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rangle","display":"\\rangle","kind":"use","occurrences":[760]},{"id":"symbol-22449108dc31ebd8","tex":"\\longmapsto","aliases":"⟼ longmapsto","display":"\\longmapsto","kind":"use","occurrences":[94,104,112,162,300,313]}],"contexts":["\\begin{definition}[Classical, shortcut, and odd maps]\\label{def:maps}\nFor a positive integer $n$, define the classical map and its shortcut by\n\\begin{equation}\\label{eq:classical}\n\\operatorname{Col}(n)=\n\\begin{cases}3n+1,&n\\text{ odd},","\\begin{definition}[Classical, shortcut, and odd maps]\\label{def:maps}\nFor a positive integer $n$, define the classical map and its shortcut by\n\\begin{equation}\\label{eq:classical}\n\\operatorname{Col}(n)=\n\\begin{cases}3n+1,&n\\text{ odd},\\\\n/2,&n\\text{ even},\\end{cases}\n\\qquad\nT(n)=\n\\begin{cases}(3n+1)/2,&n\\text{ odd},\\\\n/2,&n\\text{ even}.\\end{cases}\n\\end{equation}\nFor a positive odd integer $n$, let\n\\begin{equation}\\label{eq:oddmap}\na(n)=\\nu_2(3n+1),\\qquad A(n)=\\frac{3n+1}{2^{a(n)}}.\n\\end{equation}\nHere $\\nu_2(m)$ is the largest exponent $a$ for which $2^a$ divides the positive integer $m$. Thus $a(n)\\ge1$, and $A(n)$ is positive and odd.\n\\end{definition}","n/2,&n\\text{ even}.\\end{cases}\n\\end{equation}\nFor a positive odd integer $n$, let\n\\begin{equation}\\label{eq:oddmap}\na(n)=\\nu_2(3n+1),\\qquad A(n)=\\frac{3n+1}{2^{a(n)}}.\n\\end{equation}\nHere $\\nu_2(m)$ is the largest exponent $a$ for which $2^a$ divides the positive integer $m$. Thus $a(n)\\ge1$, and $A(n)$ is positive and odd.\n\\end{definition}","The map $A$ removes every factor of two after an odd multiplication. It is the odd acceleration, also called the Syracuse map. One application of $A$ corresponds to one odd shortcut step followed by the remaining even shortcut steps. These maps have different clocks. In particular, a statement about 37 applications of $A$ need not concern 37 applications of $T$.","More precisely, if an accelerated prefix has $r$ odd steps and cumulative valuation $S_r$, it uses $S_r$ shortcut steps and $S_r+r$ classical steps. In the shortcut map the division immediately after an odd multiplication is included in that odd step. In the classical map it is a separate step, as are all the other divisions. For example, $A(27)=41$ is realized classically by $27,82,41$. Since its valuation is 1, this has one shortcut step and two classical steps. The next acceleration $41\\mapsto31$ has valuation 2 and uses two shortcut steps or three classical steps. The combined prefix therefore has accelerated, shortcut, and classical clocks $2,3,5$.","t uses $S_r$ shortcut steps and $S_r+r$ classical steps. In the shortcut map the division immediately after an odd multiplication is included in that odd step. In the classical map it is a separate step, as are all the other divisions. For example, $A(27)=41$ is realized classically by $27,82,41$. Since its valuation is 1, this has one shortcut step and two classical steps. The next acceleration $41\\mapsto31$ has valuation 2 and uses two shortcut steps or three classical steps. The combined prefix therefore has accelerated, shortcut, and classical clocks $2,3,5$.","the classical map it is a separate step, as are all the other divisions. For example, $A(27)=41$ is realized classically by $27,82,41$. Since its valuation is 1, this has one shortcut step and two classical steps. The next acceleration $41\\mapsto31$ has valuation 2 and uses two shortcut steps or three classical steps. The combined prefix therefore has accelerated, shortcut, and classical clocks $2,3,5$.","\\begin{conjecture}[Collatz]\\label{conj:collatz}\nFor every positive odd integer $n$, there exists $r\\ge0$ such that $A^r(n)=1$.\n\\end{conjecture}","This version is equivalent to the classical assertion. Every positive integer first reaches an odd integer by division by two. Conversely, each application of $A$ is a finite portion of the classical orbit. The fixed point $A(1)=1$ represents the classical cycle $1,4,2$.","\\section{Exact valuation packets}\nFix a positive odd seed $n$, and write $x_r=A^r(n)$. Each step has the exact equation\n\\begin{equation}\\label{eq:onestep}\n2^{a_r}x_{r+1}=3x_r+1,\\qquad a_r=a(x_r).\n\\end{equation}\nThe exponent records how much dyadic information is removed at that step. It is determined by the actual state, rather than chosen independently.","\\begin{definition}[Valuation word and offset]\\label{def:word}\nA valuation word is a finite sequence $w=(a_0,\\ldots,a_{k-1})$ of positive integers. Define its partial depths and offsets by\n\\begin{equation}\\label{eq:offsetrec}\nS_0=0,\\quad S_j=\\sum_{i<j}a_i,\\qquad\nC_0=0,\\quad C_{j+1}=3C_j+2^{S_j}.\n\\end{equation}\nIts total depth is $S=S_k$ and its terminal offset is $C=C_k$. The word is realized by $n$ if these are the first $k$ valuations of the actual odd orbit from $n$.\n\\end{definition}","on word and offset]\\label{def:word}\nA valuation word is a finite sequence $w=(a_0,\\ldots,a_{k-1})$ of positive integers. Define its partial depths and offsets by\n\\begin{equation}\\label{eq:offsetrec}\nS_0=0,\\quad S_j=\\sum_{i<j}a_i,\\qquad\nC_0=0,\\quad C_{j+1}=3C_j+2^{S_j}.\n\\end{equation}\nIts total depth is $S=S_k$ and its terminal offset is $C=C_k$. The word is realized by $n$ if these are the first $k$ valuations of the actual odd orbit from $n$.\n\\end{definition}","ion word is a finite sequence $w=(a_0,\\ldots,a_{k-1})$ of positive integers. Define its partial depths and offsets by\n\\begin{equation}\\label{eq:offsetrec}\nS_0=0,\\quad S_j=\\sum_{i<j}a_i,\\qquad\nC_0=0,\\quad C_{j+1}=3C_j+2^{S_j}.\n\\end{equation}\nIts total depth is $S=S_k$ and its terminal offset is $C=C_k$. The word is realized by $n$ if these are the first $k$ valuations of the actual odd orbit from $n$.\n\\end{definition}","\\begin{theorem}[Affine packet law]\\label{thm:affine}\nIf $n$ realizes $w$, then every prefix satisfies\n\\begin{equation}\\label{eq:affine}\n2^{S_j}x_j=3^j n+C_j.\n\\end{equation}\nThe terminal offset also has the explicit expression\n\\begin{equation}\\label{eq:offsetsum}\nC_k=\\sum_{i=0}^{k-1}3^{k-1-i}2^{S_i}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAt $j=0$ the identity is $x_0=n$. If it holds at $j$, multiply \\eqref{eq:onestep} by $2^{S_j}$ to obtain\n\\[\n2^{S_{j+1}}x_{j+1}=3^{j+1}n+3C_j+2^{S_j}.\n\\]\nThis is the next identity by \\eqref{eq:offsetrec}. Iterating the same offset recurrence gives \\eqref{eq:offsetsum}.\n\\end{proof}","law]\\label{thm:affine}\nIf $n$ realizes $w$, then every prefix satisfies\n\\begin{equation}\\label{eq:affine}\n2^{S_j}x_j=3^j n+C_j.\n\\end{equation}\nThe terminal offset also has the explicit expression\n\\begin{equation}\\label{eq:offsetsum}\nC_k=\\sum_{i=0}^{k-1}3^{k-1-i}2^{S_i}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAt $j=0$ the identity is $x_0=n$. If it holds at $j$, multiply \\eqref{eq:onestep} by $2^{S_j}$ to obtain\n\\[\n2^{S_{j+1}}x_{j+1}=3^{j+1}n+3C_j+2^{S_j}.\n\\]\nThis is the next identity by \\eqref{eq:offsetrec}. Iterating the same offset recurrence gives \\eqref{eq:offsetsum}.\n\\end{proof}","}\nIf $n$ realizes $w$, then every prefix satisfies\n\\begin{equation}\\label{eq:affine}\n2^{S_j}x_j=3^j n+C_j.\n\\end{equation}\nThe terminal offset also has the explicit expression\n\\begin{equation}\\label{eq:offsetsum}\nC_k=\\sum_{i=0}^{k-1}3^{k-1-i}2^{S_i}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAt $j=0$ the identity is $x_0=n$. If it holds at $j$, multiply \\eqref{eq:onestep} by $2^{S_j}$ to obtain\n\\[\n2^{S_{j+1}}x_{j+1}=3^{j+1}n+3C_j+2^{S_j}.\n\\]\nThis is the next identity by \\eqref{eq:offsetrec}. Iterating the same offset recurrence gives \\eqref{eq:offsetsum}.\n\\end{proof}","One can see the offset being created before writing the general formula. The first step is $2^{a_0}x_1=3n+1$. Multiplying the second step by $2^{a_0}$ gives\n\\[\n2^{a_0+a_1}x_2=3(3n+1)+2^{a_0}\n=9n+3+2^{a_0}.\n\\]\nThe first addition has been multiplied by 3. The second addition has been multiplied by the dyadic depth already removed. After a third step the offset is $9+3\\cdot2^{a_0}+2^{a_0+a_1}$. Every later addition enters at its own prefix depth and receives one factor of 3 for each later odd multiplication. This explains each summand in \\eqref{eq:offsetsum}.","The coefficient $3^k/2^S$ consequently describes the transport of the original source. The offset describes the transport of the repeated additions. Neither term is a statistical correction. They are exact rational numbers, and their sum is an integer on the appropriate cylinder.","\\begin{theorem}[Exact integer cylinder]\\label{thm:cylinder}\nLet $w$ be a nonempty valuation word of length $k$ and total depth $S$. A positive odd integer $n$ realizes $w$ if and only if\n\\begin{equation}\\label{eq:cylinder}\n3^k n+C_k\\equiv 2^S\\pmod{2^{S+1}}.\n\\end{equation}\nConsequently the realizing integers form one odd residue class modulo $2^{S+1}$, whose residue is\n\\begin{equation}\\label{eq:address}\nu_w\\equiv(2^S-C_k)(3^k)^{-1}\\pmod{2^{S+1}}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nIf the word is realized, \\eqref{eq:affine} has an odd terminal quotient $x_k$. Multiplying an odd integer by $2^S$ gives exactly the congruence \\eqref{eq:cylinder}.","s $w$ if and only if\n\\begin{equation}\\label{eq:cylinder}\n3^k n+C_k\\equiv 2^S\\pmod{2^{S+1}}.\n\\end{equation}\nConsequently the realizing integers form one odd residue class modulo $2^{S+1}$, whose residue is\n\\begin{equation}\\label{eq:address}\nu_w\\equiv(2^S-C_k)(3^k)^{-1}\\pmod{2^{S+1}}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nIf the word is realized, \\eqref{eq:affine} has an odd terminal quotient $x_k$. Multiplying an odd integer by $2^S$ gives exactly the congruence \\eqref{eq:cylinder}.","\\label{eq:cylinder}\n3^k n+C_k\\equiv 2^S\\pmod{2^{S+1}}.\n\\end{equation}\nConsequently the realizing integers form one odd residue class modulo $2^{S+1}$, whose residue is\n\\begin{equation}\\label{eq:address}\nu_w\\equiv(2^S-C_k)(3^k)^{-1}\\pmod{2^{S+1}}.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nIf the word is realized, \\eqref{eq:affine} has an odd terminal quotient $x_k$. Multiplying an odd integer by $2^S$ gives exactly the congruence \\eqref{eq:cylinder}.","Conversely, define the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence give","Conversely, define the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are the","Conversely, define the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actua","Conversely, define the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^","Conversely, define the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its","Conversely, define the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because","Conversely, define the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","e the rational candidates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","idates $x_j=(3^j n+C_j)/2^{S_j}$. We show that every candidate is a positive odd integer. