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L_n^2-5F_n^2=4(-1)^n}\n{The two families share the same recurrence and have different seeds: $(F_0,F_1)=(0,1)$ and $(L_0,L_1)=(2,1)$. The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude.}\n{At $n=7$, $F_7=13$ and $L_7=29$. Thus $29^2-5\\cdot13^2=841-845=-4$. Comparing terms at different indices would destroy this identity.}\n{A recurrence operator and its initial state are separate parts of the sequence specification.}\n{OEIS A000045 and A000032, indexed from zero. Exact recurrence and common-index identity; the plotted prefix ends at $n=12$.}","act transforms}\n\\lesson{SEQUENCES / 01}{Two seeds, one Fibonacci recurrence}{S01_fibonacci_lucas}\n{F_{n+2}=F_{n+1}+F_n,\\quad L_{n+2}=L_{n+1}+L_n,\\quad L_n^2-5F_n^2=4(-1)^n}\n{The two families share the same recurrence and have different seeds: $(F_0,F_1)=(0,1)$ and $(L_0,L_1)=(2,1)$. The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude.}\n{At $n=7$, $F_7=13$ and $L_7=29$. Thus $29^2-5\\cdot13^2=841-845=-4$. Comparing terms at different indices would destroy this identity.}\n{A recurrence operator and its initial state are separate parts of the sequence specification.}\n{OEIS A000045 and A000032, indexed from zero. Exact recurrence and common-index identity; the plotted prefix ends at $n=12$.}","nsforms}\n\\lesson{SEQUENCES / 01}{Two seeds, one Fibonacci recurrence}{S01_fibonacci_lucas}\n{F_{n+2}=F_{n+1}+F_n,\\quad L_{n+2}=L_{n+1}+L_n,\\quad L_n^2-5F_n^2=4(-1)^n}\n{The two families share the same recurrence and have different seeds: $(F_0,F_1)=(0,1)$ and $(L_0,L_1)=(2,1)$. The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude.}\n{At $n=7$, $F_7=13$ and $L_7=29$. Thus $29^2-5\\cdot13^2=841-845=-4$. Comparing terms at different indices would destroy this identity.}\n{A recurrence operator and its initial state are separate parts of the sequence specification.}\n{OEIS A000045 and A000032, indexed from zero. Exact recurrence and common-index identity; the plotted prefix ends at $n=12$.}","son{SEQUENCES / 01}{Two seeds, one Fibonacci recurrence}{S01_fibonacci_lucas}\n{F_{n+2}=F_{n+1}+F_n,\\quad L_{n+2}=L_{n+1}+L_n,\\quad L_n^2-5F_n^2=4(-1)^n}\n{The two families share the same recurrence and have different seeds: $(F_0,F_1)=(0,1)$ and $(L_0,L_1)=(2,1)$. The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude.}\n{At $n=7$, $F_7=13$ and $L_7=29$. Thus $29^2-5\\cdot13^2=841-845=-4$. Comparing terms at different indices would destroy this identity.}\n{A recurrence operator and its initial state are separate parts of the sequence specification.}\n{OEIS A000045 and A000032, indexed from zero. Exact recurrence and common-index identity; the plotted prefix ends at $n=12$.}","01}{Two seeds, one Fibonacci recurrence}{S01_fibonacci_lucas}\n{F_{n+2}=F_{n+1}+F_n,\\quad L_{n+2}=L_{n+1}+L_n,\\quad L_n^2-5F_n^2=4(-1)^n}\n{The two families share the same recurrence and have different seeds: $(F_0,F_1)=(0,1)$ and $(L_0,L_1)=(2,1)$. The right panel uses the same index in both families. Its alternating defect stays exactly four in magnitude.}\n{At $n=7$, $F_7=13$ and $L_7=29$. Thus $29^2-5\\cdot13^2=841-845=-4$. Comparing terms at different indices would destroy this identity.}\n{A recurrence operator and its initial state are separate parts of the sequence specification.}\n{OEIS A000045 and A000032, indexed from zero. Exact recurrence and common-index identity; the plotted prefix ends at $n=12$.}","ernating defect stays exactly four in magnitude.}\n{At $n=7$, $F_7=13$ and $L_7=29$. Thus $29^2-5\\cdot13^2=841-845=-4$. Comparing terms at different indices would destroy this identity.}\n{A recurrence operator and its initial state are separate parts of the sequence specification.}\n{OEIS A000045 and A000032, indexed from zero. Exact recurrence and common-index identity; the plotted prefix ends at $n=12$.}","\\lesson{SEQUENCES / 02}{Longer recurrence memory}{S02_tribonacci_tetranacci}\n{T_n=T_{n-1}+T_{n-2}+T_{n-3},\\qquad Q_n=\\sum_{j=1}^{4}Q_{n-j}}\n{The seed table fixes the OEIS convention. Tribonacci starts $0,0,1$; tetranacci starts $0,0,0,1$. The curves use $\\log_{10}(1+a_n)$ so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}","CES / 02}{Longer recurrence memory}{S02_tribonacci_tetranacci}\n{T_n=T_{n-1}+T_{n-2}+T_{n-3},\\qquad Q_n=\\sum_{j=1}^{4}Q_{n-j}}\n{The seed table fixes the OEIS convention. Tribonacci starts $0,0,1$; tetranacci starts $0,0,0,1$. The curves use $\\log_{10}(1+a_n)$ so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}","ry}{S02_tribonacci_tetranacci}\n{T_n=T_{n-1}+T_{n-2}+T_{n-3},\\qquad Q_n=\\sum_{j=1}^{4}Q_{n-j}}\n{The seed table fixes the OEIS convention. Tribonacci starts $0,0,1$; tetranacci starts $0,0,0,1$. The curves use $\\log_{10}(1+a_n)$ so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}","2}+T_{n-3},\\qquad Q_n=\\sum_{j=1}^{4}Q_{n-j}}\n{The seed table fixes the OEIS convention. Tribonacci starts $0,0,1$; tetranacci starts $0,0,0,1$. The curves use $\\log_{10}(1+a_n)$ so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}",",\\qquad Q_n=\\sum_{j=1}^{4}Q_{n-j}}\n{The seed table fixes the OEIS convention. Tribonacci starts $0,0,1$; tetranacci starts $0,0,0,1$. The curves use $\\log_{10}(1+a_n)$ so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}","Q_n=\\sum_{j=1}^{4}Q_{n-j}}\n{The seed table fixes the OEIS convention. Tribonacci starts $0,0,1$; tetranacci starts $0,0,0,1$. The curves use $\\log_{10}(1+a_n)$ so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}","j=1}^{4}Q_{n-j}}\n{The seed table fixes the OEIS convention. Tribonacci starts $0,0,1$; tetranacci starts $0,0,0,1$. The curves use $\\log_{10}(1+a_n)$ so that zeros and rapidly growing terms fit together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}","together. The transform changes the plotted scale, not the integer sequence.}\n{Tribonacci gives $T_5=2+1+1=4$. Tetranacci gives $Q_7=4+2+1+1=8$. The recurrence first applies at $n=3$ for $T$ and $n=4$ for $Q$.}\n{These finite-memory recurrences have three and four feedback taps. The seed length must match the memory.}\n{OEIS A000073 and A000078, both indexed from zero. Exact integer terms through $n=16$; logarithmic ordinate explicitly labeled.}","\\lesson{SEQUENCES / 03}{Jacobsthal separates two modes}{S03_jacobsthal}\n{J_0=0,\\quad J_1=1,\\quad J_{n+2}=J_{n+1}+2J_n=\\frac{2^{n+2}-(-1)^{n+2}}{3}}\n{The rising bars are Jacobsthal numbers. The right panel removes their growing mode: $2^n-3J_n=(-1)^n$. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence.}\n{At $n=6$, $J_6=21$ and $2^6-3\\cdot21=1$. At $n=7$, $J_7=43$ and $2^7-3\\cdot43=-1$.}\n{A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction.}\n{OEIS A001045, indexed from zero. Direct closed form and exact recurrence agree for $0\\le n\\le18$.}","NCES / 03}{Jacobsthal separates two modes}{S03_jacobsthal}\n{J_0=0,\\quad J_1=1,\\quad J_{n+2}=J_{n+1}+2J_n=\\frac{2^{n+2}-(-1)^{n+2}}{3}}\n{The rising bars are Jacobsthal numbers. The right panel removes their growing mode: $2^n-3J_n=(-1)^n$. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence.}\n{At $n=6$, $J_6=21$ and $2^6-3\\cdot21=1$. At $n=7$, $J_7=43$ and $2^7-3\\cdot43=-1$.}\n{A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction.}\n{OEIS A001045, indexed from zero. Direct closed form and exact recurrence agree for $0\\le n\\le18$.}","03}{Jacobsthal separates two modes}{S03_jacobsthal}\n{J_0=0,\\quad J_1=1,\\quad J_{n+2}=J_{n+1}+2J_n=\\frac{2^{n+2}-(-1)^{n+2}}{3}}\n{The rising bars are Jacobsthal numbers. The right panel removes their growing mode: $2^n-3J_n=(-1)^n$. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence.}\n{At $n=6$, $J_6=21$ and $2^6-3\\cdot21=1$. At $n=7$, $J_7=43$ and $2^7-3\\cdot43=-1$.}\n{A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction.}\n{OEIS A001045, indexed from zero. Direct closed form and exact recurrence agree for $0\\le n\\le18$.}","l separates two modes}{S03_jacobsthal}\n{J_0=0,\\quad J_1=1,\\quad J_{n+2}=J_{n+1}+2J_n=\\frac{2^{n+2}-(-1)^{n+2}}{3}}\n{The rising bars are Jacobsthal numbers. The right panel removes their growing mode: $2^n-3J_n=(-1)^n$. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence.}\n{At $n=6$, $J_6=21$ and $2^6-3\\cdot21=1$. At $n=7$, $J_7=43$ and $2^7-3\\cdot43=-1$.}\n{A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction.}\n{OEIS A001045, indexed from zero. Direct closed form and exact recurrence agree for $0\\le n\\le18$.}","_n=(-1)^n$. What remains is an alternating mode of constant magnitude. Both modes belong to the same second-order recurrence.}\n{At $n=6$, $J_6=21$ and $2^6-3\\cdot21=1$. At $n=7$, $J_7=43$ and $2^7-3\\cdot43=-1$.}\n{A recurrence can combine growth and oscillation. Splitting the modes explains the parity correction.}\n{OEIS A001045, indexed from zero. Direct closed form and exact recurrence agree for $0\\le n\\le18$.}","\\lesson{SEQUENCES / 04}{A Pell pair shares one unit}{S04_sqrt3_pell}\n{x_n+y_n\\sqrt3=(2+\\sqrt3)^n,\\qquad x_n^2-3y_n^2=1}\n{The table keeps both coordinates at the same index. Multiplication by $2+\\sqrt3$ advances the pair using the displayed integer update. Its determinant is one, and it preserves the quadratic norm. The zero-index pair is $(1,0)$.}\n{Starting from $(x_2,y_2)=(7,4)$ gives $(x_3,y_3)=(2\\cdot7+3\\cdot4,7+2\\cdot4)=(26,15)$. The norm is $26^2-3\\cdot15^2=1$.}\n{A shared recurrence state can retain an exact invariant while both coordinates grow.}\n{OEIS A001075 and A001353, indexed from zero. Integer matrix iteration and the norm are checked through $n=18$.}","}{S04_sqrt3_pell}\n{x_n+y_n\\sqrt3=(2+\\sqrt3)^n,\\qquad x_n^2-3y_n^2=1}\n{The table keeps both coordinates at the same index. Multiplication by $2+\\sqrt3$ advances the pair using the displayed integer update. Its determinant is one, and it preserves the quadratic norm. The zero-index pair is $(1,0)$.}\n{Starting from $(x_2,y_2)=(7,4)$ gives $(x_3,y_3)=(2\\cdot7+3\\cdot4,7+2\\cdot4)=(26,15)$. The norm is $26^2-3\\cdot15^2=1$.}\n{A shared recurrence state can retain an exact invariant while both coordinates grow.}\n{OEIS A001075 and A001353, indexed from zero. Integer matrix iteration and the norm are checked through $n=18$.}","rminant is one, and it preserves the quadratic norm. The zero-index pair is $(1,0)$.}\n{Starting from $(x_2,y_2)=(7,4)$ gives $(x_3,y_3)=(2\\cdot7+3\\cdot4,7+2\\cdot4)=(26,15)$. The norm is $26^2-3\\cdot15^2=1$.}\n{A shared recurrence state can retain an exact invariant while both coordinates grow.}\n{OEIS A001075 and A001353, indexed from zero. Integer matrix iteration and the norm are checked through $n=18$.}","\\lesson{SEQUENCES / 05}{Normalize central binomial coefficients}{S05_binomial_catalan}\n{B_n=\\binom{2n}{n},\\qquad C_n=\\frac{B_n}{n+1}}\n{Each table column shows one central binomial coefficient and its Catalan normalization. The curves extend the comparison through index fourteen on a base-ten logarithmic axis. Division by $n+1$ is exact here and gives an integer at every index.}\n{For $n=4$, $B_4=\\binom84=70$ and $C_4=70/5=14$. For $n=5$, $252/6=42$. The divisor changes with the index.}\n{Normalization is an operator on coefficients. It need not be a fixed convolution or a constant tap weight.}\n{OEIS A000984 and A000108, indexed from zero. Binomial formulas and integrality checked through $n=18$.}","S / 05}{Normalize central binomial coefficients}{S05_binomial_catalan}\n{B_n=\\binom{2n}{n},\\qquad C_n=\\frac{B_n}{n+1}}\n{Each table column shows one central binomial coefficient and its Catalan normalization. The curves extend the comparison through index fourteen on a base-ten logarithmic axis. Division by $n+1$ is exact here and gives an integer at every index.}\n{For $n=4$, $B_4=\\binom84=70$ and $C_4=70/5=14$. For $n=5$, $252/6=42$. The divisor changes with the index.}\n{Normalization is an operator on coefficients. It need not be a fixed convolution or a constant tap weight.}\n{OEIS A000984 and A000108, indexed from zero. Binomial formulas and integrality checked through $n=18$.}","entral binomial coefficients}{S05_binomial_catalan}\n{B_n=\\binom{2n}{n},\\qquad C_n=\\frac{B_n}{n+1}}\n{Each table column shows one central binomial coefficient and its Catalan normalization. The curves extend the comparison through index fourteen on a base-ten logarithmic axis. Division by $n+1$ is exact here and gives an integer at every index.}\n{For $n=4$, $B_4=\\binom84=70$ and $C_4=70/5=14$. For $n=5$, $252/6=42$. The divisor changes with the index.}\n{Normalization is an operator on coefficients. It need not be a fixed convolution or a constant tap weight.}\n{OEIS A000984 and A000108, indexed from zero. Binomial formulas and integrality checked through $n=18$.}","binomial coefficients}{S05_binomial_catalan}\n{B_n=\\binom{2n}{n},\\qquad C_n=\\frac{B_n}{n+1}}\n{Each table column shows one central binomial coefficient and its Catalan normalization. The curves extend the comparison through index fourteen on a base-ten logarithmic axis. Division by $n+1$ is exact here and gives an integer at every index.}\n{For $n=4$, $B_4=\\binom84=70$ and $C_4=70/5=14$. For $n=5$, $252/6=42$. The divisor changes with the index.}\n{Normalization is an operator on coefficients. It need not be a fixed convolution or a constant tap weight.}\n{OEIS A000984 and A000108, indexed from zero. Binomial formulas and integrality checked through $n=18$.}","arithmic axis. Division by $n+1$ is exact here and gives an integer at every index.}\n{For $n=4$, $B_4=\\binom84=70$ and $C_4=70/5=14$. For $n=5$, $252/6=42$. The divisor changes with the index.}\n{Normalization is an operator on coefficients. It need not be a fixed convolution or a constant tap weight.}\n{OEIS A000984 and A000108, indexed from zero. Binomial formulas and integrality checked through $n=18$.}","\\lesson{SEQUENCES / 06}{Catalan convolution rebuilds the next term}{S06_catalan_convolution}\n{C_{n+1}=\\sum_{k=0}^{n}C_kC_{n-k},\\qquad C(z)=1+zC(z)^2}\n{The five bars are the five products contributing to $C_5$. The input indices add to four, and their sum supplies the next coefficient. Here $C(z)=\\sum_{n\\ge0}C_nz^n$ is an ordinary generating function.}\n{The contributions are $1\\cdot14$, $1\\cdot5$, $2\\cdot2$, $5\\cdot1$, and $14\\cdot1$. Their sum is $14+5+4+5+14=42=C_5$.}\n{A nonlinear generating-function equation becomes an exact convolution identity among coefficients.}\n{OEIS A000108. All convolution identities with $0\\le n\\le17$ are checked against the binomial definition.}","lesson{SEQUENCES / 06}{Catalan convolution rebuilds the next term}{S06_catalan_convolution}\n{C_{n+1}=\\sum_{k=0}^{n}C_kC_{n-k},\\qquad C(z)=1+zC(z)^2}\n{The five bars are the five products contributing to $C_5$. The input indices add to four, and their sum supplies the next coefficient. Here $C(z)=\\sum_{n\\ge0}C_nz^n$ is an ordinary generating function.}\n{The contributions are $1\\cdot14$, $1\\cdot5$, $2\\cdot2$, $5\\cdot1$, and $14\\cdot1$. Their sum is $14+5+4+5+14=42=C_5$.}\n{A nonlinear generating-function equation becomes an exact convolution identity among coefficients.}\n{OEIS A000108. All convolution identities with $0\\le n\\le17$ are checked against the binomial definition.}","ENCES / 06}{Catalan convolution rebuilds the next term}{S06_catalan_convolution}\n{C_{n+1}=\\sum_{k=0}^{n}C_kC_{n-k},\\qquad C(z)=1+zC(z)^2}\n{The five bars are the five products contributing to $C_5$. The input indices add to four, and their sum supplies the next coefficient. Here $C(z)=\\sum_{n\\ge0}C_nz^n$ is an ordinary generating function.}\n{The contributions are $1\\cdot14$, $1\\cdot5$, $2\\cdot2$, $5\\cdot1$, and $14\\cdot1$. Their sum is $14+5+4+5+14=42=C_5$.}\n{A nonlinear generating-function equation becomes an exact convolution identity among coefficients.}\n{OEIS A000108. All convolution identities with $0\\le n\\le17$ are checked against the binomial definition.}","alan convolution rebuilds the next term}{S06_catalan_convolution}\n{C_{n+1}=\\sum_{k=0}^{n}C_kC_{n-k},\\qquad C(z)=1+zC(z)^2}\n{The five bars are the five products contributing to $C_5$. The input indices add to four, and their sum supplies the next coefficient. Here $C(z)=\\sum_{n\\ge0}C_nz^n$ is an ordinary generating function.}\n{The contributions are $1\\cdot14$, $1\\cdot5$, $2\\cdot2$, $5\\cdot1$, and $14\\cdot1$. Their sum is $14+5+4+5+14=42=C_5$.}\n{A nonlinear generating-function equation becomes an exact convolution identity among coefficients.}\n{OEIS A000108. All convolution identities with $0\\le n\\le17$ are checked against the binomial definition.}","the next term}{S06_catalan_convolution}\n{C_{n+1}=\\sum_{k=0}^{n}C_kC_{n-k},\\qquad C(z)=1+zC(z)^2}\n{The five bars are the five products contributing to $C_5$. The input indices add to four, and their sum supplies the next coefficient. Here $C(z)=\\sum_{n\\ge0}C_nz^n$ is an ordinary generating function.}\n{The contributions are $1\\cdot14$, $1\\cdot5$, $2\\cdot2$, $5\\cdot1$, and $14\\cdot1$. Their sum is $14+5+4+5+14=42=C_5$.}\n{A nonlinear generating-function equation becomes an exact convolution identity among coefficients.}\n{OEIS A000108. All convolution identities with $0\\le n\\le17$ are checked against the binomial definition.}","ices add to four, and their sum supplies the next coefficient. Here $C(z)=\\sum_{n\\ge0}C_nz^n$ is an ordinary generating function.}\n{The contributions are $1\\cdot14$, $1\\cdot5$, $2\\cdot2$, $5\\cdot1$, and $14\\cdot1$. Their sum is $14+5+4+5+14=42=C_5$.}\n{A nonlinear generating-function equation becomes an exact convolution identity among coefficients.}\n{OEIS A000108. All convolution identities with $0\\le n\\le17$ are checked against the binomial definition.}","\\lesson{SEQUENCES / 07}{Bell numbers are Stirling row sums}{S07_bell_stirling}\n{B_n=\\sum_{k=0}^{n}S(n,k),\\qquad \\sum_{n\\ge0}B_n\\frac{t^n}{n!}=\\exp(e^t-1)}\n{Each matrix row contains Stirling numbers of the second kind, counting partitions into $k$ nonempty blocks. Summing across the row gives the Bell number at its right. The generating function is exponential, so its coefficient is $B_n/n!$.}\n{The fifth row contains $0,1,15,25,10,1$. Its sum is $52=B_5$. Equivalently, apply the Stirling transform to the constant input sequence $1,1,1,\\ldots$.}\n{A triangular transform uses all relevant input positions. It is not automatically a fixed-length tap filter.}\n{OEIS A000110. Stirling recurrence and an independent binomial Bell recurrence agree through $n=18$; the figure shows rows zero through seven.}","7}{Bell numbers are Stirling row sums}{S07_bell_stirling}\n{B_n=\\sum_{k=0}^{n}S(n,k),\\qquad \\sum_{n\\ge0}B_n\\frac{t^n}{n!}=\\exp(e^t-1)}\n{Each matrix row contains Stirling numbers of the second kind, counting partitions into $k$ nonempty blocks. Summing across the row gives the Bell number at its right. The generating function is exponential, so its coefficient is $B_n/n!$.}\n{The fifth row contains $0,1,15,25,10,1$. Its sum is $52=B_5$. Equivalently, apply the Stirling transform to the constant input sequence $1,1,1,\\ldots$.}\n{A triangular transform uses all relevant input positions. It is not automatically a fixed-length tap filter.}\n{OEIS A000110. Stirling recurrence and an independent binomial Bell recurrence agree through $n=18$; the figure shows rows zero through seven.}","row sums}{S07_bell_stirling}\n{B_n=\\sum_{k=0}^{n}S(n,k),\\qquad \\sum_{n\\ge0}B_n\\frac{t^n}{n!}=\\exp(e^t-1)}\n{Each matrix row contains Stirling numbers of the second kind, counting partitions into $k$ nonempty blocks. Summing across the row gives the Bell number at its right. The generating function is exponential, so its coefficient is $B_n/n!$.}\n{The fifth row contains $0,1,15,25,10,1$. Its sum is $52=B_5$. Equivalently, apply the Stirling transform to the constant input sequence $1,1,1,\\ldots$.}\n{A triangular transform uses all relevant input positions. It is not automatically a fixed-length tap filter.}\n{OEIS A000110. Stirling recurrence and an independent binomial Bell recurrence agree through $n=18$; the figure shows rows zero through seven.}","^n}{n!}=\\exp(e^t-1)}\n{Each matrix row contains Stirling numbers of the second kind, counting partitions into $k$ nonempty blocks. Summing across the row gives the Bell number at its right. The generating function is exponential, so its coefficient is $B_n/n!$.}\n{The fifth row contains $0,1,15,25,10,1$. Its sum is $52=B_5$. Equivalently, apply the Stirling transform to the constant input sequence $1,1,1,\\ldots$.}\n{A triangular transform uses all relevant input positions. It is not automatically a fixed-length tap filter.}\n{OEIS A000110. Stirling recurrence and an independent binomial Bell recurrence agree through $n=18$; the figure shows rows zero through seven.}","ential, so its coefficient is $B_n/n!$.}\n{The fifth row contains $0,1,15,25,10,1$. Its sum is $52=B_5$. Equivalently, apply the Stirling transform to the constant input sequence $1,1,1,\\ldots$.}\n{A triangular transform uses all relevant input positions. It is not automatically a fixed-length tap filter.}\n{OEIS A000110. Stirling recurrence and an independent binomial Bell recurrence agree through $n=18$; the figure shows rows zero through seven.}","\\lesson{SEQUENCES / 08}{Factorial weights order the blocks}{S08_fubini_factorial}\n{A_n=\\sum_{k=0}^{n}k!S(n,k),\\qquad \\sum_{n\\ge0}A_n\\frac{t^n}{n!}=\\frac{1}{2-e^t}}\n{The same Stirling row now receives factorial weights. Ordering $k$ blocks gives $k!$ arrangements. The upper table shows the factors at $n=5$; the bars show their products. The sum is the ordered Bell, or Fubini, number.}\n{At $n=5$, the weighted contributions are $1,30,150,240,120$, totaling $541$. The unweighted row sums to only $52$. The weights encode a different combinatorial operation.}\n{Changing the transform's input from one to $k!$ changes the output family while retaining the same triangular kernel.}\n{OEIS A000670 and A000142, indexed from zero. Integer Stirling sums and an independent ordered-partition recurrence agree through $n=18$.}","ial weights order the blocks}{S08_fubini_factorial}\n{A_n=\\sum_{k=0}^{n}k!S(n,k),\\qquad \\sum_{n\\ge0}A_n\\frac{t^n}{n!}=\\frac{1}{2-e^t}}\n{The same Stirling row now receives factorial weights. Ordering $k$ blocks gives $k!$ arrangements. The upper table shows the factors at $n=5$; the bars show their products. The sum is the ordered Bell, or Fubini, number.}\n{At $n=5$, the weighted contributions are $1,30,150,240,120$, totaling $541$. The unweighted row sums to only $52$. The weights encode a different combinatorial operation.}\n{Changing the transform's input from one to $k!$ changes the output family while retaining the same triangular kernel.}\n{OEIS A000670 and A000142, indexed from zero. Integer Stirling sums and an independent ordered-partition recurrence agree through $n=18$.}","{S08_fubini_factorial}\n{A_n=\\sum_{k=0}^{n}k!S(n,k),\\qquad \\sum_{n\\ge0}A_n\\frac{t^n}{n!}=\\frac{1}{2-e^t}}\n{The same Stirling row now receives factorial weights. Ordering $k$ blocks gives $k!$ arrangements. The upper table shows the factors at $n=5$; the bars show their products. The sum is the ordered Bell, or Fubini, number.}\n{At $n=5$, the weighted contributions are $1,30,150,240,120$, totaling $541$. The unweighted row sums to only $52$. The weights encode a different combinatorial operation.}\n{Changing the transform's input from one to $k!$ changes the output family while retaining the same triangular kernel.}\n{OEIS A000670 and A000142, indexed from zero. Integer Stirling sums and an independent ordered-partition recurrence agree through $n=18$.}","{n}k!S(n,k),\\qquad \\sum_{n\\ge0}A_n\\frac{t^n}{n!}=\\frac{1}{2-e^t}}\n{The same Stirling row now receives factorial weights. Ordering $k$ blocks gives $k!$ arrangements. The upper table shows the factors at $n=5$; the bars show their products. The sum is the ordered Bell, or Fubini, number.}\n{At $n=5$, the weighted contributions are $1,30,150,240,120$, totaling $541$. The unweighted row sums to only $52$. The weights encode a different combinatorial operation.}\n{Changing the transform's input from one to $k!$ changes the output family while retaining the same triangular kernel.}\n{OEIS A000670 and A000142, indexed from zero. Integer Stirling sums and an independent ordered-partition recurrence agree through $n=18$.}","rial weights. Ordering $k$ blocks gives $k!$ arrangements. The upper table shows the factors at $n=5$; the bars show their products. The sum is the ordered Bell, or Fubini, number.}\n{At $n=5$, the weighted contributions are $1,30,150,240,120$, totaling $541$. The unweighted row sums to only $52$. The weights encode a different combinatorial operation.}\n{Changing the transform's input from one to $k!$ changes the output family while retaining the same triangular kernel.}\n{OEIS A000670 and A000142, indexed from zero. Integer Stirling sums and an independent ordered-partition recurrence agree through $n=18$.}","contributions are $1,30,150,240,120$, totaling $541$. The unweighted row sums to only $52$. The weights encode a different combinatorial operation.}\n{Changing the transform's input from one to $k!$ changes the output family while retaining the same triangular kernel.}\n{OEIS A000670 and A000142, indexed from zero. Integer Stirling sums and an independent ordered-partition recurrence agree through $n=18$.}","\\lesson{SEQUENCES / 09}{Derangements show signed cancellation}{S09_derangements}\n{D_n=n!\\sum_{k=0}^{n}\\frac{(-1)^k}{k!},\\qquad D_n=(n-1)(D_{n-1}+D_{n-2})\\quad(n\\ge2)}\n{Positive and negative bars are the inclusion-exclusion contributions at $n=6$. Their large magnitudes cancel to a smaller integer. The dashed line records the final sum rather than an individual contribution. Here $D_0=1$ and $D_1=0$.}\n{The sum is $720-720+360-120+30-6+1=265$. The recurrence independently gives $D_6=5(44+9)=265$.}\n{Signed transforms must retain cancellation. Magnitudes alone do not determine the result.}\n{OEIS A000166. Factorial sum, variable-coefficient recurrence, and brute-force permutation counts through $n=8$ agree; exact formulas checked through $n=18$.}","/ 09}{Derangements show signed cancellation}{S09_derangements}\n{D_n=n!\\sum_{k=0}^{n}\\frac{(-1)^k}{k!},\\qquad D_n=(n-1)(D_{n-1}+D_{n-2})\\quad(n\\ge2)}\n{Positive and negative bars are the inclusion-exclusion contributions at $n=6$. Their large magnitudes cancel to a smaller integer. The dashed line records the final sum rather than an individual contribution. Here $D_0=1$ and $D_1=0$.}\n{The sum is $720-720+360-120+30-6+1=265$. The recurrence independently gives $D_6=5(44+9)=265$.}\n{Signed transforms must retain cancellation. Magnitudes alone do not determine the result.}\n{OEIS A000166. Factorial sum, variable-coefficient recurrence, and brute-force permutation counts through $n=8$ agree; exact formulas checked through $n=18$.}","D_n=n!\\sum_{k=0}^{n}\\frac{(-1)^k}{k!},\\qquad D_n=(n-1)(D_{n-1}+D_{n-2})\\quad(n\\ge2)}\n{Positive and negative bars are the inclusion-exclusion contributions at $n=6$. Their large magnitudes cancel to a smaller integer. The dashed line records the final sum rather than an individual contribution. Here $D_0=1$ and $D_1=0$.}\n{The sum is $720-720+360-120+30-6+1=265$. The recurrence independently gives $D_6=5(44+9)=265$.}\n{Signed transforms must retain cancellation. Magnitudes alone do not determine the result.}\n{OEIS A000166. Factorial sum, variable-coefficient recurrence, and brute-force permutation counts through $n=8$ agree; exact formulas checked through $n=18$.}","he dashed line records the final sum rather than an individual contribution. Here $D_0=1$ and $D_1=0$.}\n{The sum is $720-720+360-120+30-6+1=265$. The recurrence independently gives $D_6=5(44+9)=265$.}\n{Signed transforms must retain cancellation. Magnitudes alone do not determine the result.}\n{OEIS A000166. Factorial sum, variable-coefficient recurrence, and brute-force permutation counts through $n=8$ agree; exact formulas checked through $n=18$.}","than an individual contribution. Here $D_0=1$ and $D_1=0$.}\n{The sum is $720-720+360-120+30-6+1=265$. The recurrence independently gives $D_6=5(44+9)=265$.}\n{Signed transforms must retain cancellation. Magnitudes alone do not determine the result.}\n{OEIS A000166. Factorial sum, variable-coefficient recurrence, and brute-force permutation counts through $n=8$ agree; exact formulas checked through $n=18$.}","ler_updown}\n{\\sum_{n\\ge0}E_n\\frac{t^n}{n!}=\\sec t+\\tan t,\\quad E_{2m}=s_m,\\quad E_{2m+1}=\\tau_m}\n{The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients.}\n{At degree four, $E_4=5$ contributes $5t^4/4!$ to secant. At degree five, $E_5=16$ contributes $16t^5/5!$ to tangent. The first tangent entry corresponds to degree one.}\n{Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept.}\n{OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; $\\tau_m$ denotes its degree-$2m+1$ value. Exact series checked through degree eighteen.}","\\ge0}E_n\\frac{t^n}{n!}=\\sec t+\\tan t,\\quad E_{2m}=s_m,\\quad E_{2m+1}=\\tau_m}\n{The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients.}\n{At degree four, $E_4=5$ contributes $5t^4/4!$ to secant. At degree five, $E_5=16$ contributes $16t^5/5!$ to tangent. The first tangent entry corresponds to degree one.}\n{Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept.}\n{OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; $\\tau_m$ denotes its degree-$2m+1$ value. Exact series checked through degree eighteen.}","\\quad E_{2m}=s_m,\\quad E_{2m+1}=\\tau_m}\n{The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients.}\n{At degree four, $E_4=5$ contributes $5t^4/4!$ to secant. At degree five, $E_5=16$ contributes $16t^5/5!$ to tangent. The first tangent entry corresponds to degree one.}\n{Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept.}\n{OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; $\\tau_m$ denotes its degree-$2m+1$ value. Exact series checked through degree eighteen.}","d E_{2m+1}=\\tau_m}\n{The coefficient table separates even and odd positions of the Euler up/down family. Even positions are positive secant numbers. Odd positions are tangent numbers. The factorial row makes clear that these are exponential-generating-function numbers, not ordinary Taylor coefficients.}\n{At degree four, $E_4=5$ contributes $5t^4/4!$ to secant. At degree five, $E_5=16$ contributes $16t^5/5!$ to tangent. The first tangent entry corresponds to degree one.}\n{Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept.}\n{OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; $\\tau_m$ denotes its degree-$2m+1$ value. Exact series checked through degree eighteen.}","dinary Taylor coefficients.}\n{At degree four, $E_4=5$ contributes $5t^4/4!$ to secant. At degree five, $E_5=16$ contributes $16t^5/5!$ to tangent. The first tangent entry corresponds to degree one.}\n{Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept.}\n{OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; $\\tau_m$ denotes its degree-$2m+1$ value. Exact series checked through degree eighteen.}","{At degree four, $E_4=5$ contributes $5t^4/4!$ to secant. At degree five, $E_5=16$ contributes $16t^5/5!$ to tangent. The first tangent entry corresponds to degree one.}\n{Parity packets can identify familiar functions, provided the factorial normalization and sequence offset are kept.}\n{OEIS A000111, A000364, and A000182. The tangent entry has OEIS initial index one; $\\tau_m$ denotes its degree-$2m+1$ value. Exact series checked through degree eighteen.}","\\lesson{SEQUENCES / 11}{Bernoulli data need both coordinates}{S11_bernoulli_pair}\n{\\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","ES / 11}{Bernoulli data need both coordinates}{S11_bernoulli_pair}\n{\\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","{Bernoulli data need both coordinates}{S11_bernoulli_pair}\n{\\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","need both coordinates}{S11_bernoulli_pair}\n{\\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","inates}{S11_bernoulli_pair}\n{\\frac{t}{e^t-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","-1}=\\sum_{n\\ge0}B_n\\frac{t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","_n\\frac{t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","t^n}{n!},\\qquad B_n=\\frac{p_n}{q_n},\\quad q_n>0}\n{The top panel shows signed numerators on a symmetric logarithmic scale. The lower panel shows positive denominators on a logarithmic scale. Fractions are reduced. A zero Bernoulli number is stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","stored as $0/1$, and this guide uses $B_1=-1/2$.}\n{At $n=10$, $p_{10}=5$ and $q_{10}=66$, so $B_{10}=5/66$. Its Taylor coefficient is $(5/66)/10!$. At $n=12$, $B_{12}=-691/2730$.}\n{An OEIS integer numerator sequence does not by itself specify the rational family or its generating-function coefficient.}\n{OEIS A027641 and A027642, indexed from zero. Exact rational recurrence and multiplication by $(e^t-1)/t$ checked through degree eighteen.}","\\lesson{SEQUENCES / 12}{Partitions multiply reciprocal factors}{S12_partition_product}\n{P(z)=\\prod_{j\\ge1}\\frac{1}{1-z^j}=\\sum_{n\\ge0}p(n)z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","ES / 12}{Partitions multiply reciprocal factors}{S12_partition_product}\n{P(z)=\\prod_{j\\ge1}\\frac{1}{1-z^j}=\\sum_{n\\ge0}p(n)z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","12}{Partitions multiply reciprocal factors}{S12_partition_product}\n{P(z)=\\prod_{j\\ge1}\\frac{1}{1-z^j}=\\sum_{n\\ge0}p(n)z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","titions multiply reciprocal factors}{S12_partition_product}\n{P(z)=\\prod_{j\\ge1}\\frac{1}{1-z^j}=\\sum_{n\\ge0}p(n)z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","multiply reciprocal factors}{S12_partition_product}\n{P(z)=\\prod_{j\\ge1}\\frac{1}{1-z^j}=\\sum_{n\\ge0}p(n)z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","iprocal factors}{S12_partition_product}\n{P(z)=\\prod_{j\\ge1}\\frac{1}{1-z^j}=\\sum_{n\\ge0}p(n)z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","{S12_partition_product}\n{P(z)=\\prod_{j\\ge1}\\frac{1}{1-z^j}=\\sum_{n\\ge0}p(n)z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}",")z^n}\n{A factor indexed by $j$ chooses how many parts of size $j$ occur. The left table lists the five partitions of four. The right bars show the first nineteen partition numbers. This is an ordinary generating function with integer coefficients.}\n{The partitions of four are $4$, $3+1$, $2+2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","2$, $2+1+1$, and $1+1+1+1$. Thus $p(4)=5$. To find coefficients through degree eighteen, only factors with $j\\le18$ are needed.}\n{Reciprocal factors turn independent multiplicities into a coefficient-counting product. The finite prefix can be reconstructed exactly.}\n{OEIS A000041. Product coefficients computed by integer dynamic programming, independently checked by partition enumeration through $n=10$.}","\\lesson{SEQUENCES / 13}{Apéry uses taps that depend on the index}{S13_apery_recurrence}\n{(n+1)^3A_{n+1}=(2n+1)(17n^2+17n+5)A_n-n^3A_{n-1}}\n{The prefix annotation begins the Apéry sequence defined by $A_n=\\sum_{k=0}^{n}\\binom nk^2\\binom{n+k}{k}^2$. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","on{SEQUENCES / 13}{Apéry uses taps that depend on the index}{S13_apery_recurrence}\n{(n+1)^3A_{n+1}=(2n+1)(17n^2+17n+5)A_n-n^3A_{n-1}}\n{The prefix annotation begins the Apéry sequence defined by $A_n=\\sum_{k=0}^{n}\\binom nk^2\\binom{n+k}{k}^2$. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","13}{Apéry uses taps that depend on the index}{S13_apery_recurrence}\n{(n+1)^3A_{n+1}=(2n+1)(17n^2+17n+5)A_n-n^3A_{n-1}}\n{The prefix annotation begins the Apéry sequence defined by $A_n=\\sum_{k=0}^{n}\\binom nk^2\\binom{n+k}{k}^2$. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","}{S13_apery_recurrence}\n{(n+1)^3A_{n+1}=(2n+1)(17n^2+17n+5)A_n-n^3A_{n-1}}\n{The prefix annotation begins the Apéry sequence defined by $A_n=\\sum_{k=0}^{n}\\binom nk^2\\binom{n+k}{k}^2$. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","=(2n+1)(17n^2+17n+5)A_n-n^3A_{n-1}}\n{The prefix annotation begins the Apéry sequence defined by $A_n=\\sum_{k=0}^{n}\\binom nk^2\\binom{n+k}{k}^2$. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","17n+5)A_n-n^3A_{n-1}}\n{The prefix annotation begins the Apéry sequence defined by $A_n=\\sum_{k=0}^{n}\\binom nk^2\\binom{n+k}{k}^2$. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","^{n}\\binom nk^2\\binom{n+k}{k}^2$. The curves show its growth through index fourteen on a base-ten logarithmic scale. The lower diagram evaluates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","uates the recurrence at $n=2$.}\n{Using $A_1=5$ and $A_2=73$ gives $27A_3=535\\cdot73-8\\cdot5=39015$. Therefore $A_3=1445$. The recurrence applies for $n\\ge1$, with $A_0=1$.}\n{This is a three-term recurrence with polynomial coefficients. Its weights change with $n$, so it is not a constant-tap filter.}\n{OEIS A005259. Exact binomial sums and the displayed polynomial-coefficient recurrence agree for $1\\le n\\le17$.}","olution changes the address rule}{S14_divisor_convolution}\n{\\sum_{d\\mid n}\\mu(d)=\\mathbf1_{n=1},\\qquad \\sum_{d\\mid n}\\varphi(d)=n}\n{The figure places the six divisors of twelve in matching columns. Möbius contributions sum to zero; totient contributions sum to twelve. The address rule is divisibility. This differs from the additive index splitting used by ordinary power-series convolution.}\n{For $n=12$, $\\mu(1),\\mu(2),\\mu(3),\\mu(4),\\mu(6),\\mu(12)$ are $1,-1,-1,0,1,0$. The totient values are $1,1,2,2,2,4$. Their sums are zero and twelve.}\n{Choose the convolution law before identifying a transform. Divisor packets and residue packets organize different relations.}\n{OEIS A008683 and A000010, both indexed from one. Prime-factor definitions and both divisor sums checked for $1\\le n\\le60$.}","changes the address rule}{S14_divisor_convolution}\n{\\sum_{d\\mid n}\\mu(d)=\\mathbf1_{n=1},\\qquad \\sum_{d\\mid n}\\varphi(d)=n}\n{The figure places the six divisors of twelve in matching columns. Möbius contributions sum to zero; totient contributions sum to twelve. The address rule is divisibility. This differs from the additive index splitting used by ordinary power-series convolution.}\n{For $n=12$, $\\mu(1),\\mu(2),\\mu(3),\\mu(4),\\mu(6),\\mu(12)$ are $1,-1,-1,0,1,0$. The totient values are $1,1,2,2,2,4$. Their sums are zero and twelve.}\n{Choose the convolution law before identifying a transform. Divisor packets and residue packets organize different relations.}\n{OEIS A008683 and A000010, both indexed from one. Prime-factor definitions and both divisor sums checked for $1\\le n\\le60$.}","}\n{\\sum_{d\\mid n}\\mu(d)=\\mathbf1_{n=1},\\qquad \\sum_{d\\mid n}\\varphi(d)=n}\n{The figure places the six divisors of twelve in matching columns. Möbius contributions sum to zero; totient contributions sum to twelve. The address rule is divisibility. This differs from the additive index splitting used by ordinary power-series convolution.}\n{For $n=12$, $\\mu(1),\\mu(2),\\mu(3),\\mu(4),\\mu(6),\\mu(12)$ are $1,-1,-1,0,1,0$. The totient values are $1,1,2,2,2,4$. Their sums are zero and twelve.}\n{Choose the convolution law before identifying a transform. Divisor packets and residue packets organize different relations.}\n{OEIS A008683 and A000010, both indexed from one. Prime-factor definitions and both divisor sums checked for $1\\le n\\le60$.}","qquad \\sum_{d\\mid n}\\varphi(d)=n}\n{The figure places the six divisors of twelve in matching columns. Möbius contributions sum to zero; totient contributions sum to twelve. The address rule is divisibility. This differs from the additive index splitting used by ordinary power-series convolution.}\n{For $n=12$, $\\mu(1),\\mu(2),\\mu(3),\\mu(4),\\mu(6),\\mu(12)$ are $1,-1,-1,0,1,0$. The totient values are $1,1,2,2,2,4$. Their sums are zero and twelve.}\n{Choose the convolution law before identifying a transform. Divisor packets and residue packets organize different relations.}\n{OEIS A008683 and A000010, both indexed from one. Prime-factor definitions and both divisor sums checked for $1\\le n\\le60$.}","onvolution.