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","ger. The terminal congruence supplies this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","this for $j=k$. For $0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","$0\\le j<k$, split the offset sum at $j$:\n\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","\\[\nC_k=3^{k-j}C_j+2^{S_j}D_{j,k}.\n\\]\nThe integer $D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","D_{j,k}$ is odd. Its first summand is $3^{k-1-j}$, while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","while every later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","ry later summand has an additional factor of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","r of two. Reducing \\eqref{eq:cylinder} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","r} modulo $2^{S_j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","j+1}$ gives\n\\[\n3^{k-j}(3^j n+C_j)+2^{S_j}D_{j,k}\\equiv0\\pmod{2^{S_j+1}}.\n\\]\nEvery odd multiple of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","e of $2^{S_j}$ has the same residue modulo $2^{S_j+1}$. Since $3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","3^{k-j}$ is invertible, it follows that\n\\[\n3^j n+C_j\\equiv2^{S_j}\\pmod{2^{S_j+1}}.\n\\]\nThus $x_j$ is integral and odd. It is positive because $n>0$ and $C_j\\ge0$. Finally the recurrence gives $3x_j+1=2^{a_j}x_{j+1}$. Oddness of $x_{j+1}$ makes $a_j$ the exact valuation. These candidates are therefore the actual orbit. Invertibility of $3^k$ yields a unique residue class. Its parity is odd because $C_k$ is odd.\n\\end{proof}","Every such cylinder contains infinitely many positive integers. It can have no representatives in a particular finite test interval. That finite absence must not be confused with an empty arithmetic cylinder. The additional endpoint bit in modulus $2^{S+1}$ is essential. Modulus $2^S$ certifies only integrality of the terminal quotient, while exact realization requires its oddness.","\\subsection{Two ordered words with the same total depth}\\label{sec:orderedexamples}\nFor $w=(1,3)$, the recurrence gives $C_1=1$ and $C_2=5$, with total depth $S=4$. The affine expression is\n\\[\nF_{(1,3)}(n)=\\frac{9n+5}{16}.\n\\]\nThe exact cylinder requires $9n+5\\equiv16\\pmod{32}$. The inverse of 9 modulo 32 is 25, so its address is $19$ modulo 32. Every positive representative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51","\\subsection{Two ordered words with the same total depth}\\label{sec:orderedexamples}\nFor $w=(1,3)$, the recurrence gives $C_1=1$ and $C_2=5$, with total depth $S=4$. The affine expression is\n\\[\nF_{(1,3)}(n)=\\frac{9n+5}{16}.\n\\]\nThe exact cylinder requires $9n+5\\equiv16\\pmod{32}$. The inverse of 9 modulo 32 is 25, so its address is $19$ modulo 32. Every positive representative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","red words with the same total depth}\\label{sec:orderedexamples}\nFor $w=(1,3)$, the recurrence gives $C_1=1$ and $C_2=5$, with total depth $S=4$. The affine expression is\n\\[\nF_{(1,3)}(n)=\\frac{9n+5}{16}.\n\\]\nThe exact cylinder requires $9n+5\\equiv16\\pmod{32}$. The inverse of 9 modulo 32 is 25, so its address is $19$ modulo 32. Every positive representative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.",",3)$, the recurrence gives $C_1=1$ and $C_2=5$, with total depth $S=4$. The affine expression is\n\\[\nF_{(1,3)}(n)=\\frac{9n+5}{16}.\n\\]\nThe exact cylinder requires $9n+5\\equiv16\\pmod{32}$. The inverse of 9 modulo 32 is 25, so its address is $19$ modulo 32. Every positive representative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","F_{(1,3)}(n)=\\frac{9n+5}{16}.\n\\]\nThe exact cylinder requires $9n+5\\equiv16\\pmod{32}$. The inverse of 9 modulo 32 is 25, so its address is $19$ modulo 32. Every positive representative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","ires $9n+5\\equiv16\\pmod{32}$. The inverse of 9 modulo 32 is 25, so its address is $19$ modulo 32. Every positive representative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","positive representative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","sentative can be written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","e written $n=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","=32t+19$ with an integer $t\\ge0$. Direct substitution gives the full orbit segment\n\\begin{equation}\\label{eq:word13}\n32t+19\\longmapsto48t+29\\longmapsto18t+11.\n\\end{equation}\nThe first numerator is $2(48t+29)$, with an odd quotient. The next numerator is $8(18t+11)$, again with an odd quotient. Thus the valuations really are 1 and 3 for every parameter. At $t=0$ this is $19,29,11$. At $t=1$ it is $51,77,29$.","For $v=(3,1)$, the first offset is still 1, but the second is $3+2^3=11$. Its affine expression and cylinder are\n\\[\nF_{(3,1)}(n)=\\frac{9n+11}{16},\\qquad n\\equiv29\\pmod{32}.\n\\]\nThe realized segment is\n\\begin{equation}\\label{eq:word31}\n32t+29\\longmapsto12t+11\\longmapsto18t+17,\n\\qquad t\\ge0.\n\\end{equation}\nBoth words have two odd multiplications and four divisions by two. Their coefficients are the same. Their offsets, residue addresses, and intermediate states differ. The first word grows before falling. The second falls before growing. Both endpoints lie below their sources for every displayed parameter, but this common conclusion follows from two different exact budgets.","The distinction persists under composition. Starting from 29, the word $(3,1)$ gives $29,11,17$. The source 19 instead realizes $(1,3)$ and gives $19,29,11$. It is possible for one segment's endpoint to be another segment's source, yet the words still cannot be reordered at will. A permutation of valuations must be assigned its own cylinder before its transfer can be interpreted as an actual orbit.","\\subsection{The endpoint bit as an admissibility test}\nThe shorter congruence $9n+5\\equiv0\\pmod{16}$ for the first word allows $n\\equiv3\\pmod{16}$. This contains two odd classes modulo 32, namely 3 and 19. Only 19 realizes $(1,3)$. At the other address, the smallest example is\n\\[\n3\\longmapsto5\\longmapsto1.\n\\]\nThe actual word is $(1,4)$. Substitution into the proposed expression gives $(9\\cdot3+5)/16=2$, an even integer. That quotient is a legitimate result after dividing by four total powers of two, but it is not an odd accelerated state. One more division is required.","The failed address is useful because it fails in a precisely visible way. All the arithmetic expressions are integral. It is exact valuation, rather than integrality, that is missing. The congruence modulo $2^{S+1}$ supplies the missing parity bit and thereby fixes the terminal quotient as odd. The backward part of Theorem~\\ref{thm:cylinder} then propagates that exactness to every preceding step.","For a prefix $w$, further extensions refine its address. Each extension remains inside the prefix cylinder because realization of a longer word includes realization of every shorter prefix. There are infinitely many possible positive next valuations, so the full refinement tree has infinitely many branches. A finite display can preserve exact addresses for the branches shown while omitting the rest. Omitted branches are arithmetic possibilities that have not been drawn.","\\subsection{Density without a random orbit assumption}\nA length $k$ word of total depth $S$ occupies one class modulo $2^{S+1}$ among all positive integers. Its ordinary arithmetic density is therefore $2^{-S-1}$. Among the odd integers its relative density is $2^{-S}$. These two normalizations differ by a factor of two and will remain distinct in the counting proof.","For fixed $k\\ge1$, the sum of all relative odd cylinder densities is\n\\[\n\\sum_{a_0,\\ldots,a_{k-1}\\ge1}2^{-(a_0+\\cdots+a_{k-1})}\n=\\left(\\sum_{a\\ge1}2^{-a}\\right)^k=1.\n\\]\nThis identity reflects the exact partition of positive odd integers by their first $k$ valuations. It does not assert that valuations along one particular infinite orbit are independent samples. A density calculation describes how sources are distributed among finite cylinders. An orbit theorem describes the behavior of every state of a chosen trajectory. Moving between those statements needs an additional argument, which the later finite packing proof supplies under a bounded deficit hypothesis.","\\figplaceholder{collatz_residue_tree}{Valuation words give exact dyadic cylinders. The diagram labels the first two valuations by \\(r_1=a_0\\) and \\(r_2=a_1\\). Refinement reveals further valuation information. A bounded diagram displays finitely many branches of this infinite arithmetic structure.}","\\section{Composition and the descent budget}\nA realized word acts by the affine transformation\n\\begin{equation}\\label{eq:transfer}\nF_w(n)=\\lambda_w n+c_w,\\qquad\n\\lambda_w=\\frac{3^k}{2^S},\\qquad c_w=\\frac{C_k}{2^S}.\n\\end{equation}\nThe transfer law is evaluated only on the integer cylinder of the word. As a rational affine expression it is defined elsewhere, but there it need not describe an actual Collatz segment.","\\begin{proposition}[Concatenation]\\label{prop:composition}\nLet a word $w$ have data $(k,S,C)$ and a following word $v$ have data $(\\ell,U,D)$. Their concatenation has data\n\\begin{equation}\\label{eq:composition}\n(k+\\ell,S+U,3^\\ell C+2^S D).\n\\end{equation}\nOn every realizing source, its transfer is $F_v\\circ F_w$.\n\\end{proposition}\n\\begin{proof}\nSubstitute $2^S y=3^k n+C$ into $2^U z=3^\\ell y+D$. Multiplication by $2^S$ gives\n\\[\n2^{S+U}z=3^{k+\\ell}n+3^\\ell C+2^S D.\n\\]\nThe exact cylinder theorem identifies the admissible integer sources.\n\\end{proof}","n}[Concatenation]\\label{prop:composition}\nLet a word $w$ have data $(k,S,C)$ and a following word $v$ have data $(\\ell,U,D)$. Their concatenation has data\n\\begin{equation}\\label{eq:composition}\n(k+\\ell,S+U,3^\\ell C+2^S D).\n\\end{equation}\nOn every realizing source, its transfer is $F_v\\circ F_w$.\n\\end{proposition}\n\\begin{proof}\nSubstitute $2^S y=3^k n+C$ into $2^U z=3^\\ell y+D$. Multiplication by $2^S$ gives\n\\[\n2^{S+U}z=3^{k+\\ell}n+3^\\ell C+2^S D.\n\\]\nThe exact cylinder theorem identifies the admissible integer sources.\n\\end{proof}","ion]\\label{prop:composition}\nLet a word $w$ have data $(k,S,C)$ and a following word $v$ have data $(\\ell,U,D)$. Their concatenation has data\n\\begin{equation}\\label{eq:composition}\n(k+\\ell,S+U,3^\\ell C+2^S D).\n\\end{equation}\nOn every realizing source, its transfer is $F_v\\circ F_w$.