}\n{For $n=12$, $\\mu(1),\\mu(2),\\mu(3),\\mu(4),\\mu(6),\\mu(12)$ are $1,-1,-1,0,1,0$. The totient values are $1,1,2,2,2,4$. Their sums are zero and twelve.}\n{Choose the convolution law before identifying a transform. Divisor packets and residue packets organize different relations.}\n{OEIS A008683 and A000010, both indexed from one. Prime-factor definitions and both divisor sums checked for $1\\le n\\le60$.}","/ 15}{One binary statistic creates three families}{S15_binary_families}\n{w(n)=\\operatorname{popcount}(n),\\quad t(n)=w(n)\\bmod2,\\quad g(n)=2^{w(n)}}\n{Each column has a four-bit address, its binary weight, its Thue--Morse parity, and its Gould value. Reading a final zero bit preserves weight; reading a final one adds one. Parity toggles and the Gould value doubles in the latter case.}\n{The address $13=1101_2$ has weight three, so $t(13)=1$ and $g(13)=8$. Address $12=1100_2$ has weight two, parity zero, and Gould value four.}\n{The same state statistic can feed several exact outputs. Binary refinement gives simple even and odd channel recurrences.}\n{OEIS A000120, A010060, and A001316, indexed from zero. Binary formulas and even/odd recurrences checked for $0\\le n\\le63$; figure shows zero through fifteen.}","s three families}{S15_binary_families}\n{w(n)=\\operatorname{popcount}(n),\\quad t(n)=w(n)\\bmod2,\\quad g(n)=2^{w(n)}}\n{Each column has a four-bit address, its binary weight, its Thue--Morse parity, and its Gould value. Reading a final zero bit preserves weight; reading a final one adds one. Parity toggles and the Gould value doubles in the latter case.}\n{The address $13=1101_2$ has weight three, so $t(13)=1$ and $g(13)=8$. Address $12=1100_2$ has weight two, parity zero, and Gould value four.}\n{The same state statistic can feed several exact outputs. Binary refinement gives simple even and odd channel recurrences.}\n{OEIS A000120, A010060, and A001316, indexed from zero. Binary formulas and even/odd recurrences checked for $0\\le n\\le63$; figure shows zero through fifteen.}","es}{S15_binary_families}\n{w(n)=\\operatorname{popcount}(n),\\quad t(n)=w(n)\\bmod2,\\quad g(n)=2^{w(n)}}\n{Each column has a four-bit address, its binary weight, its Thue--Morse parity, and its Gould value. Reading a final zero bit preserves weight; reading a final one adds one. Parity toggles and the Gould value doubles in the latter case.}\n{The address $13=1101_2$ has weight three, so $t(13)=1$ and $g(13)=8$. Address $12=1100_2$ has weight two, parity zero, and Gould value four.}\n{The same state statistic can feed several exact outputs. Binary refinement gives simple even and odd channel recurrences.}\n{OEIS A000120, A010060, and A001316, indexed from zero. Binary formulas and even/odd recurrences checked for $0\\le n\\le63$; figure shows zero through fifteen.}","lies}\n{w(n)=\\operatorname{popcount}(n),\\quad t(n)=w(n)\\bmod2,\\quad g(n)=2^{w(n)}}\n{Each column has a four-bit address, its binary weight, its Thue--Morse parity, and its Gould value. Reading a final zero bit preserves weight; reading a final one adds one. Parity toggles and the Gould value doubles in the latter case.}\n{The address $13=1101_2$ has weight three, so $t(13)=1$ and $g(13)=8$. Address $12=1100_2$ has weight two, parity zero, and Gould value four.}\n{The same state statistic can feed several exact outputs. Binary refinement gives simple even and odd channel recurrences.}\n{OEIS A000120, A010060, and A001316, indexed from zero. Binary formulas and even/odd recurrences checked for $0\\le n\\le63$; figure shows zero through fifteen.}","s in the latter case.}\n{The address $13=1101_2$ has weight three, so $t(13)=1$ and $g(13)=8$. Address $12=1100_2$ has weight two, parity zero, and Gould value four.}\n{The same state statistic can feed several exact outputs. Binary refinement gives simple even and odd channel recurrences.}\n{OEIS A000120, A010060, and A001316, indexed from zero. Binary formulas and even/odd recurrences checked for $0\\le n\\le63$; figure shows zero through fifteen.}","\\lesson{SEQUENCES / 16}{Sums and differences form a polynomial ladder}{S16_polynomial_ladder}\n{T_n=\\sum_{j=1}^{n}j,\\quad H_n=\\sum_{j=1}^{n}T_j,\\quad \\Delta T_n=n,\\quad \\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities","dder}{S16_polynomial_ladder}\n{T_n=\\sum_{j=1}^{n}j,\\quad H_n=\\sum_{j=1}^{n}T_j,\\quad \\Delta T_n=n,\\quad \\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","16_polynomial_ladder}\n{T_n=\\sum_{j=1}^{n}j,\\quad H_n=\\sum_{j=1}^{n}T_j,\\quad \\Delta T_n=n,\\quad \\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","_ladder}\n{T_n=\\sum_{j=1}^{n}j,\\quad H_n=\\sum_{j=1}^{n}T_j,\\quad \\Delta T_n=n,\\quad \\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","\\quad H_n=\\sum_{j=1}^{n}T_j,\\quad \\Delta T_n=n,\\quad \\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","m_{j=1}^{n}T_j,\\quad \\Delta T_n=n,\\quad \\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","lta T_n=n,\\quad \\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","\\Delta H_n=T_n}\n{The upper table follows positive integers into triangular and tetrahedral numbers. Squares and cubes occupy the lower rows for comparison. The arrows reverse prefix sums by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","s by backward differences, $\\Delta f_n=f_n-f_{n-1}$. All plotted values use their displayed common index.}\n{At $n=4$, $T_4=10$ and $H_4=20$. The differences are $10-6=4$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","$ and $20-10=10$. Separately, $4^2-3^2=7$ and $4^3-3^3=37$.}\n{Prefix summation raises polynomial degree; differencing lowers it. These are exact operators with boundary values at zero; the displayed difference identities apply for $n\\ge1$.}\n{OEIS A000027, A000217, A000292, A000290, and A000578. Positive integers start at one; the other four families include index zero. Identities checked through $n=18$.}","\\[\nu_{n+1}=\\frac{\\sum_{i=0}^{n}q^{2i+1}r^{2(n-i)+1}u_i v_{n-i}}{2n+3},\\qquad\nv_{n+1}=\\frac{\\sum_{i=0}^{n}a^{2i+1}b^{2(n-i)+1}u_i v_{n-i}}{2n+3}.\n\\]","\\[\n0<u_n\\le 2^{-n},\\qquad 0<v_n\\le 2^{-n}.\n\\]","\\[\n\\left|\\sum_{n\\ge N}c_nz^{2n+1}\\right|\n\\le \\frac{|z|\\rho^N}{1-\\rho},\\qquad |z|^2<2.\n\\]","\\[\nD=k^2u_2-p^2v_2\n=\\frac{pkar}{15}(q^2-b^2)(qb-ar).\n\\]","\\[\n(x-X)(pX-ky)=0.\n\\]","\\[\n(k^2u-p^2v)(x^2-X^2)=0.\n\\]","The earlier moving-width lower argument used a fixed \\(\\alpha<1\\); it\nproved orders below \\(\\rho\\) and then recovered the order by\n\\(\\alpha\\uparrow1\\), but did not establish positive lower type at\n\\(\\rho\\). This is a limitation of that argument, not a counterexample.\nThe fixed-width comparison closes precisely that missing implication.\nThe evidence archive retains execution failures, their causes and the\nsubsequent corrected runs.","The decisive extension is from a scalar positive comparison to a\ncomparison weighted by a positive eigenvector. The fixed-width integral\ncontrols the lower type at the exact exponent. A fixed proportional\ninterval \\([\\alpha R,R]\\), with \\(\\alpha<1\\), would only give a smaller\nexponent through \\(\\alpha^sP(s)\\). Letting \\(\\alpha\\) tend to one\nrecovers exact order but does not, by itself, yield positive lower type.\nThis is a limitation of that alternative route, not a failure of the\ntheorem.","\\claim{An established diagonal antecedent}\nThe equation $f_q'(z)=1+f_q(qz)^2$, $f_q(0)=0$, is already present in\nM\\'endez~\\cite[Example~2.4]{Mendez2024}. It remains an\nimportant baseline, but its definition is not a discovery of this project.\nThe asymmetric extension keeps two separate contractions and permits a\nproved distinction from every amplitude/argument rescaling of that baseline.","\\mainmatter\n\\MonographPart{I}{Entire functions and their growth}{Contractions, coefficients and nonlinear feedback.}\n\\chapter{Contracted nonlinear feedback}\\label{ch:feedback}\n\\pageheading{FAMILY 1}{Two scales inside the function}\nFor real $0<q,r<1$, define the normalized odd solution\n\\[\n\\boxed{T_{q,r}'(z)=1+T_{q,r}(qz)T_{q,r}(rz),\\qquad T_{q,r}(0)=0.}\n\\]\nThe contractions are part of the defining feedback. The function is symmetric\nin $q,r$, but unequal scales retain left/right address information.\n\\fig{01_asymmetric_functions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}","\\mainmatter\n\\MonographPart{I}{Entire functions and their growth}{Contractions, coefficients and nonlinear feedback.}\n\\chapter{Contracted nonlinear feedback}\\label{ch:feedback}\n\\pageheading{FAMILY 1}{Two scales inside the function}\nFor real $0<q,r<1$, define the normalized odd solution\n\\[\n\\boxed{T_{q,r}'(z)=1+T_{q,r}(qz)T_{q,r}(rz),\\qquad T_{q,r}(0)=0.}\n\\]\nThe contractions are part of the defining feedback. The function is symmetric\nin $q,r$, but unequal scales retain left/right address information.\n\\fig{01_asymmetric_functions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coeff","ctions and their growth}{Contractions, coefficients and nonlinear feedback.}\n\\chapter{Contracted nonlinear feedback}\\label{ch:feedback}\n\\pageheading{FAMILY 1}{Two scales inside the function}\nFor real $0<q,r<1$, define the normalized odd solution\n\\[\n\\boxed{T_{q,r}'(z)=1+T_{q,r}(qz)T_{q,r}(rz),\\qquad T_{q,r}(0)=0.}\n\\]\nThe contractions are part of the defining feedback. The function is symmetric\nin $q,r$, but unequal scales retain left/right address information.\n\\fig{01_asymmetric_functions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral …","1}{Two scales inside the function}\nFor real $0<q,r<1$, define the normalized odd solution\n\\[\n\\boxed{T_{q,r}'(z)=1+T_{q,r}(qz)T_{q,r}(rz),\\qquad T_{q,r}(0)=0.}\n\\]\nThe contractions are part of the defining feedback. The function is symmetric\nin $q,r$, but unequal scales retain left/right address information.\n\\fig{01_asymmetric_functions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral equation\nextends it to $|z|<R/\\max(q,r)$. Repetition proves that it is entire.","r}(0)=0.}\n\\]\nThe contractions are part of the defining feedback. The function is symmetric\nin $q,r$, but unequal scales retain left/right address information.\n\\fig{01_asymmetric_functions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral equation\nextends it to $|z|<R/\\max(q,r)$. Repetition proves that it is entire.","of the defining feedback. The function is symmetric\nin $q,r$, but unequal scales retain left/right address information.\n\\fig{01_asymmetric_functions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral equation\nextends it to $|z|<R/\\max(q,r)$. Repetition proves that it is entire.","on is symmetric\nin $q,r$, but unequal scales retain left/right address information.\n\\fig{01_asymmetric_functions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral equation\nextends it to $|z|<R/\\max(q,r)$. Repetition proves that it is entire.","nctions}{Computed Taylor curves. Each entire curve uses\nterms through $z^{321}$; positive-coefficient Cauchy estimates bound the omitted\ntail on each displayed interval. The black dashed curve is ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral equation\nextends it to $|z|<R/\\max(q,r)$. Repetition proves that it is entire.","s ordinary tangent.}\nFor $T_{q,r}(z)=\\sum_{n\\ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral equation\nextends it to $|z|<R/\\max(q,r)$. Repetition proves that it is entire.","ge0}c_nz^{2n+1}$, $c_0=1$, coefficient matching gives\n\\[\nc_n=\\frac{1}{2n+1}\\sum_{i=0}^{n-1}\nq^{2i+1}r^{2(n-1-i)+1}c_ic_{n-1-i},\\qquad n\\ge1.\n\\]\nIn particular,\n\\[\nT_{q,r}(z)=z+\\frac{qr}{3}z^3+\n\\frac{q^2r^2(q^2+r^2)}{15}z^5+O(z^7).\n\\]\nThe coefficients are positive and bounded by tangent coefficients, giving a\nconvergent germ. If the solution is known on $|z|<R$, its integral equation\nextends it to $|z|<R/\\max(q,r)$. Repetition proves that it is entire.","The parameters can be recovered, up to swapping, from $c_1,c_2$. Put\n$p=3c_1$ and $u=15c_2/p^2$. Then $q,r$ are the roots of\n$x^2-\\sqrt{u+2p}\\,x+p=0$.","\\pageheading{FAMILY 1\\; /\\; GROWTH}{A scale-balance atlas}\n\\[\n\\boxed{\\rho(T_{q,r})=\\rho\\quad\\text{where}\\quad q^\\rho+r^\\rho=1.}\n\\]\nHere $M_f(R)=\\max_{|z|=R}|f(z)|$ and\n$\\rho(f)=\\limsup_{R\\to\\infty}\\log\\log M_f(R)/\\log R$ is the order of an entire function. Moreover $\\log M_{T}(R)=O(R^\\rho)$: the upper type is finite.\n\\smallfig{02_order_atlas}{The displayed values numerically solve the proved\ncharacteristic equation. Colors saturate at order 7; the upper corner contains\nlarger orders. The dashed diagonal is the previously known one-scale family.}\n\\begin{tabular}{@{}lll@{}}\\toprule\nScales & Order & Interpretation","midrule\n$q=r=1/4$ & $1/2$ & Slow entire growth","$q+r=1$ & $1$ & Many distinct functions at one order","$q=r=1/\\sqrt2$ & $2$ & Quadratic logarithmic growth scale","$q=r=3/4$ & $\\log2/\\log(4/3)$ & Approximately $2.4094208397$","\\claim{Why this equation appears}\nNonnegative coefficients give $M(R)=T_{q,r}(R)$. The integral equation implies\n\\[\nM(R)\\le R[1+M(qR)M(rR)],\n\\]\n\\[\nM(R)\\ge(1-\\alpha)R M(\\alpha qR)M(\\alpha rR),\\quad0<\\alpha<1.\n\\]\nTaking logarithms turns product feedback into a sum over contracted scales.\nComparison with $CR^\\rho$ proves the upper bound; comparison with\n$cR^{\\rho_\\alpha}$, where $(\\alpha q)^{\\rho_\\alpha}+\n(\\alpha r)^{\\rho_\\alpha}=1$, proves the lower bound. Letting $\\alpha\\uparrow1$\ngives the same order. The complete interval-induction proof is given in Appendix~\\ref{app:growth}.",")=T_{q,r}(R)$. The integral equation implies\n\\[\nM(R)\\le R[1+M(qR)M(rR)],\n\\]\n\\[\nM(R)\\ge(1-\\alpha)R M(\\alpha qR)M(\\alpha rR),\\quad0<\\alpha<1.\n\\]\nTaking logarithms turns product feedback into a sum over contracted scales.\nComparison with $CR^\\rho$ proves the upper bound; comparison with\n$cR^{\\rho_\\alpha}$, where $(\\alpha q)^{\\rho_\\alpha}+\n(\\alpha r)^{\\rho_\\alpha}=1$, proves the lower bound. Letting $\\alpha\\uparrow1$\ngives the same order. The complete interval-induction proof is given in Appendix~\\ref{app:growth}.","\\pageheading{FAMILY 1\\; /\\; SEPARATION}{More than a rescaled one-scale seed}\n\\claim{Nonrescaling theorem}\nIf $q\\ne r$, then $T_{q,r}$ cannot equal $A f_s(Bz)$ for nonzero constants\n$A,B$ and any real $0<s<1$, where $f_s'=1+f_s(sz)^2$ is the known family.","Suppose such an equality held. Normalization gives $AB=1$. Write $p=qr$,\n$u=q^2+r^2$, and let $\\rho$ be the common order. The first two nontrivial\ncoefficients and the growth law force\n\\[\nB^2s^2=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","held. Normalization gives $AB=1$. Write $p=qr$,\n$u=q^2+r^2$, and let $\\rho$ be the common order. The first two nontrivial\ncoefficients and the growth law force\n\\[\nB^2s^2=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","on gives $AB=1$. Write $p=qr$,\n$u=q^2+r^2$, and let $\\rho$ be the common order. The first two nontrivial\ncoefficients and the growth law force\n\\[\nB^2s^2=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","s $AB=1$. Write $p=qr$,\n$u=q^2+r^2$, and let $\\rho$ be the common order. The first two nontrivial\ncoefficients and the growth law force\n\\[\nB^2s^2=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","rite $p=qr$,\n$u=q^2+r^2$, and let $\\rho$ be the common order. The first two nontrivial\ncoefficients and the growth law force\n\\[\nB^2s^2=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","2+r^2$, and let $\\rho$ be the common order. The first two nontrivial\ncoefficients and the growth law force\n\\[\nB^2s^2=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","he common order. The first two nontrivial\ncoefficients and the growth law force\n\\[\nB^2s^2=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","=p,\\qquad s^2=u/2,\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","\\qquad 2s^\\rho=q^\\rho+r^\\rho=1.\n\\]\nConsequently\n\\[\n2\\left(\\frac{q^2+r^2}{2}\\right)^{\\rho/2}=q^\\rho+r^\\rho.\n\\]\nFor $\\rho\\ne2$, strict convexity or concavity of $x^{\\rho/2}$ forces $q=r$.\nFor the exceptional case $\\rho=2$, we have $u=1$, $s^2=1/2$, $B^2=2p$.\nThe two $z^7$ coefficients become\n\\[\n\\frac{p^3(3-p^2)}{315}\\quad\\text{and}\\quad\\frac{11p^3}{1260}.\n\\]\nTheir equality forces $p=1/2$ and again $(q-r)^2=u-2p=0$.\nThis completes the exceptional-order case.","\\claim{A worked order-one member}\n\\[\nT_{1/3,2/3}(z)=z+\\frac{2}{27}z^3+\n\\frac{4}{2187}z^5+\\frac{232}{11160261}z^7+O(z^9).\n\\]\nIt has order one, and $T_{1/3,2/3}(1)\\approx1.0759239897408249$.\nThe diagonal point $q=r=1/2$ gives\n$f_{1/2}(z)=\\sqrt2\\sinh(z/\\sqrt2)$. Every off-diagonal member is separated\nfrom all rescaled diagonal members by the theorem above.","\\claim{A combinatorial address formula}\nSum over full ordered binary trees with $n$ internal vertices. A vertex with\n$i$ internal vertices in its left subtree and $j$ in its right contributes\n\\[\n\\frac{q^{2i+1}r^{2j+1}}{2(i+j+1)+1}.\n\\]\nMultiply these factors over internal vertices, then sum over trees. Root\nsplitting reproduces the coefficient recurrence exactly. Unequal contractions\ntherefore record left and right path lengths separately.","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with we","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsil","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function v","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the sec","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Co","\\pageheading{FAMILY 1\\; /\\; TAPS}{Feedback banks and weak-channel sensitivity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coeffici…","vity}\\label{sec:banks}\nChoose finitely many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of t…","y many channels $j=1,\\ldots,J$, with $h_j>0$, $\\sum_jh_j=1$,\nand $0<q_j,r_j<1$. Define\n\\[\nF_\\Theta'(z)=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz),\\qquad F_\\Theta(0)=0.\n\\]\nThese taps act inside the nonlinear definition. With\n$\\mu_1=\\sum_jh_jq_jr_j$ and\n$\\mu_2=\\sum_jh_jq_jr_j(q_j^2+r_j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated conver…","j^2)$,\n\\[\nF_\\Theta(z)=z+\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","\\frac{\\mu_1}{3}z^3+\\frac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","rac{\\mu_1\\mu_2}{15}z^5+O(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","(z^7).\n\\]\nIf $q_j^{\\rho_j}+r_j^{\\rho_j}=1$, then\n\\[\n\\boxed{\\rho(F_\\Theta)=\\max_{j\\text{ active}}\\rho_j.}\n\\]\nThe positive weights alter local coefficients. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","nts. The most expansive active\nchannel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","annel determines the asymptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","symptotic order. A zero-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","o-weight channel is removed\nbefore taking the maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","he maximum.\n\\fig{07_weak_tap}{A two-channel example: scales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","ales $(1/4,1/4)$ with weight\n$1-\\varepsilon$, and $(3/4,3/4)$ with weight $\\varepsilon$.\nThe function varies continuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","ontinuously on every fixed compact set, while its growth\norder jumps when the second channel becomes active.}\nFor $0<\\varepsilon\\le1$ the order is $\\log2/\\log(4/3)$; at $\\varepsilon=0$\nit is $1/2$. Coefficients depend continuously on $\\varepsilon$ and are\ndominated by those of the entire $f_{3/4}$. Dominated convergence of the Taylor\nseries therefore proves uniform convergence on compact sets as\n$\\varepsilon\\downarrow0$.","This gives a useful design distinction: finite-window similarity does not\ncontrol entire growth order. For a bank with weights $1/4,3/4$ at these scales,\n\\[\nF(z)=z+\\frac7{48}z^3+\\frac{427}{30720}z^5+\n\\frac{176681}{188743680}z^7+O(z^9).\n\\]","\\MonographPart{II}{Arithmetic depth and signed measures}{Complex parameters, positive kernels and the boundaries of positivity.}\n\\chapter{Arithmetic at complex depth}\\label{ch:arithmetic}\n\\pageheading{FAMILY 2}{Arithmetic energy at complex depth}\nFor real $a>1/2$, $\\beta>0$, and integer $Q\\ge1$, define\n\\[\n\\boxed{\\A(s)=\\sum_{\\substack{n\\ge2","\\MonographPart{II}{Arithmetic depth and signed measures}{Complex parameters, positive kernels and the boundaries of positivity.}\n\\chapter{Arithmetic at complex depth}\\label{ch:arithmetic}\n\\pageheading{FAMILY 2}{Arithmetic energy at complex depth}\nFor real $a>1/2$, $\\beta>0$, and integer $Q\\ge1$, define\n\\[\n\\boxed{\\A(s)=\\sum_{\\substack{n\\ge2\\\\(n,Q)=1}}\n\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s},\\qquad s\\in\\mathbb C.}\n\\]\nThe complex power uses the real logarithm of $1-n^{-\\beta}>0$.\nThe atom $n=1$ is excluded because its logarithm is undefined. At positive\ninteger depths its omitted contribution would be zero. At depth zero the\ncorresponding mass-one atom is absent. In particular, $A(0)=Z_Q(2a)-1$, where\n\\[\nZ_Q(v)=\\sum_{\\substack{n\\ge1","(n,Q)=1}}\n\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s},\\qquad s\\in\\mathbb C.}\n\\]\nThe complex power uses the real logarithm of $1-n^{-\\beta}>0$.\nThe atom $n=1$ is excluded because its logarithm is undefined. At positive\ninteger depths its omitted contribution would be zero. At depth zero the\ncorresponding mass-one atom is absent. In particular, $A(0)=Z_Q(2a)-1$, where\n\\[\nZ_Q(v)=\\sum_{\\substack{n\\ge1","(n,Q)=1}}\n\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s},\\qquad s\\in\\mathbb C.}\n\\]\nThe complex power uses the real logarithm of $1-n^{-\\beta}>0$.\nThe atom $n=1$ is excluded because its logarithm is undefined. At positive\ninteger depths its omitted contribution would be zero. At depth zero the\ncorresponding mass-one atom is absent. In particular, $A(0)=Z_Q(2a)-1$, where\n\\[\nZ_Q(v)=\\sum_{\\substack{n\\ge1\\\\(n,Q)=1}}\\mu(n)^2n^{-v}\n=\\frac{\\zeta(v)}{\\zeta(2v)}\\prod_{p\\mid Q}(1+p^{-v})^{-1},\\qquad v>1.\n\\]\n\\fig{03_complex_depth}{Left: $a=\\beta=2$ at real depths, using the exact\nzeta--Newton representation at 90-digit working precision. Right: initial\nsquarefree frequencies and weights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qqua…","(n,Q)=1}}\\mu(n)^2n^{-v}\n=\\frac{\\zeta(v)}{\\zeta(2v)}\\prod_{p\\mid Q}(1+p^{-v})^{-1},\\qquad v>1.\n\\]\n\\fig{03_complex_depth}{Left: $a=\\beta=2$ at real depths, using the exact\nzeta--Newton representation at 90-digit working precision. Right: initial\nsquarefree frequencies and weights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The fir","(n,Q)=1}}\\mu(n)^2n^{-v}\n=\\frac{\\zeta(v)}{\\zeta(2v)}\\prod_{p\\mid Q}(1+p^{-v})^{-1},\\qquad v>1.\n\\]\n\\fig{03_complex_depth}{Left: $a=\\beta=2$ at real depths, using the exact\nzeta--Newton representation at 90-digit working precision. Right: initial\nsquarefree frequencies and weights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","{-v}\n=\\frac{\\zeta(v)}{\\zeta(2v)}\\prod_{p\\mid Q}(1+p^{-v})^{-1},\\qquad v>1.\n\\]\n\\fig{03_complex_depth}{Left: $a=\\beta=2$ at real depths, using the exact\nzeta--Newton representation at 90-digit working precision. Right: initial\nsquarefree frequencies and weights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","\\mid Q}(1+p^{-v})^{-1},\\qquad v>1.\n\\]\n\\fig{03_complex_depth}{Left: $a=\\beta=2$ at real depths, using the exact\nzeta--Newton representation at 90-digit working precision. Right: initial\nsquarefree frequencies and weights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","ion. Right: initial\nsquarefree frequencies and weights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","cies and weights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","ights for $Q=1$. Only finitely many atoms are\nshown; the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","the proof uses the infinite support.}\nLet $\\mathcal S_Q=\\{n\\ge2:\\mu(n)^2=1,\\ (n,Q)=1\\}$ and\n$\\lambda_n=-2\\log(1-n^{-\\beta})$. Then\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","n\n\\[\nA(s)=\\sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","sum_{n\\in\\mathcal S_Q}w_ne^{-\\lambda_ns},\\qquad w_n=n^{-2a}.\n\\]\nThe positive weights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","eights are summable, and the distinct support frequencies decrease to zero.\nIf $p_*$ is the least prime not dividing $Q$, then\n\\[\n\\tau=-2\\log(1-p_*^{-\\beta}),\\qquad M=\\Z(2a)-1,\n\\]\n\\[\n|A(s)|\\le M e^{\\tau\\max(-\\Re s,0)}.\n\\]\nThis proves local uniform convergence and entire continuation. The first atom\ndominates $A(-t)$ as $t\\to\\infty$, so the entire function has exact order one\nand exponential type $\\tau$.","\\pageheading{FAMILY 2\\; /\\; CALCULUS}{Closed shifts, infinite differential complexity}\nThe Euler product for $Z_Q$ shows that only\n$\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ matters. Expansion of the depth factor gives\n\\[\n\\boxed{\\A(s+1)=\\A(s)-2A_{a+\\beta/2,\\beta,Q}(s)\n+A_{a+\\beta,\\beta,Q}(s).}\n\\]\nThe fractional transfer and Newton formulas are\n\\[\n\\A(s+t)=\\sum_{j\\ge0}(-1)^j\\binom{2t}{j}A_{a+j\\beta/2,\\beta,Q}(s),\n\\]\n\\[\n\\A(t)=\\sum_{j\\ge0}(-1)^j\\binom{2t}{j}[\\Z(2a+j\\beta)-1].\n\\]\nThey converge absolutely and locally uniformly: binomial coefficients have\nat most polynomial growth on compact parameter sets, while the shifted sums\ndecay geometrically. The subtraction of one is essential at noninteger depth.","Y 2\\; /\\; CALCULUS}{Closed shifts, infinite differential complexity}\nThe Euler product for $Z_Q$ shows that only\n$\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ matters. Expansion of the depth factor gives\n\\[\n\\boxed{\\A(s+1)=\\A(s)-2A_{a+\\beta/2,\\beta,Q}(s)\n+A_{a+\\beta,\\beta,Q}(s).}\n\\]\nThe fractional transfer and Newton formulas are\n\\[\n\\A(s+t)=\\sum_{j\\ge0}(-1)^j\\binom{2t}{j}A_{a+j\\beta/2,\\beta,Q}(s),\n\\]\n\\[\n\\A(t)=\\sum_{j\\ge0}(-1)^j\\binom{2t}{j}[\\Z(2a+j\\beta)-1].\n\\]\nThey converge absolutely and locally uniformly: binomial coefficients have\nat most polynomial growth on compact parameter sets, while the shifted sums\ndecay geometrically. The subtraction of one is essential at noninteger depth.","\\claim{No polynomial-coefficient linear ODE}\nSuppose $\\sum_{j=0}^{r}P_j(s)A^{(j)}(s)=0$, where the $P_j$ are polynomials and are not all zero.\nAt $s=-t$, divide by the exponential of the first support atom. The remaining\ntail is exponentially smaller, so the polynomial\n$\\sum_jP_j(-t)(-\\lambda_{p_*})^j$ tends to zero exponentially and must vanish\nidentically. Remove that atom and repeat. Every support frequency therefore\nsatisfies $\\sum_jP_j(s)(-\\lambda_n)^j=0$ for all $s$.\nFor fixed $s$, this polynomial in $\\lambda$ has infinitely many distinct\nroots. All its coefficients vanish. Thus all $P_j$ are zero, a contradiction.","near ODE}\nSuppose $\\sum_{j=0}^{r}P_j(s)A^{(j)}(s)=0$, where the $P_j$ are polynomials and are not all zero.\nAt $s=-t$, divide by the exponential of the first support atom. The remaining\ntail is exponentially smaller, so the polynomial\n$\\sum_jP_j(-t)(-\\lambda_{p_*})^j$ tends to zero exponentially and must vanish\nidentically. Remove that atom and repeat. Every support frequency therefore\nsatisfies $\\sum_jP_j(s)(-\\lambda_n)^j=0$ for all $s$.\nFor fixed $s$, this polynomial in $\\lambda$ has infinitely many distinct\nroots. All its coefficients vanish. Thus all $P_j$ are zero, a contradiction.","A^{(j)}(s)=0$, where the $P_j$ are polynomials and are not all zero.\nAt $s=-t$, divide by the exponential of the first support atom. The remaining\ntail is exponentially smaller, so the polynomial\n$\\sum_jP_j(-t)(-\\lambda_{p_*})^j$ tends to zero exponentially and must vanish\nidentically. Remove that atom and repeat. Every support frequency therefore\nsatisfies $\\sum_jP_j(s)(-\\lambda_n)^j=0$ for all $s$.\nFor fixed $s$, this polynomial in $\\lambda$ has infinitely many distinct\nroots. All its coefficients vanish. Thus all $P_j$ are zero, a contradiction.","where the $P_j$ are polynomials and are not all zero.\nAt $s=-t$, divide by the exponential of the first support atom. The remaining\ntail is exponentially smaller, so the polynomial\n$\\sum_jP_j(-t)(-\\lambda_{p_*})^j$ tends to zero exponentially and must vanish\nidentically. Remove that atom and repeat. Every support frequency therefore\nsatisfies $\\sum_jP_j(s)(-\\lambda_n)^j=0$ for all $s$.\nFor fixed $s$, this polynomial in $\\lambda$ has infinitely many distinct\nroots. All its coefficients vanish. Thus all $P_j$ are zero, a contradiction.","nomials and are not all zero.\nAt $s=-t$, divide by the exponential of the first support atom. The remaining\ntail is exponentially smaller, so the polynomial\n$\\sum_jP_j(-t)(-\\lambda_{p_*})^j$ tends to zero exponentially and must vanish\nidentically. Remove that atom and repeat. Every support frequency therefore\nsatisfies $\\sum_jP_j(s)(-\\lambda_n)^j=0$ for all $s$.\nFor fixed $s$, this polynomial in $\\lambda$ has infinitely many distinct\nroots. All its coefficients vanish. Thus all $P_j$ are zero, a contradiction.","ort atom. The remaining\ntail is exponentially smaller, so the polynomial\n$\\sum_jP_j(-t)(-\\lambda_{p_*})^j$ tends to zero exponentially and must vanish\nidentically. Remove that atom and repeat. Every support frequency therefore\nsatisfies $\\sum_jP_j(s)(-\\lambda_n)^j=0$ for all $s$.\nFor fixed $s$, this polynomial in $\\lambda$ has infinitely many distinct\nroots. All its coefficients vanish. Thus all $P_j$ are zero, a contradiction.","\\claim{Strict positive complex kernel}\nFor distinct complex points $s_i$, $K_{ij}=A(s_i+\\overline{s_j})$ is strictly\npositive definite. A zero quadratic form would force a finite exponential\nsum to vanish at every $\\lambda_n$. Their accumulation at zero invokes the\nidentity theorem; derivatives at zero then give a nonsingular Vandermonde\nsystem. On the real axis, $(-1)^jA^{(j)}(s)>0$ for every $j\\ge0$.","Let\n\\[\n\\alpha=\\frac{2a-1}{\\beta},\\qquad\n\\delta_Q=\\frac1{\\zeta(2)}\\prod_{p\\mid Q}\\frac{p}{p+1}.\n\\]\nAs $|s|\\to\\infty$ in any closed sector\n$|\\arg s|\\le\\pi/2-\\epsilon$ with $0<\\epsilon<\\pi/2$,\n\\[\n\\boxed{\\A(s)=\\frac{\\delta_Q}{\\beta}B(\\alpha,2s+1)\n+O_{a,\\beta,Q,\\epsilon}\\left(|s|^{-(2a-1/2)/\\beta}\\right).}\n\\]\nIn particular,\n\\[\n\\A(s)\\sim\\frac{\\delta_Q\\Gamma(\\alpha)}{\\beta}(2s)^{-\\alpha}.\n\\]\nThe principal power is used in the right half-plane. The beta expression is\nan asymptotic continuous-density model, not an identity with the arithmetic sum.","\\claim{Proof mechanism}\nSquarefree inclusion--exclusion gives\n\\[\n\\sum_{\\substack{n\\le x\\\\(n,Q)=1}}\\mu(n)^2\n=\\delta_Q x+O_Q(\\sqrt{x}).\n\\]\nIntegrate against $g_s(x)=x^{-2a}(1-x^{-\\beta})^{2s}$ on $[1,\\infty)$.\nThe density term becomes the beta integral after $y=x^{-\\beta}$. Partial\nintegration bounds the error by\n$C_Q\\int_1^\\infty\\sqrt{x}|g_s'(x)|\\dd x$.\nThe two derivative terms are beta integrals with decay exponent\n$(2a-1/2)/\\beta$; $|s|/\\Re s$ stays bounded in the stated sector.\nThis gives the error term and then the leading asymptotic.","(n,Q)=1}}\\mu(n)^2\n=\\delta_Q x+O_Q(\\sqrt{x}).\n\\]\nIntegrate against $g_s(x)=x^{-2a}(1-x^{-\\beta})^{2s}$ on $[1,\\infty)$.\nThe density term becomes the beta integral after $y=x^{-\\beta}$. Partial\nintegration bounds the error by\n$C_Q\\int_1^\\infty\\sqrt{x}|g_s'(x)|\\dd x$.\nThe two derivative terms are beta integrals with decay exponent\n$(2a-1/2)/\\beta$; $|s|/\\Re s$ stays bounded in the stated sector.\nThis gives the error term and then the leading asymptotic.","\\claim{One complex example}\nAt $a=\\beta=2$, $Q=1$,\n\\[\nA(1/2+i/4)\\approx\n0.06033183289343401791658\n-0.007412384053785436251981\\,i.\n\\]\nDirect finite sums and a zeta--Newton evaluation agree within the recorded\nomitted-tail envelope. This is an ordinary high-precision example, not an\ninterval certificate.","\\chapter{Prime sieves and signed measures}\\label{ch:sieves}\n\\pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lam","\\chapter{Prime sieves and signed measures}\\label{ch:sieves}\n\\pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\ri","\\chapter{Prime sieves and signed measures}\\label{ch:sieves}\n\\pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac","\\chapter{Prime sieves and signed measures}\\label{ch:sieves}\n\\pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\f","\\chapter{Prime sieves and signed measures}\\label{ch:sieves}\n\\pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infini","\\chapter{Prime sieves and signed measures}\\label{ch:sieves}\n\\pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many","d measures}\\label{ch:sieves}\n\\pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix s…","pageheading{FAMILY 3}{An infinite prime shelf changes the packets}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite thro…","}\nFor a prime $p$ and $\\lambda\\ge-1$, let\n\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnor…","2})^2,\\qquad\n\\nu_{p,\\lambda}=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","^{-4}\n(1+\\lambda\\mathbf1_{p\\mid n})\\delta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","lta_{x_n}.\n\\]\nDefine monic orthogonal polynomials $P_m^{(k;p,\\lambda)}$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","$ for the measure\n$x^k\\dd\\nu_{p,\\lambda}$, $k\\ge0$ integer. The zero atom has mass one at\n$k=0$ and is annihilated when $k\\ge1$.\n\\fig{04_prime_polynomials}{Left: degree-two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","two packets for $p=2$, depth zero.\nRight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","ight: initial atom masses. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","s. Strength $-1$ deletes every even squarefree atom;\nstrength $0$ is the original measure; strength $1$ doubles that shelf.}\nThe exact moments are finite zeta stencils:\n\\[\nM_j^{p,\\lambda}=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\left(1+\\frac{\\lambda}{p^{4+2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","2r}+1}\\right).\n\\]\nAll polynomial coefficients belong to $\\mathbb Q(\\pi^{-2},\\lambda)$.\nWith $\\Delta=M_0M_2-M_1^2$,\n\\[\nP_1=x-\\frac{M_1}{M_0},\\qquad\nP_2=x^2-\\frac{M_0M_3-M_1M_2}{\\Delta}x+\n\\frac{M_1M_3-M_2^2}{\\Delta}.\n\\]\nThe infinitely many positive nodes make every Hankel matrix strictly positive\ndefinite throughout $\\lambda\\ge-1$. Hence all degrees exist, with positive\nnorms, real simple zeros in $(0,1)$, and interlacing.","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pe","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pa","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\m","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setm","\\pageheading{FAMILY 3\\; /\\; BOUNDARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","ARY}{Finite positivity can conceal a signed defect}\nLet $\\nu_{\\mathrm{non}p}$ and $\\nu_p$ be the restrictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","strictions of the\nundeformed measure $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","re $\\nu_{p,0}$ to indices not divisible by $p$ and divisible\nby $p$, respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","respectively. Then\n$\\nu_{p,\\lambda}=\\nu_{\\mathrm{non}p}+(1+\\lambda)\\nu_p$.\nLet $A_m,B_m$ be the order-$m$ Hankel matrices of these two unscaled\npositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","ositive measures. Both are positive definite.\nThus\n\\[\n\\tau_m(\\lambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","ambda)=\\det(A_m+(1+\\lambda)B_m)\n=\\det A_m\\prod_{i=1}^{m}[1+(1+\\lambda)\\eta_i],\n\\]\nwhere $\\eta_i>0$ are the eigenvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","nvalues of $A_m^{-1/2}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","}B_mA_m^{-1/2}$.\nEvery determinant root is real and below $-1$; possible coefficient poles\noccur there, although numerator cancellation is possible.\n\\fig{05_sieve_boundary}{The line is a proved bound for $p=2$; dots are\nordinary high-precision pencil roots. A defect $\\lambda=-1.001$ eventually\nviolates positivity even though it can pass low-order checks.}\nThe nearest root is $-1-\\delta_m$, with\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}\\\\\\deg f<m}}\n\\frac{\\int f^2\\dd\\nu_{\\mathrm{non}p}}{\\int f^2\\dd\\nu_p}>0,\n\\qquad\\delta_m\\downarrow0.\n\\]\nTo prove the limit, approximate a continuous bump supported at the isolated\nnode $x_p$ by polynomials on $[0,1]$. The numerator tends to zero while the\ndenominator retains the $p^{-4}$ atom. Hence every $\\lambda<-1$ eventually\nproduces a negative finite-degree quadratic form.","deg f<m}}\n\\frac{\\int f^2\\dd\\nu_{\\mathrm{non}p}}{\\int f^2\\dd\\nu_p}>0,\n\\qquad\\delta_m\\downarrow0.\n\\]\nTo prove the limit, approximate a continuous bump supported at the isolated\nnode $x_p$ by polynomials on $[0,1]$. The numerator tends to zero while the\ndenominator retains the $p^{-4}$ atom. Hence every $\\lambda<-1$ eventually\nproduces a negative finite-degree quadratic form.","For $p=2$, test with $f(x)=x(1-x)^{m-2}$, $m\\ge2$:\n\\[\n\\delta_m\\le\\frac{4096}{81}\\left(\\frac{16M_0}{17}-1\\right)\n\\left(\\frac{272}{567}\\right)^{2(m-2)},\\qquad M_0=105\\pi^{-4}.\n\\]\nThis follows from $x_2=9/16$ and $x_n\\ge64/81$ at nonzero odd nodes.\nExactly $\\delta_1=p^4$; the limit is an infinite-resolution phenomenon.","\\pageheading{FAMILY 3\\; /\\; TRANSPORT}{Classical operations with arithmetic coordinates}\nThe depth tilt $x^k$ is a Christoffel transformation. Suppressing $p,\\lambda$,\n\\[\nP_m^{(k+1)}(x)=\\frac{P_{m+1}^{(k)}(x)\n-[P_{m+1}^{(k)}(0)/P_m^{(k)}(0)]P_m^{(k)}(x)}{x}.\n\\]\nThe numerator vanishes at zero. All roots are positive, so the displayed\ndenominator coefficient is nonzero. For shifted Hankel determinants $\\tau_m^{(k)}=\\det[M_{k+i+j}]_{i,j<m}$,\nwith $\\tau_0^{(k)}=1$,\n\\[\n\\tau_m^{(k)}\\tau_m^{(k+2)}-(\\tau_m^{(k+1)})^2\n=\\tau_{m+1}^{(k)}\\tau_{m-1}^{(k+2)}>0,\\qquad m\\ge1.\n\\]\nThis is the Desnanot--Jacobi identity applied to exact arithmetic moments.\nThe construction mechanism is classical.","eheading{FAMILY 3\\; /\\; TRANSPORT}{Classical operations with arithmetic coordinates}\nThe depth tilt $x^k$ is a Christoffel transformation. Suppressing $p,\\lambda$,\n\\[\nP_m^{(k+1)}(x)=\\frac{P_{m+1}^{(k)}(x)\n-[P_{m+1}^{(k)}(0)/P_m^{(k)}(0)]P_m^{(k)}(x)}{x}.\n\\]\nThe numerator vanishes at zero. All roots are positive, so the displayed\ndenominator coefficient is nonzero. For shifted Hankel determinants $\\tau_m^{(k)}=\\det[M_{k+i+j}]_{i,j<m}$,\nwith $\\tau_0^{(k)}=1$,\n\\[\n\\tau_m^{(k)}\\tau_m^{(k+2)}-(\\tau_m^{(k+1)})^2\n=\\tau_{m+1}^{(k)}\\tau_{m-1}^{(k+2)}>0,\\qquad m\\ge1.\n\\]\nThis is the Desnanot--Jacobi identity applied to exact arithmetic moments.\nThe construction mechanism is classical.","operations with arithmetic coordinates}\nThe depth tilt $x^k$ is a Christoffel transformation. Suppressing $p,\\lambda$,\n\\[\nP_m^{(k+1)}(x)=\\frac{P_{m+1}^{(k)}(x)\n-[P_{m+1}^{(k)}(0)/P_m^{(k)}(0)]P_m^{(k)}(x)}{x}.\n\\]\nThe numerator vanishes at zero. All roots are positive, so the displayed\ndenominator coefficient is nonzero. For shifted Hankel determinants $\\tau_m^{(k)}=\\det[M_{k+i+j}]_{i,j<m}$,\nwith $\\tau_0^{(k)}=1$,\n\\[\n\\tau_m^{(k)}\\tau_m^{(k+2)}-(\\tau_m^{(k+1)})^2\n=\\tau_{m+1}^{(k)}\\tau_{m-1}^{(k+2)}>0,\\qquad m\\ge1.\n\\]\nThis is the Desnanot--Jacobi identity applied to exact arithmetic moments.\nThe construction mechanism is classical.","thmetic coordinates}\nThe depth tilt $x^k$ is a Christoffel transformation. Suppressing $p,\\lambda$,\n\\[\nP_m^{(k+1)}(x)=\\frac{P_{m+1}^{(k)}(x)\n-[P_{m+1}^{(k)}(0)/P_m^{(k)}(0)]P_m^{(k)}(x)}{x}.\n\\]\nThe numerator vanishes at zero. All roots are positive, so the displayed\ndenominator coefficient is nonzero. For shifted Hankel determinants $\\tau_m^{(k)}=\\det[M_{k+i+j}]_{i,j<m}$,\nwith $\\tau_0^{(k)}=1$,\n\\[\n\\tau_m^{(k)}\\tau_m^{(k+2)}-(\\tau_m^{(k+1)})^2\n=\\tau_{m+1}^{(k)}\\tau_{m-1}^{(k+2)}>0,\\qquad m\\ge1.