\n\\end{proposition}\n\\begin{proof}\nSubstitute $2^S y=3^k n+C$ into $2^U z=3^\\ell y+D$. Multiplication by $2^S$ gives\n\\[\n2^{S+U}z=3^{k+\\ell}n+3^\\ell C+2^S D.\n\\]\nThe exact cylinder theorem identifies the admissible integer sources.\n\\end{proof}","\\begin{theorem}[Exact descent criterion]\\label{thm:descent}\nA realized word of positive length ends below its source if and only if\n\\begin{equation}\\label{eq:descent}\n(2^S-3^k)n>C_k.\n\\end{equation}\nIf $2^S\\le3^k$, strict descent is impossible. If $2^S>3^k$, write $G=2^S-3^k$. The endpoint is at least the source if and only if\n\\begin{equation}\\label{eq:threshold}\nn\\le\\left\\lfloor\\frac{C_k}{G}\\right\\rfloor.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nThe affine identity gives $2^S(x_k-n)=C_k-(2^S-3^k)n$. Its sign proves the assertions. Since $n$ is an integer, the inequality $Gn\\le C_k$ is equivalent to \\eqref{eq:threshold}.\n\\end{proof}","positive length ends below its source if and only if\n\\begin{equation}\\label{eq:descent}\n(2^S-3^k)n>C_k.\n\\end{equation}\nIf $2^S\\le3^k$, strict descent is impossible. If $2^S>3^k$, write $G=2^S-3^k$. The endpoint is at least the source if and only if\n\\begin{equation}\\label{eq:threshold}\nn\\le\\left\\lfloor\\frac{C_k}{G}\\right\\rfloor.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nThe affine identity gives $2^S(x_k-n)=C_k-(2^S-3^k)n$. Its sign proves the assertions. Since $n$ is an integer, the inequality $Gn\\le C_k$ is equivalent to \\eqref{eq:threshold}.\n\\end{proof}","n}\\label{eq:descent}\n(2^S-3^k)n>C_k.\n\\end{equation}\nIf $2^S\\le3^k$, strict descent is impossible. If $2^S>3^k$, write $G=2^S-3^k$. The endpoint is at least the source if and only if\n\\begin{equation}\\label{eq:threshold}\nn\\le\\left\\lfloor\\frac{C_k}{G}\\right\\rfloor.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nThe affine identity gives $2^S(x_k-n)=C_k-(2^S-3^k)n$. Its sign proves the assertions. Since $n$ is an integer, the inequality $Gn\\le C_k$ is equivalent to \\eqref{eq:threshold}.\n\\end{proof}","k.\n\\end{equation}\nIf $2^S\\le3^k$, strict descent is impossible. If $2^S>3^k$, write $G=2^S-3^k$. The endpoint is at least the source if and only if\n\\begin{equation}\\label{eq:threshold}\nn\\le\\left\\lfloor\\frac{C_k}{G}\\right\\rfloor.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nThe affine identity gives $2^S(x_k-n)=C_k-(2^S-3^k)n$. Its sign proves the assertions. Since $n$ is an integer, the inequality $Gn\\le C_k$ is equivalent to \\eqref{eq:threshold}.\n\\end{proof}","Coefficient contraction is therefore necessary for descent, but the offset remains part of the budget. A coefficient crossing means $2^S>3^k$. Actual descent requires the stronger source dependent inequality \\eqref{eq:descent}. The real affine fixed point is $C_k/G$. The integer threshold is its floor. The threshold can differ from the real fixed point, and the two are equal when that fixed point is an integer.","The composition formula also explains why a local fall need not resolve a global obstruction. If an expanding segment has data $(\\lambda_g,c_g)$ and a following falling segment has $(\\lambda_f,c_f)$, the final value is\n\\[\n\\lambda_f\\lambda_g n+\\lambda_f c_g+c_f.\n\\]\nA comparison with the local peak controls only part of this expression. To prove descent below $n$, the entire expression must be compared with the same original source.","\\subsection{The threshold as a fixed point}\nWhen $G=2^S-3^k>0$, the affine transfer has one real fixed point\n\\[\nn_* = \\frac{C_k}{G}=\\frac{c_w}{1-\\lambda_w}.\n\\]\nFor a real input above $n_*$, the affine expression falls. For an input below $n_*$, it rises. Equality gives a fixed point of the affine expression. These are statements about that expression before its integer cylinder is imposed. An actual periodic state would additionally have to be positive odd and realize the word.","For $(1,3)$, the fixed point is $5/7$, and for $(3,1)$ it is $11/7$. Their exact cylinders start at 19 and 29. Every positive source in either cylinder therefore lies above the relevant threshold. This is why both entire progressions descend. The word $(2,2)$ instead has offset 7 and gap 7, so its fixed point is 1. Its cylinder is $1$ modulo 32. The smallest source stays at 1 through two accelerations, while all larger positive sources in the cylinder descend:\n\\[\n32t+1\\longmapsto24t+1\\longmapsto18t+1,\n\\qquad t\\ge0.\n\\]\nStrictness matters at $t=0$. A rule stated for every positive odd source without excluding the terminal state would fail there even though its coefficient is less than one.",". Their exact cylinders start at 19 and 29. Every positive source in either cylinder therefore lies above the relevant threshold. This is why both entire progressions descend. The word $(2,2)$ instead has offset 7 and gap 7, so its fixed point is 1. Its cylinder is $1$ modulo 32. The smallest source stays at 1 through two accelerations, while all larger positive sources in the cylinder descend:\n\\[\n32t+1\\longmapsto24t+1\\longmapsto18t+1,\n\\qquad t\\ge0.\n\\]\nStrictness matters at $t=0$. A rule stated for every positive odd source without excluding the terminal state would fail there even though its coefficient is less than one.","nder therefore lies above the relevant threshold. This is why both entire progressions descend. The word $(2,2)$ instead has offset 7 and gap 7, so its fixed point is 1. Its cylinder is $1$ modulo 32. The smallest source stays at 1 through two accelerations, while all larger positive sources in the cylinder descend:\n\\[\n32t+1\\longmapsto24t+1\\longmapsto18t+1,\n\\qquad t\\ge0.\n\\]\nStrictness matters at $t=0$. A rule stated for every positive odd source without excluding the terminal state would fail there even though its coefficient is less than one.","The finite exceptions to an individual contracting word are also explicit. Only positive odd sources in that word's cylinder at most $\\lfloor C_k/G\\rfloor$ can fail strict descent. They can be listed by intersecting one arithmetic progression with one finite interval. This observation proves a finite exceptional set for one fixed word. It does not produce a single finite exceptional set for infinitely many varying words, whose offsets and thresholds may grow.","\\subsection{A contracting prefix can still finish above its source}\nThe distinction is not limited to the state 1. Consider the actual source $n=165$ and the word\n\\[\n(4,1,1,1,1,2,2,1,2,1,1,2,1,1,1,2,3).\n\\]\nIts successive odd states are\n\\begin{align*}\n&165,31,47,71,107,161,121,91,137,103,","\\subsection{A contracting prefix can still finish above its source}\nThe distinction is not limited to the state 1. Consider the actual source $n=165$ and the word\n\\[\n(4,1,1,1,1,2,2,1,2,1,1,2,1,1,1,2,3).\n\\]\nIts successive odd states are\n\\begin{align*}\n&165,31,47,71,107,161,121,91,137,103,\\\\\n&155,233,175,263,395,593,445,167.\n\\end{align*}\nThe exact packet data are\n\\[\nk=17,\\qquad S=27,\\qquad C=1106233681,\\qquad\nG=2^{27}-3^{17}=5077565.\n\\]\nThe affine identity reads\n\\[\n2^{27}\\cdot167=3^{17}\\cdot165+1106233681=22414360576.\n\\]\nHere $G>0$, but $G\\cdot165<C$, so the endpoint 167 is above the source 165. The integer threshold is 217. This prefix is a concrete check against replacing the full descent budget with a coefficient test.","&155,233,175,263,395,593,445,167.\n\\end{align*}\nThe exact packet data are\n\\[\nk=17,\\qquad S=27,\\qquad C=1106233681,\\qquad\nG=2^{27}-3^{17}=5077565.\n\\]\nThe affine identity reads\n\\[\n2^{27}\\cdot167=3^{17}\\cdot165+1106233681=22414360576.\n\\]\nHere $G>0$, but $G\\cdot165<C$, so the endpoint 167 is above the source 165. The integer threshold is 217. This prefix is a concrete check against replacing the full descent budget with a coefficient test.","\\subsection{Composing realization as well as formulas}\nThere are two checks in a concatenation. The source must realize $w$, and its endpoint must realize $v$. Substitution then composes the transfers. Writing down $F_v(F_w(n))$ without those two checks gives a rational affine expression, which may have no connection with the orbit from $n$.","The exact cylinder theorem performs both checks at once for the concatenated word. Its modulus is $2^{S+U+1}$, so a longer exact itinerary usually requires a finer source address than either displayed short example. This is why finite precision and finite time are linked. A large depth can be compactly represented by a valuation word, but its exact cylinder still records all the source bits required by that word.","One can also retain a weaker certificate when exact final valuation is unnecessary. If the last numerator is known to equal $2^b q$ with $q>0$ integral, then the next odd state is at most $q$. Exact equality requires $q$ odd. A descent proof that only uses an upper bound can accept both parities. Example~\\ref{ex:1087} exploits this distinction. It obtains a uniform falling estimate without claiming one final exact word for every member of its coarse progression.","\\section{Why universal descent is the remaining theorem}\n\\begin{theorem}[Universal descent equivalence]\\label{thm:induction}\nConjecture~\\ref{conj:collatz} is equivalent to the assertion\n\\begin{equation}\\label{eq:universal}\n\\text{for every odd }n>1\\text{ there exists }r\\ge1\\text{ with }A^r(n)<n.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nConvergence to 1 implies descent for each $n>1$. For the converse, use strong induction on positive odd integers. The case $n=1$ is immediate. For $n>1$, choose a descendant $m=A^r(n)<n$. It is positive and odd, so the induction hypothesis gives $A^s(m)=1$. Hence $A^{r+s}(n)=1$.\n\\end{proof}","{theorem}[Universal descent equivalence]\\label{thm:induction}\nConjecture~\\ref{conj:collatz} is equivalent to the assertion\n\\begin{equation}\\label{eq:universal}\n\\text{for every odd }n>1\\text{ there exists }r\\ge1\\text{ with }A^r(n)<n.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nConvergence to 1 implies descent for each $n>1$. For the converse, use strong induction on positive odd integers. The case $n=1$ is immediate. For $n>1$, choose a descendant $m=A^r(n)<n$. It is positive and odd, so the induction hypothesis gives $A^s(m)=1$. Hence $A^{r+s}(n)=1$.\n\\end{proof}","ent equivalence]\\label{thm:induction}\nConjecture~\\ref{conj:collatz} is equivalent to the assertion\n\\begin{equation}\\label{eq:universal}\n\\text{for every odd }n>1\\text{ there exists }r\\ge1\\text{ with }A^r(n)<n.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nConvergence to 1 implies descent for each $n>1$. For the converse, use strong induction on positive odd integers. The case $n=1$ is immediate. For $n>1$, choose a descendant $m=A^r(n)<n$. It is positive and odd, so the induction hypothesis gives $A^s(m)=1$. Hence $A^{r+s}(n)=1$.\n\\end{proof}","induction}\nConjecture~\\ref{conj:collatz} is equivalent to the assertion\n\\begin{equation}\\label{eq:universal}\n\\text{for every odd }n>1\\text{ there exists }r\\ge1\\text{ with }A^r(n)<n.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nConvergence to 1 implies descent for each $n>1$. For the converse, use strong induction on positive odd integers. The case $n=1$ is immediate. For $n>1$, choose a descendant $m=A^r(n)<n$. It is positive and odd, so the induction hypothesis gives $A^s(m)=1$. Hence $A^{r+s}(n)=1$.\n\\end{proof}","\\begin{equation}\\label{eq:universal}\n\\text{for every odd }n>1\\text{ there exists }r\\ge1\\text{ with }A^r(n)<n.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nConvergence to 1 implies descent for each $n>1$. For the converse, use strong induction on positive odd integers. The case $n=1$ is immediate. For $n>1$, choose a descendant $m=A^r(n)<n$. It is positive and odd, so the induction hypothesis gives $A^s(m)=1$. Hence $A^{r+s}(n)=1$.\n\\end{proof}","abel{eq:universal}\n\\text{for every odd }n>1\\text{ there exists }r\\ge1\\text{ with }A^r(n)<n.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nConvergence to 1 implies descent for each $n>1$. For the converse, use strong induction on positive odd integers. The case $n=1$ is immediate. For $n>1$, choose a descendant $m=A^r(n)<n$. It is positive and odd, so the induction hypothesis gives $A^s(m)=1$. Hence $A^{r+s}(n)=1$.\n\\end{proof}","This floor condition concerns values $x_r\\ge n$. It does not imply that the cumulative affine coefficient remains at least one. An orbit can have a contracting coefficient and still remain above its source because of its positive offset. The coefficient split examined later is a split of the algebraic multiplier, rather than a replacement for the actual descent problem.","The induction is an ordinary induction on the source value. It does not require a bound on how long the descent takes. Once a lower odd state $m$ is reached, the induction hypothesis supplies a possibly unrelated finite termination time for $m$. Adding the two times finishes the source. This makes an unbounded adaptive descent theorem entirely compatible with a finite proof.","\\section{A complete finite example from 27}\nThe seed 27 makes the distinction between the clocks and the floors visible. Its first ten odd accelerated steps are shown in Table~\\ref{tab:27}. Each row is checked by multiplying the displayed state by $2^{S_r}$ and comparing it with $3^r\\cdot27+C_r$.","\\begin{table}[htbp]\\centering\n\\caption{Exact odd packet coordinates for the first ten steps from 27.}\\label{tab:27}\n\\begin{tabular}{rrrrr}\\toprule\n$r$&$x_r$&$a_{r-1}$&$S_r$&$C_r$","For example, the word $(1,2)$ has $S_2=3$ and $C_2=5$, so\n\\[\n8\\cdot31=9\\cdot27+5.\n\\]\nThe corresponding cylinder is $n\\equiv11\\pmod{16}$. Writing $n=16t+11$ gives\n\\[\nA(n)=24t+17,\\qquad A^2(n)=18t+13.\n\\]\nEvery member of this progression grows over the two steps, even though the second step falls from $24t+17$ to $18t+13$. For $t=1$, the local fall is $41$ to $31$, while both states remain above 27. These identities are present in the recovered Lean source \\cite{projectcollatzcore}.","The fresh finite integer audit for this edition finds the first actual descent at accelerated time 37:\n\\begin{equation}\\label{eq:27descent}\nx_{37}=23,\\qquad S_{37}=59,\\qquad\nC_{37}=1100931843921811423.\n\\end{equation}\nThus\n\\[\n2^{59}\\cdot23=3^{37}\\cdot27+1100931843921811423.\n\\]\nThe same segment has 59 shortcut steps. The historical serious audit separately records a Lean check of $T^{59}(27)=23$ \\cite{projectcollatzaudit}. The two statements use different iteration counts and the same endpoint.","\\begin{lemma}[A run of unit valuations]\\label{lem:run}\nFor integers $j\\ge0$, $q\\ge1$, and $0\\le t\\le j$, put $n=2^{j+1}q-1$. Then\n\\begin{equation}\\label{eq:run}\nA^t(n)=2^{j+1-t}3^t q-1.\n\\end{equation}\nEvery one of the first $j$ valuations is exactly 1. Moreover $A^t(n)\\ge n$ for $0\\le t\\le j$.\n\\end{lemma}\n\\begin{proof}\nFor $t<j$, the displayed state has the form $4u-1$ with $u>0$. Thus $3(4u-1)+1=2(6u-1)$, whose quotient is odd. This proves the next state and valuation by induction. The comparison with the source follows from $3^t\\ge2^t$.\n\\end{proof}","gin{equation}\\label{eq:run}\nA^t(n)=2^{j+1-t}3^t q-1.\n\\end{equation}\nEvery one of the first $j$ valuations is exactly 1. Moreover $A^t(n)\\ge n$ for $0\\le t\\le j$.\n\\end{lemma}\n\\begin{proof}\nFor $t<j$, the displayed state has the form $4u-1$ with $u>0$. Thus $3(4u-1)+1=2(6u-1)$, whose quotient is odd. This proves the next state and valuation by induction. The comparison with the source follows from $3^t\\ge2^t$.\n\\end{proof}","\\begin{theorem}[No finite universal descent horizon]\\label{thm:horizon}\nFor every integer $K\\ge0$, there exists a positive odd $n>1$ such that $A^t(n)\\ge n$ for every $0\\le t\\le K$. Consequently no finite collection of descent rules with fixed step counts can cover all odd sources above 1.\n\\end{theorem}\n\\begin{proof}\nTake $n=2^{K+2}-1$ and apply Lemma~\\ref{lem:run} with $j=K+1$ and $q=1$. For the final assertion, a finite collection has a largest step count $K$. The constructed source violates every claimed strict descent at those step counts.\n\\end{proof}","horizon}\nFor every integer $K\\ge0$, there exists a positive odd $n>1$ such that $A^t(n)\\ge n$ for every $0\\le t\\le K$. Consequently no finite collection of descent rules with fixed step counts can cover all odd sources above 1.\n\\end{theorem}\n\\begin{proof}\nTake $n=2^{K+2}-1$ and apply Lemma~\\ref{lem:run} with $j=K+1$ and $q=1$. For the final assertion, a finite collection has a largest step count $K$. The constructed source violates every claimed strict descent at those step counts.\n\\end{proof}","\\begin{theorem}[Compressed run exit]\\label{thm:exit}\nSuppose $j\\ge0$, $q\\ge1$, $b\\ge0$, and $m$ is positive odd, with\n\\begin{equation}\\label{eq:exitpremise}\n3^{j+1}q=2^b m+1.\n\\end{equation}\nFor $n=2^{j+1}q-1$, the exact exit is\n\\begin{equation}\\label{eq:exit}\nA^{j+1}(n)=m.\n\\end{equation}\nIf $m<n$, then $j+1$ is the exact first strict descent time.\n\\end{theorem}\n\\begin{proof}\nLemma~\\ref{lem:run} gives $A^j(n)=2\\cdot3^j q-1$. Its next odd numerator is\n\\[\n3(2\\cdot3^j q-1)+1=2(3^{j+1}q-1)=2^{b+1}m.\n\\]\nOddness of $m$ proves the exact final valuation $b+1$ and the exit. Earlier states are at least $n$ by the lemma.\n\\end{proof}","n{theorem}[Compressed run exit]\\label{thm:exit}\nSuppose $j\\ge0$, $q\\ge1$, $b\\ge0$, and $m$ is positive odd, with\n\\begin{equation}\\label{eq:exitpremise}\n3^{j+1}q=2^b m+1.\n\\end{equation}\nFor $n=2^{j+1}q-1$, the exact exit is\n\\begin{equation}\\label{eq:exit}\nA^{j+1}(n)=m.\n\\end{equation}\nIf $m<n$, then $j+1$ is the exact first strict descent time.\n\\end{theorem}\n\\begin{proof}\nLemma~\\ref{lem:run} gives $A^j(n)=2\\cdot3^j q-1$. Its next odd numerator is\n\\[\n3(2\\cdot3^j q-1)+1=2(3^{j+1}q-1)=2^{b+1}m.\n\\]\nOddness of $m$ proves the exact final valuation $b+1$ and the exit. Earlier states are at least $n$ by the lemma.\n\\end{proof}","pose $j\\ge0$, $q\\ge1$, $b\\ge0$, and $m$ is positive odd, with\n\\begin{equation}\\label{eq:exitpremise}\n3^{j+1}q=2^b m+1.\n\\end{equation}\nFor $n=2^{j+1}q-1$, the exact exit is\n\\begin{equation}\\label{eq:exit}\nA^{j+1}(n)=m.\n\\end{equation}\nIf $m<n$, then $j+1$ is the exact first strict descent time.\n\\end{theorem}\n\\begin{proof}\nLemma~\\ref{lem:run} gives $A^j(n)=2\\cdot3^j q-1$. Its next odd numerator is\n\\[\n3(2\\cdot3^j q-1)+1=2(3^{j+1}q-1)=2^{b+1}m.\n\\]\nOddness of $m$ proves the exact final valuation $b+1$ and the exit. Earlier states are at least $n$ by the lemma.\n\\end{proof}","th\n\\begin{equation}\\label{eq:exitpremise}\n3^{j+1}q=2^b m+1.\n\\end{equation}\nFor $n=2^{j+1}q-1$, the exact exit is\n\\begin{equation}\\label{eq:exit}\nA^{j+1}(n)=m.\n\\end{equation}\nIf $m<n$, then $j+1$ is the exact first strict descent time.\n\\end{theorem}\n\\begin{proof}\nLemma~\\ref{lem:run} gives $A^j(n)=2\\cdot3^j q-1$. Its next odd numerator is\n\\[\n3(2\\cdot3^j q-1)+1=2(3^{j+1}q-1)=2^{b+1}m.\n\\]\nOddness of $m$ proves the exact final valuation $b+1$ and the exit. Earlier states are at least $n$ by the lemma.\n\\end{proof}","ise}\n3^{j+1}q=2^b m+1.\n\\end{equation}\nFor $n=2^{j+1}q-1$, the exact exit is\n\\begin{equation}\\label{eq:exit}\nA^{j+1}(n)=m.\n\\end{equation}\nIf $m<n$, then $j+1$ is the exact first strict descent time.\n\\end{theorem}\n\\begin{proof}\nLemma~\\ref{lem:run} gives $A^j(n)=2\\cdot3^j q-1$. Its next odd numerator is\n\\[\n3(2\\cdot3^j q-1)+1=2(3^{j+1}q-1)=2^{b+1}m.\n\\]\nOddness of $m$ proves the exact final valuation $b+1$ and the exit. Earlier states are at least $n$ by the lemma.\n\\end{proof}","j+1}q-1$, the exact exit is\n\\begin{equation}\\label{eq:exit}\nA^{j+1}(n)=m.\n\\end{equation}\nIf $m<n$, then $j+1$ is the exact first strict descent time.\n\\end{theorem}\n\\begin{proof}\nLemma~\\ref{lem:run} gives $A^j(n)=2\\cdot3^j q-1$. Its next odd numerator is\n\\[\n3(2\\cdot3^j q-1)+1=2(3^{j+1}q-1)=2^{b+1}m.\n\\]\nOddness of $m$ proves the exact final valuation $b+1$ and the exit. Earlier states are at least $n$ by the lemma.\n\\end{proof}","The equation \\eqref{eq:exitpremise} is an arithmetic condition on a proposed exit. It is not asserted with a universally favorable value of $b$. Long growth can end in a strong fall, a mild fall, or continued growth. Run compression records the actual alternative without deciding the global behavior of every source.","\\subsection{Explicit terminating seeds with unbounded delay}\nThe program also constructs a closed family of sources whose first descent occurs arbitrarily late, yet whose first lower value is exactly 1 \\cite{projectcollatzseeds}. Define\n\\begin{equation}\\label{eq:seeds}\nq_j=\\frac{2^{3^j}+1}{3^{j+1}},\\qquad n_j=2^{j+1}q_j-1.\n\\end{equation}","\\begin{lemma}\\label{lem:seeddiv}\nFor every $j\\ge0$, $q_j$ is a positive integer.\n\\end{lemma}\n\\begin{proof}\nAt $j=0$ the numerator is 3. If $3^{j+1}$ divides $u+1$ for $u=2^{3^j}$, then\n\\[\n2^{3^{j+1}}+1=u^3+1=(u+1)(u^2-u+1).\n\\]\nSince $u\\equiv-1\\pmod3$, the second factor is divisible by 3. Induction proves divisibility by $3^{j+2}$. Positivity of the quotient is immediate.\n\\end{proof}","\\begin{theorem}[Exact delayed termination]\\label{thm:seeds}\nFor every $j\\ge1$, the seed $n_j$ is odd and greater than 1. It satisfies\n\\begin{equation}\\label{eq:seedresult}\nA^{j+1}(n_j)=1,\\qquad A^t(n_j)\\ge n_j\\quad(0\\le t\\le j).\n\\end{equation}\nThe seeds with positive parameters are distinct.