\n\\]\nThis is the Desnanot--Jacobi identity applied to exact arithmetic moments.\nThe construction mechanism is classical.","\\claim{An endpoint mass is numerically delicate}\nFor the undeformed depth-zero measure, set $\\nu_\\theta=\\nu+\\theta\\delta_0$,\n$\\theta\\ge-1$. Write $h_j=\\int P_j(x)^2\\dd\\nu(x)$ and\n$K_m(x,y)=\\sum_{j=0}^{m}P_j(x)P_j(y)/h_j$.\nThe Uvarov formula~\\cite{Uvarov1969} is\n\\[\nP_m^{[\\theta]}(x)=P_m(x)-\n\\frac{\\theta P_m(0)}{1+\\theta K_{m-1}(0,0)}K_{m-1}(x,0),\n\\]\nwith $K_{-1}=0$ for degree zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains t","\\claim{An endpoint mass is numerically delicate}\nFor the undeformed depth-zero measure, set $\\nu_\\theta=\\nu+\\theta\\delta_0$,\n$\\theta\\ge-1$. Write $h_j=\\int P_j(x)^2\\dd\\nu(x)$ and\n$K_m(x,y)=\\sum_{j=0}^{m}P_j(x)P_j(y)/h_j$.\nThe Uvarov formula~\\cite{Uvarov1969} is\n\\[\nP_m^{[\\theta]}(x)=P_m(x)-\n\\frac{\\theta P_m(0)}{1+\\theta K_{m-1}(0,0)}K_{m-1}(x,0),\n\\]\nwith $K_{-1}=0$ for degree zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains the conditioning","\\claim{An endpoint mass is numerically delicate}\nFor the undeformed depth-zero measure, set $\\nu_\\theta=\\nu+\\theta\\delta_0$,\n$\\theta\\ge-1$. Write $h_j=\\int P_j(x)^2\\dd\\nu(x)$ and\n$K_m(x,y)=\\sum_{j=0}^{m}P_j(x)P_j(y)/h_j$.\nThe Uvarov formula~\\cite{Uvarov1969} is\n\\[\nP_m^{[\\theta]}(x)=P_m(x)-\n\\frac{\\theta P_m(0)}{1+\\theta K_{m-1}(0,0)}K_{m-1}(x,0),\n\\]\nwith $K_{-1}=0$ for degree zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains the conditioning problem.","mass is numerically delicate}\nFor the undeformed depth-zero measure, set $\\nu_\\theta=\\nu+\\theta\\delta_0$,\n$\\theta\\ge-1$. Write $h_j=\\int P_j(x)^2\\dd\\nu(x)$ and\n$K_m(x,y)=\\sum_{j=0}^{m}P_j(x)P_j(y)/h_j$.\nThe Uvarov formula~\\cite{Uvarov1969} is\n\\[\nP_m^{[\\theta]}(x)=P_m(x)-\n\\frac{\\theta P_m(0)}{1+\\theta K_{m-1}(0,0)}K_{m-1}(x,0),\n\\]\nwith $K_{-1}=0$ for degree zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains the conditioning problem.","=\\int P_j(x)^2\\dd\\nu(x)$ and\n$K_m(x,y)=\\sum_{j=0}^{m}P_j(x)P_j(y)/h_j$.\nThe Uvarov formula~\\cite{Uvarov1969} is\n\\[\nP_m^{[\\theta]}(x)=P_m(x)-\n\\frac{\\theta P_m(0)}{1+\\theta K_{m-1}(0,0)}K_{m-1}(x,0),\n\\]\nwith $K_{-1}=0$ for degree zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains the conditioning problem.","_j$.\nThe Uvarov formula~\\cite{Uvarov1969} is\n\\[\nP_m^{[\\theta]}(x)=P_m(x)-\n\\frac{\\theta P_m(0)}{1+\\theta K_{m-1}(0,0)}K_{m-1}(x,0),\n\\]\nwith $K_{-1}=0$ for degree zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains the conditioning problem.","cite{Uvarov1969} is\n\\[\nP_m^{[\\theta]}(x)=P_m(x)-\n\\frac{\\theta P_m(0)}{1+\\theta K_{m-1}(0,0)}K_{m-1}(x,0),\n\\]\nwith $K_{-1}=0$ for degree zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains the conditioning problem.","ee zero. The mass at zero is recovered by\n\\[\n1\\le\\frac1{K_m(0,0)}\\le1+(M_0-1)(7/16)^{2m}.\n\\]\nThe lower bound comes from the zero atom, whose mass is one. The test polynomial\n$(1-x)^m$ gives the upper bound since every other node is at least $9/16$.\nAt finite degree $K_m(0,0)<1$, because no nonzero polynomial can vanish at\nall the other infinitely many nodes. Complete mass removal is valid, while\n$1-K_m(0,0)$ becomes very small. This explains the conditioning problem.","\\claim{The resolvent is non-D-finite}\nFor $\\lambda\\ge-1$, the Stieltjes function $S_{p,\\lambda}(z)=\\int(z-x)^{-1}\\dd\\nu_{p,\\lambda}(x)$\nhas infinitely many isolated poles at the positive-weight nodes $x_n\\to1$.\nA solution of a finite rational-coefficient linear ODE has only finitely many\npossible finite singular points. The germ at $z=2$ therefore is not D-finite.\nThis is a general infinite-atomic-resolvent mechanism, specialized here to\narithmetic data.","\\MonographPart{III}{Orbit spectra and exact coordinates}{Spectral continuation and the coefficient language of polyphase transforms.}\n\\chapter{Spectral functions of zero orbits}\\label{ch:spectra}\n\\pageheading{FAMILY 4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting","\\MonographPart{III}{Orbit spectra and exact coordinates}{Spectral continuation and the coefficient language of polyphase transforms.}\n\\chapter{Spectral functions of zero orbits}\\label{ch:spectra}\n\\pageheading{FAMILY 4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1…","act coordinates}{Spectral continuation and the coefficient language of polyphase transforms.}\n\\chapter{Spectral functions of zero orbits}\\label{ch:spectra}\n\\pageheading{FAMILY 4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus …","inuation and the coefficient language of polyphase transforms.}\n\\chapter{Spectral functions of zero orbits}\\label{ch:spectra}\n\\pageheading{FAMILY 4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_…","ctral functions of zero orbits}\\label{ch:spectra}\n\\pageheading{FAMILY 4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(…","ns of zero orbits}\\label{ch:spectra}\n\\pageheading{FAMILY 4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\g…","bits}\\label{ch:spectra}\n\\pageheading{FAMILY 4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z…","4}{One spectral address for each rotation orbit}\nFor integer $N\\ge2$, define the classical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=…","ical Mittag--Leffler seed~\\cite[\\S10.46]{DLMF}\n\\[\nd_N(w)=\\sum_{n\\ge0}\\frac{(-1)^nw^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive…","w^n}{(Nn)!}=E_{N,1}(-w),\n\\qquad D_N(z)=d_N(z^N).\n\\]\nLet $w_j$ be the nonzero zeros of $d_N$, counted with multiplicity. Each\ncorresponds to one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","one rotation orbit of $N$ zeros of $D_N$. Using\n$\\operatorname{Arg}w\\in(-\\pi,\\pi]$, define\n\\[\n\\boxed{\\mathcal Z_N(s)=\\sum_jw_j^{-s},\\qquad\\Re s>1/N.}\n\\]\n\\fig{06_orbit_zeros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","ros}{For $N=4$, the four real/imaginary zeros in each orbit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","bit\nshare one positive coordinate $w=z^4$. The plot shows the first six orbits.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","ts.\nMultiplicity is retained by the definition; simplicity is not assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","assumed in general.}\nThe bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","he bound $|d_N(w)|\\le e^{|w|^{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","{1/N}}$ and the finite-order zero-counting theorem\ngive absolute local uniform convergence in the stated half-plane. Since the\norder is below one and $d_N(0)=1$, the Hadamard factorization has genus zero:\n\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\]\nAt positive integers, a log derivative gives\n\\[\n-w\\frac{d_N'(w)}{d_N(w)}=\\sum_{k\\ge1}\\mathcal Z_N(k)w^k.\n\\]\nIf $a_n=(-1)^n/(Nn)!$ and $b_k=\\mathcal Z_N(k)$, then\n\\[\nb_k=-ka_k-\\sum_{j=1}^{k-1}a_jb_{k-j}.\n\\]\nThus every positive integer value is rational without numerical zero finding.","\\pageheading{FAMILY 4\\; /\\; REDUCTIONS}{Classical anchors and the open spectral frontier}\nThe first values are\n\\[\n\\begin{aligned}\n\\mathcal Z_N(1)&=\\frac1{N!},\\\\\n\\mathcal Z_N(2)&=\\frac1{(N!)^2}-\\frac2{(2N)!},\\\\\n\\mathcal Z_N(3)&=\\frac1{(N!)^3}-\\frac3{N!(2N)!}+\\frac3{(3N)!}.\n\\end{aligned}\n\\]\n\\begin{tabular}{@{}llll@{}}\\toprule\n$N$ & $\\mathcal Z_N(1)$ & $\\mathcal Z_N(2)$ & $\\mathcal Z_N(3)$","\\mathcal Z_N(3)&=\\frac1{(N!)^3}-\\frac3{N!(2N)!}+\\frac3{(3N)!}.\n\\end{aligned}\n\\]\n\\begin{tabular}{@{}llll@{}}\\toprule\n$N$ & $\\mathcal Z_N(1)$ & $\\mathcal Z_N(2)$ & $\\mathcal Z_N(3)$","midrule\n2 & $1/2$ & $1/6$ & $1/15$","3 & $1/6$ & $1/40$ & $53/13440$","4 & $1/24$ & $17/10080$ & $691/9979200$","5 & $1/120$ & $5/72576$ & $997/1743565824$","\\claim{Two exact Hurwitz-zeta reductions}\nFor $N=2$, $d_2(w)=\\cos\\sqrt w$, whose zeros are\n$w_j=\\pi^2(j+1/2)^2$, $j\\ge0$. Therefore\n\\[\n\\mathcal Z_2(s)=\\pi^{-2s}\\zeta(2s,1/2).\n\\]\nFor $N=4$,\n\\[\nD_4(z)=\\cos(z/\\sqrt2)\\cosh(z/\\sqrt2),\n\\]\nand the rotation quotient has one zero\n$w_j=(\\sqrt2\\pi)^4(j+1/2)^4$ per orbit. Hence\n\\[\n\\mathcal Z_4(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2).\n\\]\nThese anchors are classical reductions. The general spectral-zeta mechanism\nalso has substantial antecedents, including hypergeometric zero zeta functions.","\\FrontChapter{Notation and glossary}\\label{glossary}\nNotation is defined locally when each family is introduced. The same letter\ncan serve different roles in separate chapters. In particular, $a$ is an\narithmetic weight parameter in Chapter~\\ref{ch:arithmetic}, a contraction in\nChapter~\\ref{ch:coupled}, and a spectral shift in Appendix~\\ref{app:spectral}.","\\item[Contracted feedback $T_{q,r}$.] The normalized odd entire solution of\n$T'_{q,r}(z)=1+T_{q,r}(qz)T_{q,r}(rz)$, $T_{q,r}(0)=0$, for $0<q,r<1$.\nThe diagonal notation is $f_q=T_{q,q}$.\n\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Thei","\\item[Contracted feedback $T_{q,r}$.] The normalized odd entire solution of\n$T'_{q,r}(z)=1+T_{q,r}(qz)T_{q,r}(rz)$, $T_{q,r}(0)=0$, for $0<q,r<1$.\nThe diagonal notation is $f_q=T_{q,q}$.\n\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire s","\\item[Contracted feedback $T_{q,r}$.] The normalized odd entire solution of\n$T'_{q,r}(z)=1+T_{q,r}(qz)T_{q,r}(rz)$, $T_{q,r}(0)=0$, for $0<q,r<1$.\nThe diagonal notation is $f_q=T_{q,q}$.\n\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1","\\item[Contracted feedback $T_{q,r}$.] The normalized odd entire solution of\n$T'_{q,r}(z)=1+T_{q,r}(qz)T_{q,r}(rz)$, $T_{q,r}(0)=0$, for $0<q,r<1$.\nThe diagonal notation is $f_q=T_{q,q}$.\n\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $","\\item[Contracted feedback $T_{q,r}$.] The normalized odd entire solution of\n$T'_{q,r}(z)=1+T_{q,r}(qz)T_{q,r}(rz)$, $T_{q,r}(0)=0$, for $0<q,r<1$.\nThe diagonal notation is $f_q=T_{q,q}$.\n\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith","\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith $q,r,a,b\\in(0,1)$.\n\\item[Perron","\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith $q,r,a,b\\in(0,1)$.\n\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spec","\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith $q,r,a,b\\in(0,1)$.\n\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spectral radius of","\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith $q,r,a,b\\in(0,1)$.\n\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spectral radius of\na matrix. The char","\\item[Feedback bank $F_\\Theta$.] A finite collection of channels with\n$h_j\\ge0$, $\\sum_jh_j=1$ and $0<q_j,r_j<1$, entering\n$F_\\Theta'=1+\\sum_jh_jF_\\Theta(q_jz)F_\\Theta(r_jz)$.\nZero-weight channels are removed before taking an active-channel maximum.\n\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith $q,r,a,b\\in(0,1)$.\n\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spectral radius of\na matrix. The characteristic exponent satisfies\n$\\operatorname{spr}\\begin{pmatri","\\item[Maximum modulus and entire order.] $M_f(R)=\\max_{|z|=R}|f(z)|$.\nFor a nonpolynomial entire function,\n\\[\n\\rho(f)=\\limsup_{R\\to\\infty}\\frac{\\log\\log M_f(R)}{\\log R}.\n\\]\nAt positive finite order $\\rho$, the lower and upper types used here are the\nliminf and limsup of $\\log M_f(R)/R^\\rho$. Their equality is not assumed.\n\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith $q,r,a,b\\in(0,1)$.\n\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spectral radius of\na matrix. The characteristic exponent satisfies\n$\\operatorname{spr}\\begin{pmatrix}q^\\rho&r^\\rho","\\item[Coupled pair $(U,V)$.] The normalized entire solution of\n$U'=1+U(qz)V(rz)$, $V'=1+U(az)V(bz)$ and $U(0)=V(0)=0$,\nwith $q,r,a,b\\in(0,1)$.\n\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spectral radius of\na matrix. The characteristic exponent satisfies\n$\\operatorname{spr}\\begin{pmatrix}q^\\rho&r^\\rho","\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spectral radius of\na matrix. The characteristic exponent satisfies\n$\\operatorname{spr}\\begin{pmatrix}q^\\rho&r^\\rho","\\item[Perron balance.] $\\operatorname{spr}P$ denotes the spectral radius of\na matrix. The characteristic exponent satisfies\n$\\operatorname{spr}\\begin{pmatrix}q^\\rho&r^\\rho\\\\a^\\rho&b^\\rho\\end{pmatrix}=1$.\nA positive right eigenvector gives the proportions of the two type vectors.\n\\item[Arithmetic depth $A_{a,\\beta,Q}$.] The entire sum\n$\\sum_{n\\ge2,\\ (n,Q)=1}\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s}$, with\n$a>1/2$, $\\beta>0$, $Q\\ge1$ integer and the real logarithm in the complex power.\nThe atom $n=1$ is excluded at every depth.\n\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^","\\item[Arithmetic depth $A_{a,\\beta,Q}$.] The entire sum\n$\\sum_{n\\ge2,\\ (n,Q)=1}\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s}$, with\n$a>1/2$, $\\beta>0$, $Q\\ge1$ integer and the real logarithm in the complex power.\nThe atom $n=1$ is excluded at every depth.\n\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\","\\item[Arithmetic depth $A_{a,\\beta,Q}$.] The entire sum\n$\\sum_{n\\ge2,\\ (n,Q)=1}\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s}$, with\n$a>1/2$, $\\beta>0$, $Q\\ge1$ integer and the real logarithm in the complex power.\nThe atom $n=1$ is excluded at every depth.\n\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by mult","\\item[Arithmetic depth $A_{a,\\beta,Q}$.] The entire sum\n$\\sum_{n\\ge2,\\ (n,Q)=1}\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s}$, with\n$a>1/2$, $\\beta>0$, $Q\\ge1$ integer and the real logarithm in the complex power.\nThe atom $n=1$ is excluded at every depth.\n\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the ma","\\item[Arithmetic depth $A_{a,\\beta,Q}$.] The entire sum\n$\\sum_{n\\ge2,\\ (n,Q)=1}\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s}$, with\n$a>1/2$, $\\beta>0$, $Q\\ge1$ integer and the real logarithm in the complex power.\nThe atom $n=1$ is excluded at every depth.\n\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=","\\item[Arithmetic depth $A_{a,\\beta,Q}$.] The entire sum\n$\\sum_{n\\ge2,\\ (n,Q)=1}\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s}$, with\n$a>1/2$, $\\beta>0$, $Q\\ge1$ integer and the real logarithm in the complex power.\nThe atom $n=1$ is excluded at every depth.\n\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2}","\\item[Arithmetic depth $A_{a,\\beta,Q}$.] The entire sum\n$\\sum_{n\\ge2,\\ (n,Q)=1}\\mu(n)^2n^{-2a}(1-n^{-\\beta})^{2s}$, with\n$a>1/2$, $\\beta>0$, $Q\\ge1$ integer and the real logarithm in the complex power.\nThe atom $n=1$ is excluded at every depth.\n\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is posit","\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order","\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree","\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\n","\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_","\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le","\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials","\\item[Möbius function and modulus radical.] $\\mu(1)=1$.\nFor a product of $k$ distinct primes, $\\mu(n)=(-1)^k$.\nIf a prime square divides $n$, then $\\mu(n)=0$.\nThe radical $\\operatorname{rad}(Q)=\\prod_{p\\mid Q}p$ records the excluded\nprime set, with $\\operatorname{rad}(1)=1$.\n\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+","\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test theref","\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic","\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int","\\item[Squarefree Dirichlet sum.] For $v>1$,\n$Z_Q(v)=\\zeta(v)\\zeta(2v)^{-1}\\prod_{p\\mid Q}(1+p^{-v})^{-1}$.\nThe unrestricted notation is $D(v)=Z_1(v)$. Thus\n$A_{a,\\beta,Q}(0)=Z_Q(2a)-1$.\n\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive me","\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be","\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_","\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\g","\\item[Prime-sieve measure.] $\\nu_{F,\\boldsymbol\\lambda}$ is the arithmetic\natomic measure obtained by multiplying the mass at $x_n=(1-n^{-2})^2$ by\n$\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})$.\nIt is positive exactly when every real strength satisfies $\\lambda_p\\ge-1$.\n\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zer","\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ rep","\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$","\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.","\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial","\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+","\\item[Hankel order and polynomial degree.] For moments $M_j=\\int x^j\\dd\\nu$,\nthe order-$m$ matrix is $H_m=[M_{i+j}]_{0\\le i,j<m}$.\nIt tests quadratic forms of real polynomials of degree at most $m-1$.\nAt depth $k$, the matrix is $[M_{k+i+j}]_{0\\le i,j<m}$.\nAn order-five test therefore covers degree four, not degree five.\n\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1","\\item[Signed quadratic form.] For real polynomials,\n$Q_\\nu(f)=\\int f(x)^2\\dd\\nu(x)$. On a positive measure it is a squared\nnorm. On a signed measure it may be negative and is not a norm.\n\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential gene","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coeff","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite im","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) t","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act b","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncross","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz pro","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz product of\nexponential","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz product of\nexponential generating functions.\n\\item[Resolutio","\\item[Orbit spectral zeta $\\mathcal Z_N$.] Let\n$d_N(w)=\\sum_{n\\ge0}(-1)^nw^n/(Nn)!$ and $D_N(z)=d_N(z^N)$, $N\\ge2$.\nThe series $\\mathcal Z_N(s)=\\sum_jw_j^{-s}$ initially runs over the zeros\n$w_j$ of $d_N$, counted with multiplicity, for $\\Re s>1/N$.\nEach $w_j$ represents one $N$-fold rotation orbit of zeros of $D_N$.\nPowers use $\\operatorname{Arg}w\\in(-\\pi,\\pi]$.\n\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz product of\nexponential generating functions.\n\\item[Resolution certificate.] An explicit bound or rational w","\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz product of\nexponential generating functions.\n\\item[Resolution certificate.] An explicit bound or rational witness for a\nstructural distinction at a specified rad","\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz product of\nexponential generating functions.\n\\item[Resolution certificate.] An explicit bound or rational witness for a\nstructural distinction at a specified radius, precisi","\\item[Rising factorial.] $(s)_k=s(s+1)\\cdots(s+k-1)$ for $k\\ge1$, and\n$(s)_0=1$. It occurs in the shifted spectral residues.\n\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz product of\nexponential generating functions.\n\\item[Resolution certificate.] An explicit bound or rational witness for a\nstructural distinction at a specified radius, precision or polynomia","\\item[Coefficient conventions.] An ordinary generating function uses\n$\\sum_na_nz^n$. An exponential generating function uses\n$\\sum_na_nz^n/n!$. Their product and shift rules differ.\n\\item[Polyphase section and finite taps.] Polyphase splits a coefficient\nsequence by residue class. Finite impulse response (FIR) taps act by finite\nordinary convolution. A phase matrix must retain the carry delay when a tap\ncrosses a residue-class boundary. This is not the Hurwitz product of\nexponential generating functions.\n\\item[Resolution certificate.] An explicit bound or rational witness for a\nstructural distinction at a specified radius, precision or polynomial\ndegree. A sufficient threshold need not be optimal.\n\\end{description}","\\bibitem{Dascaliuc2025} Radu Dascaliuc, Tuan N. Pham, Enrique Thomann and Edward C. Waymire.\n\\emph{Transformation of Stochastic Recursions and Critical Phenomena in the\nAnalysis of the Aldous-Shields-Athreya Cascade and Related Mean Flow Equations}.\narXiv:2411.00629v2 (2025).\nThe nonlinear equation $u'=-u+u(\\alpha t)^2$ is an adjacent model, not the\nequation studied here.\n\\url{https://arxiv.org/abs/2411.00629v2}","\\bibitem{DLMF} NIST Digital Library of Mathematical Functions.\nSections 5.5 and 5.12 give the gamma reflection and beta identities.\nSection 10.46 defines the Mittag--Leffler function. Sections 16.8 and 16.21\ngive the generalized hypergeometric and Meijer $G$ differential equations.\nSection 18.2 develops general orthogonal-polynomial theory.\n\\url{https://dlmf.nist.gov/}","For real \\(a>1/2\\), real \\(\\beta>0\\), and integer \\(Q\\ge1\\), write","\\[\n A_{a,\\beta,Q}(s)=\\sum_{\\substack{n\\ge2\\\\ \\mu(n)^2=1\\\\ (n,Q)=1}}\n n^{-2a}\\exp(-\\lambda_n s),\\qquad\n \\lambda_n=-2\\log(1-n^{-\\beta}).\n\\]","The logarithm is real. The omitted \\(n=1\\) atom is not restored in any\ninverse formula. Let \\(n_1<n_2<\\cdots\\) be the support integers, and\nabbreviate \\(\\lambda_j=\\lambda_{n_j}\\), \\(w_j=n_j^{-2a}\\). Then\n\\(\\lambda_1>\\lambda_2>\\cdots>0\\), \\(\\lambda_j\\to0\\), and\n\\(\\sum_jw_j=M=Z_Q(2a)-1<\\infty\\), where","\\[\n Z_Q(v)=\\frac{\\zeta(v)}{\\zeta(2v)}\n          \\prod_{p\\mid Q}(1+p^{-v})^{-1},\\qquad v>1.\n\\]","Here a germ means the complete exact analytic germ, equivalently all\nexact derivatives at a point, rather than a finite Taylor polynomial or\nnoisy measurements. The base route proves that this germ has a unique\nentire continuation, with\n\\(|A(s)|\\le M\\exp(\\lambda_1\\max\\{-\\Re s,0\\})\\).","\\[\n a=a',\\qquad \\beta=\\beta',\\qquad\n \\operatorname{rad}(Q)=\\operatorname{rad}(Q').\n\\]","The function itself therefore encodes the weight exponent, the depth\nexponent, and the finite excluded prime set. It does not encode\nvaluations \\(v_p(Q)\\).","Starting with \\(R_1=A\\), define successively","\\[\n \\lambda_j=\\lim_{t\\to\\infty}\\frac{\\log R_j(-t)}t,\n \\qquad\n w_j=\\lim_{t\\to\\infty}e^{-\\lambda_jt}R_j(-t),\n \\qquad\n R_{j+1}(s)=R_j(s)-w_je^{-\\lambda_js}.\n\\]","Every limit exists and recovers exactly the \\(j\\)-th support frequency\nand weight. Indeed, if \\(M_{j+1}=\\sum_{k>j}w_k\\), then for every\n\\(t\\ge0\\),","\\[\n R_j(-t)=w_je^{\\lambda_jt}(1+r_j(t)),\\qquad\n 0\\le r_j(t)\\le\\frac{M_{j+1}}{w_j}\n                    e^{-(\\lambda_j-\\lambda_{j+1})t}.\n\\]","The bound follows by replacing every tail frequency by its largest value\n\\(\\lambda_{j+1}\\). Taking logarithms and dividing by \\(t\\) proves the\nfirst limit. Multiplying by \\(e^{-\\lambda_jt}\\) proves the second.\nRemoving this exact entire atom leaves the same hypotheses for the next\nstep. Thus the complete ordered atom list is intrinsic to the entire\nfunction.","\\hypertarget{step-2-recover-beta-without-knowing-the-integer-support}{%\n\\subsection{\\texorpdfstring{Step 2: recover \\(\\beta\\) without knowing\nthe integer\nsupport}{Step 2: recover \\textbackslash beta without knowing the integer support}}\\label{step-2-recover-beta-without-knowing-the-integer-support}}","\\[\n \\delta_Q=\\frac1{\\zeta(2)}\\prod_{p\\mid Q}\\frac{p}{p+1}>0.\n\\]","\\[\n N_Q(x):=\\#\\{2\\le n\\le x:\\mu(n)^2=1,(n,Q)=1\\}\n       =\\delta_Qx+O_Q(\\sqrt{x}).\n\\]","Since \\(N_Q(n_j)=j\\), it follows that","\\[\n n_j=\\frac{j}{\\delta_Q}\\left(1+O_Q(j^{-1/2})\\right),\n \\quad\n \\lambda_j=2\\delta_Q^{\\beta}j^{-\\beta}\n     \\left(1+O_{Q,\\beta}(j^{-1/2}+j^{-\\beta})\\right).\n\\]","For completeness, the first estimate follows initially from\n\\(j=\\delta_Qn_j+O_Q(\\sqrt{n_j})\\), which gives \\(n_j\\asymp_Qj\\);\nsubstituting this back improves the error to \\(O_Q(\\sqrt j)\\). The\nsecond uses \\(-\\log(1-u)=u(1+O(u))\\) as \\(u\\downarrow0\\).","\\[\n \\boxed{\\displaystyle\n \\beta=-\\lim_{j\\to\\infty}\\frac{\\log\\lambda_j}{\\log j}.}\n\\]","A rank ratio avoids the slowly decaying constant divided by \\(\\log j\\):\nfor every fixed real \\(c>1\\),","\\[\n -\\frac{\\log(\\lambda_{\\lfloor cj\\rfloor}/\\lambda_j)}{\\log c}\n =\\beta+O_{Q,\\beta,c}(j^{-1/2}+j^{-\\beta}).\n\\]","The floor contributes an additional \\(O(j^{-1})\\), which is absorbed by\nthe displayed error. These are fixed-parameter asymptotics. They do not\ngive a constant uniform in arbitrary moduli.","Once \\(\\beta\\) is known, every spectral atom gives its exact integer\nlabel:","\\[\n \\boxed{\\displaystyle\n n_j=(1-e^{-\\lambda_j/2})^{-1/\\beta},\\qquad\n a=-\\frac{\\log w_j}{2\\log n_j}.}\n\\]","The recovered real number in the first formula is an integer by the\nfamily definition; rounding is not part of the exact theorem. Since\n\\(n_j\\ge2\\), the denominator in the second formula is nonzero. Every\nrecovered atom yields the same value of \\(a\\).","A prime \\(p\\) lies in the recovered integer support exactly when\n\\(p\\nmid Q\\): primes are squarefree, and no other restriction is\nimposed. Hence","\\[\n \\boxed{\\displaystyle\n \\operatorname{rad}(Q)=\\prod_{\\substack{p\\ \\mathrm{prime}\\\\\n                                      p\\notin\\{n_j:j\\ge1\\}}}p.}\n\\]","The product is finite because the parameters are assumed to come from\nthis family. This formula is an exact reconstruction from infinite data,\nnot an algorithm that can certify completion of the prime list after\nfinitely many observations. Conversely, equal \\(a,\\beta\\) and equal\nradicals produce exactly the same terms. This proves both directions of\nAI-1.","The rank asymptotic can be made effective without an RH input. Write\n\\(\\omega(Q)=\\#\\{p:p\\mid Q\\}\\). For real \\(x\\ge4\\), inclusion and\nexclusion gives","\\[\n |N_Q(x)-\\delta_Qx|\\le C_Q\\sqrt x+1,\n \\qquad C_Q=2^{\\omega(Q)}+2\\frac{\\varphi(Q)}Q.\n\\]","To verify the constants, expand \\(\\mu(n)^2=\\sum_{d^2\\mid n}\\mu(d)\\). For\n\\((d,Q)=1\\), the number of \\(m\\le x/d^2\\) coprime to \\(Q\\) differs from\n\\((\\varphi(Q)/Q)x/d^2\\) by at most \\(2^{\\omega(Q)}\\). Summing over at\nmost \\(\\sqrt x\\) values of \\(d\\) gives the first part of \\(C_Q\\). The\nomitted main-term tail is at most\n\\((\\varphi(Q)/Q)x/\\lfloor\\sqrt x\\rfloor\\le2(\\varphi(Q)/Q)\\sqrt x\\).\nRemoving \\(n=1\\) adds one. The main-term density is\n\\((\\varphi(Q)/Q)\\sum_{(d,Q)=1}\\mu(d)d^{-2}=\\delta_Q\\) by its absolutely\nconvergent Euler product.","2=\\sum_{d^2\\mid n}\\mu(d)\\). For\n\\((d,Q)=1\\), the number of \\(m\\le x/d^2\\) coprime to \\(Q\\) differs from\n\\((\\varphi(Q)/Q)x/d^2\\) by at most \\(2^{\\omega(Q)}\\). Summing over at\nmost \\(\\sqrt x\\) values of \\(d\\) gives the first part of \\(C_Q\\). The\nomitted main-term tail is at most\n\\((\\varphi(Q)/Q)x/\\lfloor\\sqrt x\\rfloor\\le2(\\varphi(Q)/Q)\\sqrt x\\).\nRemoving \\(n=1\\) adds one. The main-term density is\n\\((\\varphi(Q)/Q)\\sum_{(d,Q)=1}\\mu(d)d^{-2}=\\delta_Q\\) by its absolutely\nconvergent Euler product.","For \\(j\\ge3\\), therefore,","\\[\n \\left(\\frac{\\sqrt{C_Q^2+4\\delta_Q(j-1)}-C_Q}{2\\delta_Q}\\right)^2\n \\le n_j\\le\n \\left(\\frac{\\sqrt{C_Q^2+4\\delta_Q(j+1)}+C_Q}{2\\delta_Q}\\right)^2.\n\\]","The lower and upper bounds are the positive solutions of the\ncorresponding quadratic inequalities in \\(\\sqrt{n_j}\\). At \\(j\\ge3\\),\nall possible supports have \\(n_j\\ge5\\), so the \\(x\\ge4\\) counting\nenvelope applies.","\\textbf{Theorem B.2 (AI-2), gap bound.} Suppose the first \\(j-1\\) atoms have\nalready been removed exactly, put\n\\(\\Delta_j=\\lambda_j-\\lambda_{j+1}>0\\), and fix \\(h>0\\). Define","\\[\n \\widehat\\lambda_j(t,h)\n =\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h.\n\\]","Then for every \\(t\\ge0\\),","\\[\n 0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n \\le\\frac1h\\log\\!\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right),\n\\]","\\[\n 0\\le e^{-\\lambda_jt}R_j(-t)-w_j\n     \\le M_{j+1}e^{-\\Delta_jt}.\n\\]","\\textbf{Proof.} The preceding positive remainder formula gives\n\\(\\widehat\\lambda_j=\\lambda_j+h^{-1}[\\log(1+r_j(t+h))-\\log(1+r_j(t))]\\).\nEach tail factor decreases, so \\(0\\le r_j(t+h)\\le r_j(t)\\). The two\ninequalities now follow from the bound on \\(r_j(t)\\). For the amplitude,\nmultiply the positive tail bound by \\(e^{-\\lambda_jt}\\).","In particular, to guarantee frequency error at most \\(\\varepsilon>0\\)\nfrom exact residual values, it suffices to take","\\[\n t\\ge\\max\\left\\{0,\\frac1{\\Delta_j}\n      \\log\\frac{M_{j+1}}{w_j(e^{h\\varepsilon}-1)}\\right\\}.\n\\]","For integer \\(L\\ge n_j\\), a direct sum through \\(n\\le L\\) at depth\n\\(-t\\), \\(t\\ge0\\), omits at most","\\[\n E_L(t)=e^{t\\lambda_{L+1}}\\frac{L^{1-2a}}{2a-1}.\n\\]","Indeed all omitted frequencies are at most \\(\\lambda_{L+1}\\), and\n\\(\\sum_{n>L}n^{-2a}\\le\\int_L^\\infty x^{-2a}\\,dx\\). Ignoring\nsquarefreeness and coprimality makes this a valid upper bound. If\n\\(r_0,r_1\\) are the two positive truncated residual sums at \\(t,t+h\\),\nthe true secant belongs to","\\[\n \\left[\n \\frac{\\log r_1-\\log(r_0+E_L(t))}h,\n \\frac{\\log(r_1+E_L(t+h))-\\log r_0}h\n \\right].\n\\]","Evaluating a residual by subtracting previous atoms is different from\nevaluating its known positive series. If an observation of \\(A(-t)\\) has\nrelative error at most \\(\\eta\\) and the previous atoms are exact, then\nthe residual error can have relative size","\\[\n \\eta\\,\\kappa_j(t),\\qquad\n \\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n \\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]","Thus the large negative depth that separates nearby atoms also worsens\ncancellation for later atoms. Uncertain previously subtracted\nfrequencies introduce further error. The computation evaluates residuals\nas positive sums and reports \\(\\log_{10}\\kappa_j\\) so this conditioning\ncost remains visible.","\\textbf{Theorem B.3 (AI-3), bounded-window instability.} Let \\(p\\) be a prime\nnot dividing \\(Q\\), and let \\(R\\ge0\\). For every integer \\(k\\ge0\\),","\\[\n \\boxed{\\displaystyle\n \\sup_{|s|\\le R}|A_{a,\\beta,Q}^{(k)}(s)-A_{a,\\beta,Qp}^{(k)}(s)|\n \\le \\lambda_p^k e^{R\\lambda_p}\n      \\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]","\\textbf{Proof.} The difference consists exactly of the squarefree\nintegers divisible by \\(p\\) and coprime to \\(Q\\), all of which satisfy\n\\(n\\ge p\\). Their frequencies are at most \\(\\lambda_p\\). The Euler\nfactor at \\(p\\) shows that their total weight is","\\[\n \\sum_{\\substack{n\\ge1\\\\\\mu(n)^2=1,(n,Q)=1\\\\p\\mid n}}n^{-2a}\n =\\frac{Z_Q(2a)}{p^{2a}+1}.\n\\]","There is no \\(n=1\\) term in this difference. Bound the \\(k\\)-th\nderivative of every exponential by \\(\\lambda_p^ke^{R\\lambda_p}\\) and\nsum. Since \\(\\lambda_p\\to0\\) and \\(p^{-2a}\\to0\\), the bound tends to\nzero. This proves the derivative convergence and the claimed\ninstability.","For \\(a=\\beta=2,Q=1\\), the mass removed by \\(p\\) is exactly\n\\(105/[\\pi^4(p^4+1)]\\). The examples \\(p=101,1009,10007\\) quantify how\nquickly large-prime distinctions disappear on \\(|s|\\le5\\).","For \\(c=(c_j)\\in\\ell^2\\), define","\\[\n (Uc)(s)=\\sum_{j\\ge1}c_j\\sqrt{w_j}\\,e^{-\\lambda_js}.\n\\]","\\[\n |(Uc)(s)|\\le\\|c\\|_{\\ell^2}\\sqrt{A(2\\Re s)}.\n\\]","The map is injective: \\(\\sum|c_j|\\sqrt{w_j}\\le\\|c\\|\\sqrt M<\\infty\\), so\nthe same spectral peeling argument, now for complex coefficients, forces\nthe first coefficient and then every coefficient to vanish whenever\n\\(Uc=0\\). Positivity was needed to recover unknown frequencies in AI-1;\ninjectivity here only needs absolute summability and the known ordered\nsupport. Specifically, after dividing by the largest remaining\nexponential, the tail is bounded by\n\\(e^{-\\Delta_jt}\\sum_{k>j}|c_k|\\sqrt{w_k}\\), which tends to zero.","M<\\infty\\), so\nthe same spectral peeling argument, now for complex coefficients, forces\nthe first coefficient and then every coefficient to vanish whenever\n\\(Uc=0\\). Positivity was needed to recover unknown frequencies in AI-1;\ninjectivity here only needs absolute summability and the known ordered\nsupport. Specifically, after dividing by the largest remaining\nexponential, the tail is bounded by\n\\(e^{-\\Delta_jt}\\sum_{k>j}|c_k|\\sqrt{w_k}\\), which tends to zero.","The image \\(\\mathcal H_{a,\\beta,Q}=U\\ell^2\\), with\n\\(\\|Uc\\|_{\\mathcal H}=\\|c\\|_{\\ell^2}\\), is therefore a complete Hilbert\nspace of entire functions. Its reproducing kernel is","\\[\n K(s,t)=A(s+\\overline t).\n\\]","Using an inner product linear in the first argument, the coefficient\nvector of \\(K(\\cdot,t)\\) is \\((\\sqrt{w_j}e^{-\\lambda_j\\overline t})_j\\),\nso \\(\\langle Uc,K(\\cdot,t)\\rangle=(Uc)(t)\\). This also verifies the sign\nand conjugation convention directly. Arbitrarily large finite Gram\nmatrices at distinct complex depths are strictly positive, as proved in\nthe base route. Thus the feature space has infinite rank, not a\nfinite-dimensional Prony model.","The monomial moment matrices also have infinite rank:\n\\(m_k=(-1)^kA^{(k)}(0)=\\sum_jw_j\\lambda_j^k\\), and for every nonzero\npolynomial \\(P\\), \\(\\sum_jw_j|P(\\lambda_j)|^2>0\\), because a nonzero\npolynomial cannot vanish at infinitely many distinct frequencies. All\nfinite derivative Hankel matrices are strictly positive definite.","Let \\(\\chi,\\psi\\) be Dirichlet characters modulo the same fixed integer\n\\(Q\\), extended by zero outside the units, and put\n\\(\\eta=\\chi\\overline\\psi\\). The base extension is","\\[\n A_{\\chi,\\psi}(s)=\\sum_{n\\ge2}\\mu(n)^2\n            \\eta(n)n^{-2a}(1-n^{-\\beta})^{2s}.\n\\]","\\[\n A_{\\chi,\\psi}(s)\n =\\sum_{j\\ge0}(-1)^j\\binom{2s}{j}\n \\left[\\frac{L(2a+j\\beta,\\eta)}\n              {L(4a+2j\\beta,\\eta^2)}-1\\right].\n\\]","Every character takes the value one at \\(n=1\\); at primes dividing\n\\(Q\\), it is zero. The Euler product identity is valid even when\n\\(\\eta\\) is imprimitive, because the actual Dirichlet character and its\nactual modulus are retained. All arguments of \\(L\\) lie in \\(\\Re v>1\\),\nwhere its Euler product converges absolutely and is nonzero. No unproved\nzero-free-region hypothesis is used.","\\textbf{Theorem B.4 (AI-4), character sector decay.} Fix \\(a>1/2,\\beta>0,Q\\),\nand a closed sector \\(|\\arg s|\\le\\pi/2-\\varepsilon\\),\n\\(0<\\varepsilon<\\pi/2\\). If \\(\\chi\\ne\\psi\\), then uniformly in that\nsector,","\\[\n A_{\\chi,\\psi}(s)\n =O_{a,\\beta,Q,\\varepsilon}\\bigl(|s|^{-(2a-1/2)/\\beta}\\bigr),\n \\qquad\n \\frac{A_{\\chi,\\psi}(s)}{A_{a,\\beta,Q}(s)}\n =O_{a,\\beta,Q,\\varepsilon}\\bigl(|s|^{-1/(2\\beta)}\\bigr).\n\\]","These statements are uniform over the finite character set of this fixed\nmodulus. They are not uniform as \\(Q\\) grows.","\\textbf{Proof.} Distinct characters give a nonprincipal \\(\\eta\\). Its\nsum over a complete period is zero, so \\(|\\sum_{m\\le y}\\eta(m)|\\le Q\\).\nHence","\\[\n \\sum_{n\\le x}\\mu(n)^2\\eta(n)\n =\\sum_{d\\le\\sqrt x}\\mu(d)\\eta(d)^2\n     \\sum_{m\\le x/d^2}\\eta(m)=O_Q(\\sqrt x).\n\\]","Use partial integration against \\(g_s(x)=x^{-2a}(1-x^{-\\beta})^{2s}\\),\nwhose endpoint values vanish when \\(\\Re s>0\\). The counting bound\nreduces the magnitude to a constant times\n\\(\\int_1^\\infty\\sqrt x\\,|g_s'(x)|\\,dx\\). Writing \\(u=\\Re s\\) and\n\\(\\gamma=(2a-1/2)/\\beta>0\\), direct substitution \\(y=x^{-\\beta}\\) bounds\nits two derivative terms by","\\[\n \\frac{2a}{\\beta}B(\\gamma,2u+1)\n +2|s|B(\\gamma+1,2u),\n\\]","which is \\(O_{a,\\beta,\\varepsilon}(|s|^{-\\gamma})\\) because\n\\(u\\ge |s|\\sin\\varepsilon\\). The principal diagonal has the nonzero\nleading term \\(\\delta_Q\\Gamma(\\alpha)\\beta^{-1}(2s)^{-\\alpha}\\),\n\\(\\alpha=(2a-1)/\\beta\\), by the base sector theorem. Its modulus is\nbounded below by a positive fixed-sector multiple of \\(|s|^{-\\alpha}\\)\nfor sufficiently large \\(|s|\\). Division gives the second bound, since\n\\(\\gamma-\\alpha=1/(2\\beta)\\).","is \\(O_{a,\\beta,\\varepsilon}(|s|^{-\\gamma})\\) because\n\\(u\\ge |s|\\sin\\varepsilon\\). The principal diagonal has the nonzero\nleading term \\(\\delta_Q\\Gamma(\\alpha)\\beta^{-1}(2s)^{-\\alpha}\\),\n\\(\\alpha=(2a-1)/\\beta\\), by the base sector theorem. Its modulus is\nbounded below by a positive fixed-sector multiple of \\(|s|^{-\\alpha}\\)\nfor sufficiently large \\(|s|\\). Division gives the second bound, since\n\\(\\gamma-\\alpha=1/(2\\beta)\\).","At real positive depth, the matrix \\((A_{\\chi,\\psi}(s))\\) on any \\(d\\)\ndistinct characters is positive semidefinite and has identical diagonal\n\\(A_{a,\\beta,Q}(s)\\). Its normalized matrix tends to \\(I_d\\), with\noperator norm error \\(O_{a,\\beta,Q,d}(s^{-1/(2\\beta)})\\). Indeed, the\noff-diagonal entry bound controls the maximum row and column sums, and\nthus the operator norm. This gives quantitative separation of a fixed\ncharacter bank. It is consistent with a complex or signed off-diagonal\nentry and makes no complete-monotonicity assertion for that entry.","Uniqueness of a positive measure from its Laplace transform is classical.\nSchilling, Song and Vondraček~\\cite[Proposition~1.2 and Theorem~1.4]{Schilling2010}\ngive the uniqueness and Bernstein representation results. Here the measure is\n$\\sum_jw_j\\delta_{\\lambda_j}$, the transform variable is real depth $s>0$,\nand finiteness follows from $\\sum_jw_j<\\infty$.\nThe exact recovery of $a$, $\\beta$ and $\\operatorname{rad}(Q)$, the rank\nasymptotic and the large-prime instability estimates are proved above.\nThey are not attributed to that reference.","\\path{checks/arithmetic_inverse.py} is deterministic. Its exact\nchecks use rational finite atomic surrogates with \\(2a=4\\), integer\n\\(\\beta=1,2,3\\), \\(Q=1,2,6,30\\), and one through five atoms. They\nindependently construct the annihilating polynomial from moments and\nfrom its known rational nodes, verify the positive Vandermonde formula\nfor Hankel determinants, reconstruct weights, recover integer labels at\nknown \\(\\beta\\), and test further exact recurrence moments. These\nsurrogates do not turn the infinite family's Hankel matrices into\nfinite-rank matrices.","metic_inverse.py} is deterministic. Its exact\nchecks use rational finite atomic surrogates with \\(2a=4\\), integer\n\\(\\beta=1,2,3\\), \\(Q=1,2,6,30\\), and one through five atoms. They\nindependently construct the annihilating polynomial from moments and\nfrom its known rational nodes, verify the positive Vandermonde formula\nfor Hankel determinants, reconstruct weights, recover integer labels at\nknown \\(\\beta\\), and test further exact recurrence moments. These\nsurrogates do not turn the infinite family's Hankel matrices into\nfinite-rank matrices.","The high-precision examples use 180 decimal digits. They show two-rank\n\\(\\beta\\) estimates at ranks 100, 1000, and 10000, conditional\nnegative-depth peeling for the first three atoms at \\(t=20,100,500\\),\nand large-prime bounded-window envelopes. Positive direct sums use\n\\(n\\le5000\\), with the analytic omitted-tail envelope above. Rank\nenumeration uses \\(n\\le100000\\). Actual cancellation factors are\nreported. Numerical values and floating-point endpoints are\nillustrations, not ball-arithmetic certificates.","\\pageheading{ARITHMETIC DEPTH\\; /\\; INVERSE}{The function encodes its parameters}\nThe complete exact analytic germ determines\n\\[\n\\boxed{a,\\quad\\beta,\\quad\\operatorname{rad}(Q).}\n\\]\nPrecisely, two admissible triples define the same germ if and only if these\nthree quantities agree. Prime powers in $Q$ are invisible because the terms\nuse only coprimality. A finite Taylor polynomial or noisy measurements are\ndifferent inputs from the complete germ in this theorem.","Let the support integers be $n_1<n_2<\\cdots$, frequencies\n$\\lambda_j=-2\\log(1-n_j^{-\\beta})$, and weights $w_j=n_j^{-2a}$.\nStarting at $R_1=A$, peel the ordered atoms:\n\\[\n\\lambda_j=\\lim_{t\\to\\infty}\\frac{\\log R_j(-t)}t,\\qquad\nw_j=\\lim_{t\\to\\infty}e^{-\\lambda_jt}R_j(-t),\n\\]\n\\[\nR_{j+1}(s)=R_j(s)-w_je^{-\\lambda_js}.