\n\\end{theorem}\n\\begin{proof}\nThe source is at least $2^{j+1}-1\\ge3$. The quotient equation is\n$3^{j+1}q_j=2^{3^j}+1$, so Theorem~\\ref{thm:exit} applies with $b=3^j$ and $m=1$. If two positive parameters gave the same seed, that seed would have two different exact first descent times. The smaller time would contradict the other parameter's absence of earlier descent.\n\\end{proof}","hm:seeds}\nFor every $j\\ge1$, the seed $n_j$ is odd and greater than 1. It satisfies\n\\begin{equation}\\label{eq:seedresult}\nA^{j+1}(n_j)=1,\\qquad A^t(n_j)\\ge n_j\\quad(0\\le t\\le j).\n\\end{equation}\nThe seeds with positive parameters are distinct.\n\\end{theorem}\n\\begin{proof}\nThe source is at least $2^{j+1}-1\\ge3$. The quotient equation is\n$3^{j+1}q_j=2^{3^j}+1$, so Theorem~\\ref{thm:exit} applies with $b=3^j$ and $m=1$. If two positive parameters gave the same seed, that seed would have two different exact first descent times. The smaller time would contradict the other parameter's absence of earlier descent.\n\\end{proof}","r every $j\\ge1$, the seed $n_j$ is odd and greater than 1. It satisfies\n\\begin{equation}\\label{eq:seedresult}\nA^{j+1}(n_j)=1,\\qquad A^t(n_j)\\ge n_j\\quad(0\\le t\\le j).\n\\end{equation}\nThe seeds with positive parameters are distinct.\n\\end{theorem}\n\\begin{proof}\nThe source is at least $2^{j+1}-1\\ge3$. The quotient equation is\n$3^{j+1}q_j=2^{3^j}+1$, so Theorem~\\ref{thm:exit} applies with $b=3^j$ and $m=1$. If two positive parameters gave the same seed, that seed would have two different exact first descent times. The smaller time would contradict the other parameter's absence of earlier descent.\n\\end{proof}","The first examples are $n_1=3$ and $n_2=151$. The latter follows the exact odd orbit $151,227,341,1$. The family does more than show that some prefixes can grow. It certifies arbitrarily large first descent times while retaining termination. Such delays must therefore be allowed in any proposed global proof.","\\subsection{Lifting a certified exit to an infinite progression}\n\\begin{theorem}[Uniform cylinder lift]\\label{thm:lift}\nUnder the hypotheses of Theorem~\\ref{thm:exit}, define for integers $s\\ge0$\n\\begin{equation}\\label{eq:lift}\nn(s)=2^{j+1}q-1+2^{j+b+2}s,\\qquad\nm(s)=m+2\\cdot3^{j+1}s.\n\\end{equation}\nThen $A^{j+1}(n(s))=m(s)$. If additionally\n\\begin{equation}\\label{eq:slopes}\nm<2^{j+1}q-1,\\qquad 3^{j+1}\\le2^{j+b+1},\n\\end{equation}\nevery member has exact first strict descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} ex","\\subsection{Lifting a certified exit to an infinite progression}\n\\begin{theorem}[Uniform cylinder lift]\\label{thm:lift}\nUnder the hypotheses of Theorem~\\ref{thm:exit}, define for integers $s\\ge0$\n\\begin{equation}\\label{eq:lift}\nn(s)=2^{j+1}q-1+2^{j+b+2}s,\\qquad\nm(s)=m+2\\cdot3^{j+1}s.\n\\end{equation}\nThen $A^{j+1}(n(s))=m(s)$. If additionally\n\\begin{equation}\\label{eq:slopes}\nm<2^{j+1}q-1,\\qquad 3^{j+1}\\le2^{j+b+1},\n\\end{equation}\nevery member has exact first strict descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} excludes earlier descent.\n\\end{proof}","iform cylinder lift]\\label{thm:lift}\nUnder the hypotheses of Theorem~\\ref{thm:exit}, define for integers $s\\ge0$\n\\begin{equation}\\label{eq:lift}\nn(s)=2^{j+1}q-1+2^{j+b+2}s,\\qquad\nm(s)=m+2\\cdot3^{j+1}s.\n\\end{equation}\nThen $A^{j+1}(n(s))=m(s)$. If additionally\n\\begin{equation}\\label{eq:slopes}\nm<2^{j+1}q-1,\\qquad 3^{j+1}\\le2^{j+b+1},\n\\end{equation}\nevery member has exact first strict descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} excludes earlier descent.\n\\end{proof}","the hypotheses of Theorem~\\ref{thm:exit}, define for integers $s\\ge0$\n\\begin{equation}\\label{eq:lift}\nn(s)=2^{j+1}q-1+2^{j+b+2}s,\\qquad\nm(s)=m+2\\cdot3^{j+1}s.\n\\end{equation}\nThen $A^{j+1}(n(s))=m(s)$. If additionally\n\\begin{equation}\\label{eq:slopes}\nm<2^{j+1}q-1,\\qquad 3^{j+1}\\le2^{j+b+1},\n\\end{equation}\nevery member has exact first strict descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} excludes earlier descent.\n\\end{proof}","otheses of Theorem~\\ref{thm:exit}, define for integers $s\\ge0$\n\\begin{equation}\\label{eq:lift}\nn(s)=2^{j+1}q-1+2^{j+b+2}s,\\qquad\nm(s)=m+2\\cdot3^{j+1}s.\n\\end{equation}\nThen $A^{j+1}(n(s))=m(s)$. If additionally\n\\begin{equation}\\label{eq:slopes}\nm<2^{j+1}q-1,\\qquad 3^{j+1}\\le2^{j+b+1},\n\\end{equation}\nevery member has exact first strict descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} excludes earlier descent.\n\\end{proof}","integers $s\\ge0$\n\\begin{equation}\\label{eq:lift}\nn(s)=2^{j+1}q-1+2^{j+b+2}s,\\qquad\nm(s)=m+2\\cdot3^{j+1}s.\n\\end{equation}\nThen $A^{j+1}(n(s))=m(s)$. If additionally\n\\begin{equation}\\label{eq:slopes}\nm<2^{j+1}q-1,\\qquad 3^{j+1}\\le2^{j+b+1},\n\\end{equation}\nevery member has exact first strict descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} excludes earlier descent.\n\\end{proof}","ad 3^{j+1}\\le2^{j+b+1},\n\\end{equation}\nevery member has exact first strict descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} excludes earlier descent.\n\\end{proof}","descent time $j+1$.\n\\end{theorem}\n\\begin{proof}\nReplace $q$ by $q(s)=q+2^{b+1}s$. The exit equation becomes\n\\[\n3^{j+1}q(s)=2^b\\bigl(m+2\\cdot3^{j+1}s\\bigr)+1.\n\\]\nThe new target is odd, and Theorem~\\ref{thm:exit} gives its exact exit. The initial target is strictly smaller than the initial source. The slope inequality gives $2\\cdot3^{j+1}\\le2^{j+b+2}$, so this strict comparison persists for every $s$. Lemma~\\ref{lem:run} excludes earlier descent.\n\\end{proof}","For the terminating seed family, the slope condition is automatic. Indeed $j+1\\le3^j$ by induction, hence\n\\[\n3^{j+1}\\le4^{j+1}=2^{2j+2}\\le2^{j+3^j+1}.\n\\]\nThus, for every $j\\ge1$, all inputs\n\\begin{equation}\\label{eq:seedprogressions}\nn_j+2^{j+3^j+2}s\n\\end{equation}\nhave exact first descent time $j+1$ and exit $1+2\\cdot3^{j+1}s$. The case $s=0$ reaches 1 at that time. For $s>0$, the theorem certifies descent rather than termination of the exit. If an exit is independently known to terminate, termination transfers to the source with $j+1$ added steps.","2^{j+3^j+1}.\n\\]\nThus, for every $j\\ge1$, all inputs\n\\begin{equation}\\label{eq:seedprogressions}\nn_j+2^{j+3^j+2}s\n\\end{equation}\nhave exact first descent time $j+1$ and exit $1+2\\cdot3^{j+1}s$. The case $s=0$ reaches 1 at that time. For $s>0$, the theorem certifies descent rather than termination of the exit. If an exit is independently known to terminate, termination transfers to the source with $j+1$ added steps.","For $j\\ge11$, both the seed and the entire progression in \\eqref{eq:seedprogressions} are congruent to $4095$ modulo $4096$. The source has a factor $2^{j+1}$ before subtraction of 1, and the progression increment is divisible by $4096$. Consequently one unresolved coarse residue contains infinitely many sources with exact successful descent certificates. Residual status describes a chosen collection of rules, rather than the behavior of every individual integer in that class \\cite{projectcollatzlifts,projectcollatzboundary}.","\\subsection{The full progression through 151}\nThe parameter $j=2$ gives $3^j=9$ and\n\\[\nq_2=\\frac{2^9+1}{3^3}=19,\\qquad n_2=8\\cdot19-1=151.\n\\]\nThe exit equation is $27\\cdot19=2^9\\cdot1+1$. Thus the first two valuations are 1 and the third is 10. Their sum is 12. The cylinder modulus is $2^{13}=8192$, rather than $2^{12}$. That last factor of two keeps the exiting quotient odd.","The lift theorem now gives, for every integer $s\\ge0$,\n\\begin{equation}\\label{eq:151lift}\n151+8192s\\longmapsto227+12288s\n\\longmapsto341+18432s\\longmapsto1+54s.\n\\end{equation}\nEach valuation can be inspected directly. The first numerator is\n\\[\n3(151+8192s)+1=2(227+12288s),\n\\]\nand its quotient is odd. The second numerator is $2(341+18432s)$, again with an odd quotient. The third is\n\\[\n3(341+18432s)+1=1024(1+54s),\n\\]\nwhose quotient is odd for every integer parameter. The large final division is consequently exact throughout the progression.","The initial two states remain above the source, since their differences are $76+4096s$ and $190+10240s$. The exit lies below it, since the source minus the exit is $150+8138s$. These three positive comparisons give exact first descent without needing a numerical search. The first few source and exit pairs are\n\\[\n(151,1),\\qquad(8343,55),\\qquad(16535,109),\\qquad(24727,163).\n\\]\nAt the base source, the exit 1 proves immediate termination. For another parameter, a separate termination certificate for its smaller exit can be composed with the three step lift. For example, $55,83,125,47$ is an exact odd prefix. The state 47 appears on the verified orbit from 27 in Appendix~\\ref{app:orbit27} and therefore terminates. That finite certificate proves termination of the particular lifte","The initial two states remain above the source, since their differences are $76+4096s$ and $190+10240s$. The exit lies below it, since the source minus the exit is $150+8138s$. These three positive comparisons give exact first descent without needing a numerical search. The first few source and exit pairs are\n\\[\n(151,1),\\qquad(8343,55),\\qquad(16535,109),\\qquad(24727,163).\n\\]\nAt the base source, the exit 1 proves immediate termination. For another parameter, a separate termination certificate for its smaller exit can be composed with the three step lift. For example, $55,83,125,47$ is an exact odd prefix. The state 47 appears on the verified orbit from 27 in Appendix~\\ref{app:orbit27} and therefore terminates. That finite certificate proves termination of the particular lifted source 8343 whe","The initial two states remain above the source, since their differences are $76+4096s$ and $190+10240s$. The exit lies below it, since the source minus the exit is $150+8138s$. These three positive comparisons give exact first descent without needing a numerical search. The first few source and exit pairs are\n\\[\n(151,1),\\qquad(8343,55),\\qquad(16535,109),\\qquad(24727,163).\n\\]\nAt the base source, the exit 1 proves immediate termination. For another parameter, a separate termination certificate for its smaller exit can be composed with the three step lift. For example, $55,83,125,47$ is an exact odd prefix. The state 47 appears on the verified orbit from 27 in Appendix~\\ref{app:orbit27} and therefore terminates. That finite certificate proves termination of the particular lifted source 8343 when composed with \\eqref{eq:151lift}. It does not establish termination of","The initial two states remain above the source, since their differences are $76+4096s$ and $190+10240s$. The exit lies below it, since the source minus the exit is $150+8138s$. These three positive comparisons give exact first descent without needing a numerical search. The first few source and exit pairs are\n\\[\n(151,1),\\qquad(8343,55),\\qquad(16535,109),\\qquad(24727,163).