\n\\]\nEvery limit follows from the isolated largest remaining frequency and the\npositive summable tail. Thus the exact function determines its spectral list.","The counting law $N_Q(x)=\\delta_Qx+O_Q(\\sqrt x)$ gives\n\\[\n\\lambda_j=2\\delta_Q^\\beta j^{-\\beta}\n[1+O_{Q,\\beta}(j^{-1/2}+j^{-\\beta})],\n\\quad\n\\beta=-\\lim_{j\\to\\infty}\\frac{\\log\\lambda_j}{\\log j}.\n\\]\nOnce $\\beta$ is recovered,\n\\[\nn_j=(1-e^{-\\lambda_j/2})^{-1/\\beta},\\qquad\na=-\\frac{\\log w_j}{2\\log n_j}.\n\\]\nPrime membership in this integer support then recovers the excluded prime\nset. The reconstruction is exact from infinite data; it does not certify\nthat a finite search has found every excluded prime.","ESOLUTION}{Separation improves while subtraction worsens}\n\\fig{10_arithmetic_inverse}{The frequency-rank estimates converge toward\nthe depth exponent. Later-atom peeling at large negative depth also has a\nrapidly growing subtraction condition factor. These are finite examples\nfrom the checked arithmetic run, not certified floating-point intervals.}\nIf earlier atoms have been removed exactly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","ubtraction worsens}\n\\fig{10_arithmetic_inverse}{The frequency-rank estimates converge toward\nthe depth exponent. Later-atom peeling at large negative depth also has a\nrapidly growing subtraction condition factor. These are finite examples\nfrom the checked arithmetic run, not certified floating-point intervals.}\nIf earlier atoms have been removed exactly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","he frequency-rank estimates converge toward\nthe depth exponent. Later-atom peeling at large negative depth also has a\nrapidly growing subtraction condition factor. These are finite examples\nfrom the checked arithmetic run, not certified floating-point intervals.}\nIf earlier atoms have been removed exactly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","peeling at large negative depth also has a\nrapidly growing subtraction condition factor. These are finite examples\nfrom the checked arithmetic run, not certified floating-point intervals.}\nIf earlier atoms have been removed exactly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","t large negative depth also has a\nrapidly growing subtraction condition factor. These are finite examples\nfrom the checked arithmetic run, not certified floating-point intervals.}\nIf earlier atoms have been removed exactly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","negative depth also has a\nrapidly growing subtraction condition factor. These are finite examples\nfrom the checked arithmetic run, not certified floating-point intervals.}\nIf earlier atoms have been removed exactly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","ervals.}\nIf earlier atoms have been removed exactly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","ctly, set\n$\\Delta_j=\\lambda_j-\\lambda_{j+1}>0$, $M_{j+1}=\\sum_{k>j}w_k$.\nThe logarithmic secant\n\\[\n\\widehat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","hat\\lambda_j(t,h)=\\frac{\\log R_j(-t-h)-\\log R_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","_j(-t)}h\n\\]\nobeys, for $t\\ge0$, $h>0$,\n\\[\n0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","0\\le\\lambda_j-\\widehat\\lambda_j(t,h)\n\\le\\frac1h\\log\\left(1+\\frac{M_{j+1}}{w_j}e^{-\\Delta_jt}\\right).\n\\]\nThis improves with negative depth. But relative error in $A(-t)$ can be\namplified after subtraction by\n\\[\n\\kappa_j(t)=\\frac{A(-t)}{R_j(-t)}\n\\sim\\frac{w_1}{w_j}e^{(\\lambda_1-\\lambda_j)t}.\n\\]\nFor the displayed $Q=1$ example, the third-atom subtraction at $t=500$\nhas a condition factor of roughly $10^{108.8}$. Exact uniqueness therefore\ndoes not promise easy recovery from ordinary precision.","\\pageheading{ARITHMETIC DEPTH\\; /\\; STABILITY}{A large prime can hide in every finite window}\nFor a prime $p\\nmid Q$, any $R\\ge0$, and integer $k\\ge0$,\n\\[\n\\boxed{\\sup_{|s|\\le R}|A_{a,\\beta,Q}^{(k)}(s)-A_{a,\\beta,Qp}^{(k)}(s)|\n\\le\\lambda_p^k e^{R\\lambda_p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radic","\\pageheading{ARITHMETIC DEPTH\\; /\\; STABILITY}{A large prime can hide in every finite window}\nFor a prime $p\\nmid Q$, any $R\\ge0$, and integer $k\\ge0$,\n\\[\n\\boxed{\\sup_{|s|\\le R}|A_{a,\\beta,Q}^{(k)}(s)-A_{a,\\beta,Qp}^{(k)}(s)|\n\\le\\lambda_p^k e^{R\\lambda_p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs fo","\\pageheading{ARITHMETIC DEPTH\\; /\\; STABILITY}{A large prime can hide in every finite window}\nFor a prime $p\\nmid Q$, any $R\\ge0$, and integer $k\\ge0$,\n\\[\n\\boxed{\\sup_{|s|\\le R}|A_{a,\\beta,Q}^{(k)}(s)-A_{a,\\beta,Qp}^{(k)}(s)|\n\\le\\lambda_p^k e^{R\\lambda_p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, w","\\pageheading{ARITHMETIC DEPTH\\; /\\; STABILITY}{A large prime can hide in every finite window}\nFor a prime $p\\nmid Q$, any $R\\ge0$, and integer $k\\ge0$,\n\\[\n\\boxed{\\sup_{|s|\\le R}|A_{a,\\beta,Q}^{(k)}(s)-A_{a,\\beta,Qp}^{(k)}(s)|\n\\le\\lambda_p^k e^{R\\lambda_p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inve","\\pageheading{ARITHMETIC DEPTH\\; /\\; STABILITY}{A large prime can hide in every finite window}\nFor a prime $p\\nmid Q$, any $R\\ge0$, and integer $k\\ge0$,\n\\[\n\\boxed{\\sup_{|s|\\le R}|A_{a,\\beta,Q}^{(k)}(s)-A_{a,\\beta,Qp}^{(k)}(s)|\n\\le\\lambda_p^k e^{R\\lambda_p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radica","\\pageheading{ARITHMETIC DEPTH\\; /\\; STABILITY}{A large prime can hide in every finite window}\nFor a prime $p\\nmid Q$, any $R\\ge0$, and integer $k\\ge0$,\n\\[\n\\boxed{\\sup_{|s|\\le R}|A_{a,\\beta,Q}^{(k)}(s)-A_{a,\\beta,Qp}^{(k)}(s)|\n\\le\\lambda_p^k e^{R\\lambda_p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radical parameter.","e^{R\\lambda_p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radical parameter.","p}\\frac{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radical parameter.","{Z_Q(2a)}{p^{2a}+1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radical parameter.","1}.}\n\\]\nThe difference consists of the $p$-divisible admissible squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radical parameter.","le squarefree atoms.\nTheir frequencies are at most $\\lambda_p$ and their total weight is the\nEuler-factor expression on the right. Termwise differentiation proves the\nbound for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radical parameter.","und for every derivative.\n\\fig{11_prime_instability}{The proved compact-window envelope shrinks as\nthe excluded prime increases. At $a=\\beta=2$, $Q=1$, $R=5$, excluding\n$p=10007$ changes the entire window by at most approximately\n$1.0749\\times10^{-16}$. The depth-zero difference is known exactly.}\nThe radical differs for every such prime, while all bounded-window derivatives\nconverge to those of $A_Q$. Without a bound on excluded primes, there is no\nuniformly stable inverse from these windows to the radical parameter.","\\pageheading{ARITHMETIC DEPTH\\; /\\; REALIZATION}{An infinite feature space with arithmetic labels}\nFor the ordered support, define $\\phi_s(j)=\\sqrt{w_j}e^{-\\lambda_js}$.\nThe kernel $A(s+\\overline t)$ is its weighted $\\ell^2$ Gram kernel.\nSummability and bounded frequencies make every feature vector well defined.","If a weighted $\\ell^2$ combination produces the zero entire function,\nCauchy--Schwarz makes its coefficients absolutely summable. Successive\ndominant-frequency peeling then forces every coefficient to vanish. This\ngives an injective Hilbert realization, and all finite derivative Hankel\nmatrices have full rank: a nonzero polynomial in $\\lambda$ cannot vanish\non the infinite frequency support.","\\claim{Fixed-modulus character resolution}\nFor characters $\\chi,\\psi$ modulo a fixed $Q$, let\n$\\eta=\\chi\\overline\\psi$. When $\\eta$ is nonprincipal, the elementary\nsquarefree character sum is $O_Q(\\sqrt x)$, with no density term.\nThe same sector partial-integration argument therefore gives\n\\[\nA_{\\chi,\\psi}(s)\n=O_{a,\\beta,Q,\\epsilon}(|s|^{-(2a-1/2)/\\beta}),\n\\qquad |\\arg s|\\le\\pi/2-\\epsilon.\n\\]\nCompared with the principal $|s|^{-(2a-1)/\\beta}$ scale, this is faster\nby $|s|^{-1/(2\\beta)}$. The modulus is fixed; the constants are not asserted uniform as $Q$ grows. Individual off-diagonal functions can be complex and are not asserted\ncompletely monotone.","acter resolution}\nFor characters $\\chi,\\psi$ modulo a fixed $Q$, let\n$\\eta=\\chi\\overline\\psi$. When $\\eta$ is nonprincipal, the elementary\nsquarefree character sum is $O_Q(\\sqrt x)$, with no density term.\nThe same sector partial-integration argument therefore gives\n\\[\nA_{\\chi,\\psi}(s)\n=O_{a,\\beta,Q,\\epsilon}(|s|^{-(2a-1/2)/\\beta}),\n\\qquad |\\arg s|\\le\\pi/2-\\epsilon.\n\\]\nCompared with the principal $|s|^{-(2a-1)/\\beta}$ scale, this is faster\nby $|s|^{-1/(2\\beta)}$. The modulus is fixed; the constants are not asserted uniform as $Q$ grows. Individual off-diagonal functions can be complex and are not asserted\ncompletely monotone.","dulo a fixed $Q$, let\n$\\eta=\\chi\\overline\\psi$. When $\\eta$ is nonprincipal, the elementary\nsquarefree character sum is $O_Q(\\sqrt x)$, with no density term.\nThe same sector partial-integration argument therefore gives\n\\[\nA_{\\chi,\\psi}(s)\n=O_{a,\\beta,Q,\\epsilon}(|s|^{-(2a-1/2)/\\beta}),\n\\qquad |\\arg s|\\le\\pi/2-\\epsilon.\n\\]\nCompared with the principal $|s|^{-(2a-1)/\\beta}$ scale, this is faster\nby $|s|^{-1/(2\\beta)}$. The modulus is fixed; the constants are not asserted uniform as $Q$ grows. Individual off-diagonal functions can be complex and are not asserted\ncompletely monotone.","entary\nsquarefree character sum is $O_Q(\\sqrt x)$, with no density term.\nThe same sector partial-integration argument therefore gives\n\\[\nA_{\\chi,\\psi}(s)\n=O_{a,\\beta,Q,\\epsilon}(|s|^{-(2a-1/2)/\\beta}),\n\\qquad |\\arg s|\\le\\pi/2-\\epsilon.\n\\]\nCompared with the principal $|s|^{-(2a-1)/\\beta}$ scale, this is faster\nby $|s|^{-1/(2\\beta)}$. The modulus is fixed; the constants are not asserted uniform as $Q$ grows. Individual off-diagonal functions can be complex and are not asserted\ncompletely monotone.","\\[\nF'(z)=1+\\sum_{j=1}^{J}h_jF(q_jz)F(r_jz),\\qquad F(0)=0,\n\\]","where \\(h_j>0\\), \\(\\sum_jh_j=1\\), and \\(0<q_j,r_j<1\\). Zero-weight\nchannels may instead be allowed if they are removed from the active bank\nbefore defining its growth exponent. Put","\\[\n\\rho=\\max_{j:h_j>0}\\rho_j,\\qquad q_j^{\\rho_j}+r_j^{\\rho_j}=1.\n\\]","The inherited coefficient recurrence gives nonnegative coefficients,\nentire extension, and \\(M(R)=\\max_{|z|=R}|F(z)|=F(R)\\ge R\\) for \\(R>0\\).","\\textbf{Theorem A.1.} There are explicit constants \\(0<c_-\\le c_+<\\infty\\)\nsuch that","\\[\nc_-\\le\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\\le c_+.\n\\]","In particular the bank has positive lower type and finite upper type at\nits exact order, including when \\(0<\\rho<1\\). Furthermore","\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]","Select any active maximizing channel and abbreviate it to \\((h,q,r)\\),\nso \\(q^\\rho+r^\\rho=1\\). Define","\\[\nL=\\frac1h,\\qquad\nC=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\qquad\nB=L+\\frac2{\\min(q,r)},\\qquad\nc_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]","\\[\n\\boxed{\\log M(R)\\ge c_-(R+C)^\\rho\\quad\\text{for every }R\\ge2.}\n\\]","Write \\(H(R)=\\log M(R)\\). On the seed interval \\([2,B]\\),\n\\(H(R)\\ge\\log R\\ge\\log2\\ge c_-(R+C)^\\rho\\). For \\(R\\ge B\\), integrate\nthe positive feedback over the fixed-width interval \\([R-L,R]\\):","\\[\n\\begin{aligned}\nM(R)&\\ge h\\int_{R-L}^{R}M(qt)M(rt)\\,dt\\\\\n&\\ge hL M(q(R-L))M(r(R-L))\\\\\n&=M(q(R-L))M(r(R-L)).\n\\end{aligned}\n\\]","The two child radii are at least two because \\(R\\ge B\\). Moreover","\\[\nq(R-L)+C\\ge q(R+C),\\qquad\nr(R-L)+C\\ge r(R+C),\n\\]","because respectively \\((1-q)C\\ge qL\\) and \\((1-r)C\\ge rL\\). If the\ncomparison bound is known at both child radii, it follows that","\\[\n\\begin{aligned}\nH(R)&\\ge H(q(R-L))+H(r(R-L))\\\\\n&\\ge c_-\\{[q(R-L)+C]^\\rho+[r(R-L)+C]^\\rho\\}\\\\\n&\\ge c_-(q^\\rho+r^\\rho)(R+C)^\\rho\n=c_-(R+C)^\\rho.\n\\end{aligned}\n\\]","For a rigorous propagation, put \\(Q_* =\\max(q,r)<1\\) and\n\\(d=(1-Q_*)B+Q_*L>0\\). Every child of every \\(R\\ge B\\) is at most\n\\(R-d\\). Induction on the intervals \\([B+kd,B+(k+1)d]\\), beginning with\nthe already controlled seed, proves the comparison for all larger radii.\nContinuity deals with common endpoints. All power comparisons use only\nmonotonicity of \\(x^\\rho\\) for positive \\(x\\); no convexity assumption\nis made, so the argument also covers \\(\\rho<1\\).","1\\) and\n\\(d=(1-Q_*)B+Q_*L>0\\). Every child of every \\(R\\ge B\\) is at most\n\\(R-d\\). Induction on the intervals \\([B+kd,B+(k+1)d]\\), beginning with\nthe already controlled seed, proves the comparison for all larger radii.\nContinuity deals with common endpoints. All power comparisons use only\nmonotonicity of \\(x^\\rho\\) for positive \\(x\\); no convexity assumption\nis made, so the argument also covers \\(\\rho<1\\).","\\[\nQ=\\max_j\\max(q_j,r_j),\\quad\nm=\\min_j\\min(q_j,r_j),\\quad\n\\Delta=\\log(1/m),\\quad\nB_+=\\max_j\\log\\frac1{q_jr_j}+\\log2.\n\\]","Choose an integer \\(N\\ge0\\) with \\(Q^{-N}\\ge3/m\\), and put","\\[\nV_N=(2^{N+1}-1)\\log2+\n       (2^{N+1}-N-2)\\log(1/Q),\n\\]","\\[\nc_+=e^{-\\rho}(V_N+1+\\Delta+B_+).\n\\]","These constants are intentionally conservative, and require no numerical\nevaluation of an unknown solution. The tangent majorant gives\n\\(M(1)<2\\). At \\(R_k=Q^{-k}\\), the integral equation gives","\\[\nM(R_k)\\le R_k[1+M(R_{k-1})^2].\n\\]","Define \\(U_0=2\\), \\(U_k=Q^{-k}(1+U_{k-1}^2)\\). Then \\(M(R_k)\\le U_k\\),\n\\(U_k\\ge1\\), and \\(\\log U_k\\le k\\log(1/Q)+\\log2+2\\log U_{k-1}\\). Summing\nthis elementary recurrence gives \\(\\log U_N\\le V_N\\). Since \\(e<3\\), the\nlogarithmic-radius seed \\(1\\le t\\le1+\\Delta\\) has\n\\(e^t\\le e/m<3/m\\le R_N\\), and therefore \\(\\log M(e^t)\\le V_N\\).","Let \\(a_j=\\log(1/q_j)\\), \\(b_j=\\log(1/r_j)\\), and\n\\(G(t)=\\log M(e^t)+t+B_+\\). For \\(t\\ge1+\\Delta\\), all child moduli are\nat least one, so the positive integral and weight normalization give","\\[\n\\log M(e^t)\\le t+\\log2+\n\\max_j\\{\\log M(e^{t-a_j})+\\log M(e^{t-b_j})\\}.\n\\]","The definition of \\(B_+\\) implies","\\[\nG(t)\\le\\max_j\\{G(t-a_j)+G(t-b_j)\\}.\n\\]","On the seed, \\(G(t)\\le V_N+1+\\Delta+B_+\\le c_+e^{\\rho t}\\). Each channel\nsatisfies \\(q_j^\\rho+r_j^\\rho\\le1\\). Induction in intervals of length\n\\(\\min_j\\min(a_j,b_j)>0\\) therefore propagates the seed comparison,\ngiving the explicit upper bound","\\[\n\\boxed{\\log M(R)\\le c_+R^\\rho-\\log R-B_+\\quad(R\\ge e).}\n\\]","Dividing the two bounds by \\(R^\\rho\\), then taking logarithms of their\npositive logarithmic-growth bounds, proves all assertions of Theorem A.1.\nThe theorem leaves open whether \\(\\log M(R)/R^\\rho\\) has a limit.\nLattice resonances or logarithmic oscillations are possible research\ndirections, not results here.","Fix the number \\(J\\) of channels. Allow complex weights \\(h_j\\), without\nnormalization, and complex contractions satisfying \\(|q_j|,|r_j|<1\\).\nThe same formal recurrence defines coefficients","\\[\nc_n(h,q,r)\\in\\mathbb Q[h_1,\\ldots,h_J,q_1,r_1,\\ldots,q_J,r_J].\n\\]","Each is homogeneous of degree \\(n\\) in the weights. If a parameter set\nhas \\(|q_j|,|r_j|\\le Q<1\\) and \\(\\sum_j|h_j|\\le W\\), with \\(W>0\\),\ncoefficient induction gives","\\[\n|c_n(h,q,r)|\\le W^n d_n(Q),\n\\]","where \\(f_Q(z)=\\sum d_n(Q)z^{2n+1}\\) is the established diagonal\nfunction. Consequently, on every disk \\(|z|\\le R\\),","\\[\n\\sum_{n\\ge0}|c_n(h,q,r)|R^{2n+1}\n\\le W^{-1/2}f_Q(\\sqrt W R)<\\infty.\n\\]","Every compact subset of the complex parameter domain has such \\(Q,W\\).\nThe series therefore converges normally on compact sets of the joint\ndomain, and its sum is jointly holomorphic in the weights, contractions,\nand \\(z\\). This proves parameter continuity at zero weights, although\nthe growth order can jump there. Complex or signed parameters do not\ninherit the positive-bank growth theorem merely from holomorphic\ndependence.","Keep fixed real contractions \\(0<q_j,r_j\\le Q<1\\), and let \\(h,g\\) be\nnonnegative normalized weight vectors. Zero weights are permitted. Let\n\\(d=\\|h-g\\|_1\\). Then for every \\(R\\ge0\\),","\\[\n\\boxed{\\sup_{|z|\\le R}|F_h(z)-F_g(z)|\n\\le\\frac d4\\{R f_Q'(R)-f_Q(R)\\}.}\n\\]","\\textbf{Proof.} Regard \\(c_n\\) as a polynomial in the real weight\nvariables with the contractions fixed. Its coefficients are nonnegative.\nInduction in the recurrence gives the coefficientwise polynomial bound","\\[\n0\\preceq c_n(h)\\preceq d_n(Q)\\Bigl(\\sum_jh_j\\Bigr)^n.\n\\]","Here \\(\\preceq\\) compares every monomial coefficient. Differentiation\npreserves this comparison, so on the normalized nonnegative simplex","\\[\n0\\le\\partial_{h_j}c_n\\le n d_n(Q).\n\\]","Put \\(v=h-g\\). Its positive and negative parts both have sum \\(d/2\\),\nsince \\(\\sum_jv_j=0\\). At every point of the segment between \\(g\\) and\n\\(h\\), all partial derivatives lie in the interval \\([0,n d_n(Q)]\\).\nHence","\\[\n|v\\cdot\\nabla c_n|\\le\\frac d2 n d_n(Q),\\qquad\n|c_n(h)-c_n(g)|\\le\\frac d2 n d_n(Q).\n\\]","Multiply by \\(R^{2n+1}\\) and sum. Entire convergence justifies the sum,\nand\n\\[R f_Q'(R)-f_Q(R)=2\\sum_{n\\ge0}n d_n(Q)R^{2n+1},\\]\nproving the\nclaim. The same argument applies to the closed simplex and needs no\npositive minimum weight.","Take the two diagonal channels \\((q,r)=(1/4,1/4)\\) and \\((3/4,3/4)\\),\nwith weights \\(1-\\varepsilon\\) and \\(\\varepsilon\\), respectively. For\n\\(0\\le\\varepsilon\\le1\\), write the resulting function as\n\\(F_\\varepsilon\\). The quantitative bound gives","\\[\n\\sup_{|z|\\le R}|F_\\varepsilon(z)-f_{1/4}(z)|\n\\le\\frac{\\varepsilon}{2}\\{R f_{3/4}'(R)-f_{3/4}(R)\\}.\n\\]","Thus convergence to \\(f_{1/4}\\) is uniform on every fixed disk, at an\nexplicit linear rate in \\(\\varepsilon\\). Yet","\\[\n\\rho(F_0)=\\frac12,\\qquad\n\\rho(F_\\varepsilon)=\\frac{\\log2}{\\log(4/3)}\n\\quad(0<\\varepsilon\\le1).\n\\]","For the faster active channel the lower constants become\n\\(L=1/\\varepsilon\\), \\(C=3/\\varepsilon\\), \\(B=1/\\varepsilon+8/3\\). Its\ncertified lower-type constant is","\\[\nc_-(\\varepsilon)=\\frac{\\log2}{(4/\\varepsilon+8/3)^{\\log2/\\log(4/3)}}>0.\n\\]","\\pageheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot i","\\pageheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of co","\\pageheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant","\\pageheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are reco","\\pageheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\cl","\\pageheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponen","\\pageheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $","eheading{DILATION\\; /\\; GROWTH THEOREM}{The exact order has positive lower type}\nThe single-channel result extends to finite positive feedback banks. Here a\nbank has weights $h_j>0$ with $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $(h,q,…","h $\\sum_jh_j=1$, contractions\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $(h,q,r)$ and integrate over a fixed-width\nterminal interval. Let\n\\[\nL=1/h,\\quad C=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\quad\nB=L+\\frac{2}{\\min(q,r)},\\quad c_-=\\frac{\\log2}{(B+C)^\\r…","ons\n$0<q_j,r_j<1$, and normalized solution\n\\[\nF'(z)=1+\\sum_jh_jF(q_jz)F(r_jz),\\qquad F(0)=0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $(h,q,r)$ and integrate over a fixed-width\nterminal interval. Let\n\\[\nL=1/h,\\quad C=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\quad\nB=L+\\frac{2}{\\min(q,r)},\\quad c_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]\nThe result is the …","0.\n\\]\nWrite $M(R)=M_F(R)$ and let $q_j^{\\rho_j}+r_j^{\\rho_j}=1$.\nThen, with $\\rho=\\max_j\\rho_j$,\n\\[\n\\boxed{0<\\liminf_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}\n\\le\\limsup_{R\\to\\infty}\\frac{\\log M(R)}{R^\\rho}<\\infty.}\n\\]\nThus logarithmic growth is trapped between two positive multiples of\n$R^\\rho$. In particular,\n\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M(R)}{\\log R}=\\rho.\n\\]\nThe theorem does not assert that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $(h,q,r)$ and integrate over a fixed-width\nterminal interval. Let\n\\[\nL=1/h,\\quad C=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\quad\nB=L+\\frac{2}{\\min(q,r)},\\quad c_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]\nThe result is the explicit all-radius bound\n\\[\n\\log M(R)\\ge c_-(R+C)^\\rho,\\qquad R\\ge2.\n\\]\nUnlike a fixed fr…","that $\\log M(R)/R^\\rho$ has a limit.\n\\fig{08_growth_type}{Finite-radius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $(h,q,r)$ and integrate over a fixed-width\nterminal interval. Let\n\\[\nL=1/h,\\quad C=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\quad\nB=L+\\frac{2}{\\min(q,r)},\\quad c_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]\nThe result is the explicit all-radius bound\n\\[\n\\log M(R)\\ge c_-(R+C)^\\rho,\\qquad R\\ge2.\n\\]\nUnlike a fixed fractional window, the terminal window loses only an additive\nradius. The shift $C$ absorbs that loss without changing the exponent.","dius logarithmic growth and its proved lower\ncomparison. The normalized curves illustrate scale dependence; the plot is\nnot a proof of convergence to a type constant. Analytic truncation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $(h,q,r)$ and integrate over a fixed-width\nterminal interval. Let\n\\[\nL=1/h,\\quad C=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\quad\nB=L+\\frac{2}{\\min(q,r)},\\quad c_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]\nThe result is the explicit all-radius bound\n\\[\n\\log M(R)\\ge c_-(R+C)^\\rho,\\qquad R\\ge2.\n\\]\nUnlike a fixed fractional window, the terminal window loses only an additive\nradius. The shift $C$ absorbs that loss without changing the exponent.","runcation envelopes\nand ordinary high-precision rounding are recorded separately.}\n\\claim{A lower bound at the exact exponent}\nChoose a maximizing channel $(h,q,r)$ and integrate over a fixed-width\nterminal interval. Let\n\\[\nL=1/h,\\quad C=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\quad\nB=L+\\frac{2}{\\min(q,r)},\\quad c_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]\nThe result is the explicit all-radius bound\n\\[\n\\log M(R)\\ge c_-(R+C)^\\rho,\\qquad R\\ge2.\n\\]\nUnlike a fixed fractional window, the terminal window loses only an additive\nradius. The shift $C$ absorbs that loss without changing the exponent.","izing channel $(h,q,r)$ and integrate over a fixed-width\nterminal interval. Let\n\\[\nL=1/h,\\quad C=\\max\\left(\\frac{qL}{1-q},\\frac{rL}{1-r}\\right),\\quad\nB=L+\\frac{2}{\\min(q,r)},\\quad c_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]\nThe result is the explicit all-radius bound\n\\[\n\\log M(R)\\ge c_-(R+C)^\\rho,\\qquad R\\ge2.\n\\]\nUnlike a fixed fractional window, the terminal window loses only an additive\nradius. The shift $C$ absorbs that loss without changing the exponent.","\\pageheading{DILATION\\; /\\; PROOF}{A translated power comparison}\nWrite $H(R)=\\log M(R)$. Positivity gives\n\\[\nM(R)\\ge h\\int_{R-L}^{R}M(qt)M(rt)\\dd t\n\\ge M(q(R-L))M(r(R-L)),\\qquad R\\ge B.\n\\]\nThe choice of $C$ ensures\n\\[\nq(R-L)+C\\ge q(R+C),\\qquad r(R-L)+C\\ge r(R+C).\n\\]\nOn the seed interval $[2,B]$, $H(R)\\ge\\log R\\ge\\log2\\ge c_-(R+C)^\\rho$.\nIf this comparison holds at the two child radii, then\n\\[\n\\begin{aligned}\nH(R)&\\ge H(q(R-L))+H(r(R-L))","\\pageheading{DILATION\\; /\\; PROOF}{A translated power comparison}\nWrite $H(R)=\\log M(R)$. Positivity gives\n\\[\nM(R)\\ge h\\int_{R-L}^{R}M(qt)M(rt)\\dd t\n\\ge M(q(R-L))M(r(R-L)),\\qquad R\\ge B.\n\\]\nThe choice of $C$ ensures\n\\[\nq(R-L)+C\\ge q(R+C),\\qquad r(R-L)+C\\ge r(R+C).\n\\]\nOn the seed interval $[2,B]$, $H(R)\\ge\\log R\\ge\\log2\\ge c_-(R+C)^\\rho$.\nIf this comparison holds at the two child radii, then\n\\[\n\\begin{aligned}\nH(R)&\\ge H(q(R-L))+H(r(R-L))\\\\\n&\\ge c_-\\{[q(R-L)+C]^\\rho+[r(R-L)+C]^\\rho\\}\\\\\n&\\ge c_-(q^\\rho+r^\\rho)(R+C)^\\rho=c_-(R+C)^\\rho.\n\\end{aligned}\n\\]\nLet $Q_* =\\max(q,r)$ and $d=(1-Q_*)B+Q_*L>0$. Each child radius is at\nmost $R-d$ and at least two. Induction on successive intervals of length\n$d$ propagates the seed comparison to every $R\\ge B$. The argument uses\nmonotonicity of positive powers and works also when $\\rho<1$.","&\\ge c_-(q^\\rho+r^\\rho)(R+C)^\\rho=c_-(R+C)^\\rho.\n\\end{aligned}\n\\]\nLet $Q_* =\\max(q,r)$ and $d=(1-Q_*)B+Q_*L>0$. Each child radius is at\nmost $R-d$ and at least two. Induction on successive intervals of length\n$d$ propagates the seed comparison to every $R\\ge B$. The argument uses\nmonotonicity of positive powers and works also when $\\rho<1$.","\\claim{Explicit upper constants are available too}\nSet $Q=\\max_j\\max(q_j,r_j)$, $m=\\min_j\\min(q_j,r_j)$,\n$\\Delta=\\log(1/m)$, and $B_+=\\max_j\\log(1/(q_jr_j))+\\log2$.\nChoose an integer $K$ with $Q^{-K}\\ge3/m$ and put\n\\[\nV_K=(2^{K+1}-1)\\log2+(2^{K+1}-K-2)\\log(1/Q),\n\\]\n\\[\nc_+=e^{-\\rho}(V_K+1+\\Delta+B_+).\n\\]\nThen $\\log M(R)\\le c_+R^\\rho-\\log R-B_+$ for $R\\ge e$.\nThe tangent majorant $M(1)<2$, followed by the integral recurrence on radii\n$Q^{-k}$, bounds the initial seed by $V_K$. The logarithmic-radius comparison\nthen supplies the displayed upper constant. The full induction is in Appendix A.","vailable too}\nSet $Q=\\max_j\\max(q_j,r_j)$, $m=\\min_j\\min(q_j,r_j)$,\n$\\Delta=\\log(1/m)$, and $B_+=\\max_j\\log(1/(q_jr_j))+\\log2$.\nChoose an integer $K$ with $Q^{-K}\\ge3/m$ and put\n\\[\nV_K=(2^{K+1}-1)\\log2+(2^{K+1}-K-2)\\log(1/Q),\n\\]\n\\[\nc_+=e^{-\\rho}(V_K+1+\\Delta+B_+).\n\\]\nThen $\\log M(R)\\le c_+R^\\rho-\\log R-B_+$ for $R\\ge e$.\nThe tangent majorant $M(1)<2$, followed by the integral recurrence on radii\n$Q^{-k}$, bounds the initial seed by $V_K$. The logarithmic-radius comparison\nthen supplies the displayed upper constant. The full induction is in Appendix A.","_j)$, $m=\\min_j\\min(q_j,r_j)$,\n$\\Delta=\\log(1/m)$, and $B_+=\\max_j\\log(1/(q_jr_j))+\\log2$.\nChoose an integer $K$ with $Q^{-K}\\ge3/m$ and put\n\\[\nV_K=(2^{K+1}-1)\\log2+(2^{K+1}-K-2)\\log(1/Q),\n\\]\n\\[\nc_+=e^{-\\rho}(V_K+1+\\Delta+B_+).\n\\]\nThen $\\log M(R)\\le c_+R^\\rho-\\log R-B_+$ for $R\\ge e$.\nThe tangent majorant $M(1)<2$, followed by the integral recurrence on radii\n$Q^{-k}$, bounds the initial seed by $V_K$. The logarithmic-radius comparison\nthen supplies the displayed upper constant. The full induction is in Appendix A.","\\pageheading{DILATION\\; /\\; TAPS}{A quantitative compact-continuity law}\nFix the contractions and let $h,g$ be two normalized nonnegative tap vectors.\nSet $Q=\\max_j\\max(q_j,r_j)<1$ and $d=\\|h-g\\|_1$. Then\n\\[\n\\boxed{\\sup_{|z|\\le R}|F_h(z)-F_g(z)|\n\\le\\frac{d}{4}\\,[R f_Q'(R)-f_Q(R)].}\n\\]\nThe right side is an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd","\\pageheading{DILATION\\; /\\; TAPS}{A quantitative compact-continuity law}\nFix the contractions and let $h,g$ be two normalized nonnegative tap vectors.\nSet $Q=\\max_j\\max(q_j,r_j)<1$ and $d=\\|h-g\\|_1$. Then\n\\[\n\\boxed{\\sup_{|z|\\le R}|F_h(z)-F_g(z)|\n\\le\\frac{d}{4}\\,[R f_Q'(R)-f_Q(R)].}\n\\]\nThe right side is an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[","\\pageheading{DILATION\\; /\\; TAPS}{A quantitative compact-continuity law}\nFix the contractions and let $h,g$ be two normalized nonnegative tap vectors.\nSet $Q=\\max_j\\max(q_j,r_j)<1$ and $d=\\|h-g\\|_1$. Then\n\\[\n\\boxed{\\sup_{|z|\\le R}|F_h(z)-F_g(z)|\n\\le\\frac{d}{4}\\,[R f_Q'(R)-f_Q(R)].}\n\\]\nThe right side is an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe p","\\pageheading{DILATION\\; /\\; TAPS}{A quantitative compact-continuity law}\nFix the contractions and let $h,g$ be two normalized nonnegative tap vectors.\nSet $Q=\\max_j\\max(q_j,r_j)<1$ and $d=\\|h-g\\|_1$. Then\n\\[\n\\boxed{\\sup_{|z|\\le R}|F_h(z)-F_g(z)|\n\\le\\frac{d}{4}\\,[R f_Q'(R)-f_Q(R)].}\n\\]\nThe right side is an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d","x_j\\max(q_j,r_j)<1$ and $d=\\|h-g\\|_1$. Then\n\\[\n\\boxed{\\sup_{|z|\\le R}|F_h(z)-F_g(z)|\n\\le\\frac{d}{4}\\,[R f_Q'(R)-f_Q(R)].}\n\\]\nThe right side is an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.",". Then\n\\[\n\\boxed{\\sup_{|z|\\le R}|F_h(z)-F_g(z)|\n\\le\\frac{d}{4}\\,[R f_Q'(R)-f_Q(R)].}\n\\]\nThe right side is an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","le\\frac{d}{4}\\,[R f_Q'(R)-f_Q(R)].}\n\\]\nThe right side is an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","an explicitly defined entire majorant. It measures local\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","ocal\nsensitivity without asserting continuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","inuity of the asymptotic growth order.\n\\fig{09_tap_continuity}{Computed weak-tap differences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","fferences lie below the proved\ncompact bound. For a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","a slow/faster two-channel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","nnel bank, $d=2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","2\\varepsilon$, so the\nenvelope is $\\varepsilon[R f_{3/4}'(R)-f_{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","{3/4}(R)]/2$.}\nEach $c_n(h)$ is a positive polynomial homogeneous of degree $n$ in the\nweights. Coefficientwise comparison gives\n$c_n(h)\\le d_n(Q)(\\sum_jh_j)^n$, where $f_Q=\\sum_nd_n(Q)z^{2n+1}$.\nEvery weight partial derivative on the simplex lies in $[0,nd_n(Q)]$.\nThe positive and negative parts of $h-g$ each have mass $d/2$, giving\n$|c_n(h)-c_n(g)|\\le(d/2)nd_n(Q)$.\nSumming this inequality uses\n$\\sum_nnd_n(Q)R^{2n+1}=[Rf_Q'(R)-f_Q(R)]/2$.","\\pageheading{DILATION\\; /\\; PARAMETERS}{An analytic family with a discontinuous invariant}\nFor fixed finite $J$, allow complex $h_j,q_j,r_j$, with\n$|q_j|,|r_j|<1$. The coefficient recurrence defines polynomials in all these\nparameters. On a compact parameter set choose $Q<1$ and $W>0$ such that\n$|q_j|,|r_j|\\le Q$ and $\\sum_j|h_j|\\le W$. Then\n\\[\n|c_n(h,q,r)|\\le W^n d_n(Q),\n\\qquad\n\\sum_n|c_n|R^{2n+1}\\le W^{-1/2}f_Q(\\sqrt W R).\n\\]\nNormal convergence proves joint holomorphic dependence on parameters and\nargument throughout this domain. Signed and complex banks have entire\nsolutions, but the positive-bank growth theorem is not asserted for them.","\\claim{The weak-tap example now has both sides quantified}\nFor the bank with scales $(1/4,1/4)$ and $(3/4,3/4)$, weights\n$1-\\varepsilon,\\varepsilon$, the functions converge on every compact set\nas $\\varepsilon\\downarrow0$. The preceding envelope gives a linear bound\nin $\\varepsilon$. At every $\\varepsilon>0$, the order is\n$\\log2/\\log(4/3)$, while at zero it is $1/2$.","The new lower-type theorem does not contradict this: its lower constant\ndepends on the active faster weight and can tend to zero as that channel\nvanishes. For $\\varepsilon=10^{-3}$ one constructive lower constant is\nabout $1.4496\\times10^{-9}$; a conservative upper constant is about $90.4392$.","\\pageheading{PRIME SIEVES\\; /\\; MULTIPLE PARAMETERS}{Overlap is part of the arithmetic}\nFor a finite set $F$ of distinct primes, define\n\\[\n\\nu_{F,\\boldsymbol\\lambda}\n=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})\\delta_{x_n},\n\\quad x_n=(1-n^{-2})^2.\n\\]\nFor real strengths, the exact positivity domain is\n\\[\n\\boxed{\\lambda_p\\ge-1\\quad\\text{for every }p\\in F.}\n\\]\nSufficiency is termwise positivity. If one strength is below $-1$, the\natom $n=p$ isolates its negative factor. The other strengths do not occur\nat that atom. At every simultaneous deletion endpoint, primes outside\n$F$ still supply infinitely many positive nodes, so all Hankel matrices\nremain strictly positive definite.","}{Overlap is part of the arithmetic}\nFor a finite set $F$ of distinct primes, define\n\\[\n\\nu_{F,\\boldsymbol\\lambda}\n=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})\\delta_{x_n},\n\\quad x_n=(1-n^{-2})^2.\n\\]\nFor real strengths, the exact positivity domain is\n\\[\n\\boxed{\\lambda_p\\ge-1\\quad\\text{for every }p\\in F.}\n\\]\nSufficiency is termwise positivity. If one strength is below $-1$, the\natom $n=p$ isolates its negative factor. The other strengths do not occur\nat that atom. At every simultaneous deletion endpoint, primes outside\n$F$ still supply infinitely many positive nodes, so all Hankel matrices\nremain strictly positive definite.","rt of the arithmetic}\nFor a finite set $F$ of distinct primes, define\n\\[\n\\nu_{F,\\boldsymbol\\lambda}\n=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n})\\delta_{x_n},\n\\quad x_n=(1-n^{-2})^2.\n\\]\nFor real strengths, the exact positivity domain is\n\\[\n\\boxed{\\lambda_p\\ge-1\\quad\\text{for every }p\\in F.}\n\\]\nSufficiency is termwise positivity. If one strength is below $-1$, the\natom $n=p$ isolates its negative factor. The other strengths do not occur\nat that atom. At every simultaneous deletion endpoint, primes outside\n$F$ still supply infinitely many positive nodes, so all Hankel matrices\nremain strictly positive definite.","F}(1+\\lambda_p\\mathbf1_{p\\mid n})\\delta_{x_n},\n\\quad x_n=(1-n^{-2})^2.\n\\]\nFor real strengths, the exact positivity domain is\n\\[\n\\boxed{\\lambda_p\\ge-1\\quad\\text{for every }p\\in F.}\n\\]\nSufficiency is termwise positivity. If one strength is below $-1$, the\natom $n=p$ isolates its negative factor. The other strengths do not occur\nat that atom. At every simultaneous deletion endpoint, primes outside\n$F$ still supply infinitely many positive nodes, so all Hankel matrices\nremain strictly positive definite.","The exact coordinate formula is\n\\[\nM_j(\\boldsymbol\\lambda)=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\prod_{p\\in F}\\left(1+\\frac{\\lambda_p}{p^{4+2r}+1}\\right).\n\\]\nCoefficients of the monic families lie in\n$\\mathbb Q(\\pi^{-2},\\lambda_p:p\\in F)$.\nFor two primes, the overlap term is visible:\n\\[\nM_j=M_j(0,0)+\\lambda_pU_{p,j}+\\lambda_qU_{q,j}\n+\\lambda_p\\lambda_qU_{pq,j}.\n\\]\nHere $U_{pq,j}>0$ is the $pq$-divisible submeasure moment. The family\ndoes not reduce to adding separate prime responses without their overlap.","exact coordinate formula is\n\\[\nM_j(\\boldsymbol\\lambda)=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\prod_{p\\in F}\\left(1+\\frac{\\lambda_p}{p^{4+2r}+1}\\right).\n\\]\nCoefficients of the monic families lie in\n$\\mathbb Q(\\pi^{-2},\\lambda_p:p\\in F)$.\nFor two primes, the overlap term is visible:\n\\[\nM_j=M_j(0,0)+\\lambda_pU_{p,j}+\\lambda_qU_{q,j}\n+\\lambda_p\\lambda_qU_{pq,j}.\n\\]\nHere $U_{pq,j}>0$ is the $pq$-divisible submeasure moment. The family\ndoes not reduce to adding separate prime responses without their overlap.","rmula is\n\\[\nM_j(\\boldsymbol\\lambda)=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n\\frac{\\zeta(4+2r)}{\\zeta(8+4r)}\n\\prod_{p\\in F}\\left(1+\\frac{\\lambda_p}{p^{4+2r}+1}\\right).\n\\]\nCoefficients of the monic families lie in\n$\\mathbb Q(\\pi^{-2},\\lambda_p:p\\in F)$.\nFor two primes, the overlap term is visible:\n\\[\nM_j=M_j(0,0)+\\lambda_pU_{p,j}+\\lambda_qU_{q,j}\n+\\lambda_p\\lambda_qU_{pq,j}.\n\\]\nHere $U_{pq,j}>0$ is the $pq$-divisible submeasure moment. The family\ndoes not reduce to adding separate prime responses without their overlap.","\\pageheading{PRIME SIEVES\\; /\\; FINITE TESTS}{Moment positivity is a weaker requirement}\n\\fig{12_multi_prime_domain}{The exact positive-measure domain and the\norder-one moment test have different parameter geometry. Regions passing\na finite test outside the safe orthant still contain negative atoms.}\nFor primes two and three,\n\\[\nM_0=105\\pi^{-4}(1+\\lambda_2/17)(1+\\lambda_3/82).\n\\]\nBoth $(0,0)$ and $(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","pageheading{PRIME SIEVES\\; /\\; FINITE TESTS}{Moment positivity is a weaker requirement}\n\\fig{12_multi_prime_domain}{The exact positive-measure domain and the\norder-one moment test have different parameter geometry. Regions passing\na finite test outside the safe orthant still contain negative atoms.}\nFor primes two and three,\n\\[\nM_0=105\\pi^{-4}(1+\\lambda_2/17)(1+\\lambda_3/82).\n\\]\nBoth $(0,0)$ and $(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","sitivity is a weaker requirement}\n\\fig{12_multi_prime_domain}{The exact positive-measure domain and the\norder-one moment test have different parameter geometry. Regions passing\na finite test outside the safe orthant still contain negative atoms.}\nFor primes two and three,\n\\[\nM_0=105\\pi^{-4}(1+\\lambda_2/17)(1+\\lambda_3/82).\n\\]\nBoth $(0,0)$ and $(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","o and three,\n\\[\nM_0=105\\pi^{-4}(1+\\lambda_2/17)(1+\\lambda_3/82).\n\\]\nBoth $(0,0)$ and $(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","\\[\nM_0=105\\pi^{-4}(1+\\lambda_2/17)(1+\\lambda_3/82).\n\\]\nBoth $(0,0)$ and $(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","1+\\lambda_3/82).\n\\]\nBoth $(0,0)$ and $(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.",")$ and $(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","(-34,-164)$ have positive total mass. Their midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","midpoint\n$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","$(-17,-82)$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.",")$ has zero mass. Thus the parameter set passing the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","g the order-one\npositive-definiteness test is nonconvex. This particular midpoint argument\ndoes not establish nonconvexity of the semidefinite test, which permits zero.\nFor that test, take the two points $(0,0)$ and $(-34,-246)$.\nThe normalized masses $M_0/(105\\pi^{-4})$ there are $1$ and $2$.\nTheir convex combination with weights $3/5$ and $2/5$ is\n$(-68/5,-492/5)$, where the normalized mass is $-1/25$.\nThus the order-one semidefinite feasible set is also nonconvex,\nby a separate counterexample.","\\claim{A separate additive family has different geometry}\nFor weights $1+\\sum_{p\\in F}\\gamma_p\\mathbf1_{p\\mid n}$, positivity is\nequivalent to\n\\[\n1+\\sum_{\\gamma_p<0}\\gamma_p\\ge0.\n\\]\nEvery prime subset appears at its squarefree product, proving necessity.\nChecking each atom proves sufficiency. Its finite Hankel matrices are affine\nin $\\gamma$, so the semidefinite feasible sets are convex spectrahedra.\nThe multiplicative family has genuine overlap terms and polynomial parameter\ndependence. These are two distinct deformations, with distinct domains.","\\pageheading{PRIME SIEVES\\; /\\; CONSTRUCTIVE WITNESS}{Every negative prime atom has a finite detector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever","\\pageheading{PRIME SIEVES\\; /\\; CONSTRUCTIVE WITNESS}{Every negative prime atom has a finite detector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs n","\\pageheading{PRIME SIEVES\\; /\\; CONSTRUCTIVE WITNESS}{Every negative prime atom has a finite detector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ re","\\pageheading{PRIME SIEVES\\; /\\; CONSTRUCTIVE WITNESS}{Every negative prime atom has a finite detector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/","\\pageheading{PRIME SIEVES\\; /\\; CONSTRUCTIVE WITNESS}{Every negative prime atom has a finite detector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","prime atom has a finite detector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","tom has a finite detector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","etector}\nFor an integer $a\\ge2$, let\n\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","{a-1}\\frac{x-x_n}{x_a-x_n},\\quad\nC_a=L_a(1),\\quad r_a=\\frac{1-x_{a+1}}{1-x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","x_a}<1,\n\\]\n\\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m.\n\\]\nThis polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","This polynomial kills every lower node, equals one at $x_a$, and has\nabsolute value at most $C_ar_a^m$ at every higher node.\n\\fig{13_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","3_signed_detector}{A rational detector for $p=3$, $\\lambda_3=-1.1$,\n$\\lambda_2=1$. The first exact bound proving negativity is $m=5$, so a\nHankel matrix of order eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","er eight detects the signed defect.}\nSet $C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","od_{q\\in F}\\max(1,|1+\\lambda_q|)$. The summable tail gives\n\\[\n\\int f_{p,m}^2\\dd\\nu\n\\le\\frac{1+\\lambda_p}{p^4}+\n\\frac{C_FC_p^2}{3p^3}r_p^{2m}.\n\\]\nHence the quadratic form is negative whenever\n$pC_FC_p^2r_p^{2m}/3<-(1+\\lambda_p)$.\nFor rational strengths this is an exact rational certificate. It needs no\nnumerical eigenvalue calculation. A depth tilt $x^k$ replaces the negative\natom contribution by $(1+\\lambda_p)x_p^k/p^4$.","\\pageheading{PRIME SIEVES\\; /\\; FIBERS}{Real singularities survive multiple deletions}\nFix all strengths except $\\lambda_p$, with the others at least $-1$.\nThe remaining moment matrix has the form $A_m+(1+\\lambda_p)B_m$, where\n$A_m$ is the $p$-free part and $B_m$ the $p$-divisible part. Both are\npositive definite, even at other deletion endpoints: primes outside $F$\nsupply the first support, and integers $p\\ell$ the second.","ing{PRIME SIEVES\\; /\\; FIBERS}{Real singularities survive multiple deletions}\nFix all strengths except $\\lambda_p$, with the others at least $-1$.\nThe remaining moment matrix has the form $A_m+(1+\\lambda_p)B_m$, where\n$A_m$ is the $p$-free part and $B_m$ the $p$-divisible part. Both are\npositive definite, even at other deletion endpoints: primes outside $F$\nsupply the first support, and integers $p\\ell$ the second.","Thus every one-variable determinant fiber has exact degree $m$ and all\nroots real below $-1$. Its nearest margin satisfies\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}","Thus every one-variable determinant fiber has exact degree $m$ and all\nroots real below $-1$. Its nearest margin satisfies\n\\[\n\\delta_m=\\min_{\\substack{f\\in\\mathbb R[x]\\setminus\\{0\\}\\\\\\deg f<m}}\n\\frac{\\int f^2\\dd\\nu_{p\\text{-free}}}\n{\\int f^2\\dd\\nu_{p\\text{-divisible}}},\\qquad\\delta_m\\downarrow0.\n\\]\nThe detector yields, for $C_\\kappa=\\prod_{q\\ne p}\\max(1,1+\\lambda_q)$,\n\\[\n\\delta_{p+m}\\le\\frac{pC_\\kappa C_p^2}{3}r_p^{2m}.\n\\]\nThe denominator retains the atom $p$; all lower nodes in the numerator are\nkilled. This extends the prime-two bound to every prime and to simultaneous\npositive or deletion strengths. It does not assert simple roots or global\nmultivariate real stability.","deg f<m}}\n\\frac{\\int f^2\\dd\\nu_{p\\text{-free}}}\n{\\int f^2\\dd\\nu_{p\\text{-divisible}}},\\qquad\\delta_m\\downarrow0.\n\\]\nThe detector yields, for $C_\\kappa=\\prod_{q\\ne p}\\max(1,1+\\lambda_q)$,\n\\[\n\\delta_{p+m}\\le\\frac{pC_\\kappa C_p^2}{3}r_p^{2m}.\n\\]\nThe denominator retains the atom $p$; all lower nodes in the numerator are\nkilled. This extends the prime-two bound to every prime and to simultaneous\npositive or deletion strengths. It does not assert simple roots or global\nmultivariate real stability.","\\[\n\\boxed{19{,}871\\text{ exact assertion statements}\\quad+\n5{,}576\\text{ numerical assertion statements passed}.}\n\\]\nThe exact total comprises 19,331 principal checks and 540 separately recorded\nhelper checks in the inverse reconstruction. Operational path and process\nguards are excluded. Counts are assertion statements, not counts of theorems\nor every scalar operation inside a grouped assertion.","\\[\nT_{q,r}(z)=\\sum_{n\\ge0}c_n(q,r)z^{2n+1}\n          =\\sum_{n\\ge0}A_n(q,r)\\frac{z^{2n+1}}{(2n+1)!}.\n\\]","The two coefficient systems are related by \\(A_n=(2n+1)!c_n\\). In\nparticular, \\(A_n\\) is the derivative of order \\(2n+1\\) at zero. The\ncoefficient \\(c_n\\), the derivative \\(A_n\\), and the full EGF sequence\n\\(a_m\\) with \\(a_{2n+1}=A_n, a_{2n}=0\\) are different objects. None of\nthe code silently substitutes one for another.","\\[\n\\boxed{\nA_0=1,\\qquad\nA_n=\\sum_{i=0}^{n-1}\\binom{2n}{2i+1}\n q^{2i+1}r^{2(n-1-i)+1}A_i A_{n-1-i},\\quad n\\ge1.}\n\\]","\\textbf{Proof.} Multiply the ordinary Taylor recurrence by \\((2n+1)!\\).\nThe factor attached to \\(A_iA_j\\), where \\(i+j=n-1\\), is\n\\((2n)!/[(2i+1)!(2j+1)!]=\\binom{2n}{2i+1}\\). This also proves the\nrecurrence directly in the EGF convention. \\(\\square\\)","\\textbf{Polynomial theorem.} For every \\(n\\), \\(A_n\\in\\mathbb Z[q,r]\\),\nall its nonzero coefficients are positive integers, and\n\\(A_n(q,r)=A_n(r,q)\\). Every monomial \\(q^u r^v\\) satisfies","\\[\nu,v\\ge n,\\qquad u\\equiv v\\equiv n\\pmod2,\\qquad u+v\\le n(n+1).\n\\]","\\textbf{Proof.} The recurrence is an induction using positive integer\nbinomial coefficients. Swapping \\(q,r\\) and replacing \\(i\\) by \\(n-1-i\\)\nleaves the sum unchanged. If child indices are \\(i,j\\), their exponent\npairs receive the shifts \\((2i+1,2j+1)\\). The lower bounds follow from\n\\(i+j+1=n\\). The two parities are the parity of \\(i+j+1\\), also \\(n\\).\nFor the degree bound, a recurrence term has total degree at most","\\[\n2n+i(i+1)+j(j+1)\\le n(n+1),\\qquad i+j=n-1,\n\\]","since \\(i(i+1)+j(j+1)\\) is largest at \\((i,j)=(0,n-1)\\) or its swap. The\nbase \\(A_0=1\\) satisfies every assertion. \\(\\square\\)","\\[\n\\begin{aligned}\nA_0={}&1,\\\\\nA_1={}&2qr,\\\\\nA_2={}&8q^2r^2(q^2+r^2),\\\\\nA_3={}&q^3r^3\\bigl[48(q^2+r^2)(q^4+r^4)+80q^2r^2\\bigr].\n\\end{aligned}\n\\]","Consequently each integer pair \\((q,r)\\) gives an integer sequence for\nall \\(n\\), independently of whether its formal Taylor series converges.\nFor example, the first values of \\(A_n(1,2)\\) are\n\\(1,4,160,35200,38041600\\).","For rational channel tuples \\((h_t,q_t,r_t)\\), use","\\[\nF'(z)=1+\\sum_t h_tF(q_tz)F(r_tz),\\qquad F(0)=0.\n\\]","The preceding recurrence is summed over channels with the factor\n\\(h_t\\). In the universal polynomial over the weights and contractions,\nevery term of \\(A_n\\) has weight degree exactly \\(n\\) and total\ncontraction degree at most \\(n(n+1)\\). The weight statement follows from\none new weight at each recurrence step and \\(1+i+j=n\\); the contraction\nbound uses the same induction.","Let \\(H\\) be a common denominator for all channel weights, and \\(D\\) a\ncommon denominator for all channel contractions. Then","\\[\n\\boxed{B_n=H^nD^{n(n+1)}A_n\\in\\mathbb Z\\quad\\hbox{for all }n.}\n\\]","This is a proved normalization, even with signed rational weights. It is\nusually not the smallest possible denominator clearing. The catalog\nrecords both \\(A_n\\) and \\(B_n\\), and the exact multiplier, so that a\nproposed sequence lookup can state its normalization honestly.","The entire-function theorem uses \\(0<q,r<1\\). Specializations with large\npositive integers produce valid formal and integer sequences but do not\nnecessarily produce analytic functions near zero.","\\textbf{Zero-radius theorem.} If \\(q>0\\) and \\(r>1\\), the formal series\nfor \\(T_{q,r}\\) has radius of convergence zero. The same holds after\nswapping the parameters.","\\textbf{Proof.} Every coefficient is positive. In the ordinary\nrecurrence keep only the term \\(i=0\\). This gives","\\[\nc_n\\ge\\frac{q r^{2n-1}}{2n+1}c_{n-1}\n\\quad\\Longrightarrow\\quad\nc_n\\ge\\frac{q^n r^{n^2}}{3\\cdot5\\cdots(2n+1)}.\n\\]","Since the denominator is at most \\((2n+1)^n\\),","\\[\nc_n^{1/(2n+1)}\\ge\nq^{n/(2n+1)}r^{n^2/(2n+1)}(2n+1)^{-n/(2n+1)}\\longrightarrow\\infty.\n\\]","The Cauchy--Hadamard criterion therefore gives radius zero. \\(\\square\\)","More generally, a positive feedback bank with a channel satisfying\n\\(h>0,q>0,r>1\\) has the same obstruction, replacing \\(q^n\\) by\n\\((hq)^n\\). Positivity prevents cancellation. No conclusion of this form\nis asserted for a bank with signed weights.",">{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\nParameters or bank\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\nFirst five \\(A_n\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\nMathematical status\n\\end{minipage}","\\midrule\\noalign{}\n\\endhead\n\\bottomrule\\noalign{}\n\\endlastfoot\n\\((1,1)\\) & \\(1\\), \\(2\\), \\(16\\), \\(272\\), \\(7936\\) & Known tangent derivatives","\\((0,1)\\) & \\(1\\), \\(0\\), \\(0\\), \\(0\\), \\(0\\) & Known linear boundary","\\((1,-1)\\) & \\(1\\), \\(-2\\), \\(16\\), \\(-272\\), \\(7936\\) & Known hyperbolic tangent\nderivatives","\\((1,2)\\) & \\(1\\), \\(4\\), \\(160\\), \\(35200\\), \\(38041600\\) & Integer sequence, radius\nzero","\\((1/2,1/2)\\) & \\(1\\), \\(1/2\\), \\(1/4\\), \\(1/8\\), \\(1/16\\) & Known scaled hyperbolic sine","\\((1/4,3/4)\\) & \\(1\\), \\(3/8\\), \\(45/256\\), \\(21465/262144\\), \\(39833775/1073741824\\) &\nEntire asymmetric family, order one","The complete catalog contains \\textbf{18 rows}, each through \\(n=12\\):\nthree classical or degenerate integer anchors, eight positive integer\nformal specializations, four rational contractions, one entire edge case\nof infinite order, and two normalized feedback banks. Integer\nintegrality and rational clearing are established by the preceding\nproofs, not merely inferred from the finite sample.","\\textbf{Boundary theorem.} For \\(0<q<1\\), the normalized formal solution\nof","\\[\nT'(z)=1+T(qz)T(z),\\qquad T(0)=0\n\\]","\\textbf{Proof of entire extension.} The tangent-coefficient majorant\ngives an initial holomorphic germ. Suppose \\(T\\) is known on \\(|z|<R\\).\nThe function \\(T(qz)\\) is known on \\(|z|<R/q\\). On this larger disk\nsolve the linear ODE \\(U'=1+T(qz)U\\), \\(U(0)=0\\). Explicitly,","\\[\nU(z)=e^{B(z)}\\int_0^z e^{-B(w)}\\,dw,\n\\qquad B(z)=\\int_0^zT(qw)\\,dw.\n\\]","The integrals are path independent on the disk. Linear-ODE uniqueness\nshows \\(U=T\\) on the previous disk, and there \\(T(qz)\\) equals the newly\nextended function at \\(qz\\). Thus the same nonlinear equation holds on\nthe enlarged disk. Iterating \\(R\\mapsto R/q\\) exhausts the plane.","\\textbf{Proof of infinite order.} Positive coefficients imply\n\\(T(x)\\ge x\\) for \\(x>0\\) and \\(M(x)=T(x)\\). The logarithmic derivative\nobeys","\\[\n\\frac{d}{dx}\\log T(x)=\\frac1{T(x)}+T(qx)\\ge qx.\n\\]","Integrating from one, with \\(T(1)\\ge1\\), gives\n\\(T(x)\\ge\\exp(q(x^2-1)/2)\\) for \\(x\\ge1\\). For \\(R\\ge2/q\\), integrate\nthe same logarithmic derivative over \\([R/2,R]\\). Since \\(T(R/2)\\ge1\\),","\\[\n\\boxed{\n\\log M(R)\\ge\\frac R2\\exp\\left(\\frac{q^3R^2}{8}-\\frac q2\\right).}\n\\]","Therefore \\(\\log\\log M(R)/\\log R\\to\\infty\\), which proves infinite\norder. \\(\\square\\)","\\midrule\\noalign{}\n\\endhead\n\\bottomrule\\noalign{}\n\\endlastfoot\n\\(0<q,r<1\\) & Entire, finite order defined by \\(q^\\rho+r^\\rho=1\\)","One parameter equals one, the other lies in \\((0,1)\\) & Entire, infinite\norder","\\(q=r=1\\) & Tangent, meromorphic with its usual poles","\\(q>0,r>0,\\max(q,r)>1\\) & Formal solution with radius zero","One parameter equals zero & Linear solution \\(z\\), regardless of the\nother scale","For an EGF \\(G(z)=\\sum_{m\\ge0}a_m z^m/m!\\), an integer base \\(b\\ge1\\), a\nresidue \\(0\\le j<b\\), and \\(\\omega=e^{2\\pi i/b}\\), define","\\[\n\\boxed{\n\\mathcal P_{b,j}G(z)=\\frac1b\\sum_{k=0}^{b-1}\\omega^{-jk}G(\\omega^kz)\n=\\sum_{m\\equiv j\\,(b)}a_m\\frac{z^m}{m!}.}\n\\]","This holds as a formal series and wherever the original series\nconverges. Indeed the coefficient multiplier is the geometric sum\n\\(b^{-1}\\sum_{k=0}^{b-1}\\omega^{k(m-j)}\\), equal to one or zero\naccording to the residue. It proves all the projector identities","\\[\n\\sum_{j=0}^{b-1}\\mathcal P_{b,j}=I,\\qquad\n\\mathcal P_{b,j}\\mathcal P_{b,l}=\\delta_{jl}\\mathcal P_{b,j}.\n\\]","The full-position section and the decimated lane are distinct arrays. A\nsection keeps the original positions, inserting zeros outside the\nresidue. A lane \\(x_j[n]=a_{bn+j}\\) changes the index to \\(n\\).","For an \\textbf{ordinary} generating function \\(X(z)=\\sum_ma_m z^m\\),\ndefine \\(X_j(w)=\\sum_na_{bn+j}w^n\\). Then","\\[\nX(z)=\\sum_{j=0}^{b-1}z^jX_j(z^b).\n\\]","\\[\nG(z)=\\sum_{j=0}^{b-1}z^j\n \\sum_{n\\ge0}\\frac{a_{bn+j}}{(bn+j)!}(z^b)^n.\n\\]","One must retain \\((bn+j)!\\). Replacing it by \\(n!\\) gives a different\nfunction. The root-filter section identity has no such factorial\nambiguity.","Let \\(h_0,\\ldots,h_{L-1}\\) be a finite tap vector and let \\(a_m=0\\) for\n\\(m<0\\). Ordinary taps form","\\[\nb_m=\\sum_{t=0}^{L-1}h_ta_{m-t}.\n\\]","For OGFs this is multiplication by \\(H(z)=\\sum_th_tz^t\\):\n\\(B(z)=H(z)X(z)\\). For EGFs it is a sum of repeated zero-constant\nantiderivatives:","\\[\n\\boxed{B_{\\rm EGF}(z)=\\sum_{t=0}^{L-1}h_t I^tG(z),\\qquad\nIG(z)=\\int_0^zG(w)\\,dw.}\n\\]","\\textbf{Proof.} The EGF of the delayed sequence is\n\\(\\sum_{m\\ge t}a_{m-t}z^m/m!=I^tG(z)\\). Summing taps gives the formula.\nConversely, \\(G^{(t)}\\) has coefficient sequence \\(a_{m+t}\\), so\nderivatives sample future coefficients and do not implement these past\ndelays. \\(\\square\\)","For two EGFs \\(G_a,G_h\\), ordinary function multiplication instead gives\nthe \\textbf{Hurwitz product}","\\[\nG_a(z)G_h(z)=\\sum_m\\left[\n\\sum_{t=0}^m\\binom mt h_ta_{m-t}\\right]\\frac{z^m}{m!}.\n\\]","This is a binomial convolution. For example, with \\(a_m=1\\) and taps\n\\(h=(1,1)\\), ordinary taps give \\(b_2=2\\), while the Hurwitz product\ngives \\(b_2=3\\). This explicit negative control catches a common\nnormalization error.","\\textbf{Intrinsic nonlinear feedback taps} are the parameters\n\\((h_t,q_t,r_t)\\) inside \\(F'=1+\\sum_th_tF(q_tz)F(r_tz)\\). They change\nthe function's defining equation and its derivative recurrence. Ordinary\nFIR taps are a linear transformation of an already defined coefficient\nsequence. Both are useful, but they operate on different objects. The\nintrinsic-bank growth theorem is proved in Appendix~\\ref{app:growth}.","For an OGF sequence \\(a_m\\), define lanes \\(x_l[n]=a_{bn+l}\\) and\n\\(X_l(w)=\\sum_nx_l[n]w^n\\). Apply finite ordinary taps and obtain\n\\(y_j[n]=b_{bn+j}\\). Then the exact matrix identity is","\\[\n\\boxed{Y_j(w)=\\sum_{l=0}^{b-1}H_{jl}(w)X_l(w),\\qquad\nH_{jl}(w)=\\sum_{\\substack{0\\le t<L\\\\t+l\\equiv j\\,(b)}}\nh_t w^{(t+l-j)/b}.}\n\\]","\\textbf{Proof.} A term \\(h_t a_{bn+j-t}\\) is in phase \\(l\\) exactly when\n\\(t+l\\equiv j\\pmod b\\). Write \\(d=(t+l-j)/b\\). That term is\n\\(h_t x_l[n-d]\\), whose OGF is \\(h_tw^dX_l(w)\\) by zero extension.\nSumming over taps and phases gives the matrix. \\(\\square\\)","\\[\n\\begin{pmatrix}Y_0\\\\Y_1\\end{pmatrix}\n=\\begin{pmatrix}h_0&h_1w\\\\h_1&h_0\\end{pmatrix}\n\\begin{pmatrix}X_0\\\\X_1\\end{pmatrix}.\n\\]","The upper-right factor \\(w\\) is necessary: a delay from odd phase to\neven phase crosses one block. A diagonal-only phase operation cannot\nimplement this filter. The code constructs the sparse matrix for any\nfinite base and tap vector and checks that recombination gives the full\ndirect convolution.","\\begin{itemize}\n\\tightlist\n\\item\n  18 catalog rows through \\(A_{12}\\), independently compared with dense\n  formal Picard iteration through degree 25.\n\\item\n  Bivariate polynomials through \\(A_8\\), with 540 nonzero monomials\n  checked for integer positivity, symmetry, parity and the degree bound.\n\\item\n  Tangent coefficients independently obtained by division of sine and\n  cosine, plus hyperbolic tangent, scaled hyperbolic sine and linear anchors.\n\\item\n  Root-filter multipliers reduced exactly in cyclotomic fields for bases\n  one through seven and indices zero through 31.\n\\item\n  Direct FIR convolution compared with phase-matrix convolution for 26\n  input lengths, seven bases and four tap vectors, including signed\n  rational taps.\n\\item\n  The antiderivative","\\begin{itemize}\n\\tightlist\n\\item\n  18 catalog rows through \\(A_{12}\\), independently compared with dense\n  formal Picard iteration through degree 25.\n\\item\n  Bivariate polynomials through \\(A_8\\), with 540 nonzero monomials\n  checked for integer positivity, symmetry, parity and the degree bound.\n\\item\n  Tangent coefficients independently obtained by division of sine and\n  cosine, plus hyperbolic tangent, scaled hyperbolic sine and linear anchors.\n\\item\n  Root-filter multipliers reduced exactly in cyclotomic fields for bases\n  one through seven and indices zero through 31.\n\\item\n  Direct FIR convolution compared with phase-matrix convolution for 26\n  input lengths, seven bases and four tap vectors, including signed\n  rational taps.\n\\item\n  The antiderivative interpretation of EGF taps and the factorial\n  conversion of Hurwitz products, independently compared with ordinary\n  Taylor","and run time. It\nreports no failed assertion in this run. Independent audit identified\ntwo API input-validation defects: a fractional residue was accepted by\nthe section operator, and an empty phase matrix accepted base zero. Both\nare now rejected and have passing regression checks. These defects\naffected invalid inputs, not the displayed identities. A draft prose\nexample initially transcribed \\(A_3(1,2)\\) and \\(A_4(1,2)\\) incorrectly;\nthe exact catalog comparison exposed the mismatch and corrected them to\n\\(35200\\) and \\(38041600\\). The code result did not change. Larger\nnumerical evidence is unnecessary for these algebraic identities; an\nindependent mathematical audit should examine the reductions.","reports no failed assertion in this run. Independent audit identified\ntwo API input-validation defects: a fractional residue was accepted by\nthe section operator, and an empty phase matrix accepted base zero. Both\nare now rejected and have passing regression checks. These defects\naffected invalid inputs, not the displayed identities. A draft prose\nexample initially transcribed \\(A_3(1,2)\\) and \\(A_4(1,2)\\) incorrectly;\nthe exact catalog comparison exposed the mismatch and corrected them to\n\\(35200\\) and \\(38041600\\). The code result did not change. Larger\nnumerical evidence is unnecessary for these algebraic identities; an\nindependent mathematical audit should examine the reductions.","fects: a fractional residue was accepted by\nthe section operator, and an empty phase matrix accepted base zero. Both\nare now rejected and have passing regression checks. These defects\naffected invalid inputs, not the displayed identities. A draft prose\nexample initially transcribed \\(A_3(1,2)\\) and \\(A_4(1,2)\\) incorrectly;\nthe exact catalog comparison exposed the mismatch and corrected them to\n\\(35200\\) and \\(38041600\\). The code result did not change. Larger\nnumerical evidence is unnecessary for these algebraic identities; an\nindependent mathematical audit should examine the reductions.","ional residue was accepted by\nthe section operator, and an empty phase matrix accepted base zero. Both\nare now rejected and have passing regression checks. These defects\naffected invalid inputs, not the displayed identities. A draft prose\nexample initially transcribed \\(A_3(1,2)\\) and \\(A_4(1,2)\\) incorrectly;\nthe exact catalog comparison exposed the mismatch and corrected them to\n\\(35200\\) and \\(38041600\\). The code result did not change. Larger\nnumerical evidence is unnecessary for these algebraic identities; an\nindependent mathematical audit should examine the reductions.","\\pageheading{EXACT LABORATORY\\; /\\; COEFFICIENTS}{Integer coordinates for the nonlinear family}\nNormalize the Taylor coefficients by\n\\[\nA_n(q,r)=(2n+1)!c_n(q,r),\\qquad A_0=1.\n\\]\nThe recurrence becomes\n\\[\n\\boxed{A_n=\\sum_{i=0}^{n-1}\\binom{2n}{2i+1}\nq^{2i+1}r^{2(n-1-i)+1}A_iA_{n-1-i},\\quad n\\ge1.}\n\\]\nTherefore $A_n\\in\\mathbb Z[q,r]$ has nonnegative coefficients and is\nsymmetric in its two variables. The first coordinates are\n\\[\n\\begin{aligned}\nA_1&=2qr,","\\; /\\; COEFFICIENTS}{Integer coordinates for the nonlinear family}\nNormalize the Taylor coefficients by\n\\[\nA_n(q,r)=(2n+1)!c_n(q,r),\\qquad A_0=1.\n\\]\nThe recurrence becomes\n\\[\n\\boxed{A_n=\\sum_{i=0}^{n-1}\\binom{2n}{2i+1}\nq^{2i+1}r^{2(n-1-i)+1}A_iA_{n-1-i},\\quad n\\ge1.}\n\\]\nTherefore $A_n\\in\\mathbb Z[q,r]$ has nonnegative coefficients and is\nsymmetric in its two variables. The first coordinates are\n\\[\n\\begin{aligned}\nA_1&=2qr,\\\\\nA_2&=8q^2r^2(q^2+r^2),\\\\\nA_3&=q^3r^3[48(q^2+r^2)(q^4+r^4)+80q^2r^2].\n\\end{aligned}\n\\]\nAt $(q,r)=(1,1)$ these are the tangent numbers\n$1,2,16,272,7936,\\ldots$, the known OEIS A000182 anchor.\nAt $(1,2)$ the exact sequence begins\n$1,4,160,35200,38041600,\\ldots$.\n\\smallfig{17_tree_addresses}{Ordered trees retain left/right path information.\nThe two displayed shapes contribute different monomials to the same\nfactorial-normalized coefficient. Their labels are exact tree data.}\nAt rational scales, an integer denominator clearing is available. If $D$\nclears every contraction denominator and $H$ clears every bank weight,\n\\[\nH^nD^{n(n+1)}A_n\\in\\mathbb Z.\n\\]\nThe maximum roo…","A_3&=q^3r^3[48(q^2+r^2)(q^4+r^4)+80q^2r^2].\n\\end{aligned}\n\\]\nAt $(q,r)=(1,1)$ these are the tangent numbers\n$1,2,16,272,7936,\\ldots$, the known OEIS A000182 anchor.\nAt $(1,2)$ the exact sequence begins\n$1,4,160,35200,38041600,\\ldots$.\n\\smallfig{17_tree_addresses}{Ordered trees retain left/right path information.\nThe two displayed shapes contribute different monomials to the same\nfactorial-normalized coefficient. Their labels are exact tree data.}\nAt rational scales, an integer denominator clearing is available. If $D$\nclears every contraction denominator and $H$ clears every bank weight,\n\\[\nH^nD^{n(n+1)}A_n\\in\\mathbb Z.\n\\]\nThe maximum root-path exponent sum over a full binary tree is $n(n+1)$.\nThe clearing provides exact sequence exports without implying a new OEIS","A_3&=q^3r^3[48(q^2+r^2)(q^4+r^4)+80q^2r^2].\n\\end{aligned}\n\\]\nAt $(q,r)=(1,1)$ these are the tangent numbers\n$1,2,16,272,7936,\\ldots$, the known OEIS A000182 anchor.\nAt $(1,2)$ the exact sequence begins\n$1,4,160,35200,38041600,\\ldots$.\n\\smallfig{17_tree_addresses}{Ordered trees retain left/right path information.\nThe two displayed shapes contribute different monomials to the same\nfactorial-normalized coefficient. Their labels are exact tree data.}\nAt rational scales, an integer denominator clearing is available. If $D$\nclears every contraction denominator and $H$ clears every bank weight,\n\\[\nH^nD^{n(n+1)}A_n\\in\\mathbb Z.\n\\]\nThe maximum root-path exponent sum over a full binary tree is $n(n+1)$.\nThe clearing provides exact sequence exports without implying a new OEIS entry.","936,\\ldots$, the known OEIS A000182 anchor.\nAt $(1,2)$ the exact sequence begins\n$1,4,160,35200,38041600,\\ldots$.\n\\smallfig{17_tree_addresses}{Ordered trees retain left/right path information.\nThe two displayed shapes contribute different monomials to the same\nfactorial-normalized coefficient. Their labels are exact tree data.}\nAt rational scales, an integer denominator clearing is available. If $D$\nclears every contraction denominator and $H$ clears every bank weight,\n\\[\nH^nD^{n(n+1)}A_n\\in\\mathbb Z.\n\\]\nThe maximum root-path exponent sum over a full binary tree is $n(n+1)$.\nThe clearing provides exact sequence exports without implying a new OEIS entry.","t $(1,2)$ the exact sequence begins\n$1,4,160,35200,38041600,\\ldots$.\n\\smallfig{17_tree_addresses}{Ordered trees retain left/right path information.\nThe two displayed shapes contribute different monomials to the same\nfactorial-normalized coefficient. Their labels are exact tree data.}\nAt rational scales, an integer denominator clearing is available. If $D$\nclears every contraction denominator and $H$ clears every bank weight,\n\\[\nH^nD^{n(n+1)}A_n\\in\\mathbb Z.\n\\]\nThe maximum root-path exponent sum over a full binary tree is $n(n+1)$.\nThe clearing provides exact sequence exports without implying a new OEIS entry.","gins\n$1,4,160,35200,38041600,\\ldots$.\n\\smallfig{17_tree_addresses}{Ordered trees retain left/right path information.\nThe two displayed shapes contribute different monomials to the same\nfactorial-normalized coefficient. Their labels are exact tree data.}\nAt rational scales, an integer denominator clearing is available. If $D$\nclears every contraction denominator and $H$ clears every bank weight,\n\\[\nH^nD^{n(n+1)}A_n\\in\\mathbb Z.\n\\]\nThe maximum root-path exponent sum over a full binary tree is $n(n+1)$.\nThe clearing provides exact sequence exports without implying a new OEIS entry.","t path information.\nThe two displayed shapes contribute different monomials to the same\nfactorial-normalized coefficient. Their labels are exact tree data.}\nAt rational scales, an integer denominator clearing is available. If $D$\nclears every contraction denominator and $H$ clears every bank weight,\n\\[\nH^nD^{n(n+1)}A_n\\in\\mathbb Z.\n\\]\nThe maximum root-path exponent sum over a full binary tree is $n(n+1)$.\nThe clearing provides exact sequence exports without implying a new OEIS entry.","\\pageheading{EXACT LABORATORY\\; /\\; REGIMES}{A formal sequence is not always an analytic germ}\n\\smallfig[0.68]{18_parameter_regimes}{The full nonnegative real parameter map separates\nfinite-order entire functions, infinite-order entire edges, tangent, and\nzero-radius formal series. Zero scales give the linear function.}\nIf $q>0$, $r>1$, the recurrence includes its $i=0$ term and gives\n\\[\nc_n\\ge\\frac{q^nr^{n^2}}{3\\cdot5\\cdots(2n+1)}.\n\\]\nThe root test proves zero radius. By symmetry this covers either scale above\none, provided the other is positive. The exported positive integer asymmetric\nrows belong to this formal regime.","If $0<q<1$, $r=1$, the equation is linear in the uncontracted value:\n$T'=1+T(qz)T(z)$. A known coefficient on $|z|<R/q$ gives a linear ODE\nthere, so disk bootstrapping proves entire extension. On the positive ray,\n$(\\log T)'=T(qR)+1/T(R)\\ge qR$, giving an exponential lower bound for\n$T$. Substitution back gives, for $R\\ge2/q$,\n\\[\n\\log T(R)\\ge\\frac R2\\exp(q^3R^2/8-q/2).\n\\]\nThus the entire order is infinite. The corner $(1,1)$ is tangent with poles.\nThese exact boundaries protect the analytic interpretation of the catalog.","equation is linear in the uncontracted value:\n$T'=1+T(qz)T(z)$. A known coefficient on $|z|<R/q$ gives a linear ODE\nthere, so disk bootstrapping proves entire extension. On the positive ray,\n$(\\log T)'=T(qR)+1/T(R)\\ge qR$, giving an exponential lower bound for\n$T$. Substitution back gives, for $R\\ge2/q$,\n\\[\n\\log T(R)\\ge\\frac R2\\exp(q^3R^2/8-q/2).\n\\]\nThus the entire order is infinite. The corner $(1,1)$ is tangent with poles.\nThese exact boundaries protect the analytic interpretation of the catalog.","\\pageheading{EXACT LABORATORY\\; /\\; PROJECTORS}{Polyphase sections preserve the full sequence}\nFor an ordinary generating function $G(z)=\\sum_{m\\ge0}a_mz^m$ and\ninteger $b\\ge1$, write\n\\[\nG(z)=\\sum_{j=0}^{b-1}z^jG_j(z^b),\\qquad\nG_j(t)=\\sum_{n\\ge0}a_{bn+j}t^n.\n\\]\nThe root projector is\n\\[\nP_{b,j}G(z)=\\frac1b\\sum_{k=0}^{b-1}\\omega^{-jk}G(\\omega^kz)\n=z^jG_j(z^b),\\qquad\\omega=e^{2\\pi i/b}.\n\\]\nThe same projector works on an EGF, but the section retains its original\nfactorials: $\\sum_na_{bn+j}z^{bn+j}/(bn+j)!$.\nRemoving those denominators changes the representation and requires an\nexplicit transport. Root filtering does not do it automatically.","the full sequence}\nFor an ordinary generating function $G(z)=\\sum_{m\\ge0}a_mz^m$ and\ninteger $b\\ge1$, write\n\\[\nG(z)=\\sum_{j=0}^{b-1}z^jG_j(z^b),\\qquad\nG_j(t)=\\sum_{n\\ge0}a_{bn+j}t^n.\n\\]\nThe root projector is\n\\[\nP_{b,j}G(z)=\\frac1b\\sum_{k=0}^{b-1}\\omega^{-jk}G(\\omega^kz)\n=z^jG_j(z^b),\\qquad\\omega=e^{2\\pi i/b}.\n\\]\nThe same projector works on an EGF, but the section retains its original\nfactorials: $\\sum_na_{bn+j}z^{bn+j}/(bn+j)!$.\nRemoving those denominators changes the representation and requires an\nexplicit transport. Root filtering does not do it automatically.","For causal FIR taps $h_\\ell$, $b_m=\\sum_\\ell h_\\ell a_{m-\\ell}$ with\nzero extension to negative indices, split $A(z)$ and $B(z)$ into $b$ phases.\nThen $B_i(t)=\\sum_jH_{ij}(t)A_j(t)$, where\n\\[\n\\boxed{H_{ij}(t)=\\sum_{\\substack{\\ell\\ge0","For causal FIR taps $h_\\ell$, $b_m=\\sum_\\ell h_\\ell a_{m-\\ell}$ with\nzero extension to negative indices, split $A(z)$ and $B(z)$ into $b$ phases.\nThen $B_i(t)=\\sum_jH_{ij}(t)A_j(t)$, where\n\\[\n\\boxed{H_{ij}(t)=\\sum_{\\substack{\\ell\\ge0\\\\j+\\ell\\equiv i\\pmod b}}\nh_\\ell t^{(j+\\ell-i)/b}.}\n\\]\nThe exponent is the nonnegative carry to the next phase block. This is an\nexact coefficient law, with no approximation or loss of terms.","\\pageheading{EXACT LABORATORY\\; /\\; TAPS}{The carry matrix is part of the filter}\n\\smallfig{16_polyphase}{A three-phase filter with taps $[1,2,-1]$.\nThe matrix carries terms across block boundaries. Interleaving its output\nreproduces the direct scalar FIR calculation exactly.}\nWith these taps and $b=3$,\n\\[\nH(t)=\\begin{pmatrix}\n1&-t&2t","\\pageheading{EXACT LABORATORY\\; /\\; TAPS}{The carry matrix is part of the filter}\n\\smallfig{16_polyphase}{A three-phase filter with taps $[1,2,-1]$.\nThe matrix carries terms across block boundaries. Interleaving its output\nreproduces the direct scalar FIR calculation exactly.}\nWith these taps and $b=3$,\n\\[\nH(t)=\\begin{pmatrix}\n1&-t&2t\\\\2&1&-t\\\\-1&2&1\n\\end{pmatrix}.\n\\]\nThe entries arise from the same congruence condition. For instance an input\nin phase two shifted by one lands in phase zero of the next block, contributing\n$2t$ to $H_{0,2}$.","-1&2&1\n\\end{pmatrix}.\n\\]\nThe entries arise from the same congruence condition. For instance an input\nin phase two shifted by one lands in phase zero of the next block, contributing\n$2t$ to $H_{0,2}$.","\\claim{Three operations that must be kept distinct}\nFor an EGF $E_a(z)=\\sum_ma_mz^m/m!$, differentiation transports future\ncoefficients: $E_a^{(k)}=\\sum_ma_{m+k}z^m/m!$.\nCausal ordinary taps instead act by repeated zero-constant integration:\n\\[\nE_b=\\sum_\\ell h_\\ell I^\\ell E_a.\n\\]\nMultiplication of two EGFs gives Hurwitz convolution,\n$b_m=\\sum_{k=0}^m\\binom mk a_kh_{m-k}$.\nIt is not ordinary sequence FIR convolution. Intrinsic nonlinear feedback\ntaps alter the defining family and are a fourth operation with their own\ncoefficient recurrence.","midrule\n$(1,1)$ & $1,2,16,272,7936$ & Known tangent anchor","$(1,2)$ & $1,4,160,35200,38041600$ & Integer, formal radius zero","$(1/2,1/2)$ & $1,1/2,1/4,1/8,1/16$ & Scaled sinh, entire order one","$(1/3,2/3)$ & $1,4/9,160/729,18560/177147,\\ldots$ & Asymmetric entire order one","Let \\(F\\) be a finite set of distinct primes, let\n\\(\\boldsymbol\\lambda=(\\lambda_p)_{p\\in F}\\), and put","\\[\nx_n=(1-n^{-2})^2,\\qquad\n\\kappa_F(n;\\boldsymbol\\lambda)=\\prod_{p\\in F}(1+\\lambda_p\\mathbf1_{p\\mid n}),\n\\] \\[\n\\nu_{F,\\boldsymbol\\lambda}\n=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\\kappa_F(n;\\boldsymbol\\lambda)\\delta_{x_n}.\n\\]","The mass at \\(x_1=0\\) is exactly 1. A strength \\(\\lambda_p=-1\\) deletes\nevery squarefree atom divisible by \\(p\\). Several deletions commute. The\ndepth-\\(k\\) family uses \\(x^k\\,d\\nu\\), for integers \\(k\\ge0\\).","Write \\(D(s)=\\zeta(s)/\\zeta(2s)\\). For arbitrary finite complex\nstrengths and \\(\\Re s>1\\), absolute convergence permits expansion of the\nfinite prime product. The local squarefree Euler factors give","\\[\nZ_F(s;\\boldsymbol\\lambda)\n=\\sum_{n\\ge1}\\mu(n)^2n^{-s}\\kappa_F(n;\\boldsymbol\\lambda)\n=D(s)\\prod_{p\\in F}\\left(1+\\frac{\\lambda_p}{p^s+1}\\right).\n\\]","Equivalently, expansion in subsets \\(S\\subseteq F\\) gives the\ncoefficient \\(D(s)/\\prod_{p\\in S}(p^s+1)\\) for\n\\(\\prod_{p\\in S}\\lambda_p\\). This is an independently useful coordinate\nformula for prime overlap.","\\[\nM_j(\\boldsymbol\\lambda)=\\int x^j\\,d\\nu_{F,\\boldsymbol\\lambda}\n=\\sum_{r=0}^{2j}(-1)^r\\binom{2j}{r}\n D(4+2r)\\prod_{p\\in F}\n \\left(1+\\frac{\\lambda_p}{p^{4+2r}+1}\\right).\n\\tag{1}\n\\]","Depth-\\(k\\) moments are \\(M_{k+j}\\). With \\(v=\\pi^{-2}\\), the even-zeta\nformula makes every \\(D(4+2r)\\) a rational multiple of \\(v^{r+2}\\). Thus\nthe moments belong to \\(\\mathbb Q[v,\\lambda_p:p\\in F]\\), and monic\northogonal-polynomial coefficients belong to\n\\(\\mathbb Q(v,\\lambda_p:p\\in F)\\). This is an exact field statement,\nwith no decimal recognition step.","\\[\nM_j(\\lambda_p,\\lambda_q)=M_j(0,0)+\\lambda_p U_{p,j}\n+\\lambda_q U_{q,j}+\\lambda_p\\lambda_q U_{pq,j},\n\\]","where \\(U_{pq,j}=\\sum_{pq\\mid n}\\mu(n)^2n^{-4}x_n^j>0\\). More generally,\nwith all other strengths at least \\(-1\\), every mixed moment derivative\nin a set of distinct primes is strictly positive. It is the integral\nover atoms divisible by those primes, with the remaining nonnegative\nprime factors. Infinitely many such atoms remain after any finite\ncollection of deletions. Pure second derivatives in any one strength\nvanish because the moments are separately affine.","\\textbf{Theorem C.1.} For real strengths, \\(\\nu_{F,\\boldsymbol\\lambda}\\)\nis a positive measure if and only if \\(\\lambda_p\\ge-1\\) for every\n\\(p\\in F\\). On this closed orthant, every finite Hankel matrix of every\nnonnegative integer depth is strictly positive definite.","\\textbf{Proof.} The stated inequalities make each atom weight\nnonnegative. The primes outside \\(F\\) supply infinitely many positive\natoms, even when all strengths are deletion endpoints. A nonzero\npolynomial cannot vanish on infinitely many distinct nodes; therefore\nits squared norm is strictly positive. Conversely, if \\(\\lambda_p<-1\\),\nthe squarefree atom \\(n=p\\) has mass \\((1+\\lambda_p)p^{-4}<0\\): no other\nstrength appears at that atom. The same argument works after every depth\ntilt because \\(x_p>0\\). \\(\\square\\)","supply infinitely many positive\natoms, even when all strengths are deletion endpoints. A nonzero\npolynomial cannot vanish on infinitely many distinct nodes; therefore\nits squared norm is strictly positive. Conversely, if \\(\\lambda_p<-1\\),\nthe squarefree atom \\(n=p\\) has mass \\((1+\\lambda_p)p^{-4}<0\\): no other\nstrength appears at that atom. The same argument works after every depth\ntilt because \\(x_p>0\\). \\(\\square\\)","nitely many positive\natoms, even when all strengths are deletion endpoints. A nonzero\npolynomial cannot vanish on infinitely many distinct nodes; therefore\nits squared norm is strictly positive. Conversely, if \\(\\lambda_p<-1\\),\nthe squarefree atom \\(n=p\\) has mass \\((1+\\lambda_p)p^{-4}<0\\): no other\nstrength appears at that atom. The same argument works after every depth\ntilt because \\(x_p>0\\). \\(\\square\\)","The positive monic families \\(P_m^{(k;F,\\boldsymbol\\lambda)}\\)\nconsequently exist at all degrees throughout the closed orthant. Their\nHankel determinants never vanish there, including at simultaneous\ndeletion endpoints. Each finite-degree coefficient is analytic in a\nneighborhood of every point of that orthant. This neighborhood is a\nfinite-degree statement; the next theorem explains why no common\nenlargement of the positivity domain works at all degrees.","Together, Theorems C.1 and C.2 also prove the exact equivalence: every\nfinite Hankel matrix is positive semidefinite if and only if every\nstrength is at least \\(-1\\). Thus this is a complete positivity-domain\nresult, not only a sufficient condition.","For an integer \\(a\\ge2\\), define","\\[\nL_a(x)=\\prod_{n=1}^{a-1}\\frac{x-x_n}{x_a-x_n},\\qquad\nC_a=L_a(1)=\\prod_{n=1}^{a-1}\\frac{1-x_n}{x_a-x_n}>0,\n\\] \\[\nr_a=\\frac{1-x_{a+1}}{1-x_a}\n=\\frac{a^4\\bigl(2(a+1)^2-1\\bigr)}{(a+1)^4(2a^2-1)}\\in(0,1),\n\\] \\[\nf_{a,m}(x)=L_a(x)\\left(\\frac{1-x}{1-x_a}\\right)^m,\n\\qquad m=0,1,2,\\ldots.\n\\tag{2}\n\\]","The degree is \\(a-1+m\\); all lower nodes are killed exactly;\n\\(f_{a,m}(x_a)=1\\). At every higher node,\n\\(|f_{a,m}(x_n)|\\le C_a r_a^m\\), because each factor of \\(L_a\\)\nincreases on \\([x_a,1]\\) and \\(1-x_n\\le1-x_{a+1}\\).","Set \\(C_F=\\prod_{q\\in F}\\max(1,|1+\\lambda_q|)\\). Then\n\\(|\\kappa_F(n)|\\le C_F\\), and the elementary decreasing-series bound\ngives","\\[\n\\sum_{n>a}|\\mu(n)^2n^{-4}\\kappa_F(n)|\n\\le\\frac{C_F}{3a^3}.\n\\tag{3}\n\\]","\\textbf{Theorem C.2.} If \\(\\lambda_p<-1\\), then","\\[\n\\int f_{p,m}(x)^2\\,d\\nu_{F,\\boldsymbol\\lambda}(x)\n\\le\\frac{1+\\lambda_p}{p^4}\n+\\frac{C_F C_p^2}{3p^3}r_p^{2m}.\n\\tag{4}\n\\]","Hence the Hankel matrix of order \\(p+m\\) is not positive semidefinite\nwhenever","\\[\n\\boxed{\\frac{p C_F C_p^2}{3}r_p^{2m}<-(1+\\lambda_p).}\n\\tag{5}\n\\]","All quantities in (5) are rational when the strengths are rational. It\nprovides an exact finite certificate without calculating Hankel\neigenvalues or relying on floating-point root finding. A suitable finite\n\\(m\\) exists because \\(r_p<1\\).","\\textbf{Proof.} The polynomial kills all atoms below \\(p\\), equals 1 at\nthe negative atom \\(p\\), and has the stated uniform bound above \\(p\\).\nReplacing all higher signed contributions by their absolute values and\napplying (3) proves (4). The coefficients of \\(f_{p,m}\\) then supply a\nnegative quadratic form for \\([M_{i+j}]_{i,j<p+m}\\) when (5) holds.\n\\(\\square\\)","For depth \\(k\\), the same polynomial satisfies","\\[\n\\int x^k f_{p,m}(x)^2\\,d\\nu\n\\le\\frac{(1+\\lambda_p)x_p^k}{p^4}\n+\\frac{C_F C_p^2}{3p^3}r_p^{2m}.\n\\]","Replace the right side of (5) by \\(-(1+\\lambda_p)x_p^k\\) to obtain an\nexplicit depth-\\(k\\) detector. This proof extends to any finite signed\natomic measure with these nodes and the same summable weight envelope.\nIts mechanism is an isolating polynomial followed by exponential\nsuppression of the higher nodes.","Fix \\(p\\in F\\), with all other strengths at least \\(-1\\), and write\n\\(\\kappa(n)=\\prod_{q\\in F\\setminus\\{p\\}}(1+\\lambda_q\\mathbf1_{q\\mid n})\\).\nSeparate the positive measure into its \\(p\\)-free and \\(p\\)-divisible\nparts:","\\[\nA_m=[\\sum_{p\\nmid n}\\mu(n)^2n^{-4}\\kappa(n)x_n^{i+j}]_{i,j<m},\n\\quad\nB_m=[\\sum_{p\\mid n}\\mu(n)^2n^{-4}\\kappa(n)x_n^{i+j}]_{i,j<m}.\n\\]","Both matrices are positive definite. For \\(A_m\\), use infinitely many\nprimes outside \\(F\\); for \\(B_m\\), use infinitely many squarefree\nintegers \\(p\\ell\\), with primes \\(\\ell\\) outside \\(F\\). This remains\ntrue at every deletion endpoint of the other strengths.","\\[\n\\tau_m(\\lambda_p)=\\det\\bigl(A_m+(1+\\lambda_p)B_m\\bigr)\n\\]","has exact degree \\(m\\), and all its roots, with multiplicity, are real\nand strictly less than \\(-1\\). The root nearest the positivity boundary\nis \\(-1-\\delta_m\\), with","\\[\n\\delta_m=\\min_{0\\ne f\\in\\mathbb R[x],\\ \\deg f<m}\n\\frac{\\int f^2\\,d\\nu_{p\\text{-free}}}{\\int f^2\\,d\\nu_{p\\text{-divisible}}}>0.\n\\tag{6}\n\\]","The sequence \\(\\delta_m\\) is nonincreasing, and, with\n\\(C_\\kappa=\\prod_{q\\ne p}\\max(1,1+\\lambda_q)\\),","\\[\n\\boxed{\\delta_{p+m}\\le\\frac{p C_\\kappa C_p^2}{3}r_p^{2m}.}\n\\tag{7}\n\\]","\\textbf{Proof.} The definite pencil is congruent to\n\\(I+(1+\\lambda_p)A_m^{-1/2}B_mA_m^{-1/2}\\), whose eigenvalues have\npositive slopes. Its roots are \\(-1-1/\\eta_i\\), with \\(\\eta_i>0\\). This\nproves the degree and root statements, and the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","tbf{Proof.