\n\\]\nAt the base source, the exit 1 proves immediate termination. For another parameter, a separate termination certificate for its smaller exit can be composed with the three step lift. For example, $55,83,125,47$ is an exact odd prefix. The state 47 appears on the verified orbit from 27 in Appendix~\\ref{app:orbit27} and therefore terminates. That finite certificate proves termination of the particular lifted source 8343 when composed with \\eqref{eq:151lift}. It does not establish termination of all the other exits $1+54s$.","$. These three positive comparisons give exact first descent without needing a numerical search. The first few source and exit pairs are\n\\[\n(151,1),\\qquad(8343,55),\\qquad(16535,109),\\qquad(24727,163).\n\\]\nAt the base source, the exit 1 proves immediate termination. For another parameter, a separate termination certificate for its smaller exit can be composed with the three step lift. For example, $55,83,125,47$ is an exact odd prefix. The state 47 appears on the verified orbit from 27 in Appendix~\\ref{app:orbit27} and therefore terminates. That finite certificate proves termination of the particular lifted source 8343 when composed with \\eqref{eq:151lift}. It does not establish termination of all the other exits $1+54s$.","for its smaller exit can be composed with the three step lift. For example, $55,83,125,47$ is an exact odd prefix. The state 47 appears on the verified orbit from 27 in Appendix~\\ref{app:orbit27} and therefore terminates. That finite certificate proves termination of the particular lifted source 8343 when composed with \\eqref{eq:151lift}. It does not establish termination of all the other exits $1+54s$.","\\begin{example}[An entire progression modulo 4096]\\label{ex:1087}\nFor $n=4096t+1087$, with integer $t\\ge0$, six exact applications of $A$ give\n\\begin{align*}\n4096t+1087&\\longmapsto6144t+1631\\longmapsto9216t+2447","\\begin{example}[An entire progression modulo 4096]\\label{ex:1087}\nFor $n=4096t+1087$, with integer $t\\ge0$, six exact applications of $A$ give\n\\begin{align*}\n4096t+1087&\\longmapsto6144t+1631\\longmapsto9216t+2447\\\\\n&\\longmapsto13824t+3671\\longmapsto20736t+5507\\\\\n&\\longmapsto31104t+8261\\longmapsto5832t+1549.\n\\end{align*}\nThe valuations are $1,1,1,1,1,4$. The next numerator is\n\\[\n3(5832t+1549)+1=2^3(2187t+581).\n\\]\nTherefore\n\\[\nA^7(n)=\\operatorname{oddPart}(2187t+581)\n\\le2187t+581<4096t+1087.\n\\]\nThe final valuation may exceed 3 for some parameters. The inequality remains valid because removing further factors of two only reduces the positive quotient. This proof is represented in the recovered deep residue certificate source.\n\\end{example}","&\\longmapsto31104t+8261\\longmapsto5832t+1549.\n\\end{align*}\nThe valuations are $1,1,1,1,1,4$. The next numerator is\n\\[\n3(5832t+1549)+1=2^3(2187t+581).\n\\]\nTherefore\n\\[\nA^7(n)=\\operatorname{oddPart}(2187t+581)\n\\le2187t+581<4096t+1087.\n\\]\nThe final valuation may exceed 3 for some parameters. The inequality remains valid because removing further factors of two only reduces the positive quotient. This proof is represented in the recovered deep residue certificate source.\n\\end{example}","\\section{Permanent coefficient noncrossing}\nThe next result concerns a particular hypothetical global regime. Its hypothesis is stronger than absence of actual descent. Throughout this section $x_r=A^r(n)$, and the valuations and depths come from the actual orbit.","\\begin{definition}[Coefficient ratio and deficit]\\label{def:noncross}\nDefine\n\\begin{equation}\\label{eq:rho}\n\\rho_r=\\frac{2^{S_r}}{3^r},\\qquad\nd_r=\\lfloor r\\log_2 3\\rfloor-S_r.\n\\end{equation}\nPermanent coefficient noncrossing means $\\rho_r\\le1$ for every $r\\ge0$. Under that hypothesis $d_r$ is a nonnegative integer.\n\\end{definition}","The reciprocal $1/\\rho_r$ is the homogeneous coefficient of the prefix transfer. The deficit records distance below the critical dyadic depth. Outside the stated hypothesis, the displayed subtraction can be negative. The historical Lean deficit is a natural number subtraction, which truncates there. We use \\eqref{eq:rho} only in its noncrossing domain.","\\begin{lemma}[Normalized offset and height]\\label{lem:height}\nUnder permanent noncrossing,\n\\begin{equation}\\label{eq:normalized}\n\\rho_r x_r=n+\\frac13\\sum_{i<r}\\rho_i\\le n+\\frac r3.\n\\end{equation}\nAt every time with $d_r\\le D$, where $D\\ge0$ is integral,\n\\begin{equation}\\label{eq:height}\nx_r<2^{D+1}\\left(n+\\frac r3\\right).\n\\end{equation}\n\\end{lemma}\n\\begin{proof}\nDivide the affine law by $3^r$ and use the offset sum. Noncrossing gives $\\rho_i\\le1$ for every preceding time, which is essential for the upper bound. From $d_r\\le D$ we have $S_r\\ge\\lfloor r\\log_2 3\\rfloor-D>r\\log_2 3-D-1$. Therefore $\\rho_r>2^{-D-1}$, proving the height estimate.\n\\end{proof}","ght]\\label{lem:height}\nUnder permanent noncrossing,\n\\begin{equation}\\label{eq:normalized}\n\\rho_r x_r=n+\\frac13\\sum_{i<r}\\rho_i\\le n+\\frac r3.\n\\end{equation}\nAt every time with $d_r\\le D$, where $D\\ge0$ is integral,\n\\begin{equation}\\label{eq:height}\nx_r<2^{D+1}\\left(n+\\frac r3\\right).\n\\end{equation}\n\\end{lemma}\n\\begin{proof}\nDivide the affine law by $3^r$ and use the offset sum. Noncrossing gives $\\rho_i\\le1$ for every preceding time, which is essential for the upper bound. From $d_r\\le D$ we have $S_r\\ge\\lfloor r\\log_2 3\\rfloor-D>r\\log_2 3-D-1$. Therefore $\\rho_r>2^{-D-1}$, proving the height estimate.\n\\end{proof}","ized}\n\\rho_r x_r=n+\\frac13\\sum_{i<r}\\rho_i\\le n+\\frac r3.\n\\end{equation}\nAt every time with $d_r\\le D$, where $D\\ge0$ is integral,\n\\begin{equation}\\label{eq:height}\nx_r<2^{D+1}\\left(n+\\frac r3\\right).\n\\end{equation}\n\\end{lemma}\n\\begin{proof}\nDivide the affine law by $3^r$ and use the offset sum. Noncrossing gives $\\rho_i\\le1$ for every preceding time, which is essential for the upper bound. From $d_r\\le D$ we have $S_r\\ge\\lfloor r\\log_2 3\\rfloor-D>r\\log_2 3-D-1$. Therefore $\\rho_r>2^{-D-1}$, proving the height estimate.\n\\end{proof}","\\frac13\\sum_{i<r}\\rho_i\\le n+\\frac r3.\n\\end{equation}\nAt every time with $d_r\\le D$, where $D\\ge0$ is integral,\n\\begin{equation}\\label{eq:height}\nx_r<2^{D+1}\\left(n+\\frac r3\\right).\n\\end{equation}\n\\end{lemma}\n\\begin{proof}\nDivide the affine law by $3^r$ and use the offset sum. Noncrossing gives $\\rho_i\\le1$ for every preceding time, which is essential for the upper bound. From $d_r\\le D$ we have $S_r\\ge\\lfloor r\\log_2 3\\rfloor-D>r\\log_2 3-D-1$. Therefore $\\rho_r>2^{-D-1}$, proving the height estimate.\n\\end{proof}","ry time with $d_r\\le D$, where $D\\ge0$ is integral,\n\\begin{equation}\\label{eq:height}\nx_r<2^{D+1}\\left(n+\\frac r3\\right).\n\\end{equation}\n\\end{lemma}\n\\begin{proof}\nDivide the affine law by $3^r$ and use the offset sum. Noncrossing gives $\\rho_i\\le1$ for every preceding time, which is essential for the upper bound. From $d_r\\le D$ we have $S_r\\ge\\lfloor r\\log_2 3\\rfloor-D>r\\log_2 3-D-1$. Therefore $\\rho_r>2^{-D-1}$, proving the height estimate.\n\\end{proof}","\\begin{lemma}[Injectivity]\\label{lem:injective}\nA permanently noncrossing positive odd orbit never repeats a state.\n\\end{lemma}\n\\begin{proof}\nSuppose $x_u=x_v$ with $v>u$. Put $k=v-u$ and $B=S_v-S_u$. The affine law for this positive length block gives\n\\[\n(2^B-3^k)x_u=C_{u,v}>0.\n\\]\nHence $2^B/3^k>1$. Determinism makes the same states and valuations repeat, so\n$\\rho_{u+hk}=\\rho_u(2^B/3^k)^h$. This exceeds 1 for sufficiently large $h$, contrary to permanent noncrossing.\n\\end{proof}","em:injective}\nA permanently noncrossing positive odd orbit never repeats a state.\n\\end{lemma}\n\\begin{proof}\nSuppose $x_u=x_v$ with $v>u$. Put $k=v-u$ and $B=S_v-S_u$. The affine law for this positive length block gives\n\\[\n(2^B-3^k)x_u=C_{u,v}>0.\n\\]\nHence $2^B/3^k>1$. Determinism makes the same states and valuations repeat, so\n$\\rho_{u+hk}=\\rho_u(2^B/3^k)^h$. This exceeds 1 for sufficiently large $h$, contrary to permanent noncrossing.\n\\end{proof}","\\begin{theorem}[No eventually bounded noncrossing deficit]\\label{thm:boundeddeficit}\nFor a positive odd actual orbit with $\\rho_r\\le1$ at every time, the deficit is unbounded on every tail. More precisely, for every pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","entually bounded noncrossing deficit]\\label{thm:boundeddeficit}\nFor a positive odd actual orbit with $\\rho_r\\le1$ at every time, the deficit is unbounded on every tail. More precisely, for every pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","ddeficit}\nFor a positive odd actual orbit with $\\rho_r\\le1$ at every time, the deficit is unbounded on every tail. More precisely, for every pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","e odd actual orbit with $\\rho_r\\le1$ at every time, the deficit is unbounded on every tail. More precisely, for every pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","o_r\\le1$ at every time, the deficit is unbounded on every tail. More precisely, for every pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","nbounded on every tail. More precisely, for every pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","l. More precisely, for every pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","very pair of integers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","gers $D,N\\ge0$ there exists $r\\ge N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","e N$ with\n\\begin{equation}\\label{eq:unboundeddeficit}\n2^{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","{D+1}2^{S_r}\\le3^r.\n\\end{equation}\n\\end{theorem}\n\\begin{proof}\nAssume instead that $d_r\\le D$ for every $r\\ge N$. At any such starting time, a local block of $k$ valuations has sum $s=S_{r+k}-S_r$ satisfying\n\\[\ns<k\\log_2 3+D+1.\n\\]\nThis follows from $\\rho_{r+k}\\le1$ and $\\rho_r>2^{-D-1}$. Take $k=5\\ell$, where $\\ell\\ge1$ is an integer. Since $3^5<2^8$, every local word has total valuation at most $h=8\\ell+D+1$.","Let $E_\\ell$ be the union of integer cylinders for all length $5\\ell$ words of total valuation at most $h$. Every state $x_r$ with $r\\ge N$ belongs to this finite union. A word of total $s$ has one residue class modulo $2^{s+1}$ by Theorem~\\ref{thm:cylinder}. Its number of positive representatives at most $H$ is bounded by $H/2^{s+1}+1$.","There are $\\binom{s-1}{5\\ell-1}$ positive words of total $s$. If $W_\\ell$ denotes the finite total number of selected words, summing the cylinder counts gives\n\\[\n\\#(E_\\ell\\cap[1,H])\\le\\varepsilon_\\ell H+W_\\ell,\n\\qquad\n\\varepsilon_\\ell=\\frac12\\sum_{s=5\\ell}^{h}\n\\binom{s-1}{5\\ell-1}2^{-s}.