} The definite pencil is congruent to\n\\(I+(1+\\lambda_p)A_m^{-1/2}B_mA_m^{-1/2}\\), whose eigenvalues have\npositive slopes. Its roots are \\(-1-1/\\eta_i\\), with \\(\\eta_i>0\\). This\nproves the degree and root statements, and the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","definite pencil is congruent to\n\\(I+(1+\\lambda_p)A_m^{-1/2}B_mA_m^{-1/2}\\), whose eigenvalues have\npositive slopes. Its roots are \\(-1-1/\\eta_i\\), with \\(\\eta_i>0\\). This\nproves the degree and root statements, and the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","congruent to\n\\(I+(1+\\lambda_p)A_m^{-1/2}B_mA_m^{-1/2}\\), whose eigenvalues have\npositive slopes. Its roots are \\(-1-1/\\eta_i\\), with \\(\\eta_i>0\\). This\nproves the degree and root statements, and the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","lues have\npositive slopes. Its roots are \\(-1-1/\\eta_i\\), with \\(\\eta_i>0\\). This\nproves the degree and root statements, and the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","\\(-1-1/\\eta_i\\), with \\(\\eta_i>0\\). This\nproves the degree and root statements, and the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","ot statements, and the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","the reciprocal largest\neigenvalue is exactly (6). Polynomial-space inclusion proves\nmonotonicity. Test (6) at order \\(p+m\\) with (2). The denominator\nincludes the atom \\(p\\), of mass \\(p^{-4}\\), because \\(\\kappa(p)=1\\).\nThe numerator kills every lower atom, has no atom at \\(p\\), and its\nhigher tail is bounded by \\(C_\\kappa C_p^2r_p^{2m}/(3p^3)\\). This proves\n(7) and therefore \\(\\delta_m\\to0\\). \\(\\square\\)","\\[\nA_{\\rm tail}=D(4)\\frac{p^4}{p^4+1}\n\\prod_{q\\ne p}\\left(1+\\frac{\\lambda_q}{q^4+1}\\right)\n-\\sum_{1\\le n<p}\\mu(n)^2n^{-4}\\kappa(n)>0,\n\\]","and replaces (7) by \\(\\delta_{p+m}\\le p^4 C_p^2 A_{\\rm tail}r_p^{2m}\\).\nEvery finite-degree coefficient can have poles only at the fiber's\nHankel roots; numerator cancellations can remove a possible pole. The\ntheorem does not assert simple roots or global multivariate real\nstability.","At depth zero, the initial margin is exactly \\(\\delta_1=p^4\\),\nindependently of all other safe strengths: the total \\(p\\)-free and\n\\(p\\)-divisible masses have ratio \\(p^4\\), since all the other Euler\nfactors cancel. The higher-order margins contain the arithmetic overlap\ninformation.","The same argument applies to every nonnegative depth, replacing each\nintegral by its \\(x^k\\) tilt. The denominator's explicit atom\ncontribution becomes \\(p^{-4}x_p^k\\), and (7) acquires the factor\n\\(x_p^{-k}\\).","For \\(F=\\{2,3\\}\\), the order-one Hankel matrix is the scalar","\\[\nM_0(\\lambda_2,\\lambda_3)=105\\pi^{-4}\n\\left(1+\\frac{\\lambda_2}{17}\\right)\n\\left(1+\\frac{\\lambda_3}{82}\\right).\n\\]","Thus \\((0,0)\\) and \\((-34,-164)\\) have the same positive order-one\nHankel matrix. Their midpoint \\((-17,-82)\\) has zero total mass. The\nunrestricted real parameter set passing the order-one\npositive-definiteness test is therefore nonconvex. This is a concrete\nwarning about a finite test, while the true positivity domain in Theorem\nC.1 is a convex closed orthant.","The positive-semidefinite feasible set is nonconvex as well: the points\n\\((0,0)\\) and \\((-34,-246)\\) have positive mass factors 1 and 2, while\ntheir convex combination with weights \\(3/5\\) and \\(2/5\\), namely\n\\((-68/5,-492/5)\\), has mass factor \\(-1/25\\).","\\[\n\\widetilde\\nu_{F,\\boldsymbol\\gamma}\n=\\sum_{n\\ge1}\\mu(n)^2n^{-4}\n\\left(1+\\sum_{p\\in F}\\gamma_p\\mathbf1_{p\\mid n}\\right)\\delta_{x_n}.\n\\]","\\[\n1+\\sum_{p\\in S}\\gamma_p\\ge0\\quad(S\\subseteq F),\n\\quad\\text{equivalently}\\quad\n1+\\sum_{\\gamma_p<0}\\gamma_p\\ge0.\n\\]","Necessity uses the squarefree integer \\(n=\\prod_{p\\in S}p\\); sufficiency\nchecks every atom's prime subset. Its finite Hankel matrices are affine\nfunctions of the strengths, so their positive-semidefinite feasible sets\nare convex spectrahedra. In contrast, the multiplicative family includes\nthe arithmetic prime-overlap terms and has polynomial parameter\ndependence. For example, additive strengths \\(\\gamma_2=\\gamma_3=-3/4\\)\nleave both individual prime weights positive but make the \\(n=6\\) weight\nnegative; multiplicative strengths \\(\\lambda_2=\\lambda_3=-3/4\\) remain\nsafely positive.","ecessity uses the squarefree integer \\(n=\\prod_{p\\in S}p\\); sufficiency\nchecks every atom's prime subset. Its finite Hankel matrices are affine\nfunctions of the strengths, so their positive-semidefinite feasible sets\nare convex spectrahedra. In contrast, the multiplicative family includes\nthe arithmetic prime-overlap terms and has polynomial parameter\ndependence. For example, additive strengths \\(\\gamma_2=\\gamma_3=-3/4\\)\nleave both individual prime weights positive but make the \\(n=6\\) weight\nnegative; multiplicative strengths \\(\\lambda_2=\\lambda_3=-3/4\\) remain\nsafely positive.","atom's prime subset. Its finite Hankel matrices are affine\nfunctions of the strengths, so their positive-semidefinite feasible sets\nare convex spectrahedra. In contrast, the multiplicative family includes\nthe arithmetic prime-overlap terms and has polynomial parameter\ndependence. For example, additive strengths \\(\\gamma_2=\\gamma_3=-3/4\\)\nleave both individual prime weights positive but make the \\(n=6\\) weight\nnegative; multiplicative strengths \\(\\lambda_2=\\lambda_3=-3/4\\) remain\nsafely positive.","e affine\nfunctions of the strengths, so their positive-semidefinite feasible sets\nare convex spectrahedra. In contrast, the multiplicative family includes\nthe arithmetic prime-overlap terms and has polynomial parameter\ndependence. For example, additive strengths \\(\\gamma_2=\\gamma_3=-3/4\\)\nleave both individual prime weights positive but make the \\(n=6\\) weight\nnegative; multiplicative strengths \\(\\lambda_2=\\lambda_3=-3/4\\) remain\nsafely positive.","If at least two strengths in a positive multiplicative family are\nnonzero, its entire depth-zero monic polynomial sequence cannot equal\nany one-prime sequence on the same variable and node system. A direct\nproof avoids importing a determinacy theorem. Write\n\\(P_m=x^m+\\sum_{j<m}c_{mj}x^j\\). The equations \\(\\int P_m\\,d\\nu=0\\)\nrecursively determine every normalized moment from the monic sequence.\nEqual sequences therefore give equal normalized Stieltjes series for\n\\(|z|>1\\). Analytic continuation on the connected complement of the\ncommon closed node set makes their Stieltjes functions equal, and\nresidues at each isolated node give equal normalized atom weights. The\nmass 1 at zero fixes the normalization constant. Finally, the atom at\neach prime \\(p\\) reads off \\(1+\\lambda_p\\), whereas a one-prime\ndeformation changes at most one such atom. This is a precise separator\nfrom the one-prime family. It\ndoes not address arbitrary affine variable changes.","nzero, its entire depth-zero monic polynomial sequence cannot equal\nany one-prime sequence on the same variable and node system. A direct\nproof avoids importing a determinacy theorem. Write\n\\(P_m=x^m+\\sum_{j<m}c_{mj}x^j\\). The equations \\(\\int P_m\\,d\\nu=0\\)\nrecursively determine every normalized moment from the monic sequence.\nEqual sequences therefore give equal normalized Stieltjes series for\n\\(|z|>1\\). Analytic continuation on the connected complement of the\ncommon closed node set makes their Stieltjes functions equal, and\nresidues at each isolated node give equal normalized atom weights. The\nmass 1 at zero fixes the normalization constant. Finally, the atom at\neach prime \\(p\\) reads off \\(1+\\lambda_p\\), whereas a one-prime\ndeformation changes at most one such atom. This is a precise separator\nfrom the one-prime family. It\ndoes not address arbitrary affine variable changes.","normalized moment from the monic sequence.\nEqual sequences therefore give equal normalized Stieltjes series for\n\\(|z|>1\\). Analytic continuation on the connected complement of the\ncommon closed node set makes their Stieltjes functions equal, and\nresidues at each isolated node give equal normalized atom weights. The\nmass 1 at zero fixes the normalization constant. Finally, the atom at\neach prime \\(p\\) reads off \\(1+\\lambda_p\\), whereas a one-prime\ndeformation changes at most one such atom. This is a precise separator\nfrom the one-prime family. It\ndoes not address arbitrary affine variable changes.","t from the monic sequence.\nEqual sequences therefore give equal normalized Stieltjes series for\n\\(|z|>1\\). Analytic continuation on the connected complement of the\ncommon closed node set makes their Stieltjes functions equal, and\nresidues at each isolated node give equal normalized atom weights. The\nmass 1 at zero fixes the normalization constant. Finally, the atom at\neach prime \\(p\\) reads off \\(1+\\lambda_p\\), whereas a one-prime\ndeformation changes at most one such atom. This is a precise separator\nfrom the one-prime family. It\ndoes not address arbitrary affine variable changes.","The multi-prime Stieltjes kernel also retains infinitely many isolated\npoles at primes outside \\(F\\), so the preceding one-prime proof of\nnon-D-finiteness applies unchanged. Orthogonal-polynomial existence,\ndefinite-pencil spectral theory, and compact moment determinacy are\nclassical mechanisms. The specific arithmetic coordinates, parameter\ngeometry, and explicit finite-resolution detector are the objects\ndeveloped here.","For \\(F=\\{2,3\\}\\), the following coefficients are rounded high-precision\nevaluations of (1) and the Hankel formulas. They are not exact decimal\nconstants.","\\begin{longtable}[]{@{}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\n\\((\\lambda_2,\\lambda_3)\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\n\\(P_1(x)\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\n\\(P_2(x)\\)\n\\end{minipage}","@{}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\n\\((\\lambda_2,\\lambda_3)\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\n\\(P_1(x)\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\n\\(P_2(x)\\)\n\\end{minipage}","al{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3333}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\n\\((\\lambda_2,\\lambda_3)\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\n\\(P_1(x)\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedright\n\\(P_2(x)\\)\n\\end{minipage}","\\midrule\\noalign{}\n\\endhead\n\\bottomrule\\noalign{}\n\\endlastfoot\n\\((-1,-1)\\) & \\(x-0.00200014863265043361\\) &\n\\(x^2-0.933902261393426053x+0.000000921271332319859170\\)","\\((-1/2,1)\\) & \\(x-0.0377167661732468061\\) &\n\\(x^2-0.725293001442481628x+0.000932648146085806112\\)","\\((1,1)\\) & \\(x-0.0822907496962840844\\) &\n\\(x^2-0.655873200757636623x+0.00211153000026086129\\)","Simultaneously deleting \\(2,3,5\\) gives\n\\(P_1(x)=x-0.000528721422047782375\\) and\n\\(P_2(x)=x^2-0.966491641778821472x+0.0000000795531413796910514\\), again\nrounded. This removes the low prime shelves while retaining the fixed\nmass at zero and an infinite tail.","For the variable strength at \\(p=2\\), the nearest conditional margins\nare:","{@{}\n  >{\\raggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.2727}}\n  >{\\raggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}\n  >{\\raggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\nHankel order\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nOther strength \\(\\lambda_3=-1\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nOther strength \\(\\lambda_3=3\\)\n\\end{minipage}","ggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}\n  >{\\raggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\nHankel order\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nOther strength \\(\\lambda_3=-1\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nOther strength \\(\\lambda_3=3\\)\n\\end{minipage}","\\midrule\\noalign{}\n\\endhead\n\\bottomrule\\noalign{}\n\\endlastfoot\n1 & \\(16\\), exact & \\(16\\), exact","2 & \\(0.0939109919936561\\) & \\(1.43829172807931\\)","3 & \\(0.000302450514004102\\) & \\(0.0346655628067259\\)","4 & \\(0.000000528145808944195\\) & \\(0.000223749375519984\\)","The noninteger entries are approximate symmetric-pencil calculations.\nThey show an actual prime-overlap effect beyond the mass test. In these\ncalculations the defect \\(\\lambda_2=-1.001\\) is visible at order 3 when\nthe 3-shelf is deleted, but only at order 4 when that shelf is\nstrengthened to 3. This comparison is a numerical example, not a\nuniversal monotonicity theorem.","The separate rational certificates (5) do not depend on those\napproximate roots. The script constructs negative detectors for primes\n\\(2,3,5,7\\), at Hankel orders \\(8,11,19,40\\), respectively. These\nsufficient orders are not asserted to be minimal. In the \\(p=7\\)\nexample, the other strengths are\n\\(\\lambda_2=1/2,\\lambda_3=-1,\\lambda_5=2\\), and \\(\\lambda_7=-1.0001\\);\nthe exact rational upper bound on the detector quadratic form is approximately\n\\(-9.74271\\times10^{-9}\\).","end on those\napproximate roots. The script constructs negative detectors for primes\n\\(2,3,5,7\\), at Hankel orders \\(8,11,19,40\\), respectively. These\nsufficient orders are not asserted to be minimal. In the \\(p=7\\)\nexample, the other strengths are\n\\(\\lambda_2=1/2,\\lambda_3=-1,\\lambda_5=2\\), and \\(\\lambda_7=-1.0001\\);\nthe exact rational upper bound on the detector quadratic form is approximately\n\\(-9.74271\\times10^{-9}\\).","The passing run records 1,019 exact rational assertions, 102 positive\nparameter/depth cases, 1,395 lower-monomial orthogonality conditions,\n381 recurrence identities, 40 conditional-pencil cases, and four exact\nnegative-atom certificates. Working precision is 320 decimal digits.\nMaximum absolute orthogonality residual was approximately\n\\(6.33\\times10^{-322}\\), maximum recurrence residual\n\\(2.23\\times10^{-303}\\), and maximum scale-normalized pencil determinant\nresidual \\(5.06\\times10^{-321}\\). These are finite numerical consistency\nresults, not interval certificates. No execution failure occurred in\nthis route's run.","eter/depth cases, 1,395 lower-monomial orthogonality conditions,\n381 recurrence identities, 40 conditional-pencil cases, and four exact\nnegative-atom certificates. Working precision is 320 decimal digits.\nMaximum absolute orthogonality residual was approximately\n\\(6.33\\times10^{-322}\\), maximum recurrence residual\n\\(2.23\\times10^{-303}\\), and maximum scale-normalized pencil determinant\nresidual \\(5.06\\times10^{-321}\\). These are finite numerical consistency\nresults, not interval certificates. No execution failure occurred in\nthis route's run.","\\pageheading{ORBIT SPECTRA\\; /\\; CONTINUATION THEOREM}{Global meromorphic continuation}\nFor every integer $N\\ge2$, the orbit zeta has a meromorphic continuation to\nthe whole complex plane with exactly one pole:\n\\[\n\\boxed{s=1/N,\\qquad\n\\operatorname{Res}_{s=1/N}\\mathcal Z_N(s)=\\frac{\\sin(\\pi/N)}{N\\pi}.}\n\\]\nIts exact absolute-convergence abscissa is $1/N$. At nonpositive integers,\n\\[\n\\boxed{\\mathcal Z_N(-m)=0\\quad(m\\ge0),\\qquad\n\\mathcal Z_N'(0)=-\\log N.}\n\\]\nThe zeta-regularized product of the orbit coordinates is consequently\n$\\exp[-\\mathcal Z_N'(0)]=N$.\n\\fig{15_spectral_poles}{Pole and zero locations follow from the proved\ncontinuation. The second panel shows the shifted extension, where a pole\nladder appears. These are exact structural locations rather than fitted roots.}\nThe continuation uses classica","\\pageheading{ORBIT SPECTRA\\; /\\; CONTINUATION THEOREM}{Global meromorphic continuation}\nFor every integer $N\\ge2$, the orbit zeta has a meromorphic continuation to\nthe whole complex plane with exactly one pole:\n\\[\n\\boxed{s=1/N,\\qquad\n\\operatorname{Res}_{s=1/N}\\mathcal Z_N(s)=\\frac{\\sin(\\pi/N)}{N\\pi}.}\n\\]\nIts exact absolute-convergence abscissa is $1/N$. At nonpositive integers,\n\\[\n\\boxed{\\mathcal Z_N(-m)=0\\quad(m\\ge0),\\qquad\n\\mathcal Z_N'(0)=-\\log N.}\n\\]\nThe zeta-regularized product of the orbit coordinates is consequently\n$\\exp[-\\mathcal Z_N'(0)]=N$.\n\\fig{15_spectral_poles}{Pole and zero locations follow from the proved\ncontinuation. The second panel shows the shifted extension, where a pole\nladder appears. These are exact structural locations rather than fitted roots.}\nThe continuation uses classical Mellin regularization. The contribution here is a complete local application with multiplicities,\nprincipal branches, and all convergence hypotheses checked.","has a meromorphic continuation to\nthe whole complex plane with exactly one pole:\n\\[\n\\boxed{s=1/N,\\qquad\n\\operatorname{Res}_{s=1/N}\\mathcal Z_N(s)=\\frac{\\sin(\\pi/N)}{N\\pi}.}\n\\]\nIts exact absolute-convergence abscissa is $1/N$. At nonpositive integers,\n\\[\n\\boxed{\\mathcal Z_N(-m)=0\\quad(m\\ge0),\\qquad\n\\mathcal Z_N'(0)=-\\log N.}\n\\]\nThe zeta-regularized product of the orbit coordinates is consequently\n$\\exp[-\\mathcal Z_N'(0)]=N$.\n\\fig{15_spectral_poles}{Pole and zero locations follow from the proved\ncontinuation. The second panel shows the shifted extension, where a pole\nladder appears. These are exact structural locations rather than fitted roots.}\nThe continuation uses classical Mellin regularization. The contribution here is a complete local application with multiplicities,\nprincipal branches, and all convergence hypotheses checked.","\\pageheading{ORBIT SPECTRA\\; /\\; LARGE RADIUS}{A finite exponential sum controls the tail}\nPut $E_N(t)=d_N(-t)=\\sum_{n\\ge0}t^n/(Nn)!$ and\n$\\omega=e^{2\\pi i/N}$. For $x\\ge0$,\n\\[\nE_N(x^N)=\\frac1N\\sum_{j=0}^{N-1}e^{\\omega^jx}.\n\\]\nRoot filtering proves this identity term by term: only powers divisible by\n$N$ remain. The positive real exponential is dominant, so\n\\[\n\\log E_N(t)=t^{1/N}-\\log N+O(e^{-\\kappa_N t^{1/N}}),\n\\quad \\kappa_N=1-\\cos(2\\pi/N)>0,\n\\]\nwith corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative","\\pageheading{ORBIT SPECTRA\\; /\\; LARGE RADIUS}{A finite exponential sum controls the tail}\nPut $E_N(t)=d_N(-t)=\\sum_{n\\ge0}t^n/(Nn)!$ and\n$\\omega=e^{2\\pi i/N}$. For $x\\ge0$,\n\\[\nE_N(x^N)=\\frac1N\\sum_{j=0}^{N-1}e^{\\omega^jx}.\n\\]\nRoot filtering proves this identity term by term: only powers divisible by\n$N$ remain. The positive real exponential is dominant, so\n\\[\n\\log E_N(t)=t^{1/N}-\\log N+O(e^{-\\kappa_N t^{1/N}}),\n\\quad \\kappa_N=1-\\cos(2\\pi/N)>0,\n\\]\nwith corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(","\\pageheading{ORBIT SPECTRA\\; /\\; LARGE RADIUS}{A finite exponential sum controls the tail}\nPut $E_N(t)=d_N(-t)=\\sum_{n\\ge0}t^n/(Nn)!$ and\n$\\omega=e^{2\\pi i/N}$. For $x\\ge0$,\n\\[\nE_N(x^N)=\\frac1N\\sum_{j=0}^{N-1}e^{\\omega^jx}.\n\\]\nRoot filtering proves this identity term by term: only powers divisible by\n$N$ remain. The positive real exponential is dominant, so\n\\[\n\\log E_N(t)=t^{1/N}-\\log N+O(e^{-\\kappa_N t^{1/N}}),\n\\quad \\kappa_N=1-\\cos(2\\pi/N)>0,\n\\]\nwith corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ ne","\\pageheading{ORBIT SPECTRA\\; /\\; LARGE RADIUS}{A finite exponential sum controls the tail}\nPut $E_N(t)=d_N(-t)=\\sum_{n\\ge0}t^n/(Nn)!$ and\n$\\omega=e^{2\\pi i/N}$. For $x\\ge0$,\n\\[\nE_N(x^N)=\\frac1N\\sum_{j=0}^{N-1}e^{\\omega^jx}.\n\\]\nRoot filtering proves this identity term by term: only powers divisible by\n$N$ remain. The positive real exponential is dominant, so\n\\[\n\\log E_N(t)=t^{1/N}-\\log N+O(e^{-\\kappa_N t^{1/N}}),\n\\quad \\kappa_N=1-\\cos(2\\pi/N)>0,\n\\]\nwith corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential domin","\\pageheading{ORBIT SPECTRA\\; /\\; LARGE RADIUS}{A finite exponential sum controls the tail}\nPut $E_N(t)=d_N(-t)=\\sum_{n\\ge0}t^n/(Nn)!$ and\n$\\omega=e^{2\\pi i/N}$. For $x\\ge0$,\n\\[\nE_N(x^N)=\\frac1N\\sum_{j=0}^{N-1}e^{\\omega^jx}.\n\\]\nRoot filtering proves this identity term by term: only powers divisible by\n$N$ remain. The positive real exponential is dominant, so\n\\[\n\\log E_N(t)=t^{1/N}-\\log N+O(e^{-\\kappa_N t^{1/N}}),\n\\quad \\kappa_N=1-\\cos(2\\pi/N)>0,\n\\]\nwith corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\ns","\\pageheading{ORBIT SPECTRA\\; /\\; LARGE RADIUS}{A finite exponential sum controls the tail}\nPut $E_N(t)=d_N(-t)=\\sum_{n\\ge0}t^n/(Nn)!$ and\n$\\omega=e^{2\\pi i/N}$. For $x\\ge0$,\n\\[\nE_N(x^N)=\\frac1N\\sum_{j=0}^{N-1}e^{\\omega^jx}.\n\\]\nRoot filtering proves this identity term by term: only powers divisible by\n$N$ remain. The positive real exponential is dominant, so\n\\[\n\\log E_N(t)=t^{1/N}-\\log N+O(e^{-\\kappa_N t^{1/N}}),\n\\quad \\kappa_N=1-\\cos(2\\pi/N)>0,\n\\]\nwith corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray bec…","cos(2\\pi/N)>0,\n\\]\nwith corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray because $d_N(-t)>0$ for $t\\ge0$.\nThus all zero arguments have a uniform gap from the branch cut.\nIn particular $|t+w_j|\\ge c_N(t+|w_j|)$ for $t\\ge0$.","h corresponding differentiated estimates on the positive ray.\n\\fig{14_spectral_tail}{Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray because $d_N(-t)>0$ for $t\\ge0$.\nThus all zero arguments have a uniform gap from the branch cut.\nIn particular $|t+w_j|\\ge c_N(t+|w_j|)$ for $t\\ge0$.","Computed logarithmic derivatives of the finite\nexponential sum approach one in the root-radius variable. The decaying\nremainder supplies the Mellin continuation's large-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray because $d_N(-t)>0$ for $t\\ge0$.\nThus all zero arguments have a uniform gap from the branch cut.\nIn particular $|t+w_j|\\ge c_N(t+|w_j|)$ for $t\\ge0$.","ge-radius input.}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray because $d_N(-t)>0$ for $t\\ge0$.\nThus all zero arguments have a uniform gap from the branch cut.\nIn particular $|t+w_j|\\ge c_N(t+|w_j|)$ for $t\\ge0$.","}\n\\claim{Why complex zero arguments do not break the integral}\nIn a sufficiently narrow sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray because $d_N(-t)>0$ for $t\\ge0$.\nThus all zero arguments have a uniform gap from the branch cut.\nIn particular $|t+w_j|\\ge c_N(t+|w_j|)$ for $t\\ge0$.","sector around negative $w$, choose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray because $d_N(-t)>0$ for $t\\ge0$.\nThus all zero arguments have a uniform gap from the branch cut.\nIn particular $|t+w_j|\\ge c_N(t+|w_j|)$ for $t\\ge0$.","hoose the root\n$x=(-w)^{1/N}$ near the positive real ray. The first exponential dominates\nthe other $N-1$ terms by a uniform real-part gap at large radius, so this\nsector is eventually zero free. The finitely many remaining zeros avoid\nthe negative real ray because $d_N(-t)>0$ for $t\\ge0$.\nThus all zero arguments have a uniform gap from the branch cut.\nIn particular $|t+w_j|\\ge c_N(t+|w_j|)$ for $t\\ge0$.","\\pageheading{ORBIT SPECTRA\\; /\\; MELLIN PROOF}{The two ends have different jobs}\nFor $1/N<\\Re s<1$, the genus-zero log derivative and the angular gap give\nabsolute interchange in\n\\[\n\\mathcal Z_N(s)=\\frac{\\sin\\pi s}{\\pi}\n\\int_0^\\infty t^{-s}\\frac{d}{dt}\\log E_N(t)\\dd t.\n\\]\nEach term uses the principal-branch beta integral\n$\\int_0^\\infty t^{-s}(t+w)^{-1}\\dd t=\\pi w^{-s}/\\sin\\pi s$.\nThe denominator estimate bounds the absolute integral by a constant times\n$|w|^{-\\Re s}$, whose sum converges in this strip.","s have different jobs}\nFor $1/N<\\Re s<1$, the genus-zero log derivative and the angular gap give\nabsolute interchange in\n\\[\n\\mathcal Z_N(s)=\\frac{\\sin\\pi s}{\\pi}\n\\int_0^\\infty t^{-s}\\frac{d}{dt}\\log E_N(t)\\dd t.\n\\]\nEach term uses the principal-branch beta integral\n$\\int_0^\\infty t^{-s}(t+w)^{-1}\\dd t=\\pi w^{-s}/\\sin\\pi s$.\nThe denominator estimate bounds the absolute integral by a constant times\n$|w|^{-\\Re s}$, whose sum converges in this strip.","Near zero write $f(t)=(\\log E_N)'(t)=\\sum_{k\\ge1}\\alpha_kt^{k-1}$,\nwhere $\\alpha_k=(-1)^{k+1}\\mathcal Z_N(k)$.\nFor any integer $K\\ge1$, subtract $K$ terms on $[0,1]$:\n\\[\n\\begin{aligned}\nI_K(s)={}&\\int_0^1t^{-s}\\left[f(t)-\\sum_{k=1}^K\\alpha_kt^{k-1}\\right]\\dd t\n+\\sum_{k=1}^K\\frac{\\alpha_k}{k-s}","Near zero write $f(t)=(\\log E_N)'(t)=\\sum_{k\\ge1}\\alpha_kt^{k-1}$,\nwhere $\\alpha_k=(-1)^{k+1}\\mathcal Z_N(k)$.\nFor any integer $K\\ge1$, subtract $K$ terms on $[0,1]$:\n\\[\n\\begin{aligned}\nI_K(s)={}&\\int_0^1t^{-s}\\left[f(t)-\\sum_{k=1}^K\\alpha_kt^{k-1}\\right]\\dd t\n+\\sum_{k=1}^K\\frac{\\alpha_k}{k-s}\\\\\n&+\\int_1^\\infty t^{-s}\\left[f(t)-\\frac1N t^{1/N-1}\\right]\\dd t\n+\\frac{1}{N(s-1/N)}.\n\\end{aligned}\n\\]\nThe first integral is holomorphic for $\\Re s<K+1$. The second is entire because\nits remainder decays exponentially in $t^{1/N}$. The factor $\\sin\\pi s$ cancels\nthe positive-integer poles. Increasing $K$ covers the plane, leaving only\n$1/N$. At nonpositive integers $I_K$ is regular, so the sine factor gives zero.","&+\\int_1^\\infty t^{-s}\\left[f(t)-\\frac1N t^{1/N-1}\\right]\\dd t\n+\\frac{1}{N(s-1/N)}.\n\\end{aligned}\n\\]\nThe first integral is holomorphic for $\\Re s<K+1$. The second is entire because\nits remainder decays exponentially in $t^{1/N}$. The factor $\\sin\\pi s$ cancels\nthe positive-integer poles. Increasing $K$ covers the plane, leaving only\n$1/N$. At nonpositive integers $I_K$ is regular, so the sine factor gives zero.","At zero the finite parts give\n\\(I_K(0)=\\lim_{T\\to\\infty}[\\log E_N(T)-T^{1/N}]=-\\log N.\\)\nSince $(\\sin\\pi s/\\pi)'_{s=0}=1$, this equals $\\mathcal Z_N'(0)$.\nAbsolute convergence at a smaller real exponent would give a holomorphic\nseries across $1/N$, contradicting the pole and proving the exact abscissa.","\\pageheading{ORBIT SPECTRA\\; /\\; SHIFTS}{A Hurwitz-type deformation and its determinant}\nFor real $a\\ge0$, define initially\n\\[\n\\mathcal Z_N(s;a)=\\sum_j(w_j+a)^{-s},\\qquad\\Re s>1/N.\n\\]\nThe rightward shift preserves avoidance of the negative branch cut.\nNormalize the product with $E_N(a)$ and use\n$\\log E_N(a+t)$ in the Mellin argument.\nFor $a>0$ there are genuine simple poles at $s=1/N-k$, $k\\ge0$:\n\\[\n\\boxed{\\operatorname{Res}_{s=1/N-k}\\mathcal Z_N(s;a)\n=\\frac{\\sin(\\pi/N)}{N\\pi}\\frac{(1-1/N)_k}{k!}a^k.}\n\\]\nIndeed the large-radius derivative expands as\n\\[\n\\frac1N(a+t)^{1/N-1}\n=\\frac1N\\sum_{k\\ge0}\\frac{(-1)^k(1-1/N)_k}{k!}a^kt^{1/N-1-k}\n\\]\nas an asymptotic series. The alternating sign is canceled by\n$\\sin(\\pi(1/N-k))=(-1)^k\\sin(\\pi/N)$.\nAt $a=0$ all secondary residues vanish.","ageheading{ORBIT SPECTRA\\; /\\; SHIFTS}{A Hurwitz-type deformation and its determinant}\nFor real $a\\ge0$, define initially\n\\[\n\\mathcal Z_N(s;a)=\\sum_j(w_j+a)^{-s},\\qquad\\Re s>1/N.\n\\]\nThe rightward shift preserves avoidance of the negative branch cut.\nNormalize the product with $E_N(a)$ and use\n$\\log E_N(a+t)$ in the Mellin argument.\nFor $a>0$ there are genuine simple poles at $s=1/N-k$, $k\\ge0$:\n\\[\n\\boxed{\\operatorname{Res}_{s=1/N-k}\\mathcal Z_N(s;a)\n=\\frac{\\sin(\\pi/N)}{N\\pi}\\frac{(1-1/N)_k}{k!}a^k.}\n\\]\nIndeed the large-radius derivative expands as\n\\[\n\\frac1N(a+t)^{1/N-1}\n=\\frac1N\\sum_{k\\ge0}\\frac{(-1)^k(1-1/N)_k}{k!}a^kt^{1/N-1-k}\n\\]\nas an asymptotic series. The alternating sign is canceled by\n$\\sin(\\pi(1/N-k))=(-1)^k\\sin(\\pi/N)$.\nAt $a=0$ all secondary residues vanish.","^{-s},\\qquad\\Re s>1/N.\n\\]\nThe rightward shift preserves avoidance of the negative branch cut.\nNormalize the product with $E_N(a)$ and use\n$\\log E_N(a+t)$ in the Mellin argument.\nFor $a>0$ there are genuine simple poles at $s=1/N-k$, $k\\ge0$:\n\\[\n\\boxed{\\operatorname{Res}_{s=1/N-k}\\mathcal Z_N(s;a)\n=\\frac{\\sin(\\pi/N)}{N\\pi}\\frac{(1-1/N)_k}{k!}a^k.}\n\\]\nIndeed the large-radius derivative expands as\n\\[\n\\frac1N(a+t)^{1/N-1}\n=\\frac1N\\sum_{k\\ge0}\\frac{(-1)^k(1-1/N)_k}{k!}a^kt^{1/N-1-k}\n\\]\nas an asymptotic series. The alternating sign is canceled by\n$\\sin(\\pi(1/N-k))=(-1)^k\\sin(\\pi/N)$.\nAt $a=0$ all secondary residues vanish.","a+t)$ in the Mellin argument.\nFor $a>0$ there are genuine simple poles at $s=1/N-k$, $k\\ge0$:\n\\[\n\\boxed{\\operatorname{Res}_{s=1/N-k}\\mathcal Z_N(s;a)\n=\\frac{\\sin(\\pi/N)}{N\\pi}\\frac{(1-1/N)_k}{k!}a^k.}\n\\]\nIndeed the large-radius derivative expands as\n\\[\n\\frac1N(a+t)^{1/N-1}\n=\\frac1N\\sum_{k\\ge0}\\frac{(-1)^k(1-1/N)_k}{k!}a^kt^{1/N-1-k}\n\\]\nas an asymptotic series. The alternating sign is canceled by\n$\\sin(\\pi(1/N-k))=(-1)^k\\sin(\\pi/N)$.\nAt $a=0$ all secondary residues vanish.","here are genuine simple poles at $s=1/N-k$, $k\\ge0$:\n\\[\n\\boxed{\\operatorname{Res}_{s=1/N-k}\\mathcal Z_N(s;a)\n=\\frac{\\sin(\\pi/N)}{N\\pi}\\frac{(1-1/N)_k}{k!}a^k.}\n\\]\nIndeed the large-radius derivative expands as\n\\[\n\\frac1N(a+t)^{1/N-1}\n=\\frac1N\\sum_{k\\ge0}\\frac{(-1)^k(1-1/N)_k}{k!}a^kt^{1/N-1-k}\n\\]\nas an asymptotic series. The alternating sign is canceled by\n$\\sin(\\pi(1/N-k))=(-1)^k\\sin(\\pi/N)$.\nAt $a=0$ all secondary residues vanish.","The negative-integer values remain zero, and\n\\[\n\\mathcal Z_N'(0;a)=-\\log[N E_N(a)],\\qquad\n\\boxed{\\prod_j^{\\rm reg}(w_j+a)=N E_{N,1}(a).}\n\\]\nThe finite part $-\\log N$ and normalization $-\\log E_N(a)$ give a\nzeta-regularized product, not an ordinary product. At $N=2$ it is\n$2\\cosh\\sqrt a$, the classical half-integer spectral determinant reproduced\nby the four shifted benchmarks. Functional equations and derivative-weighted\nzero sums remain open here.","Let \\(N\\ge2\\) be an integer, and put \\(\\alpha=1/N\\). Define","\\[\nd_N(w)=\\sum_{n=0}^{\\infty}\\frac{(-1)^n w^n}{(Nn)!}\n      =E_{N,1}(-w).\n\\]","List its nonzero zeros \\(w_j\\) with multiplicity, and use the principal\nlogarithm, with \\(-\\pi<\\operatorname{Arg}w_j<\\pi\\). No zero lies on the\nomitted negative ray. The orbit zeta function","\\[\n\\mathcal Z_N(s)=\\sum_j e^{-s\\operatorname{Log}w_j}\n\\]","converges absolutely exactly when \\(\\Re s>1/N\\). It has a meromorphic\ncontinuation to the whole plane, with exactly one pole: a simple pole at\n\\(s=1/N\\), of residue","\\[\n\\boxed{\\operatorname{Res}_{s=1/N}\\mathcal Z_N(s)\n      =\\frac{\\sin(\\pi/N)}{N\\pi}.}\n\\]","\\[\n\\boxed{\\mathcal Z_N(-m)=0\\quad(m=0,1,2,\\ldots).}\n\\]","\\[\n\\boxed{\\mathcal Z_N'(0)=-\\log N,\\qquad\n       \\exp[-\\mathcal Z_N'(0)]=N.}\n\\]","Let \\(\\omega=e^{2\\pi i/N}\\). The finite root-of-unity filter gives, for\nevery \\(x\\in\\mathbb C\\),","\\[\nF_N(x):=d_N(-x^N)\n=\\frac1N\\sum_{j=0}^{N-1}e^{\\omega^j x}\n=\\sum_{n=0}^{\\infty}\\frac{x^{Nn}}{(Nn)!}.\n\\tag{S1}\n\\]","This equality follows by expanding the exponentials and using\n\\(N^{-1}\\sum_j\\omega^{jm}=1\\) when \\(N\\mid m\\), and zero otherwise. For\nreal \\(x\\ge0\\), the power series makes \\(F_N(x)>0\\). Put","\\[\nh_N(x)=\\sum_{j=1}^{N-1}e^{(\\omega^j-1)x},\n\\qquad c_N=1-\\cos(2\\pi/N)>0.\n\\]","\\[\nF_N(x)=\\frac{e^x}{N}(1+h_N(x)),\\qquad\n|h_N(x)|\\le(N-1)e^{-c_Nx},\\qquad\n|h_N'(x)|\\le2(N-1)e^{-c_Nx}.\n\\tag{S2}\n\\]","Since \\(1+h_N(x)=Ne^{-x}F_N(x)>0\\), the real logarithm is well defined\nfor all \\(x\\ge0\\). At infinity,","\\[\n\\log F_N(x)=x-\\log N+O(e^{-c_Nx}),\n\\qquad g_N(x):=\\frac{F_N'(x)}{F_N(x)}\n=1+O(e^{-c_Nx}).\n\\tag{S3}\n\\]","At the origin \\(g_N(x)=O(x^{N-1})\\). Its only Taylor powers are\n\\(x^{Nm-1}\\), with \\(m\\ge1\\).","\\[\nd_N(w)=\\prod_j(1-w/w_j).\n\\tag{S4}\n\\]","For completeness, the series bound \\(|d_N(w)|\\le\\exp(|w|^{1/N})\\) gives\norder at most \\(1/N<1\\). Hadamard factorization therefore has genus zero\nand a constant exponential factor, which equals one because\n\\(d_N(0)=1\\). The usual zero-count consequence gives\n\\(\\sum_j|w_j|^{-\\sigma}<\\infty\\) for every \\(\\sigma>1/N\\). Along the\nnegative real axis (S3) also shows the order is exactly \\(1/N\\).","For \\(|\\arg(-w)|\\le\\pi/2\\), choose \\(x=(-w)^{1/N}\\) with\n\\(|\\arg x|\\le\\pi/(2N)\\). For every \\(j\\ne0\\),","\\[\n\\Re x-\\Re(\\omega^j x)\n\\ge \\left[\\cos\\frac{\\pi}{2N}-\\cos\\frac{3\\pi}{2N}\\right]|x|\n=:\\gamma_N|x|>0.\n\\tag{S5}\n\\]","Indeed, the nearest competing root is separated from the positive\ndirection by at least \\(3\\pi/(2N)\\), while the distinguished root is\nwithin \\(\\pi/(2N)\\). Cosine decreases on \\([0,\\pi]\\). Thus, once\n\\((N-1)e^{-\\gamma_N|x|}<1\\), the distinguished term \\(e^x\\) in (S1) is\nlarger in modulus than the sum of all competitors, and \\(d_N(w)\\ne0\\).\nConsequently, sufficiently large zeros avoid the entire closed left\nhalf-plane. There are only finitely many remaining zeros in a bounded\ndisk. None lies on the negative ray, since \\(d_N(-t)>0\\) for \\(t\\ge0\\).\nIt follows that some \\(\\eta_N>0\\) satisfies","d root is\nwithin \\(\\pi/(2N)\\). Cosine decreases on \\([0,\\pi]\\). Thus, once\n\\((N-1)e^{-\\gamma_N|x|}<1\\), the distinguished term \\(e^x\\) in (S1) is\nlarger in modulus than the sum of all competitors, and \\(d_N(w)\\ne0\\).\nConsequently, sufficiently large zeros avoid the entire closed left\nhalf-plane. There are only finitely many remaining zeros in a bounded\ndisk. None lies on the negative ray, since \\(d_N(-t)>0\\) for \\(t\\ge0\\).\nIt follows that some \\(\\eta_N>0\\) satisfies","(\\pi/(2N)\\). Cosine decreases on \\([0,\\pi]\\). Thus, once\n\\((N-1)e^{-\\gamma_N|x|}<1\\), the distinguished term \\(e^x\\) in (S1) is\nlarger in modulus than the sum of all competitors, and \\(d_N(w)\\ne0\\).\nConsequently, sufficiently large zeros avoid the entire closed left\nhalf-plane. There are only finitely many remaining zeros in a bounded\ndisk. None lies on the negative ray, since \\(d_N(-t)>0\\) for \\(t\\ge0\\).\nIt follows that some \\(\\eta_N>0\\) satisfies","\\([0,\\pi]\\). Thus, once\n\\((N-1)e^{-\\gamma_N|x|}<1\\), the distinguished term \\(e^x\\) in (S1) is\nlarger in modulus than the sum of all competitors, and \\(d_N(w)\\ne0\\).\nConsequently, sufficiently large zeros avoid the entire closed left\nhalf-plane. There are only finitely many remaining zeros in a bounded\ndisk. None lies on the negative ray, since \\(d_N(-t)>0\\) for \\(t\\ge0\\).\nIt follows that some \\(\\eta_N>0\\) satisfies","\\[\n|\\operatorname{Arg}w_j|\\le\\pi-\\eta_N\\qquad\\text{for every }j.\n\\tag{S6}\n\\]","With \\(c=\\sin(\\eta_N/2)>0\\), this implies","\\[\n|t+w_j|\\ge c(t+|w_j|),\\qquad t\\ge0.\n\\tag{S7}\n\\]","To check the elementary estimate, write \\(w=re^{i\\theta}\\):\n\\(|t+w|^2=(t+r)^2-4tr\\sin^2(\\theta/2)\\ge(t+r)^2\\cos^2(\\theta/2)\\).","Let \\(H_N(t)=d_N(-t)=E_{N,1}(t)\\), and \\(G_N(t)=H_N'(t)/H_N(t)\\).\nLogarithmic differentiation of (S4), locally uniformly off the zeros,\ngives","\\[\nG_N(t)=\\sum_j\\frac1{t+w_j},\\qquad t\\ge0.\n\\tag{S8}\n\\]","For \\(1/N<\\sigma<1\\), (S7) yields","\\[\n\\sum_j\\int_0^\\infty\\frac{t^{-\\sigma}}{|t+w_j|}\\,dt\n\\le\\frac{\\pi}{c\\sin(\\pi\\sigma)}\\sum_j|w_j|^{-\\sigma}<\\infty.\n\\tag{S9}\n\\]","This is the absolute interchange needed for complex \\(s\\) with\n\\(\\Re s=\\sigma\\). Euler's beta integral and the reflection identity give","\\[\n\\int_0^\\infty\\frac{t^{-s}}{t+w}\\,dt\n=\\frac{\\pi}{\\sin(\\pi s)}e^{-s\\operatorname{Log}w},\n\\quad 0<\\Re s<1,\\quad |\\arg w|<\\pi.\n\\tag{S10}\n\\]","First prove this for positive \\(w\\) by substitution and the beta\nintegral, then extend to the slit \\(w\\)-plane by holomorphic dependence.\nOn compact subsets of that slit plane the denominator estimate supplies\nthe needed local domination.","Combining (S8)--(S10), and substituting \\(t=x^N\\), proves","\\[\n\\boxed{\\mathcal Z_N(s)=\\frac{\\sin(\\pi s)}{\\pi}\n\\int_0^\\infty x^{-Ns}g_N(x)\\,dx,\n\\qquad \\frac1N<\\Re s<1.}\n\\tag{S11}\n\\]","For \\(x>0\\), \\(x^{-Ns}=e^{-Ns\\log x}\\) uses the real logarithm.\nMultiplicities are already incorporated in (S4), so no simple-zero\nassumption entered the proof.","One can see directly that \\(d_N\\) has no zero on \\(|w|\\le1\\):","\\[\n|d_N(w)-1|\\le\\sum_{n\\ge1}\\frac1{(Nn)!}\n\\le\\cosh(1)-1<1.\n\\]","The strict inequality also gives a slightly larger zero-free disk by\ncontinuity. Thus \\(G_N(t)\\) has a Taylor series converging uniformly on\n\\([0,1]\\),","\\[\nG_N(t)=\\sum_{k\\ge0}a_{N,k}t^k,\n\\qquad g_N(x)=\\sum_{m\\ge1}N a_{N,m-1}x^{Nm-1}.\n\\tag{S12}\n\\]","For any integer \\(K\\ge0\\), define","\\[\n\\begin{aligned}\nB_{N,K}(s)={}&\n\\int_0^1x^{-Ns}\\left[g_N(x)-\\sum_{m=1}^{K}N a_{N,m-1}x^{Nm-1}\\right]dx\\\\\n&+\\sum_{m=1}^{K}\\frac{a_{N,m-1}}{m-s}\n +\\int_1^\\infty x^{-Ns}\\bigl[g_N(x)-1\\bigr]dx\n +\\frac1{Ns-1}.\n\\end{aligned}\n\\tag{S13}\n\\]","The first integral is holomorphic for \\(\\Re s<K+1\\). This includes local\nuniform domination, since the Taylor remainder is \\(O(x^{N(K+1)-1})\\).\nThe last integral is entire in \\(s\\), because (S3) gives exponential\ndecay against every power of \\(x\\), uniformly on compact \\(s\\)-sets,\nincluding after any number of \\(s\\)-derivatives. The rational terms\nexhibit the only possible poles: positive integers \\(1,\\ldots,K\\) and\n\\(1/N\\).","first integral is holomorphic for \\(\\Re s<K+1\\). This includes local\nuniform domination, since the Taylor remainder is \\(O(x^{N(K+1)-1})\\).\nThe last integral is entire in \\(s\\), because (S3) gives exponential\ndecay against every power of \\(x\\), uniformly on compact \\(s\\)-sets,\nincluding after any number of \\(s\\)-derivatives. The rational terms\nexhibit the only possible poles: positive integers \\(1,\\ldots,K\\) and\n\\(1/N\\).","holomorphic for \\(\\Re s<K+1\\). This includes local\nuniform domination, since the Taylor remainder is \\(O(x^{N(K+1)-1})\\).\nThe last integral is entire in \\(s\\), because (S3) gives exponential\ndecay against every power of \\(x\\), uniformly on compact \\(s\\)-sets,\nincluding after any number of \\(s\\)-derivatives. The rational terms\nexhibit the only possible poles: positive integers \\(1,\\ldots,K\\) and\n\\(1/N\\).","\\[\n\\mathcal Z_N(s)=\\frac{\\sin(\\pi s)}{\\pi}B_{N,K}(s)\n\\tag{S14}\n\\]","provides compatible meromorphic continuations covering the whole plane\nas \\(K\\to\\infty\\). At each positive integer \\(m\\), the apparent simple\npole \\(a_{N,m-1}/(m-s)\\) cancels with the simple zero of\n\\(\\sin(\\pi s)\\). The term \\(1/(Ns-1)\\) has residue \\(1/N\\), and its\nmultiplier at \\(s=1/N\\) is nonzero. This proves the stated unique simple\npole and its residue.","It also proves that the original series cannot converge absolutely at\n\\(s=1/N\\). If it did, its absolute terms would dominate the series for\nreal \\(s\\downarrow1/N\\), except for finitely many zeros of modulus below\none. Dominated convergence would then keep that real limit bounded,\ncontradicting the nonzero pole. Absolute convergence at any smaller real\npart would imply convergence at \\(1/N\\). The abscissa is consequently\nexactly \\(1/N\\).","ries cannot converge absolutely at\n\\(s=1/N\\). If it did, its absolute terms would dominate the series for\nreal \\(s\\downarrow1/N\\), except for finitely many zeros of modulus below\none. Dominated convergence would then keep that real limit bounded,\ncontradicting the nonzero pole. Absolute convergence at any smaller real\npart would imply convergence at \\(1/N\\). The abscissa is consequently\nexactly \\(1/N\\).","For \\(\\Re s<1\\) we may use \\(K=0\\):","\\[\nB_{N,0}(s)=\\int_0^1x^{-Ns}g_N(x)\\,dx\n+\\int_1^\\infty x^{-Ns}(g_N(x)-1)\\,dx+\\frac1{Ns-1}.\n\\tag{S15}\n\\]","This expression is holomorphic at every nonpositive integer. Hence (S14)\nproves \\(\\mathcal Z_N(-m)=0\\). These statements do not assert that every\nsuch zero is simple.","At a positive integer \\(m\\), taking the removable limit in (S14) gives","\\[\n\\mathcal Z_N(m)=(-1)^{m+1}a_{N,m-1}.\n\\tag{S16}\n\\]","If \\(c_n=(-1)^n/(Nn)!\\), the previous logarithmic-product law therefore\nremains","\\[\nb_m:=\\mathcal Z_N(m)\n=-m c_m-\\sum_{j=1}^{m-1}c_jb_{m-j}.\n\\tag{S17}\n\\]","At \\(s=0\\), the elementary antiderivatives in (S15) give","\\[\n\\begin{aligned}\nB_{N,0}(0)\n&=\\log F_N(1)\n +\\lim_{R\\to\\infty}\\bigl[\\log F_N(R)-R-\\log F_N(1)+1\\bigr]-1\\\\\n&=-\\log N.\n\\end{aligned}\n\\]","Since \\(\\sin(\\pi s)/\\pi=s+O(s^3)\\), (S14) implies \\(\\mathcal Z_N(0)=0\\)\nand \\(\\mathcal Z_N'(0)=-\\log N\\).","\\[\n\\mathcal Z_2(s)=\\pi^{-2s}\\zeta(2s,1/2),\\qquad\n\\mathcal Z_4(s)=(\\sqrt2\\pi)^{-4s}\\zeta(4s,1/2)\n\\]","now hold as meromorphic identities, by equality in the original\nhalf-plane and uniqueness of continuation. Their pole residues are\nrespectively \\(1/(2\\pi)\\) and \\(1/(4\\sqrt2\\pi)\\), agreeing with the\ngeneral theorem. Their zero values at nonpositive integers follow from\nthe Hurwitz formula, and their regularized products are respectively 2\nand 4.","The companion script performs 411 exact rational assertions: independent\nformal-logarithm and quotient calculations for \\(2\\le N\\le10\\) through\ndegree 40; three explicit integer values for each \\(N\\); and 12\nBernoulli/Hurwitz anchors each for \\(N=2,4\\). It then evaluates (S13)\nusing a truncated local Taylor sum and high-precision exponential-tail\nquadrature. Sixteen \\(N=2,4\\) benchmarks sample both sides of the\noriginal convergence boundary, negative real values, and complex values.\nValues for \\(N=3,5\\) and checks of \\(\\mathcal Z_N'(0)\\) are recorded in\nthe JSON evidence.","t calculations for \\(2\\le N\\le10\\) through\ndegree 40; three explicit integer values for each \\(N\\); and 12\nBernoulli/Hurwitz anchors each for \\(N=2,4\\). It then evaluates (S13)\nusing a truncated local Taylor sum and high-precision exponential-tail\nquadrature. Sixteen \\(N=2,4\\) benchmarks sample both sides of the\noriginal convergence boundary, negative real values, and complex values.\nValues for \\(N=3,5\\) and checks of \\(\\mathcal Z_N'(0)\\) are recorded in\nthe JSON evidence.","e N\\le10\\) through\ndegree 40; three explicit integer values for each \\(N\\); and 12\nBernoulli/Hurwitz anchors each for \\(N=2,4\\). It then evaluates (S13)\nusing a truncated local Taylor sum and high-precision exponential-tail\nquadrature. Sixteen \\(N=2,4\\) benchmarks sample both sides of the\noriginal convergence boundary, negative real values, and complex values.\nValues for \\(N=3,5\\) and checks of \\(\\mathcal Z_N'(0)\\) are recorded in\nthe JSON evidence.","There is a concrete shifted family. For a real shift \\(a\\ge0\\), define","\\[\n\\mathcal Z_N(s;a)=\\sum_j(w_j+a)^{-s},\\qquad \\Re s>1/N,\n\\tag{T1}\n\\]","again with the principal logarithm and multiplicities. A rightward shift\npreserves the angular gap from the negative ray: for each nonreal \\(w\\),\n\\(|\\arg(w+a)|\\) is nonincreasing in \\(a\\), and positive real zeros\nremain positive. The zero sequence is unchanged at infinity up to a\nbounded translation, so the previous absolute convergence argument\napplies. The normalized determinant is exactly","\\[\n\\frac{H_N(a+t)}{H_N(a)}\n=\\prod_j\\left(1+\\frac{t}{w_j+a}\\right),\\qquad\nH_N(t)=E_{N,1}(t)>0\\quad(t\\ge0).\n\\tag{T2}\n\\]","This follows by taking the ratio of the two convergent products in (S4),\nnot by introducing an unverified exponential factor. Its logarithmic\nderivative is \\(G_{N,a}(t)=H_N'(a+t)/H_N(a+t)\\). The same dominated\nbeta-integral argument gives","\\[\n\\mathcal Z_N(s;a)=\\frac{\\sin(\\pi s)}{\\pi}\n\\int_0^\\infty t^{-s}G_{N,a}(t)\\,dt,\n\\qquad 1/N<\\Re s<1.\n\\tag{T3}\n\\]","At infinity, (S2)--(S3), with \\(x=(a+t)^{1/N}\\), imply","\\[\nG_{N,a}(t)=\\alpha(t+a)^{\\alpha-1}\n+O\\!\\left((t+a)^{\\alpha-1}e^{-c_N(t+a)^\\alpha}\\right).\n\\tag{T4}\n\\]","For fixed \\(a\\) and any integer \\(K\\ge0\\), the binomial expansion\ntherefore gives","\\[\nG_{N,a}(t)=\\sum_{k=0}^{K}\\alpha\\binom{\\alpha-1}{k}a^k t^{\\alpha-k-1}\n+O(t^{\\alpha-K-2}).\n\\tag{T5}\n\\]","To be explicit about continuation, choose a finite splitting point\n\\(T>2a+1\\). Subtract the first \\(K+1\\) terms of (T5) inside the integral\nfrom \\(T\\) to infinity, and add their integrals back as the rational\nterms","\\[\n\\frac{\\alpha\\binom{\\alpha-1}{k}a^k\n T^{\\alpha-k-s}}{s-\\alpha+k}.\n\\tag{T6}\n\\]","The remainder integral is holomorphic for \\(\\Re s>\\alpha-K-1\\). The\nintegral on any compact interval away from zero is entire. At zero,\n\\(G_{N,a}\\) is analytic since \\(H_N(a)>0\\); subtracting any finite\nTaylor polynomial there extends the other side, with apparent poles only\nat positive integers. Such Taylor poles cancel against \\(\\sin(\\pi s)\\),\njust as in Section~\\ref{global-continuation-and-the-only-pole}. Choosing arbitrarily large subtractions at both\nends covers every \\(s\\in\\mathbb C\\). The resulting continuations agree\non overlaps because all added terms are exactly the integrals of the\nterms removed.","any compact interval away from zero is entire. At zero,\n\\(G_{N,a}\\) is analytic since \\(H_N(a)>0\\); subtracting any finite\nTaylor polynomial there extends the other side, with apparent poles only\nat positive integers. Such Taylor poles cancel against \\(\\sin(\\pi s)\\),\njust as in Section~\\ref{global-continuation-and-the-only-pole}. Choosing arbitrarily large subtractions at both\nends covers every \\(s\\in\\mathbb C\\). The resulting continuations agree\non overlaps because all added terms are exactly the integrals of the\nterms removed.","For \\(a>0\\), the complete pole set is consequently","\\[\n\\boxed{s=\\frac1N-k,\\quad k=0,1,2,\\ldots,}\n\\]","\\[\n\\begin{aligned}\n\\operatorname{Res}_{s=\\alpha-k}\\mathcal Z_N(s;a)\n&=\\frac{\\alpha\\sin(\\pi(\\alpha-k))}{\\pi}\n  \\binom{\\alpha-1}{k}a^k\\\\\n&=\\boxed{\\frac{\\sin(\\pi/N)}{N\\pi}\n          \\frac{(1-1/N)_k}{k!}a^k.}\n\\end{aligned}\n\\tag{T7}\n\\]","Here \\((b)_0=1\\) and \\((b)_k=b(b+1)\\cdots(b+k-1)\\). The sign calculation\nuses both \\(\\sin(\\pi(\\alpha-k))=(-1)^k\\sin(\\pi\\alpha)\\) and\n\\(\\binom{\\alpha-1}{k}=(-1)^k(1-\\alpha)_k/k!\\); their signs cancel. All\nresidues are positive when \\(a>0\\). At \\(a=0\\), the secondary\ncoefficients vanish, so the only remaining pole is the unshifted pole at\n\\(1/N\\). As before, its nonzero residue makes the absolute convergence\nabscissa exactly \\(1/N\\).","b(b+1)\\cdots(b+k-1)\\). The sign calculation\nuses both \\(\\sin(\\pi(\\alpha-k))=(-1)^k\\sin(\\pi\\alpha)\\) and\n\\(\\binom{\\alpha-1}{k}=(-1)^k(1-\\alpha)_k/k!\\); their signs cancel. All\nresidues are positive when \\(a>0\\). At \\(a=0\\), the secondary\ncoefficients vanish, so the only remaining pole is the unshifted pole at\n\\(1/N\\). As before, its nonzero residue makes the absolute convergence\nabscissa exactly \\(1/N\\).","None of the pole locations \\(\\alpha-k\\) is an integer. Thus the sine\nfactor vanishes at every nonpositive integer without an opposing pole:","\\[\n\\boxed{\\mathcal Z_N(-m;a)=0\\quad(m\\ge0).}\n\\tag{T8}\n\\]","Finally, the one-term subtraction is already convergent at \\(s=0\\),\nsince \\(G_{N,a}(t)-\\alpha t^{\\alpha-1}=O(t^{\\alpha-2})\\) and\n\\(\\alpha<1\\). Let","\\[\nB_{N,a}(0)=\\int_0^1G_{N,a}(t)dt\n +\\int_1^\\infty[G_{N,a}(t)-\\alpha t^{\\alpha-1}]dt-1.\n\\]","Integrating exactly and using \\(\\log H_N(a+t)-t^\\alpha\\to-\\log N\\) gives","\\[\nB_{N,a}(0)=-\\log N-\\log H_N(a).\n\\]","The multiplier \\(\\sin(\\pi s)/\\pi\\) then gives","\\[\n\\boxed{\\mathcal Z_N'(0;a)=-\\log[N E_{N,1}(a)],\\qquad\n       \\exp[-\\mathcal Z_N'(0;a)]=N E_{N,1}(a).}\n\\tag{T9}\n\\]","For \\(N=2\\), the product reduces to \\(2\\cosh\\sqrt a\\). Four direct\nquadrature checks at \\(a=0,1/4,2,9\\) agree with \\(-\\log(2\\cosh\\sqrt a)\\)\nat 80-digit working precision. They use a local integral, an exactly\nintegrated algebraic leading tail, and an exponentially decaying\nremainder; the quadrature does not sample numerical zero locations.\nThese are four additional numerical benchmarks, with the original 411\nexact assertions unchanged. Results are in\n\\path{evidence/spectral_continuation_checks.json} under\n\\path{shifted_N2_product_benchmarks}.","\\[\n0<q,r,a,b<1.\n\\]","\\[\n\\boxed{\\begin{aligned}\nU'(z)&=1+U(qz)V(rz),\\\\\nV'(z)&=1+U(az)V(bz),\\\\\nU(0)&=V(0)=0.\n\\end{aligned}}\n\\]","\\[\nU(z)=\\sum_{n\\ge0}u_nz^{2n+1},\\qquad\nV(z)=\\sum_{n\\ge0}v_nz^{2n+1}.\n\\]","Then \\(u_0=v_0=1\\), and, for \\(n\\ge1\\),","\\[\n\\begin{aligned}\nu_n&=\\frac1{2n+1}\\sum_{i=0}^{n-1}\n q^{2i+1}r^{2(n-1-i)+1}u_i v_{n-1-i},\\\\\nv_n&=\\frac1{2n+1}\\sum_{i=0}^{n-1}\n a^{2i+1}b^{2(n-1-i)+1}u_i v_{n-1-i}.\n\\end{aligned}\n\\]","\\textbf{Proof.} Comparing successive coefficients in the differential\nequations determines the coefficient of degree \\(k+1\\) from coefficients\nof degrees at most \\(k\\). Thus the formal pair is unique. The displayed\nrecurrence constructs an odd formal pair satisfying both equations.\nFormal uniqueness excludes any additional even coefficients. Strict\npositivity follows by induction.","\\[\n0<u_n\\le2^{-n},\\qquad 0<v_n\\le2^{-n}\n\\]","hold by simultaneous induction. If they hold below \\(n\\), each product\nin either recurrence is at most \\(2^{-(n-1)}\\), and there are \\(n\\)\nproducts. The resulting bound is \\(n2^{-(n-1)}/(2n+1)<2^{-n}\\). Hence\nthe two series converge on \\(|z|<\\sqrt2\\), and define a holomorphic\nsolution there.","Put \\(Q=\\max(q,r,a,b)<1\\). If the pair is already defined on \\(|z|<R\\),\nthen","\\[\n\\int_0^z[1+U(qw)V(rw)]\\,dw,\\qquad\n\\int_0^z[1+U(aw)V(bw)]\\,dw\n\\]","define holomorphic extensions on \\(|z|<R/Q\\). The integrands are\nholomorphic on that disk. On the old disk the new functions agree with\nthe old pair because their derivatives and values at zero agree. On the\nlarger disk, every contracted argument still lies in the old disk, so\nthe extensions satisfy the same equations. Iterating exhausts the\ncomplex plane. Any other entire solution agrees with the unique germ and\ntherefore everywhere. \\(\\square\\)","Q\\). The integrands are\nholomorphic on that disk. On the old disk the new functions agree with\nthe old pair because their derivatives and values at zero agree. On the\nlarger disk, every contracted argument still lies in the old disk, so\nthe extensions satisfy the same equations. Iterating exhausts the\ncomplex plane. Any other entire solution agrees with the unique germ and\ntherefore everywhere. \\(\\square\\)","\\[\n\\begin{aligned}\nU(z)&=z+\\frac{qr}{3}z^3\n +\\frac{qr(qr\\,q^2+ab\\,r^2)}{15}z^5+O(z^7),\\\\\nV(z)&=z+\\frac{ab}{3}z^3\n +\\frac{ab(qr\\,a^2+ab\\,b^2)}{15}z^5+O(z^7).\n\\end{aligned}\n\\]","The factorial coefficients \\(A_n=(2n+1)!u_n\\) and \\(B_n=(2n+1)!v_n\\)\nsatisfy","\\[\n\\begin{aligned}\nA_n&=\\sum_{i=0}^{n-1}\\binom{2n}{2i+1}\n q^{2i+1}r^{2(n-1-i)+1}A_iB_{n-1-i},\\\\\nB_n&=\\sum_{i=0}^{n-1}\\binom{2n}{2i+1}\n a^{2i+1}b^{2(n-1-i)+1}A_iB_{n-1-i}.\n\\end{aligned}\n\\]","Thus \\(A_n,B_n\\) are polynomials with nonnegative integer coefficients\nin the four scales. These formulas are exact algebraic constructions.\nFinite checks of them do not replace the analytic proof.","For \\(s>0\\), let","\\[\nP(s)=\\begin{pmatrix}q^s&r^s\\\\a^s&b^s\\end{pmatrix}.\n\\]","There is a unique \\(\\rho>0\\) with","\\[\n\\boxed{(1-q^\\rho)(1-b^\\rho)=(ar)^\\rho.}\n\\]","Indeed, the left side strictly increases continuously from zero to one,\nwhile the right side strictly decreases continuously from one to zero.\nThis equation is precisely \\(\\det(I-P(\\rho))=0\\). Define","\\[\n\\widetilde w=(r^\\rho,1-q^\\rho)^T,\n\\qquad\nw=\\frac{\\widetilde w}{\\max(r^\\rho,1-q^\\rho)}.\n\\]","Both coordinates of \\(w\\) are positive, \\(\\max(w_1,w_2)=1\\), and direct\nmultiplication gives","\\[\nP(\\rho)w=w.\n\\]","The other eigenvalue is \\(q^\\rho+b^\\rho-1\\), whose absolute value is\nstrictly less than one. Consequently","\\[\n\\boxed{\\operatorname{spr}P(\\rho)=1.}\n\\]","\\[\n\\ell=(a^\\rho,1-q^\\rho)^T,\\qquad \\ell^TP(\\rho)=\\ell^T.\n\\]","The notation \\(w\\) for the growth vector is deliberately different from\n\\(v_n\\), the coefficients of \\(V\\).","For \\(R>0\\), positivity gives","\\[\nM_U(R)=\\max_{|z|=R}|U(z)|=U(R),\\qquad\nM_V(R)=V(R).\n\\]","Set \\(F_1=U\\), \\(F_2=V\\), and \\(H_i(R)=\\log F_i(R)\\). Since\n\\(F_i(R)\\ge R\\), these logarithms are positive for \\(R>1\\). The type\nconvention used here is","\\[\n\\underline\\tau_i=\\liminf_{R\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho},\n\\qquad\n\\overline\\tau_i=\\limsup_{R\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho}.\n\\]","\\textbf{Theorem F.2.} Both functions have order \\(\\rho\\), positive lower\ntype and finite upper type. More precisely, the explicit constants below\nsatisfy","\\[\n0<c_-w_i\\le\\underline\\tau_i\\le\\overline\\tau_i\\le c_+w_i<\\infty,\n\\qquad i=1,2.\n\\]","\\[\n\\lim_{R\\to\\infty}\\frac{\\log\\log M_{F_i}(R)}{\\log R}=\\rho.\n\\]","\\[\nm=\\min(q,r,a,b),\\qquad Q=\\max(q,r,a,b),\n\\]","\\[\nC=\\frac{Q}{1-Q},\\qquad B=1+\\frac2m,\n\\qquad c_-=\\frac{\\log2}{(B+C)^\\rho}.\n\\]","\\[\n\\boxed{H_i(R)\\ge c_-w_i(R+C)^\\rho\\quad(R\\ge2,\\ i=1,2).}\n\\]","On \\(2\\le R\\le B\\), its right side is at most \\(\\log2\\), because\n\\(w_i\\le1\\), whereas \\(H_i(R)\\ge\\log R\\ge\\log2\\).","For the first equation, integrate only over the final unit interval to\nobtain, when \\(R\\ge B\\),","\\[\nU(R)\\ge\\int_{R-1}^R U(qt)V(rt)\\,dt\n\\ge U(q(R-1))V(r(R-1)).\n\\]","The second equation gives the same inequality with \\((q,r)\\) replaced by\n\\((a,b)\\) and \\(U(R)\\) replaced by \\(V(R)\\). All four child radii are at\nleast two. Every scale \\(p\\in\\{q,r,a,b\\}\\) satisfies","\\[\np(R-1)+C\\ge p(R+C),\n\\]","because \\((1-p)C\\ge p\\). If the comparison already holds at the child\nradii, then","\\[\n\\begin{aligned}\nH_1(R)&\\ge c_-\\{w_1[q(R-1)+C]^\\rho\n                     +w_2[r(R-1)+C]^\\rho\\}\\\\\n&\\ge c_-(q^\\rho w_1+r^\\rho w_2)(R+C)^\\rho\n=c_-w_1(R+C)^\\rho.\n\\end{aligned}\n\\]","The second row of \\(P(\\rho)w=w\\) proves the second component in exactly\nthe same way.","\\[\nd=(1-Q)B+Q>0.\n\\]","Every child of \\(R\\ge B\\) is at most \\(Q(R-1)\\le R-d\\). Simultaneous\ninduction over the intervals \\([B+kd,B+(k+1)d]\\), starting with the\nestablished seed interval, therefore proves both comparisons for every\n\\(R\\ge2\\). Continuity handles shared endpoints. The argument uses only\nmonotonicity of positive powers and works unchanged when \\(0<\\rho<1\\).","\\[\n\\Delta=\\log(1/m),\\qquad\nD=\\max\\left(\\log\\frac1{qr},\\log\\frac1{ab}\\right)+\\log2.\n\\]","Choose any integer \\(N\\ge0\\) satisfying \\(Q^{-N}\\ge3/m\\), and define","\\[\nW_N=(2^{N+1}-1)\\log2\n +(2^{N+1}-N-2)\\log(1/Q),\n\\]","\\[\nS=W_N+1+\\Delta+D,\\qquad\nc_+=\\frac{e^{-\\rho}S}{\\min(w_1,w_2)}.\n\\]","These constants require only the parameters and the uniquely defined\nroot \\(\\rho\\). They require no evaluation of an unknown solution.","To bound a seed interval, the coefficient majorant from Theorem F.1 gives\n\\(F_i(1)\\le2\\). Put \\(R_k=Q^{-k}\\), \\(Z_0=2\\), and","\\[\nZ_k=Q^{-k}(1+Z_{k-1}^2).\n\\]","Monotonicity of the positive functions and the integral equations give\n\\(F_i(R_k)\\le Z_k\\) simultaneously. As \\(Z_k\\ge1\\),","\\[\n\\log Z_k\\le k\\log(1/Q)+\\log2+2\\log Z_{k-1}.\n\\]","Summing this elementary recurrence gives \\(\\log Z_N\\le W_N\\). Therefore\n\\(H_i(e^t)\\le W_N\\) throughout \\(1\\le t\\le1+\\Delta\\), since\n\\(e^t\\le e/m<3/m\\le R_N\\).","\\[\nG_i(t)=H_i(e^t)+t+D.\n\\]","For \\(t\\ge1+\\Delta\\), all child radii are at least \\(e\\), so their\nfunction values are at least one. The full integral gives","\\[\nH_1(e^t)\\le t+\\log2\n +H_1(e^{t-\\log(1/q)})+H_2(e^{t-\\log(1/r)}).\n\\]","The definition of \\(D\\) converts this into","\\[\nG_1(t)\\le G_1(t-\\log(1/q))+G_2(t-\\log(1/r)).\n\\]","\\[\nG_2(t)\\le G_1(t-\\log(1/a))+G_2(t-\\log(1/b)).\n\\]","On the seed interval, \\(G_i(t)\\le S\\le c_+w_i e^{\\rho t}\\). The Perron\nequation propagates this comparison: substituting the bounds at the\nchildren in the first inequality gives\n\\(c_+e^{\\rho t}(q^\\rho w_1+r^\\rho w_2)=c_+w_1e^{\\rho t}\\), and the\nsecond row works likewise. All logarithmic delays belong to\n\\([\\log(1/Q),\\Delta]\\). Induction in intervals of length\n\\(\\log(1/Q)>0\\), after the seed interval, therefore proves","\\[\n\\boxed{H_i(R)\\le c_+w_iR^\\rho-\\log R-D\\quad(R\\ge e).}\n\\]","The lower and upper comparisons imply the asserted positive, finite\ntypes. Taking logarithms of these bounds for sufficiently large \\(R\\)\ngives the claimed radial exponent limits and hence exact orders.\n\\(\\square\\)","\\textbf{Theorem F.3.} There exist positive finite constants\n\\(\\underline\\tau\\le\\overline\\tau\\) such that","\\[\n\\boxed{\\begin{aligned}\n(\\underline\\tau_1,\\underline\\tau_2)^T&=\\underline\\tau\\,w,\\\\\n(\\overline\\tau_1,\\overline\\tau_2)^T&=\\overline\\tau\\,w.\n\\end{aligned}}\n\\]","\\[\n\\frac{\\underline\\tau_1}{\\underline\\tau_2}\n=\\frac{\\overline\\tau_1}{\\overline\\tau_2}\n=\\frac{r^\\rho}{1-q^\\rho}\n=\\frac{1-b^\\rho}{a^\\rho}.\n\\]","Either both ordinary type limits exist or neither does. When they exist,\ntheir vector is a positive multiple of \\(w\\).","\\textbf{Proof.} Use the final-unit integral inequality from the lower\nproof and divide by \\(R^\\rho\\). Taking the liminf gives","\\[\n\\underline t\\ge P(\\rho)\\underline t,\n\\qquad\n\\underline t=(\\underline\\tau_1,\\underline\\tau_2)^T.\n\\]","Here \\(p(R-1)\\) tends to infinity continuously and\n\\([p(R-1)]^\\rho/R^\\rho\\to p^\\rho\\), which justifies each scaled liminf.\nThe liminf of a sum is at least the sum of the liminfs. All quantities\nare positive and finite by Theorem F.2.","\\[\nH_1(R)\\le\\log R+\\log2+H_1(qR)+H_2(rR)\n\\]","\\[\n\\overline t\\le P(\\rho)\\overline t.\n\\]","The terms \\((\\log R+\\log2)/R^\\rho\\) vanish. Dotting either componentwise\ninequality with the strictly positive left eigenvector \\(\\ell\\) from\nSection~\\ref{v11-coupled_growth_full-the-characteristic-exponent-and-its-positive-vector} shows that its nonnegative difference vector has zero positive\nweighted sum. Every coordinate of that difference is therefore zero.\nThus both type vectors are fixed by \\(P(\\rho)\\). Its eigenvalue-one\nspace is one dimensional, so each is a positive multiple of \\(w\\).","e terms \\((\\log R+\\log2)/R^\\rho\\) vanish. Dotting either componentwise\ninequality with the strictly positive left eigenvector \\(\\ell\\) from\nSection~\\ref{v11-coupled_growth_full-the-characteristic-exponent-and-its-positive-vector} shows that its nonnegative difference vector has zero positive\nweighted sum. Every coordinate of that difference is therefore zero.\nThus both type vectors are fixed by \\(P(\\rho)\\). Its eigenvalue-one\nspace is one dimensional, so each is a positive multiple of \\(w\\).","trictly positive left eigenvector \\(\\ell\\) from\nSection~\\ref{v11-coupled_growth_full-the-characteristic-exponent-and-its-positive-vector} shows that its nonnegative difference vector has zero positive\nweighted sum. Every coordinate of that difference is therefore zero.\nThus both type vectors are fixed by \\(P(\\rho)\\). Its eigenvalue-one\nspace is one dimensional, so each is a positive multiple of \\(w\\).","If one component has an ordinary type limit, its lower and upper types\nare equal. Because that coordinate of \\(w\\) is positive, this forces\n\\(\\underline\\tau=\\overline\\tau\\), and the other component also has a\ntype limit. \\(\\square\\)","This theorem does not assert a limit for \\(\\log U(R)/\\log V(R)\\) when\ntype oscillations persist. It also does not identify the scalar values\nof either type or prove their equality.","\\[\nU=V\\quad\\Longleftrightarrow\\quad\\{q,r\\}=\\{a,b\\}.\n\\]","For sufficiency, the scalar solution \\(T_{q,r}\\) from\n\\path{routes/asymmetric_dilation.md} makes \\((T_{q,r},T_{q,r})\\)\nsolve the coupled equations. Theorem F.1 gives uniqueness. For necessity,\nequality at degree three gives \\(qr=ab=p>0\\). Equality at degree five\nthen gives \\(q^2+r^2=a^2+b^2\\). The two positive unordered pairs have\nthe same product and the same sum \\(\\sqrt{q^2+r^2+2p}\\), so they\ncoincide. This characterizes equality of the normalized components. A\nstronger assertion about all possible changes of variables requires its\nown nonreduction theorem.","On either scalar reduction locus, the characteristic equation is\nequivalent to \\(q^\\rho+r^\\rho=1\\), recovering the scalar order theorem.\nAt \\(q=r=a=b=1/2\\),","\\[\nU(z)=V(z)=\\sqrt2\\sinh(z/\\sqrt2),\\qquad \\rho=1.\n\\]","\\[\n(q,r,a,b)=\\left(\\frac15,\\frac25,\\frac35,\\frac7{10}\\right).\n\\]","The determinant identity at \\(\\rho=1\\) is\n\\((1-1/5)(1-7/10)=(3/5)(2/5)=6/25\\). Thus the exact order is one, and","\\[\nw=(1/2,1)^T.\n\\]","\\[\n\\begin{aligned}\nU(z)&=z+\\frac2{75}z^3+\\frac{88}{234375}z^5\n +\\frac{18814}{8544921875}z^7+O(z^9),\\\\\nV(z)&=z+\\frac7{50}z^3+\\frac{8211}{1250000}z^5\n +\\frac{85664601}{625000000000}z^7+O(z^9).\n\\end{aligned}\n\\]","Both lower and upper exponential types of \\(V\\) are exactly twice the\ncorresponding types of \\(U\\). The row sums of \\(P(1)\\) are \\(3/5\\) and\n\\(13/10\\). Consequently neither an unweighted average row condition nor\nthe maximum of the two scalar row orders yields this example's coupled\norder. The positive eigenvector accounts for the exchange of growth\nbetween the components.","For this example the elementary lower constants are \\(m=1/5\\),\n\\(Q=7/10\\), \\(C=7/3\\), \\(B=11\\), and \\(c_-=3\\log2/40\\). Hence, for every\n\\(R\\ge2\\),","\\[\n\\log U(R)\\ge\\frac{3\\log2}{80}\\left(R+\\frac73\\right),\\qquad\n\\log V(R)\\ge\\frac{3\\log2}{40}\\left(R+\\frac73\\right).\n\\]","\\textbf{Dependence of the order on the scales.} The determinant equation\nshows that the order depends on the self scales \\(q,b\\) and the product\n\\(ar\\) of the two cross scales. Changing \\(a\\) and \\(r\\) while keeping\ntheir product fixed leaves the order fixed, provided both remain in\n\\((0,1)\\). The Perron vector and the normalized functions generally\nchange.","For any prescribed \\(s>0\\) and any \\(q,r\\in(0,1)\\), choices satisfying","\\[\na^s r^s=(1-q^s)(1-b^s),\\qquad 0<a,b<1,\n\\]","lie on the order-\\(s\\) surface. For instance, one may select \\(q,r,b\\)\nwith \\((1-q^s)(1-b^s)<r^s\\), then define","\\[\na=\\frac{[(1-q^s)(1-b^s)]^{1/s}}{r}.\n\\]","Throughout, \\(q,r,a,b\\in(0,1)\\). Let the normalized coupled solution be","\\[\nU'=1+U(qz)V(rz),\\qquad V'=1+U(az)V(bz),\\qquad U(0)=V(0)=0.\n\\]","Write \\(U(z)=z+u_1z^3+u_2z^5+u_3z^7+\\cdots\\) and\n\\(V(z)=z+v_1z^3+v_2z^5+v_3z^7+\\cdots\\). Put \\(p=qr\\), \\(k=ab\\).\nCoefficient matching gives","\\[\nu_1=\\frac p3,\\qquad v_1=\\frac k3,\n\\] \\[\nu_2=\\frac p{15}(p q^2+k r^2),\\qquad\nv_2=\\frac k{15}(p a^2+k b^2),\n\\] \\[\nu_3=\\frac p7(q^4u_2+p^2u_1v_1+r^4v_2),\\qquad\nv_3=\\frac k7(a^4u_2+k^2u_1v_1+b^4v_2).\n\\]","\\[\n\\boxed{D=k^2u_2-p^2v_2\n=\\frac{pk ar}{15}(q^2-b^2)(qb-ar).}\n\\]","\\begin{enumerate}\n\\def\\labelenumi{\\arabic{enumi}.}\n\\tightlist\n\\item\n  There are nonzero complex constants \\(A,B\\) such that \\(V(z)=A U(Bz)\\)\n  for every complex \\(z\\).\n\\item\n  \\(D=0\\).\n\\item\n  \\(q=b\\) or \\(qb=ar\\).\n\\end{enumerate}","If these conditions hold, put \\(s=\\sqrt{k/p}>0\\), and let \\(T_{x,y}\\)\ndenote the normalized scalar solution \\(T'=1+T(xz)T(yz)\\), \\(T(0)=0\\).\nThen","\\[\n\\boxed{\nU(z)=\\sqrt{s}\\,T_{q,sr}(z/\\sqrt{s}),\\qquad\nV(z)=s^{-1/2}T_{q,sr}(\\sqrt{s}\\,z).\n}\n\\]","The effective scales belong to \\((0,1)\\): if \\(q=b\\), then\n\\(sr=\\sqrt{ar}\\), and if \\(qb=ar\\), then \\(s=b/r\\) and \\(sr=b\\).","\\textbf{Proof.} Normalization gives \\(AB=1\\). Comparing cubic\ncoefficients gives \\(B^2=k/p\\), and comparing fifth coefficients gives\n\\(v_2=(k/p)^2u_2\\). Thus component rescaling implies \\(D=0\\), even when\ncomplex rescalings are allowed. The displayed factorization, strict\npositivity, and \\(q,b>0\\) show that \\(D=0\\) is equivalent to \\(q=b\\) or\n\\(qb=ar\\).","For the converse define the displayed pair using \\(T=T_{q,sr}\\). Direct\ndifferentiation gives","\\[\nU'(z)=1+T(qz/\\sqrt{s})T(r\\sqrt{s}\\,z)\n=1+U(qz)V(rz).\n\\]","\\[\n\\{q\\sqrt{s},\\;s^{3/2}r\\}\n=\\{a/\\sqrt{s},\\;b\\sqrt{s}\\}.\n\\]","When \\(qb=ar\\), one has \\(a=sq\\) and \\(b=sr\\), giving equality in the\ndisplayed order. When \\(q=b\\), one has \\(a=s^2r\\), giving equality after\ninterchanging the factors. The pair also has the required initial values\nand derivatives. Uniqueness of the formal coefficient recursion\nidentifies it with the coupled solution. Finally, \\(V(z)=s^{-1}U(sz)\\),\nproving rescaling.","\\textbf{Theorem G.2.} If \\(D\\ne0\\), the six coefficients\n\\((u_1,v_1,u_2,v_2,u_3,v_3)\\) uniquely determine the ordered tuple\n\\((q,r,a,b)\\) within \\((0,1)^4\\). Consequently, equality of the two\nlabeled normalized germs implies equality of these parameters in this\nregime.","\\textbf{Proof.} First recover \\(p=3u_1\\), \\(k=3v_1\\). If \\(x=q^2\\), then\n\\(r^2=p^2/x\\), and the equation for \\(u_2\\) gives","\\[\np x^2-\\frac{15u_2}{p}x+kp^2=0.\n\\]","This leaves at most two positive candidates. If there are two distinct\npositive roots \\(x,x'\\), their product is \\(xx'=kp\\). The corresponding\nseventh-coefficient values differ by","\\[\nu_3(x)-u_3(x')\n=\\frac p7(x^2-x'^2)\n\\left(u_2-\\frac{p^2}{k^2}v_2\\right)\n=\\frac{pD}{7k^2}(x^2-x'^2).\n\\]","The difference is nonzero because \\(D\\ne0\\) and the distinct positive\nroots have distinct squares. Thus \\(u_3\\) selects the unique root\n\\(q^2\\), giving \\(q=\\sqrt{x}>0\\) and \\(r=p/q\\). A repeated quadratic\nroot already gives a unique value and requires no selection.","Likewise \\(y=a^2\\) obeys","\\[\np y^2-\\frac{15v_2}{k}y+k^3=0,\n\\]","with root product \\(yy'=k^3/p\\). The two candidate seventh coefficients\nhave difference","\\[\nv_3(y)-v_3(y')\n=\\frac k7(y^2-y'^2)\n\\left(u_2-\\frac{p^2}{k^2}v_2\\right)\n=\\frac{D}{7k}(y^2-y'^2),\n\\]","which is again nonzero for distinct positive candidates. Recover\n\\(a=\\sqrt{y}\\) and \\(b=k/a\\). All selections must also respect the\ncontraction domain. Since the supplied coefficients arise from an\nadmissible pair, at least one admissible selection exists. The preceding\ndifferences give uniqueness.","This is an exact-data inverse theorem. The displayed differences become\nsmall as \\(D\\to0\\), so the argument does not supply a uniform stable\ninverse near the reduction surfaces. It also does not identify unlabeled\ncomponents without the additional component-permutation equivalence.","\\[\n(q,r,a,b)=\\left(\\frac15,\\frac25,\\frac35,\\frac7{10}\\right),\\qquad\np=\\frac2{25},\\quad k=\\frac{21}{50}.\n\\]","At \\(\\rho=1\\), the characteristic matrix has positive eigenvector\n\\((1/2,1)^T\\) with eigenvalue one. Its other eigenvalue is \\(-1/10\\).\nThe growth theorem therefore gives exact order one for both components.\nHere \\(q\\ne b\\) and \\(qb\\ne ar\\), so \\(D\\ne0\\). The pair is outside both\nreduction surfaces and its four labeled parameters are recoverable from\nthe coefficients through degree seven.","\\[\nU(z)=\\sum_{n\\ge0}A_n\\frac{z^{2n+1}}{(2n+1)!},\\qquad\nV(z)=\\sum_{n\\ge0}B_n\\frac{z^{2n+1}}{(2n+1)!}.\n\\]","\\[\nA_n=\\sum_{i+j=n-1}\\binom{2n}{2i+1}q^{2i+1}r^{2j+1}A_iB_j,\n\\qquad\nB_n=\\sum_{i+j=n-1}\\binom{2n}{2i+1}a^{2i+1}b^{2j+1}A_iB_j,\n\\]","with \\(A_0=B_0=1\\). Coefficients with even derivative index vanish.\nSwapping the components sends the parameter tuple \\((q,r,a,b)\\) to\n\\((b,a,r,q)\\). Identical rows \\((a,b)=(q,r)\\), or exchanged rows\n\\((a,b)=(r,q)\\), yield the scalar reduction \\(U=V=T_{q,r}\\).","Both derivative arrays are polynomials in four scale variables with\nnonnegative integer coefficients. Their total degree is at most\n\\(n(n+1)\\). Indeed, a recurrence term with \\(i+j=n-1\\) has degree at\nmost \\(2n+i(i+1)+j(j+1)\\le n(n+1)\\). If all four denominators divide\n\\(D\\), then \\(D^{n(n+1)}A_n\\) and \\(D^{n(n+1)}B_n\\) are integers. The\nclearing factor is a convenient uniform choice and need not be minimal.","This argument uses \\(0<q,r,a,b<1\\) and does not require the order\ntheorem. Let \\(Q=\\max(q,r,a,b)\\). All ordinary odd Taylor coefficients\nare positive and at most one: induction in the ordinary recurrence\nbounds the coefficient of \\(z^{2n+1}\\) by \\(n/(2n+1)\\) for \\(n\\ge1\\).\nThus both series converge on the unit disk and, for \\(0\\le x<1\\),","\\[\n\\max(U(x),V(x))\\le \\frac{x}{1-x^2}.\n\\]","Analytic continuation by the integral equations expands each known disk\nby the factor \\(1/Q\\), so both solutions are entire. Positive\ncoefficients give \\(\\max_{|z|\\le x}|F(z)|=F(x)\\) for either component.\nThe integral equations therefore imply","\\[\nM(x):=\\max(U(x),V(x))\\le x\\bigl(1+M(Qx)^2\\bigr).\n\\]","Start at the first member \\(Q^kx<1\\) of this finite contraction ladder,\nuse \\(x/(1-x^2)\\) there, and repeatedly apply \\(B(x)=x(1+B(Qx)^2)\\)\nbackward. This defines the rational \\path{disk_majorant} bound for\nrational parameters and radius. The implementation has explicit\nrecursion and integer size guards, which raise errors rather than\nreturning a weakened result.","If \\(F_N\\) retains odd indices \\(0,\\ldots,N\\), then for \\(S>R\\ge0\\),","\\[\n\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.\n\\]","For every omitted exponent \\(m\\ge2N+3\\), use \\(R^m\\le (R/S)^{2N+3}S^m\\),\nsum positive terms, and bound their sum by \\(F(S)\\le B(S)\\). No extra\ngeometric denominator is needed. For rational \\(x\\ge0\\),\n\\path{certified_evaluation} returns the exact interval\n\\([F_N(x),F_N(x)+B(S)(x/S)^{2N+3}]\\) for each component.","At \\(x=1\\), with \\(N=30\\) and \\(S=2\\), the common tail bound is below\n\\(1.600\\times10^{-13}\\). The partial values are approximately\n\\(U(1)=1.027044341827337\\) and \\(V(1)=1.1467074957737624\\). These\ndisplayed decimals are rounded illustrations. The returned fractions are\nthe actual certificate.","Let \\(q,r,a,b>0\\), with contraction \\(<1\\) imposed when discussing the\nnormalized entire solution. For an arbitrary analytic vector input\n\\(X=(u,v)\\) with \\(u(0)=v(0)=0\\), define the functional vector field","\\[\n\\mathcal F[X](z)=\n\\begin{pmatrix}1+u(qz)v(rz)\\\\1+u(az)v(bz)\\end{pmatrix}.\n\\tag{S1}\n\\]","A nonzero constant row \\(c=(c_1,c_2)\\in\\mathbb C^2\\) is an\n\\textbf{autonomous scalar projection} if some scalar functional\n\\(\\mathcal H\\) satisfies","\\[\nc\\mathcal F[X]=\\mathcal H[cX]\n\\quad\\text{for every such analytic input }X.\n\\tag{S2}\n\\]","The functional \\(\\mathcal H\\) may use the whole scalar input. Thus this\ndefinition allows any scalar dilation equation, not only a prescribed\nquadratic ansatz. Equality (S2) means that two vector inputs with the\nsame scalar projection have the same projected vector field. It is an\nidentity of functional differential operators. It does not assert an\nidentity only along the distinguished normalized solution.","A constant invertible change \\(W=PX\\) completely decouples the operator\nif both rows of \\(P\\) are autonomous scalar projections. No assumption\nabout the form of their scalar functionals is needed.","\\textbf{Theorem H.1.} For positive nonzero scales, a nonzero row \\(c\\)\nsatisfies (S2) if and only if","\\[\nq=a,\\qquad r=b,\\qquad c=\\gamma(1,-1),\\qquad\\gamma\\ne0.\n\\tag{S3}\n\\]","In that exceptional case the projected vector field is identically zero.\nConsequently, no invertible constant linear change completely decouples\n(S1), including when some or all scales coincide. If \\((q,r)\\ne(a,b)\\),\nthere is not even one nonzero autonomous scalar projection. Four\npairwise distinct contraction scales are a special case of this stronger\nresult.","\\[\nQ_c[u,v](z)=c_1u(qz)v(rz)+c_2u(az)v(bz).\n\\]","If (S2) holds, \\(Q_c\\) is constant on every fiber of\n\\((u,v)\\mapsto c_1u+c_2v\\). Fix \\(z_0\\ne0\\). Polynomial interpolation\ngives a scalar polynomial \\(f\\) with \\(f(0)=0\\) and \\(f(sz_0)=1\\) at\nevery distinct scale \\(s\\in\\{q,r,a,b\\}\\). For inputs \\(u=xf,v=yf\\),\nwhere \\(x,y\\in\\mathbb C\\),","\\[\nQ_c[xf,yf](z_0)=(c_1+c_2)xy.\n\\tag{S4}\n\\]","It must depend only on \\(c_1x+c_2y\\). If \\(c_1+c_2\\ne0\\),\ndifferentiating the polynomial \\(xy\\) along the fiber direction\n\\((c_2,-c_1)\\) would give","\\[\nc_2y-c_1x=0\\quad\\text{for every }x,y,\n\\]","which forces \\(c=0\\), a contradiction. Therefore \\(c_2=-c_1\\), and\nscaling the row reduces the question to \\(c=(1,-1)\\).","For this row, put \\(u=w+g,v=g\\). This input and \\((w,0)\\) have the same\nprojection \\(w\\). The second input has \\(Q_c[w,0]=0\\), so autonomy\nforces","\\[\nw(qz)g(rz)-w(az)g(bz)\n+g(qz)g(rz)-g(az)g(bz)=0\n\\tag{S5}\n\\]","for all zero-constant analytic \\(w,g\\). Setting \\(w=0\\) and subtracting\nthat instance of (S5) leaves","\\[\nw(qz)g(rz)=w(az)g(bz).\n\\tag{S6}\n\\]","Taking successively \\((w,g)=(z,z),(z^2,z),(z,z^2)\\) yields \\(qr=ab\\),\n\\(q^2r=a^2b\\), and \\(qr^2=ab^2\\). Nonzero scales then give \\(q=a\\) and\n\\(r=b\\). Conversely these equalities make the two components of (S1)\nidentical on every input, so \\((1,-1)\\mathcal F[X]=0\\). The only\npossible autonomous rows therefore span a one-dimensional space, which\ncannot be the two rows of an invertible matrix. \\(\\square\\)","The same proof works on odd analytic inputs for positive scales: choose\n\\(f(z)=z p(z^2)\\) by interpolation at the distinct positive scale\nsquares, and use \\((z,z),(z^3,z),(z,z^3)\\) in (S6). Positivity then\nresolves the equalities of squares. The theorem does not depend on\nallowing even inputs.","The preceding proof also makes a coordinate invariant obstruction\nexplicit. Under an invertible state change, write the original\ncoordinates as \\(u=\\ell_1(W),v=\\ell_2(W)\\), where \\(\\ell_1,\\ell_2\\) are\nindependent linear forms. At the pair of distinct nodes \\((qz,rz)\\), the\nfirst quadratic term has bilinear coefficient","\\[\nB=\\ell_1\\otimes\\ell_2.\n\\]","It has rank one, but is not symmetric:\n\\(B-B^{\\mathsf T}=\\ell_1\\wedge\\ell_2\\ne0\\). At four distinct nonzero\nnodes, polynomial interpolation lets their vector values vary\nindependently, so terms supported at \\((az,bz)\\) cannot cancel this\ncoefficient. A scalar autonomous channel at the same node pair would\nrequire \\(B=\\kappa\\ell\\otimes\\ell\\), a symmetric rank-one tensor, after\nprojecting the output onto that channel. For any output row with a\nnonzero coefficient on this term, the alternating part remains nonzero.\nThe other node pair gives the same obstruction if the first output\ncoefficient vanishes.","This is not a rank-only argument: the tensor \\(\\ell_1\\otimes\\ell_2\\)\nalready has rank one. Its nonzero alternating part is the decisive\ninvariant at distinct nodes. When the nodes of a product coincide, the\nsymmetric quadratic form \\(uv\\) has matrix","\\[\n\\begin{pmatrix}0&1/2\\\\1/2&0\\end{pmatrix},\n\\]","For contraction scales, the normalized analytic germ is unique. If\n\\((a,b)=(q,r)\\), its components coincide because their derivatives are\nidentical and their initial values agree. The common component is the\nscalar family \\(T'=1+T(qz)T(rz),T(0)=0\\). The same coincidence holds if\n\\((a,b)=(r,q)\\): \\((T,T)\\) solves both equations because multiplication\nis commutative, and uniqueness identifies the normalized solution. For\n\\(q\\ne r\\), this reversed-pair case has no autonomous scalar operator\nprojection by Theorem H.1, despite having \\(U=V\\) along its normalized\nsolution. This is a concrete reason the two notions must remain\ndistinct.","he normalized analytic germ is unique. If\n\\((a,b)=(q,r)\\), its components coincide because their derivatives are\nidentical and their initial values agree. The common component is the\nscalar family \\(T'=1+T(qz)T(rz),T(0)=0\\). The same coincidence holds if\n\\((a,b)=(r,q)\\): \\((T,T)\\) solves both equations because multiplication\nis commutative, and uniqueness identifies the normalized solution. For\n\\(q\\ne r\\), this reversed-pair case has no autonomous scalar operator\nprojection by Theorem H.1, despite having \\(U=V\\) along its normalized\nsolution. This is a concrete reason the two notions must remain\ndistinct.","identical and their initial values agree. The common component is the\nscalar family \\(T'=1+T(qz)T(rz),T(0)=0\\). The same coincidence holds if\n\\((a,b)=(r,q)\\): \\((T,T)\\) solves both equations because multiplication\nis commutative, and uniqueness identifies the normalized solution. For\n\\(q\\ne r\\), this reversed-pair case has no autonomous scalar operator\nprojection by Theorem H.1, despite having \\(U=V\\) along its normalized\nsolution. This is a concrete reason the two notions must remain\ndistinct.","\\[\nU=V\\quad\\Longleftrightarrow\\quad\\{q,r\\}=\\{a,b\\}.\n\\tag{S7}\n\\]","To prove necessity, write \\(U=V=T=z+c_1z^3+\\cdots\\). Then\n\\(3c_1=qr=ab>0\\). Comparing the \\(z^4\\) coefficients of\n\\(T(qz)T(rz)=T(az)T(bz)\\) gives \\(c_1qr(q^2+r^2)=c_1ab(a^2+b^2)\\), so\nboth products and sums of squares agree. Positive scales then have equal\nsums and are the roots of the same quadratic. Sufficiency is the\nuniqueness argument above. For example, \\((q,r,a,b)=(1/2,1/3,1/3,1/2)\\)\nhas coincident normalized components and no nonzero autonomous operator\nprojection.","At \\(q=r=a=b=s>0\\), the variables \\(w=u+v,d=u-v\\) give","\\[\nw'=2+\\tfrac12\\bigl(w(sz)^2-d(sz)^2\\bigr),\\qquad d'=0.\n\\]","This triangular system is not a completely decoupled operator. The\nnormalized solution lies in the invariant subspace \\(d=0\\).","Zero scales lie outside Theorem H.1. On zero-constant inputs, a product\nvanishes identically if either of its scales is zero. If both products\nvanish, both equations are already \\(u'=v'=1\\). If only the first\nproduct vanishes, the projection \\(u\\) is autonomous, regardless of\nwhether the second pair agrees with the first. Thus extending Theorem H.1\nto zero scales would be false. Scale one and positive scales greater\nthan one do not affect the algebraic operator theorem, although the\ncontraction-domain entire growth theorems do not automatically extend to\nthem. Negative or complex scales and odd-input coincidences caused by\nsigns are not classified here.","\\pageheading{FAMILY 5}{Two functions exchange growth}\nFor four real contractions $0<q,r,a,b<1$, define the normalized pair\n\\[\n\\boxed{\\begin{aligned}\nU'(z)&=1+U(qz)V(rz),","\\pageheading{FAMILY 5}{Two functions exchange growth}\nFor four real contractions $0<q,r,a,b<1$, define the normalized pair\n\\[\n\\boxed{\\begin{aligned}\nU'(z)&=1+U(qz)V(rz),\\\\\nV'(z)&=1+U(az)V(bz),\\qquad U(0)=V(0)=0.\n\\end{aligned}}\n\\]\nEach equation combines one value of $U$ with one value of $V$. The four\nscales specify where that exchange occurs. Both functions begin with the\nsame unit slope, while later coefficients retain the two equations' different\nscale addresses.","V'(z)&=1+U(az)V(bz),\\qquad U(0)=V(0)=0.\n\\end{aligned}}\n\\]\nEach equation combines one value of $U$ with one value of $V$. The four\nscales specify where that exchange occurs. Both functions begin with the\nsame unit slope, while later coefficients retain the two equations' different\nscale addresses.","\\vfig[0.96]{coupled_01_matrix}{The worked pair\n$(q,r,a,b)=(1/5,2/5,3/5,7/10)$ has exact order one. The two displayed\nidentities give the positive growth vector $(1/2,1)^T$.}","There is a unique entire solution pair. Both functions are odd, and every\nodd Taylor coefficient is positive. The common order is governed by the\npositive matrix\n\\[\nP(s)=\\begin{pmatrix}q^s&r^s\\\\a^s&b^s\\end{pmatrix},\n\\qquad \\operatorname{spr}P(\\rho)=1.\n\\]\nThe eigenvector supplies the relative weights that a row sum misses. In\nthe example, the row sums at $s=1$ are $3/5$ and $13/10$. The coupled\nsystem nevertheless has order exactly one in both components.","a^s&b^s\\end{pmatrix},\n\\qquad \\operatorname{spr}P(\\rho)=1.\n\\]\nThe eigenvector supplies the relative weights that a row sum misses. In\nthe example, the row sums at $s=1$ are $3/5$ and $13/10$. The coupled\nsystem nevertheless has order exactly one in both components.","\\pageheading{FAMILY 5 / CONSTRUCTION}{The coefficients preserve both addresses}\nWrite $U(z)=\\sum_{n\\ge0}u_nz^{2n+1}$ and\n$V(z)=\\sum_{n\\ge0}v_nz^{2n+1}$, with $u_0=v_0=1$.\nThe first equation gives the exact recursion\n\\[\nu_n=\\frac1{2n+1}\\sum_{i+j=n-1}\nq^{2i+1}r^{2j+1}u_i v_j,\\qquad n\\ge1.\n\\]\nThe second replaces $(q,r)$ by $(a,b)$ while retaining the same products\n$u_i v_j$. Thus each new coefficient uses the shared history of the pair.","\\vfig[0.97]{coupled_02_functions}{For\n$(q,r,a,b)=(1/3,2/3,1/2,3/4)$, the components separate despite their\ncommon initial slope. These numerical curves use odd indices through 240.