\n\\]\nThis count is arithmetic. It does not assume that orbit valuations are independent random variables.","Set $z=3/4$. For $s\\le h$, $2^{-s}\\le z^{-h}(z/2)^s$. Summing over all positive words bounds the truncated sum by\n\\[\n\\varepsilon_\\ell\\le\\frac12z^{-h}\n\\left(\\sum_{a\\ge1}(z/2)^a\\right)^{5\\ell}\n=\\frac12\\left(\\frac43\\right)^{D+1}\n\\left(\\frac{65536}{84375}\\right)^\\ell.\n\\]\nThe exact rational factor is less than 1, so $\\varepsilon_\\ell$ tends to zero.","For $R>N$, Lemmas~\\ref{lem:height} and \\ref{lem:injective} place the $R-N$ distinct states $x_N,\\ldots,x_{R-1}$ in $E_\\ell$ below\n$H_R=\\lceil2^{D+1}(n+R/3)\\rceil$. Thus\n\\[\nR-N\\le\\varepsilon_\\ell H_R+W_\\ell.\n\\]\nFix $\\ell$ and let $R$ tend to infinity. Division by $R$ yields\n$1\\le\\varepsilon_\\ell2^{D+1}/3$. Choosing $\\ell$ sufficiently large contradicts this inequality. No eventual bounded deficit strip exists.","Finally, unbounded deficit on each tail supplies a time with $d_r\\ge D+1$. Since $2^{\\lfloor r\\log_2 3\\rfloor}\\le3^r$, this gives \\eqref{eq:unboundeddeficit}.\n\\end{proof}","The quantifier on the conclusion matters. Unbounded deficit on every tail does not imply $d_r\\to\\infty$. The deficit could return to small values between increasingly deep excursions. Even convergence $d_r\\to\\infty$ would only imply $\\rho_r\\to0$, which is compatible with growing positive states. The present theorem closes every bounded strip, rather than the entire noncrossing branch.","\\subsection{The two resources in the packing contradiction}\nThe proof joins a restriction on words to a restriction on heights. Their roles are easiest to distinguish by temporarily keeping all parameters visible. Fix the proposed eventual deficit bound $D$ and its starting time $N$. The global noncrossing hypothesis is used at every earlier time in the normalized offset estimate. The eventual bound is used at the starting time of each local word. It supplies $\\rho_r>2^{-D-1}$ for $r\\ge N$. Those two facts give a height envelope for every selected tail state.","roof joins a restriction on words to a restriction on heights. Their roles are easiest to distinguish by temporarily keeping all parameters visible. Fix the proposed eventual deficit bound $D$ and its starting time $N$. The global noncrossing hypothesis is used at every earlier time in the normalized offset estimate. The eventual bound is used at the starting time of each local word. It supplies $\\rho_r>2^{-D-1}$ for $r\\ge N$. Those two facts give a height envelope for every selected tail state.","on on words to a restriction on heights. Their roles are easiest to distinguish by temporarily keeping all parameters visible. Fix the proposed eventual deficit bound $D$ and its starting time $N$. The global noncrossing hypothesis is used at every earlier time in the normalized offset estimate. The eventual bound is used at the starting time of each local word. It supplies $\\rho_r>2^{-D-1}$ for $r\\ge N$. Those two facts give a height envelope for every selected tail state.","For a local block beginning at $r$, its valuation sum $s$ satisfies\n\\[\n\\frac{2^s}{3^k}=\\frac{\\rho_{r+k}}{\\rho_r}<2^{D+1}.\n\\]\nThis inequality is the source of the word restriction. Taking logarithms gives $s<k\\log_2 3+D+1$. No independence of valuations is used. Both ratios come from one actual orbit, and the restriction holds at each of its tail starting times.","The block length $5\\ell$ is chosen because the elementary comparison $3^5=243<256=2^8$ leaves a small positive dyadic margin. The bound $s\\le8\\ell+D+1$ is deliberately a little loose. Its virtue is that it keeps the argument integral and gives a clean geometric decay after weighting words. The number of positive words of length $k$ and total $s$ is $\\binom{s-1}{k-1}$: place $k-1$ separators in the $s-1$ gaps between $s$ unit marks. Each resulting run of marks gives one positive valuation. This is a count of words, before any orbit is selected.","e block length $5\\ell$ is chosen because the elementary comparison $3^5=243<256=2^8$ leaves a small positive dyadic margin. The bound $s\\le8\\ell+D+1$ is deliberately a little loose. Its virtue is that it keeps the argument integral and gives a clean geometric decay after weighting words. The number of positive words of length $k$ and total $s$ is $\\binom{s-1}{k-1}$: place $k-1$ separators in the $s-1$ gaps between $s$ unit marks. Each resulting run of marks gives one positive valuation. This is a count of words, before any orbit is selected.","ll$ is chosen because the elementary comparison $3^5=243<256=2^8$ leaves a small positive dyadic margin. The bound $s\\le8\\ell+D+1$ is deliberately a little loose. Its virtue is that it keeps the argument integral and gives a clean geometric decay after weighting words. The number of positive words of length $k$ and total $s$ is $\\binom{s-1}{k-1}$: place $k-1$ separators in the $s-1$ gaps between $s$ unit marks. Each resulting run of marks gives one positive valuation. This is a count of words, before any orbit is selected.","For a modulus $M$, any one residue class contributes at most $H/M+1$ positive integers up to height $H$. The $+1$ accounts for an incomplete terminal period. Applying that estimate separately to every selected word yields the finite term $W_\\ell$. One may write it exactly as\n\\[\nW_\\ell=\\sum_{s=5\\ell}^{8\\ell+D+1}\\binom{s-1}{5\\ell-1}\n=\\binom{8\\ell+D+1}{5\\ell}.\n\\]\nIt can be enormous. Its size does not invalidate the argument, because $\\ell$ is fixed before the number of orbit states tends to infinity. A growing finite constant is not being silently discarded while its defining parameter changes.","class contributes at most $H/M+1$ positive integers up to height $H$. The $+1$ accounts for an incomplete terminal period. Applying that estimate separately to every selected word yields the finite term $W_\\ell$. One may write it exactly as\n\\[\nW_\\ell=\\sum_{s=5\\ell}^{8\\ell+D+1}\\binom{s-1}{5\\ell-1}\n=\\binom{8\\ell+D+1}{5\\ell}.\n\\]\nIt can be enormous. Its size does not invalidate the argument, because $\\ell$ is fixed before the number of orbit states tends to infinity. A growing finite constant is not being silently discarded while its defining parameter changes.","The weight $z=3/4$ tilts the word sum toward the low total valuations permitted by the hypothetical strip. Since $0<z<1$ and $s\\le h$, one has $z^{-s}\\le z^{-h}$. The identity\n\\[\n\\sum_{a\\ge1}(z/2)^a=\\frac{z}{2-z}=\\frac35\n\\]\nthen accounts for all possible positive valuations in each of the $5\\ell$ positions. Combining this factor with $z^{-8\\ell}$ gives\n\\[\n\\left(\\frac43\\right)^{8\\ell}\\left(\\frac35\\right)^{5\\ell}\n=\\left(\\frac{65536}{84375}\\right)^\\ell.\n\\]\nThis exact ratio is the source of rarity. It is a bound for a finite union of residue classes, expressed through an infinite convergent generating sum. It is not an empirical frequency measured on an orbit.","The contradiction can also be written without a limit. Set $B=2^{D+1}$ and choose $\\ell$ so that $\\alpha=\\varepsilon_\\ell B/3<1$. Since $H_R\\le B(n+R/3)+1$, the packing inequality would force\n\\begin{equation}\\label{eq:finitepacking}\n(1-\\alpha)R\\le N+\\varepsilon_\\ell(Bn+1)+W_\\ell\n\\end{equation}\nfor every integer $R>N$. The right side is fixed after $D,N,n,\\ell$ are fixed. Selecting an integer $R$ above that finite bound contradicts \\eqref{eq:finitepacking}. The limit in the theorem's proof is a compact way of expressing the same comparison.","For example, a hypothetical eventual bound $D=3$ gives $B=16$. Choosing $\\ell=9$ allows block length 45 and valuation cap 76. The displayed bound on $\\varepsilon_\\ell$ gives $\\alpha<0.868$, and the exact rational comparison is below 1. There is no claim that the resulting finite threshold is small. The proof needs its existence, because an infinite injective orbit supplies arbitrarily many distinct states.","\\subsection{What the proof does not transfer to an unbounded strip}\nIf the deficit bound is replaced by a varying number $D_R$, the height multiplier $2^{D_R+1}$ varies with the same orbit window. A smaller cylinder density may be offset by a much larger permitted height. The finite family correction may also grow if its block length is allowed to vary. The preceding proof gives no uniform comparison between those competing quantities.","The deficit itself can be handled entirely with integers. Let $H_r$ be the least nonnegative integer $h$ such that $3^r<2^h$. Then $\\lfloor r\\log_2 3\\rfloor=H_r-1$, including $r=0$. For $r>0$, equality of a positive power of 3 with a power of 2 is impossible, so the strict threshold identifies the floor exactly. The displayed logarithm is a convenient notation for this integer comparison. A certificate implementation need not approximate an irrational logarithm to evaluate the deficit.","\\section{The global obligations and the failed shortcuts}\nThere are two coefficient cases for an infinite source floor candidate. If some $\\rho_r>1$, it has a contracting prefix. A successful argument must still overcome the offset and prove eventual descent below the original seed. If $\\rho_r\\le1$ for every time, the preceding theorem supplies injectivity and unbounded deficit, but a further arithmetic contradiction is needed. These are the main unresolved branches in the corrected gap register \\cite{projectcollatzgap}.","\\begin{conjecture}[Global packet descent]\\label{conj:packetdescent}\nFor every positive odd $n>1$, at least one actual prefix satisfies\n\\[\n\\left(2^{S_r}-3^r\\right)n>C_r.