\nThe two truncations at indices 120 and 240 agree at $x=8$ to displayed\ndouble precision.}","The first distinction already appears at degree three:\n\\[\nU(z)=z+\\frac{qr}{3}z^3+O(z^5),\\qquad\nV(z)=z+\\frac{ab}{3}z^3+O(z^5).\n\\]\nInduction gives $0<u_n,v_n\\le2^{-n}$, which supplies a convergent germ.\nIf both functions are known on a disk of radius $R$, their integral\nequations extend them to radius $R/Q$, where $Q=\\max(q,r,a,b)<1$.\nRepeating this step reaches every finite disk.","\\pageheading{FAMILY 5 / GROWTH}{One order, two exact type proportions}\nThe entire functions have the same order $\\rho$, with positive lower type\nand finite upper type. Positivity places the maximum modulus on the positive\nreal axis, so $M_U(R)=U(R)$ and $M_V(R)=V(R)$.","For $F_1=U$ and $F_2=V$, define\n\\[\n\\underline\\tau_i=\\liminf_{R\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho},\n\\qquad\n\\overline\\tau_i=\\limsup_{R\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho}.\n\\]\nIf $P(\\rho)w=w$ with $w>0$, then both type vectors have precisely these\nPerron proportions:\n\\[\n\\boxed{\\underline{\\boldsymbol\\tau}=\\underline\\tau\\,w,\n\\qquad\\overline{\\boldsymbol\\tau}=\\overline\\tau\\,w,\n\\qquad0<\\underline\\tau\\le\\overline\\tau<\\infty.}\n\\]\nFor the worked pair, both the lower type and the upper type of $V$ are\ntwice the corresponding type of $U$. Either both ordinary type limits\nexist or neither does. Equality of the lower and upper types remains open.\nThe finite-radius slope picture does not decide that question","For $F_1=U$ and $F_2=V$, define\n\\[\n\\underline\\tau_i=\\liminf_{R\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho},\n\\qquad\n\\overline\\tau_i=\\limsup_{R\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho}.\n\\]\nIf $P(\\rho)w=w$ with $w>0$, then both type vectors have precisely these\nPerron proportions:\n\\[\n\\boxed{\\underline{\\boldsymbol\\tau}=\\underline\\tau\\,w,\n\\qquad\\overline{\\boldsymbol\\tau}=\\overline\\tau\\,w,\n\\qquad0<\\underline\\tau\\le\\overline\\tau<\\infty.}\n\\]\nFor the worked pair, both the lower type and the upper type of $V$ are\ntwice the corresponding type of $U$. Either both ordinary type limits\nexist or neither does. Equality of the lower and upper types remains open.\nThe finite-radius slope picture does not decide that question.","\\rho},\n\\qquad\n\\overline\\tau_i=\\limsup_{R\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho}.\n\\]\nIf $P(\\rho)w=w$ with $w>0$, then both type vectors have precisely these\nPerron proportions:\n\\[\n\\boxed{\\underline{\\boldsymbol\\tau}=\\underline\\tau\\,w,\n\\qquad\\overline{\\boldsymbol\\tau}=\\overline\\tau\\,w,\n\\qquad0<\\underline\\tau\\le\\overline\\tau<\\infty.}\n\\]\nFor the worked pair, both the lower type and the upper type of $V$ are\ntwice the corresponding type of $U$. Either both ordinary type limits\nexist or neither does. Equality of the lower and upper types remains open.\nThe finite-radius slope picture does not decide that question.","\\to\\infty}\\frac{\\log M_{F_i}(R)}{R^\\rho}.\n\\]\nIf $P(\\rho)w=w$ with $w>0$, then both type vectors have precisely these\nPerron proportions:\n\\[\n\\boxed{\\underline{\\boldsymbol\\tau}=\\underline\\tau\\,w,\n\\qquad\\overline{\\boldsymbol\\tau}=\\overline\\tau\\,w,\n\\qquad0<\\underline\\tau\\le\\overline\\tau<\\infty.}\n\\]\nFor the worked pair, both the lower type and the upper type of $V$ are\ntwice the corresponding type of $U$. Either both ordinary type limits\nexist or neither does. Equality of the lower and upper types remains open.\nThe finite-radius slope picture does not decide that question.","\\pageheading{FAMILY 5 / PARAMETER SPACE}{Growth follows a spectral surface}\nThe order is the unique positive solution of\n\\[\n\\boxed{(1-q^\\rho)(1-b^\\rho)=(ar)^\\rho.}\n\\]\nThe left side increases from zero to one as $\\rho$ increases. The right\nside decreases from one to zero. Their unique crossing determines the\ngrowth exponent without fitting a truncated Taylor curve.","\\vfig[0.80]{coupled_04_landscape}{Numerical solutions of the proved\ncharacteristic equation, with $r=2/5$ and $a=3/5$. The highlighted pair\n$(q,b)=(1/5,7/10)$ has order exactly one by rational arithmetic.}","The order depends on the self scales $q,b$ and on the product $ar$ of\nthe cross scales. Holding $q,b,ar$ fixed preserves the order, while\nchanging the cross scales separately generally changes the functions and\ntheir type proportions:\n\\[\n\\frac{w_1}{w_2}=\\frac{r^\\rho}{1-q^\\rho}\n=\\frac{1-b^\\rho}{a^\\rho}.\n\\]\nConsequently, equal order does not determine the family member. It fixes\none growth exponent within a larger parameter space. The coefficients\ncarry enough additional information to recover all four ordered scales\naway from the reduction surfaces described below.","\\pageheading{FAMILY 5 / PRECISION}{A coefficient list with a certified tail}\nA finite Taylor list becomes an exact approximation certificate when it\nis paired with a bound on a larger disk. Let $F$ be either component,\nand let $F_N$ retain the odd indices $0,\\ldots,N$.","Put $Q=\\max(q,r,a,b)$. One elementary majorant starts with\n$B(x)=x/(1-x^2)$ for $0\\le x<1$ and works outward through the finite\ncontraction ladder using\n\\[\nB(x)=x\\bigl(1+B(Qx)^2\\bigr).\n\\]\nThe integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal roun","Put $Q=\\max(q,r,a,b)$. One elementary majorant starts with\n$B(x)=x/(1-x^2)$ for $0\\le x<1$ and works outward through the finite\ncontraction ladder using\n\\[\nB(x)=x\\bigl(1+B(Qx)^2\\bigr).\n\\]\nThe integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal rounding.","through the finite\ncontraction ladder using\n\\[\nB(x)=x\\bigl(1+B(Qx)^2\\bigr).\n\\]\nThe integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal rounding.","the finite\ncontraction ladder using\n\\[\nB(x)=x\\bigl(1+B(Qx)^2\\bigr).\n\\]\nThe integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal rounding.","der using\n\\[\nB(x)=x\\bigl(1+B(Qx)^2\\bigr).\n\\]\nThe integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal rounding.","g\n\\[\nB(x)=x\\bigl(1+B(Qx)^2\\bigr).\n\\]\nThe integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal rounding.","x\\bigl(1+B(Qx)^2\\bigr).\n\\]\nThe integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal rounding.","The integral equations show that this bounds both components on the disk\n$|z|\\le x$. For $S>R\\ge0$, positivity then gives\n\\[\n\\boxed{\\max_{|z|\\le R}|F(z)-F_N(z)|\n\\le B(S)\\left(\\frac RS\\right)^{2N+3}.}\n\\]\nThe exponent is the first omitted odd degree. Every later term has an\neven smaller ratio between its values at radii $R$ and $S$.\nFor the worked pair, $N=30$, $R=1$ and $S=2$ give an error below\n$1.600\\times10^{-13}$ for both functions. The laboratory returns\nrational interval endpoints, so the certificate survives decimal rounding.","\\pageheading{FAMILY 5 / SEQUENCES}{The phase calculus transports the pair}\nUse the derivative normalization\n$A_n=(2n+1)!u_n$, $B_n=(2n+1)!v_n$. Its recurrence has integer\nbinomial weights:\n\\[\nA_n=\\sum_{i+j=n-1}\\binom{2n}{2i+1}\nq^{2i+1}r^{2j+1}A_iB_j,\\qquad n\\ge1.\n\\]\nThe second component again replaces $(q,r)$ by $(a,b)$.","\\vfig[0.91]{coupled_06_polyphase}{The full derivative array of $V$ is split\ninto three residue lanes. Applying taps $(1,-2,1)$ with the correct block\ncarries, then interleaving, agrees exactly with direct convolution.}","These derivative arrays are polynomials with nonnegative integer\ncoefficients in the four scale variables. Their total degree is at most\n$n(n+1)$. If every parameter denominator divides $D_0$, then\n\\[\nD_0^{n(n+1)}A_n,\\qquad D_0^{n(n+1)}B_n\n\\quad\\hbox{are integers}.\n\\]\nThis uniform clearing factor makes exact rational families available to\ninteger sequence experiments. It need not be the smallest possible factor.","\\pageheading{FAMILY 5 / REDUCTION}{Two surfaces where one component rescales to the other}\nSet $p=qr$, $k=ab$, and use the ordinary Taylor coefficients $u_2,v_2$\nat degree five. The exact separation quantity factors as\n\\[\n\\boxed{D=k^2u_2-p^2v_2\n=\\frac{pk ar}{15}(q^2-b^2)(qb-ar).}\n\\]","\\vfig[0.88]{coupled_07_reduction_geometry}{The two reduction surfaces are\n$q=b$ and $qb=ar$. Their union is exactly the zero set of $D$ in the\npositive contraction domain. Outside this union, the components cannot\nbe obtained from each other by amplitude and argument rescaling.}","There exist nonzero complex constants $A,B$ with $V(z)=A\\,U(Bz)$\nif and only if $D=0$. On either surface, put $s=\\sqrt{k/p}>0$.\nThe scalar family is defined by\n$T_{x,y}'(z)=1+T_{x,y}(xz)T_{x,y}(yz)$, $T_{x,y}(0)=0$.\nIt gives the explicit reduction\n\\[\nU(z)=\\sqrt{s}\\,T_{q,sr}(z/\\sqrt{s}),\\qquad\nV(z)=s^{-1/2}T_{q,sr}(\\sqrt{s}\\,z).\n\\]\nBoth effective scales lie in $(0,1)$. Equality $U=V$ is more restrictive:\nit occurs exactly when $\\{q,r\\}=\\{a,b\\}$.","stants $A,B$ with $V(z)=A\\,U(Bz)$\nif and only if $D=0$. On either surface, put $s=\\sqrt{k/p}>0$.\nThe scalar family is defined by\n$T_{x,y}'(z)=1+T_{x,y}(xz)T_{x,y}(yz)$, $T_{x,y}(0)=0$.\nIt gives the explicit reduction\n\\[\nU(z)=\\sqrt{s}\\,T_{q,sr}(z/\\sqrt{s}),\\qquad\nV(z)=s^{-1/2}T_{q,sr}(\\sqrt{s}\\,z).\n\\]\nBoth effective scales lie in $(0,1)$. Equality $U=V$ is more restrictive:\nit occurs exactly when $\\{q,r\\}=\\{a,b\\}$.","\\pageheading{FAMILY 5 / RECOVERY}{Degree seven identifies the four scales}\nAway from $D=0$, the six coefficients\n$(u_1,v_1,u_2,v_2,u_3,v_3)$ uniquely recover the ordered parameters\n$(q,r,a,b)$ in $(0,1)^4$. These are the coefficients at degrees three,\nfive and seven of the two labeled functions.","\\vfig[0.92]{coupled_08_inverse}{Cubic terms recover the two scale products.\nFifth-degree terms leave at most two candidates for each remaining scale.\nThe seventh-degree terms select the unique admissible candidates when\n$D\\ne0$. Repeated quadratic roots are already unique.}","First $p=3u_1$ and $k=3v_1$. The possible values $x=q^2$ and $y=a^2$\nsatisfy\n\\[\np x^2-\\frac{15u_2}{p}x+kp^2=0,\\qquad\np y^2-\\frac{15v_2}{k}y+k^3=0.\n\\]\nFor two distinct positive candidates $x,x'$, the predicted seventh\ncoefficients differ by\n\\[\nu_3(x)-u_3(x')=\\frac{pD}{7k^2}(x^2-x'^2)\\ne0.\n\\]\nThe second component supplies the analogous separation for $y$.\nFinally take the positive square roots and recover $r=p/q$, $b=k/a$.","=3u_1$ and $k=3v_1$. The possible values $x=q^2$ and $y=a^2$\nsatisfy\n\\[\np x^2-\\frac{15u_2}{p}x+kp^2=0,\\qquad\np y^2-\\frac{15v_2}{k}y+k^3=0.\n\\]\nFor two distinct positive candidates $x,x'$, the predicted seventh\ncoefficients differ by\n\\[\nu_3(x)-u_3(x')=\\frac{pD}{7k^2}(x^2-x'^2)\\ne0.\n\\]\nThe second component supplies the analogous separation for $y$.\nFinally take the positive square roots and recover $r=p/q$, $b=k/a$.","d $k=3v_1$. The possible values $x=q^2$ and $y=a^2$\nsatisfy\n\\[\np x^2-\\frac{15u_2}{p}x+kp^2=0,\\qquad\np y^2-\\frac{15v_2}{k}y+k^3=0.\n\\]\nFor two distinct positive candidates $x,x'$, the predicted seventh\ncoefficients differ by\n\\[\nu_3(x)-u_3(x')=\\frac{pD}{7k^2}(x^2-x'^2)\\ne0.\n\\]\nThe second component supplies the analogous separation for $y$.\nFinally take the positive square roots and recover $r=p/q$, $b=k/a$.","This is an exact-data inverse theorem. Near a reduction surface, small\n$D$ can make candidate predictions close, so a uniform stable inverse\ndoes not follow. The degree-seven recovery and the reduction classification\nare proved statements. Historical priority of these particular results\nis addressed separately by the bounded literature comparison.","\\pageheading{COUPLED FAMILIES}{Coupled growth and the cost of resolution}\nThe coupled family consists of two entire functions whose derivatives depend on\ncontracted values of both components. Its growth exponent is the unique\npositive number $\\rho$ for which\n\\[\n\\operatorname{spr}\\begin{pmatrix}q^\\rho&r^\\rho","\\pageheading{COUPLED FAMILIES}{Coupled growth and the cost of resolution}\nThe coupled family consists of two entire functions whose derivatives depend on\ncontracted values of both components. Its growth exponent is the unique\npositive number $\\rho$ for which\n\\[\n\\operatorname{spr}\\begin{pmatrix}q^\\rho&r^\\rho\\\\a^\\rho&b^\\rho\\end{pmatrix}=1,\n\\qquad 0<q,r,a,b<1.\n\\]\nBoth functions have positive lower type and finite upper type at this order.\nThe positive eigenvector fixes the proportions of both type vectors. The\nexistence of an ordinary type limit remains open.","Component reduction & Rescaling occurs exactly on $q=b$ or $qb=ar$","Generic exact inverse & Coefficients through degree seven identify all four ordered scales when $D\\ne0$","\\[\nF_\\varepsilon'(z)=1+(1-\\varepsilon)F_\\varepsilon(z/4)^2\n +\\varepsilon F_\\varepsilon(3z/4)^2,\n\\qquad F_\\varepsilon(0)=0,\n\\qquad 0\\le\\varepsilon\\le1.\n\\]","The base theorem gives a unique entire function with nonnegative odd\ncoefficients. At zero weight, the order is \\(1/2\\). At every positive\nweight it is","\\[\n\\rho=\\frac{\\log2}{\\log(4/3)}\\approx2.40942084.\n\\]","Write \\(f=f_{3/4}\\) for the normalized diagonal function, and\n\\(f(z)=\\sum_{n\\ge0}d_nz^{2n+1}\\). The inherited tap continuity theorem\ngives","\\[\n\\sup_{|z|\\le R}|F_\\varepsilon(z)-F_0(z)|\n\\le \\varepsilon L(R),\\qquad\nL(R)=\\sum_{n\\ge1}n d_nR^{2n+1}\n=\\frac{Rf'(R)-f(R)}2.\n\\tag{RA-1}\n\\]","The following elementary certificate makes the unit disk example fully\ncomputable. Positive coefficient domination by tangent, followed by\n\\(\\sin x\\le x\\) and \\(\\cos x\\ge1-x^2/2\\), gives","\\[\nf(4/3)\\le\\tan(4/3)\\le12.\n\\]","For \\(N\\ge1\\) and \\(0\\le R\\le1\\), put \\(t=3R/4\\). If\n\\((N+2)t^2\\le N+1\\), then \\(n t^{2n+1}\\) is nonincreasing for\n\\(n\\ge N+1\\), and hence","\\[\n0\\le L(R)-\\sum_{n=1}^{N}n d_nR^{2n+1}\n\\le12(N+1)t^{2N+3}.\n\\tag{RA-2}\n\\]","Indeed the left side is a sum of \\(d_n(4/3)^{2n+1}\\) multiplied by\n\\(n t^{2n+1}\\). Bound the multiplier by its first value, then use the\npositive total mass \\(f(4/3)\\le12\\). The coefficients \\(d_n\\) are\nrational. With \\(N=40\\), the exact rational upper bound is less than","\\[\nL(1)<0.241890267.\n\\]","\\[\n\\boxed{\\sup_{|z|\\le1}|F_{10^{-12}}(z)-F_0(z)|<2.41890267\\times10^{-13},}\n\\]","\\textbf{How to read the figure.} The left curves divide the difference\nby the tap weight, so the relevant small quantity is the plotted height\nmultiplied by \\(\\varepsilon\\). The right panel records the proved order,\nwhich is determined by arbitrarily large radii. The curve computation\nuses powers through \\(z^{201}\\), with a normalized analytic omitted-tail\nbound below \\(5.26\\times10^{-23}\\) on the unit interval. Ordinary\nfloating rounding is separate from that analytic tail bound.","The base lower-type theorem supplies, for every \\(0<\\varepsilon\\le1\\),","\\[\n\\log F_\\varepsilon(R)\\ge c_\\varepsilon(R+3/\\varepsilon)^\\rho\n\\ge c_\\varepsilon R^\\rho\\qquad(R\\ge2),\n\\] \\[\nc_\\varepsilon=\\frac{\\log2}{(4/\\varepsilon+8/3)^\\rho}.\n\\]","For the slow function, the base upper-type construction with\n\\(q=r=1/4\\), \\(N=2\\), yields","\\[\n\\log F_0(R)\\le C_0\\sqrt R\\qquad(R\\ge e),\n\\qquad C_0=e^{-1/2}(1+22\\log2)\\approx9.85566103.\n\\]","The value of \\(C_0\\) comes directly from the earlier explicit constants:\n\\(V_2=15\\log2\\), \\(\\Delta=2\\log2\\), and \\(B_+=5\\log2\\). Thus\n\\(V_2+1+\\Delta+B_+=1+22\\log2\\).","\\[\nR_2(\\varepsilon)=\\max\\left\\{\n e,\\left(\\frac{2C_0}{c_\\varepsilon}\\right)^{1/(\\rho-1/2)},\n \\left(\\frac{2\\log2}{c_\\varepsilon}\\right)^{1/\\rho}\\right\\}.\n\\tag{RA-3}\n\\]","Then, for every \\(R\\ge R_2(\\varepsilon)\\),","\\[\n\\boxed{F_\\varepsilon(R)\\ge2F_0(R).}\n\\]","To prove this, the last two constraints imply that\n\\(c_\\varepsilon R^\\rho\\) is at least both \\(2C_0\\sqrt R\\) and\n\\(2\\log2\\). It is therefore at least their half-sum\n\\(C_0\\sqrt R+\\log2\\). Exponentiating proves the claim.","\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}\n  >{\\raggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\nTap weight\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nDifference on the unit disk is at most\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nSufficient radius \\(R_2\\), rounded\n\\end{minipage}","\\midrule\\noalign{}\n\\endhead\n\\bottomrule\\noalign{}\n\\endlastfoot\n\\(10^{-3}\\) & \\(2.419\\times10^{-4}\\) & \\(2.029\\times10^5\\)","\\(10^{-6}\\) & \\(2.419\\times10^{-7}\\) & \\(1.237\\times10^9\\)","\\(10^{-9}\\) & \\(2.419\\times10^{-10}\\) & \\(7.549\\times10^{12}\\)","\\(10^{-12}\\) & \\(2.419\\times10^{-13}\\) & \\(4.608\\times10^{16}\\)","The radius column displays the analytic formula approximately. For a\ncertificate use the formula itself or a rigorously rounded upper value.\nThese large numbers reflect the conservatism of the global comparison.\nThey do not mean that observations at all smaller radii fail. The\nformula grows proportionally to \\(\\varepsilon^{-\\rho/(\\rho-1/2)}\\) as\nthe tap vanishes.","Fix \\(a=\\beta=2\\), and use the arithmetic depth family","\\[\nA_Q(s)=\\sum_{\\substack{n\\ge2\\\\\\mu(n)^2=1\\\\(n,Q)=1}}\n n^{-4}(1-n^{-2})^{2s}.\n\\]","For a prime \\(p\\), let \\(D_p=A_1-A_p\\). This positive difference\nconsists exactly of the squarefree atoms divisible by \\(p\\). Its largest\nfrequency is","\\[\n\\lambda_p=-2\\log(1-p^{-2}),\n\\]","and its total mass is \\(D(4)/(p^4+1)\\), where\n\\(D(4)=\\zeta(4)/\\zeta(8)=105/\\pi^4\\).","For every integer \\(k\\ge0\\) and \\(R\\ge0\\),","\\[\n\\sup_{|s|\\le R}|D_p^{(k)}(s)|\n\\le \\lambda_p^k e^{R\\lambda_p}\\frac{D(4)}{p^4+1}.\n\\tag{RA-4}\n\\]","This is the inherited arithmetic inverse theorem, now read as a\nprecision requirement. For fixed \\(R\\), the envelope is of order\n\\(p^{-4-2k}\\). Each additional derivative suppresses the large-prime\ndifference even further in absolute units. This does not rule out\nmethods that rescale derivatives and account for their corresponding\nnoise amplification.","At integer radius \\(R=5\\), an especially simple rational certificate\nfollows from \\(D(4)<\\zeta(4)<4/3\\):","\\[\n\\boxed{\\sup_{|s|\\le5}|D_p(s)|\n <\\frac{4}{3(p^4+1)}(1-p^{-2})^{-10}.}\n\\tag{RA-5}\n\\]","For derivatives, multiply this rational bound by \\([2/(p^2-1)]^k\\),\nusing \\(-\\log(1-u)\\le u/(1-u)\\). Thus this part of the certificate needs\nno transcendental evaluation at all.","\\begin{longtable}[]{@{}lr@{}}\n\\toprule\\noalign{}\nExcluded prime & Rational upper bound on \\(|s|\\le5\\), rounded upward","\\midrule\\noalign{}\n\\endhead\n\\bottomrule\\noalign{}\n\\endlastfoot\n101 & \\(1.283\\times10^{-8}\\)","1009 & \\(1.287\\times10^{-12}\\)","10007 & \\(1.330\\times10^{-16}\\)","\\[\n\\boxed{p^{-4}e^{\\lambda_pT}\\le D_p(-T)\n\\le\\frac{D(4)}{p^4+1}e^{\\lambda_pT}\\qquad(T\\ge0).}\n\\tag{RA-6}\n\\]","Choose an absolute tolerance \\(\\eta>D(4)/(p^4+1)\\). The strict\ninequality makes the initial difference smaller than the tolerance.\nDefine","\\[\nT_-=\\frac{\\log\\bigl(\\eta(p^4+1)/D(4)\\bigr)}{\\lambda_p},\n\\qquad T_+=\\frac{\\log(\\eta p^4)}{\\lambda_p}.\n\\tag{RA-7}\n\\]","Then \\(D_p(-T)\\le\\eta\\) for \\(0\\le T\\le T_-\\), while \\(D_p(-T)\\ge\\eta\\)\nfor \\(T\\ge T_+\\). The true crossing exists and is unique because\n\\(D_p(-T)\\) is continuous and strictly increasing. Therefore it belongs\nto \\([T_-,T_+]\\). If the tolerance hypothesis fails, the interval\nstarting at zero is not certified by this formula. The executable record\nrepresents an absent lower interval by \\texttt{null}.","\\[\nT_+-T_-=\\frac{\\log\\bigl(D(4)p^4/(p^4+1)\\bigr)}{\\lambda_p},\n\\qquad\nT_+=\\frac{p^2}{2}\\bigl(4\\log p+\\log\\eta\\bigr)(1+O(p^{-2})).\n\\]","For \\(\\eta=10^{-6}\\), rounded values are:","aggedright\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.2727}}\n  >{\\raggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}\n  >{\\raggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\nPrime\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nNo larger than tolerance through \\(T_-\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nAt least tolerance by \\(T_+\\)\n\\end{minipage}","arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}\n  >{\\raggedleft\\arraybackslash}P{(\\columnwidth - 4\\tabcolsep) * \\real{0.3636}}@{}}\n\\toprule\\noalign{}\n\\begin{minipage}[b]{\\linewidth}\\raggedright\nPrime\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nNo larger than tolerance through \\(T_-\\)\n\\end{minipage} & \\begin{minipage}[b]{\\linewidth}\\raggedleft\nAt least tolerance by \\(T_+\\)\n\\end{minipage}","The positive series for \\(D_p\\) can be evaluated directly if the\nexcluded prime is already specified. Recovering it by subtracting\nindependently evaluated full functions can be much harder. The atom\n\\(n=2\\) in \\(A_1\\) gives","\\[\n\\boxed{\\frac{A_1(-T)}{D_p(-T)}\n\\ge\\frac{p^4+1}{16D(4)}e^{(\\lambda_2-\\lambda_p)T}.}\n\\tag{RA-8}\n\\]","At the sufficient endpoint \\(T_+\\), the base-ten logarithm of this lower\nbound is about 5,924 for \\(p=101\\), 1,761,864 for \\(p=1009\\), and\n288,119,991 for \\(p=10007\\). These are conditioning scales for\nsubtracting full signals. They are not running times, required precision\nfor every possible algorithm, or claimed numerical evaluations of the\nfull functions at those enormous depths. The logarithmic bound is\ncomputed directly without forming those functions.","Consider the one-prime signed sieve at \\(p=2\\), with\n\\(\\lambda_2=-1-\\delta\\), \\(\\delta>0\\). Its nodes are\n\\(x_n=(1-n^{-2})^2\\), and its quadratic form is","\\[\nQ_\\delta(f)=\\sum_{\\substack{n\\ge1\\\\\\mu(n)^2=1\\\\n\\text{ odd}}}n^{-4}f(x_n)^2\n-\\delta\\sum_{\\substack{n\\ge1\\\\\\mu(n)^2=1\\\\n\\text{ even}}}n^{-4}f(x_n)^2.\n\\]","Every even squarefree atom has negative mass. We will certify that all nonzero real\npolynomials through degree four nevertheless have positive quadratic form for\n\\(\\delta=10^{-12}\\), and then construct a degree-seven polynomial of\nnegative quadratic form. This is an assertion about every polynomial in a\nfinite-dimensional space, not a plot of a few test vectors.","Choose \\(d+1\\) distinct odd squarefree labels \\(n_0,\\ldots,n_d\\),\nincluding 1, and set \\(y_i=x_{n_i}\\). Their Lagrange polynomials are","\\[\n\\ell_i(x)=\\prod_{j\\ne i}\\frac{x-y_j}{y_i-y_j},\n\\quad\nb_i=\\prod_{j\\ne i}\\frac{\\max(y_j,1-y_j)}{|y_i-y_j|},\n\\quad K_d=\\sum_{i=0}^{d}n_i^4b_i^2.\n\\]","For \\(x\\in[0,1]\\), we have \\(|\\ell_i(x)|\\le b_i\\). Interpolation and\nweighted Cauchy--Schwarz give, for every real polynomial of degree at\nmost \\(d\\),","\\[\n|f(x)|^2\\le\n\\left(\\sum_{i=0}^{d}n_i^{-4}f(y_i)^2\\right)\n\\left(\\sum_{i=0}^{d}n_i^4\\ell_i(x)^2\\right)\n\\le K_d\\sum_{i=0}^{d}n_i^{-4}f(y_i)^2.\n\\]","The total even mass satisfies\n\\(\\sum_{n\\text{ even}}\\mu(n)^2n^{-4}<\\frac1{16}(1+\\frac13)=1/12\\).\nKeeping the selected nodes in the positive part yields the explicit\nbound","\\[\n\\boxed{Q_\\delta(f)\\ge\n\\left(1-\\frac{\\delta K_d}{12}\\right)\n\\sum_{i=0}^{d}n_i^{-4}f(y_i)^2.}\n\\tag{RA-9}\n\\]","A nonzero polynomial of degree at most \\(d\\) cannot vanish at all\n\\(d+1\\) nodes. Thus \\(\\delta<12/K_d\\) proves strict positive\ndefiniteness of the Hankel matrix of order \\(d+1\\).","For \\(d=4\\), use labels \\(1,3,5,7,11\\). Exact rational arithmetic gives","\\[\nK_4\\approx1.50393441996\\times10^{11},\\qquad\n12/K_4\\approx7.9790713217\\times10^{-11}>10^{-12}.\n\\]","Let \\(S\\) be exactly the set of all odd squarefree labels \\(n\\le K\\),\nwhere \\(K\\) is an odd squarefree integer, and let \\(n_+\\) be the next\nsuch label. Assume \\(n_+>2\\), and define","\\[\nP_{S,m}(x)=\\prod_{n\\in S}\\frac{x-x_n}{x_2-x_n}\n\\left(\\frac{1-x}{1-x_2}\\right)^m,\n\\] \\[\nC_S=\\prod_{n\\in S}\\frac{1-x_n}{|x_2-x_n|},\n\\qquad r_S=\\frac{1-x_{n_+}}{1-x_2}<1.\n\\]","This polynomial has degree \\(|S|+m\\), equals one at \\(x_2\\), and kills\nevery positive atom through the cutoff. At each remaining positive node,\nthe product of the absolute root factors is at most \\(C_S\\), and the\nlast factor is at most \\(r_S^m\\). Dropping all other negative atoms and\nbounding the positive tail by the integral\n\\(\\sum_{n>K}n^{-4}\\le1/(3K^3)\\) gives","\\[\n\\boxed{Q_\\delta(P_{S,m})\\le-\\frac\\delta{16}\n +\\frac{C_S^2}{3K^3}r_S^{2m}.}\n\\tag{RA-10}\n\\]","For any chosen \\(S\\), a computable sufficient exponent is the first\nnonnegative integer \\(m\\) for which","\\[\n r_S^{2m}<\\frac{3K^3\\delta}{16C_S^2}.\n\\]","It can be found by multiplying rational numbers, without logarithms or\neigenvalues. Searching finitely many choices of \\(S\\) improves the\nsufficient degree but does not prove its optimality.","For the worked defect \\(\\delta=10^{-12}\\), take","\\[\nS=\\{1,3,5,7\\},\\quad K=7,\\quad n_+=11,\\quad m=3.\n\\]","\\[\nC_S=\\frac{756482048}{20777538825},\\qquad\nr_S=\\frac{3856}{102487},\n\\]","and \\(P_{S,3}\\) has degree seven. The exact rational right side of\n(RA-10) is strictly negative and is approximately","\\[\n\\boxed{Q_{10^{-12}}(P_{S,3})\\le-5.8845708899\\times10^{-14}.}\n\\]","The negative atom contributes \\(-6.25\\times10^{-14}\\). The full positive\ntail contributes at most \\(3.6542911008\\times10^{-15}\\), a bound that\ncovers every omitted atom. The exact certificate is stored as numerator\nand denominator. The original generic isolating detector needed degree\n21 for this defect. Using the known sign pattern and killing later\npositive shelves reduces the exhibited degree to seven.","One preliminary presentation issue was corrected during development: a\nnegative formula for \\(T_-\\) was initially clamped to zero. That would\nmisleadingly label an already visible difference as invisible at zero.\nThe finished formula requires its stated tolerance hypothesis, and the\nexecutable record marks an absent interval by \\texttt{null}. The exact\nassertion run passed. This correction did not affect any of the three\nlisted prime examples.","\\pageheading{RESOLUTION ATLAS\\; /\\; WEAK TAPS}{A small change can alter the global order}\nA feedback channel may be almost invisible on a fixed disk and still determine\nthe eventual growth of the function. Consider the normalized entire solution\n\\[\nF_\\varepsilon'(z)=1+(1-\\varepsilon)F_\\varepsilon(z/4)^2\n +\\varepsilon F_\\varepsilon(3z/4)^2,\n\\qquad F_\\varepsilon(0)=0.\n\\]\nHere $0\\le\\varepsilon\\le1$. At $\\varepsilon=0$, its entire order is $1/2$.\nAt every positive weight, the order becomes\n$\\rho=\\log2/\\log(4/3)\\approx2.40942084$.","eheading{RESOLUTION ATLAS\\; /\\; WEAK TAPS}{A small change can alter the global order}\nA feedback channel may be almost invisible on a fixed disk and still determine\nthe eventual growth of the function. Consider the normalized entire solution\n\\[\nF_\\varepsilon'(z)=1+(1-\\varepsilon)F_\\varepsilon(z/4)^2\n +\\varepsilon F_\\varepsilon(3z/4)^2,\n\\qquad F_\\varepsilon(0)=0.\n\\]\nHere $0\\le\\varepsilon\\le1$. At $\\varepsilon=0$, its entire order is $1/2$.\nAt every positive weight, the order becomes\n$\\rho=\\log2/\\log(4/3)\\approx2.40942084$.","TAPS}{A small change can alter the global order}\nA feedback channel may be almost invisible on a fixed disk and still determine\nthe eventual growth of the function. Consider the normalized entire solution\n\\[\nF_\\varepsilon'(z)=1+(1-\\varepsilon)F_\\varepsilon(z/4)^2\n +\\varepsilon F_\\varepsilon(3z/4)^2,\n\\qquad F_\\varepsilon(0)=0.\n\\]\nHere $0\\le\\varepsilon\\le1$. At $\\varepsilon=0$, its entire order is $1/2$.\nAt every positive weight, the order becomes\n$\\rho=\\log2/\\log(4/3)\\approx2.40942084$.","feedback channel may be almost invisible on a fixed disk and still determine\nthe eventual growth of the function. Consider the normalized entire solution\n\\[\nF_\\varepsilon'(z)=1+(1-\\varepsilon)F_\\varepsilon(z/4)^2\n +\\varepsilon F_\\varepsilon(3z/4)^2,\n\\qquad F_\\varepsilon(0)=0.\n\\]\nHere $0\\le\\varepsilon\\le1$. At $\\varepsilon=0$, its entire order is $1/2$.\nAt every positive weight, the order becomes\n$\\rho=\\log2/\\log(4/3)\\approx2.40942084$.","\\vfig{resolution_01_weak_tap}{Left: divide the compact difference by the tap\nweight to compare its shape. Right: the proved order jumps at zero. The dashed\ncurve is an analytic envelope. The function curves use 100-digit arithmetic\nand have a normalized omitted-series bound below $5.26\\times10^{-23}$ on\n$0\\le R\\le1$. Floating rounding is separate from that tail bound.}","\\claim{The unit disk comes with an exact error certificate}\nLet $f_{3/4}=F_1$, with $f_{3/4}(z)=\\sum_{n\\ge0}d_nz^{2n+1}$.\nPositive coefficients give\n\\[\n\\sup_{|z|\\le R}|F_\\varepsilon(z)-F_0(z)|\n\\le\\varepsilon\\sum_{n\\ge1}nd_nR^{2n+1}.\n\\]\nA rational coefficient sum and a positive-tail estimate prove that, on the\nunit disk, the multiplier is less than $0.241890267$.\nFor $\\varepsilon=10^{-12}$, the entire disk therefore changes by less than\n$2.41890267\\times10^{-13}$.","error certificate}\nLet $f_{3/4}=F_1$, with $f_{3/4}(z)=\\sum_{n\\ge0}d_nz^{2n+1}$.\nPositive coefficients give\n\\[\n\\sup_{|z|\\le R}|F_\\varepsilon(z)-F_0(z)|\n\\le\\varepsilon\\sum_{n\\ge1}nd_nR^{2n+1}.\n\\]\nA rational coefficient sum and a positive-tail estimate prove that, on the\nunit disk, the multiplier is less than $0.241890267$.\nFor $\\varepsilon=10^{-12}$, the entire disk therefore changes by less than\n$2.41890267\\times10^{-13}$.","\\pageheading{RESOLUTION ATLAS\\; /\\; OBSERVATION RADIUS}{How far is far enough to guarantee separation?}\nThe growth theorem supplies positive lower type for the weak-tap function\nand finite upper type for the slow function. For $0<\\varepsilon\\le1$ and $R\\ge e$,\n\\[\n\\log F_\\varepsilon(R)\\ge c_\\varepsilon R^\\rho,\n\\qquad \\log F_0(R)\\le C_0\\sqrt R,\n\\]\n\\[\nc_\\varepsilon=\\frac{\\log2}{(4/\\varepsilon+8/3)^\\rho},\n\\qquad C_0=e^{-1/2}(1+22\\log2).\n\\]","\\claim{An explicit sufficient radius}\nPut\n\\[\nR_2(\\varepsilon)=\\max\\!\\left\\{\n e,\\left(\\frac{2C_0}{c_\\varepsilon}\\right)^{1/(\\rho-1/2)},\n \\left(\\frac{2\\log2}{c_\\varepsilon}\\right)^{1/\\rho}\\right\\}.\n\\]\nThen $F_\\varepsilon(R)\\ge2F_0(R)$ for every $R\\ge R_2(\\varepsilon)$.\nThe formula makes $c_\\varepsilon R^\\rho$ at least both\n$2C_0\\sqrt R$ and $2\\log2$, and therefore at least their half-sum.\nExponentiating gives the factor of two.","For $\\varepsilon=10^{-6}$, the formula is approximately\n$1.237\\times10^9$. This large radius is a guaranteed sufficient choice.\nIt is not the first radius at which a useful difference appears, and it does\nnot establish failure of every observation at smaller radii.\nThe exact formula carries the guarantee. Rounded figures illustrate its scale.","\\pageheading{RESOLUTION ATLAS\\; /\\; ARITHMETIC PRECISION}{A large excluded prime hides in a fixed window}\nUse the squarefree arithmetic family\n\\[\nA_Q(s)=\\sum_{\\substack{n\\ge2\\\\\\mu(n)^2=1\\\\(n,Q)=1}}\n n^{-4}(1-n^{-2})^{2s},\n\\qquad D_p(s)=A_1(s)-A_p(s),\n\\]\nwhere $p$ is prime. The difference contains exactly the squarefree atoms\ndivisible by $p$. Its largest frequency is\n$\\lambda_p=-2\\log(1-p^{-2})$, and its total mass is\n$D(4)/(p^4+1)$, with $D(4)=105/\\pi^4$.","(n,Q)=1}}\n n^{-4}(1-n^{-2})^{2s},\n\\qquad D_p(s)=A_1(s)-A_p(s),\n\\]\nwhere $p$ is prime. The difference contains exactly the squarefree atoms\ndivisible by $p$. Its largest frequency is\n$\\lambda_p=-2\\log(1-p^{-2})$, and its total mass is\n$D(4)/(p^4+1)$, with $D(4)=105/\\pi^4$.","\\vfig{resolution_03_prime_precision}{Left: analytic bounds for the\nfunction and three derivatives on $|s|\\le5$. The connecting curves show scale,\nwhile the arithmetic statement applies at primes. Right: rational bounds\nfor three particular excluded primes. The bar height is the negative\nbase-ten logarithm of the bound, not a count of recovered digits.}","\\claim{The full complex disk has a rational certificate}\nFor integer $k\\ge0$,\n\\[\n\\sup_{|s|\\le5}|D_p^{(k)}(s)|\n<\\frac{4(1-p^{-2})^{-10}}{3(p^4+1)}\n \\left(\\frac{2}{p^2-1}\\right)^k.\n\\]\nThe inequality uses $D(4)<4/3$ and\n$\\lambda_p\\le2/(p^2-1)$. Every quantity on the right is rational.\nFor $k=0$, upward decimal bounds are $1.283\\times10^{-8}$ at $p=101$,\n$1.287\\times10^{-12}$ at $p=1009$, and $1.330\\times10^{-16}$ at $p=10007$.","mplex disk has a rational certificate}\nFor integer $k\\ge0$,\n\\[\n\\sup_{|s|\\le5}|D_p^{(k)}(s)|\n<\\frac{4(1-p^{-2})^{-10}}{3(p^4+1)}\n \\left(\\frac{2}{p^2-1}\\right)^k.\n\\]\nThe inequality uses $D(4)<4/3$ and\n$\\lambda_p\\le2/(p^2-1)$. Every quantity on the right is rational.\nFor $k=0$, upward decimal bounds are $1.283\\times10^{-8}$ at $p=101$,\n$1.287\\times10^{-12}$ at $p=1009$, and $1.330\\times10^{-16}$ at $p=10007$.","\\pageheading{RESOLUTION ATLAS\\; /\\; NEGATIVE DEPTH}{Seeing the prime can make subtraction harder}\nAt negative real depth, the difference grows. The prime atom alone supplies\na lower bound, while the entire removed shelf supplies an upper bound:\n\\[\np^{-4}e^{\\lambda_pT}\\le D_p(-T)\n\\le\\frac{D(4)}{p^4+1}e^{\\lambda_pT},\\qquad T\\ge0.\n\\]\nChoose a tolerance $\\eta>D(4)/(p^4+1)$. Its unique crossing occurs between\n\\[\nT_- =\\frac{\\log\\!\\bigl(\\eta(p^4+1)/D(4)\\bigr)}{\\lambda_p},\n\\qquad\nT_+ =\\frac{\\log(\\eta p^4)}{\\lambda_p}.\n\\]","ng{RESOLUTION ATLAS\\; /\\; NEGATIVE DEPTH}{Seeing the prime can make subtraction harder}\nAt negative real depth, the difference grows. The prime atom alone supplies\na lower bound, while the entire removed shelf supplies an upper bound:\n\\[\np^{-4}e^{\\lambda_pT}\\le D_p(-T)\n\\le\\frac{D(4)}{p^4+1}e^{\\lambda_pT},\\qquad T\\ge0.\n\\]\nChoose a tolerance $\\eta>D(4)/(p^4+1)$. Its unique crossing occurs between\n\\[\nT_- =\\frac{\\log\\!\\bigl(\\eta(p^4+1)/D(4)\\bigr)}{\\lambda_p},\n\\qquad\nT_+ =\\frac{\\log(\\eta p^4)}{\\lambda_p}.\n\\]","\\vfig{resolution_04_prime_depth}{At tolerance $10^{-6}$, the left panel\nshows the analytic crossing brackets after rescaling each row by $T_+$.\nThe right panel shows a lower bound on the decimal cancellation scale when\nsubtracting full functions. Displayed endpoints and labels are ordinary\nfloating evaluations of exact formulas.}","For $p=101$, the difference is no larger than $10^{-6}$ through\n$T_-=23307.789\\ldots$, and at least that large by\n$T_+=23690.516\\ldots$. The decimals illustrate the formulas and are not\nrounded interval certificates.","\\claim{The direct positive difference and subtraction have different costs}\nThe $n=2$ atom in the full function proves\n\\[\n\\frac{A_1(-T)}{D_p(-T)}\n\\ge\\frac{p^4+1}{16D(4)}e^{(\\lambda_2-\\lambda_p)T}.\n\\]\nAt $T_+$ for $p=101$, the base-ten logarithm of this lower bound is about\n5924. Subtracting independently approximated full functions can therefore\namplify relative errors enormously. Evaluating the known positive difference\nseries is a different procedure. These conditioning figures impose no\nuniversal precision requirement on every possible recovery algorithm.","\\pageheading{RESOLUTION ATLAS\\; /\\; NEGATIVE ATOMS}{A degree-seven witness exposes a signed defect}\nLet $x_n=(1-n^{-2})^2$, and change the prime-two sieve strength to\n$\\lambda_2=-1-\\delta$. For a real polynomial $f$, the associated quadratic form is\n\\[\nQ_\\delta(f)=\\sum_{\\substack{n\\ge1","\\pageheading{RESOLUTION ATLAS\\; /\\; NEGATIVE ATOMS}{A degree-seven witness exposes a signed defect}\nLet $x_n=(1-n^{-2})^2$, and change the prime-two sieve strength to\n$\\lambda_2=-1-\\delta$. For a real polynomial $f$, the associated quadratic form is\n\\[\nQ_\\delta(f)=\\sum_{\\substack{n\\ge1\\\\\\mu(n)^2=1\\\\n\\text{ odd}}}\n \\frac{f(x_n)^2}{n^4}\n-\\delta\\sum_{\\substack{n\\ge1\\\\\\mu(n)^2=1\\\\n\\text{ even}}}\n \\frac{f(x_n)^2}{n^4}.\n\\]\nEvery even squarefree atom has negative mass. At $\\delta=10^{-12}$,\nthe polynomial below certifies that this is detectable at finite degree.","n\\text{ even}}}\n \\frac{f(x_n)^2}{n^4}.\n\\]\nEvery even squarefree atom has negative mass. At $\\delta=10^{-12}$,\nthe polynomial below certifies that this is detectable at finite degree.","\\vfig{resolution_05_signed_detector}{The detector vanishes on the first\nfour positive shelves and equals one at the negative atom $n=2$.\nThe positive bar is an upper bound for the entire remaining positive tail.\nThe total upper bound is negative, so no omitted atom can reverse the\nconclusion.}","\\claim{The certificate uses rational arithmetic throughout}\nTake $S=\\{1,3,5,7\\}$ and define\n\\[\nP(x)=\\prod_{n\\in S}\\frac{x-x_n}{x_2-x_n}\n \\left(\\frac{1-x}{1-x_2}\\right)^3.\n\\]\nIts degree is seven. All remaining positive nodes begin at $n=11$.\nWith\n$C=756482048/20777538825$ and $r=3856/102487$, a complete tail bound gives\n\\[\nQ_{10^{-12}}(P)\\le-\\frac{10^{-12}}{16}\n +\\frac{C^2r^6}{3\\cdot7^3}\n <-5.8845708899\\times10^{-14}.\n\\]\nThe inequality is certified as a comparison of exact rational numbers.\nThe first four zeros remove the dominant positive contributions.\nThe factor $(1-x)^3$ suppresses the remaining cluster near one.\nThe construction exhibits a detecting degree, without asserting that degree\nseven is the smallest possible choice.","]\nIts degree is seven. All remaining positive nodes begin at $n=11$.\nWith\n$C=756482048/20777538825$ and $r=3856/102487$, a complete tail bound gives\n\\[\nQ_{10^{-12}}(P)\\le-\\frac{10^{-12}}{16}\n +\\frac{C^2r^6}{3\\cdot7^3}\n <-5.8845708899\\times10^{-14}.\n\\]\nThe inequality is certified as a comparison of exact rational numbers.\nThe first four zeros remove the dominant positive contributions.\nThe factor $(1-x)^3$ suppresses the remaining cluster near one.\nThe construction exhibits a detecting degree, without asserting that degree\nseven is the smallest possible choice.","\\pageheading{RESOLUTION ATLAS\\; /\\; ALL LOW DEGREES}{Passing every coarse test can still miss the defect}\nThe previous witness would be less informative without a converse guarantee\nat lower resolution. For the same defect $\\delta=10^{-12}$, every nonzero\npolynomial of degree at most four has strictly positive quadratic form. This quantifies\nthe information that the coarse test actually misses.","\\vfig{resolution_06_degree_gap}{The blue boundary certifies positivity\nfor every polynomial through that degree. The orange boundary gives an\nexplicit detecting polynomial. The shaded gap remains unresolved by these\ncertificates. At $\\delta=10^{-12}$, degrees through four pass, degree seven\nfails, and degrees five and six are undecided here.}","\\claim{Interpolation controls the whole polynomial space}\nChoose the odd squarefree labels $n_i=1,3,5,7,11$, write $y_i=x_{n_i}$,\nand put\n\\[\nb_i=\\prod_{j\\ne i}\\frac{\\max(y_j,1-y_j)}{|y_i-y_j|},\n\\qquad K_4=\\sum_{i=0}^{4}n_i^4b_i^2.\n\\]\nLagrange interpolation and weighted Cauchy--Schwarz show that, for every\npolynomial $f$ of degree at most four,\n\\[\nQ_\\delta(f)\\ge\n\\left(1-\\frac{\\delta K_4}{12}\\right)\n\\sum_{i=0}^{4}n_i^{-4}f(y_i)^2.\n\\]\nExact rational comparison gives $12/K_4>7.9790\\times10^{-11}>10^{-12}$.\nA nonzero polynomial cannot vanish at all five selected nodes, so the right\nside is strictly positive. Thus the order-five Hankel matrix is positive\ndefinite, while the degree-seven witness makes the order-eight matrix fail\npositive semidefiniteness.","odd squarefree labels $n_i=1,3,5,7,11$, write $y_i=x_{n_i}$,\nand put\n\\[\nb_i=\\prod_{j\\ne i}\\frac{\\max(y_j,1-y_j)}{|y_i-y_j|},\n\\qquad K_4=\\sum_{i=0}^{4}n_i^4b_i^2.\n\\]\nLagrange interpolation and weighted Cauchy--Schwarz show that, for every\npolynomial $f$ of degree at most four,\n\\[\nQ_\\delta(f)\\ge\n\\left(1-\\frac{\\delta K_4}{12}\\right)\n\\sum_{i=0}^{4}n_i^{-4}f(y_i)^2.\n\\]\nExact rational comparison gives $12/K_4>7.9790\\times10^{-11}>10^{-12}$.\nA nonzero polynomial cannot vanish at all five selected nodes, so the right\nside is strictly positive. Thus the order-five Hankel matrix is positive\ndefinite, while the degree-seven witness makes the order-eight matrix fail\npositive semidefiniteness."]}