\n\\]\n\\end{conjecture}\nBy Theorems~\\ref{thm:descent} and \\ref{thm:induction}, this statement is exactly sufficient to prove Collatz. A completed certificate system must force such a prefix or provide a logically equivalent global argument. Its soundness alone is insufficient. Soundness says that a supplied certificate proves what it states. Completeness would say that every relevant source eventually supplies the needed certificate.","The historical half contraction lane studies full source floor packets with homogeneous coefficient at most $1/2$. It restricts a least possible failure to a particular primitive annulus and supplies finite correction rules. The general exclusion remains open. Even that auxiliary exclusion would not automatically settle the whole range $1/2<\\lambda<1$ or the permanent noncrossing branch \\cite{projectcollatzaudit,projectcollatzgap}.","The bounded deficit proof passes that last gate. It names the common height $H_R$, places $R-N$ distinct states below it, and proves a smaller admissible capacity. Once bounded deficit is removed, its linear height estimate is no longer available. Reusing only the cylinder rarity would omit the decisive half of the argument.","\\subsection{Soundness, search, and adaptive completeness}\nA finite descent certificate for a positive odd source $n>1$ can consist of a length $k\\ge1$ and a positive valuation word $w$. Its verifier computes $S_k,C_k$, sets $G=2^{S_k}-3^k$, checks the exact cylinder congruence, and checks $Gn>C_k$. All of these are finite integer operations. The cylinder theorem proves that the supplied word is realized. The descent theorem then proves that its endpoint is lower. This is soundness of the verifier.","Producing the certificate is a different task. One can follow the actual orbit from $n$ and stop the first time it becomes smaller than $n$. If that event occurs, the recorded valuations form a correct certificate. The assertion that this search halts for every positive odd $n>1$ is exactly the universal descent statement. Describing the search therefore isolates the open theorem. It does not establish that theorem by giving an executable loop.","\\begin{proposition}[A same word merger obstruction]\\label{prop:merger}\nSuppose two positive sources $n>m$ realize the same affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","oposition}[A same word merger obstruction]\\label{prop:merger}\nSuppose two positive sources $n>m$ realize the same affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","[A same word merger obstruction]\\label{prop:merger}\nSuppose two positive sources $n>m$ realize the same affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","merger obstruction]\\label{prop:merger}\nSuppose two positive sources $n>m$ realize the same affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","Suppose two positive sources $n>m$ realize the same affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","ive sources $n>m$ realize the same affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","s $n>m$ realize the same affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","ame affine word with data $(k,S,C)$ and endpoints $y>x$. Then $n-m$ is a positive multiple of $2^S$. In particular, if $2^S\\ge n$, no smaller positive source can share that word.\n\\end{proposition}\n\\begin{proof}\nSubtract the endpoint equations to obtain $3^k(n-m)=2^S(y-x)$. Coprimality of $3^k$ and $2^S$ forces $2^S\\mid n-m$. A positive multiple of $2^S$ is at least $2^S$, so $m=n-(n-m)\\le0$ when $2^S\\ge n$.\n\\end{proof}","This is the obstruction behind pulling a smaller endpoint backward through an unchanged late prefix. A useful smaller source merger must eventually change the word. For two different transfers ending at the same $y$, elimination gives\n\\[\nm=\\frac{2^{S'}3^k n+2^{S'}C-2^S C'}{2^S3^{k'}}.\n\\]\nThe remaining task is to force a realizing alternative with $0<m<n$. The formula does not itself supply such an alternative. The saved route notes develop this direction while retaining its unproved universal existence step \\cite{projectcollatzroutes}.","Cycles have their own exact requirement. A positive cycle represented by a nonempty odd word must satisfy\n\\[\n(2^S-3^k)n=C_k>0.\n\\]\nThe positive gap is therefore necessary, but its size alone does not eliminate the divisibility equation. The ordered offset must also be controlled. Repeating a contracting cycle reproduces this equality. An argument about an expanding word cannot be applied to it without changing the hypotheses.","\\appendix\n\\section{Complete odd coordinates for the source 27}\\label{app:orbit27}\nTable~\\ref{tab:27complete} records the entire accelerated orbit through its first arrival at 1. The column $a$ is the valuation used to arrive at the state in that row. The column $S$ is both cumulative dyadic depth and cumulative shortcut time. The column $L=S+r$ is cumulative classical time. The offset is computed by $C_{r+1}=3C_r+2^{S_r}$, starting from zero. Each row satisfies the exact identity $2^Sx=3^r\\cdot27+C$.","endix\n\\section{Complete odd coordinates for the source 27}\\label{app:orbit27}\nTable~\\ref{tab:27complete} records the entire accelerated orbit through its first arrival at 1. The column $a$ is the valuation used to arrive at the state in that row. The column $S$ is both cumulative dyadic depth and cumulative shortcut time. The column $L=S+r$ is cumulative classical time. The offset is computed by $C_{r+1}=3C_r+2^{S_r}$, starting from zero. Each row satisfies the exact identity $2^Sx=3^r\\cdot27+C$.","e~\\ref{tab:27complete} records the entire accelerated orbit through its first arrival at 1. The column $a$ is the valuation used to arrive at the state in that row. The column $S$ is both cumulative dyadic depth and cumulative shortcut time. The column $L=S+r$ is cumulative classical time. The offset is computed by $C_{r+1}=3C_r+2^{S_r}$, starting from zero. Each row satisfies the exact identity $2^Sx=3^r\\cdot27+C$.","\\toprule\n$r$&$x_r$&$a$&$S$&$L$&$C_r$","Every row before $r=37$ has $x_r\\ge27$, and row 37 has $x_r=23$. Within an acceleration, an odd classical multiplication first raises the state, and each following division by two lowers it monotonically to the odd endpoint. Thus no skipped intermediate state can lie below both odd endpoints. The first odd descent in the table is consequently also the first shortcut and classical descent. It has clocks $37,59,96$, respectively.","ow before $r=37$ has $x_r\\ge27$, and row 37 has $x_r=23$. Within an acceleration, an odd classical multiplication first raises the state, and each following division by two lowers it monotonically to the odd endpoint. Thus no skipped intermediate state can lie below both odd endpoints. The first odd descent in the table is consequently also the first shortcut and classical descent. It has clocks $37,59,96$, respectively.","The final arrival at 1 has clocks $41,70,111$. The largest odd state is 3077 at row 32. Its classical successor is $3\\cdot3077+1=9232$, the largest classical state on the path. Its shortcut successor is 4616, the largest shortcut state. These different peak values are another consequence of the different maps, rather than conflicting measurements of one clock.","This appendix is finite evidence with an unusually transparent verification rule. A reader can check any row using integer multiplication and exponentiation, and can check consecutive rows using $3x+1=2^a x'$. The large offsets are retained because deleting them would obscure the exact descent budget. Their size has no bearing on a universal theorem beyond this certified source.","\\item[Odd state $x_r$]\nA positive odd integer obtained by $r$ actual applications of $A$. A rational candidate from an affine formula becomes an odd state only after realization is proved.\n\\item[Valuation $a_r$]\nThe exact number of factors of two in $3x_r+1$. A proposed positive exponent is a word symbol. Equality with the actual valuation is an admissibility claim.\n\\item[Word and packet]\nA word is an ordered finite sequence of positive valuations. A realized packet retains that word together with its source cylinder and affine transfer. Equal length and equal total depth do not imply equal packets.\n\\item[Depth $S_r$ and offset $C_r$]\nThe depth sums the removed dyadic powers. The offset sums the transported add","\\item[Odd state $x_r$]\nA positive odd integer obtained by $r$ actual applications of $A$. A rational candidate from an affine formula becomes an odd state only after realization is proved.\n\\item[Valuation $a_r$]\nThe exact number of factors of two in $3x_r+1$. A proposed positive exponent is a word symbol. Equality with the actual valuation is an admissibility claim.\n\\item[Word and packet]\nA word is an ordered finite sequence of positive valuations. A realized packet retains that word together with its source cylinder and affine transfer. Equal length and equal total depth do not imply equal packets.\n\\item[Depth $S_r$ and offset $C_r$]\nThe depth sums the removed dyadic powers. The offset sums the transported additions. The first determines the multipl","\\item[Odd state $x_r$]\nA positive odd integer obtained by $r$ actual applications of $A$. A rational candidate from an affine formula becomes an odd state only after realization is proved.\n\\item[Valuation $a_r$]\nThe exact number of factors of two in $3x_r+1$. A proposed positive exponent is a word symbol. Equality with the actual valuation is an admissibility claim.\n\\item[Word and packet]\nA word is an ordered finite sequence of positive valuations. A realized packet retains that word together with its source cylinder and affine transfer. Equal length and equal total depth do not imply equal packets.\n\\item[Depth $S_r$ and offset $C_r$]\nThe depth sums the removed dyadic powers. The offset sums the transported additions. The first determines the multiplier, while both determine t","\\item[Valuation $a_r$]\nThe exact number of factors of two in $3x_r+1$. A proposed positive exponent is a word symbol. Equality with the actual valuation is an admissibility claim.\n\\item[Word and packet]\nA word is an ordered finite sequence of positive valuations. A realized packet retains that word together with its source cylinder and affine transfer. Equal length and equal total depth do not imply equal packets.\n\\item[Depth $S_r$ and offset $C_r$]\nThe depth sums the removed dyadic powers. The offset sums the transported additions. The first determines the multiplier, while both determine the endpoint.\n\\item[Cylinder]\nThe exact odd residue class modulo $2^{S+1}$ that realizes a word. It is infinite as an arith","\\item[Valuation $a_r$]\nThe exact number of factors of two in $3x_r+1$. A proposed positive exponent is a word symbol. Equality with the actual valuation is an admissibility claim.\n\\item[Word and packet]\nA word is an ordered finite sequence of positive valuations. A realized packet retains that word together with its source cylinder and affine transfer. Equal length and equal total depth do not imply equal packets.\n\\item[Depth $S_r$ and offset $C_r$]\nThe depth sums the removed dyadic powers. The offset sums the transported additions. The first determines the multiplier, while both determine the endpoint.\n\\item[Cylinder]\nThe exact odd residue class modulo $2^{S+1}$ that realizes a word. It is infinite as an arithmetic progression, even when a chosen finite tes","\\item[Depth $S_r$ and offset $C_r$]\nThe depth sums the removed dyadic powers. The offset sums the transported additions. The first determines the multiplier, while both determine the endpoint.\n\\item[Cylinder]\nThe exact odd residue class modulo $2^{S+1}$ that realizes a word. It is infinite as an arithmetic progression, even when a chosen finite test interval misses it.\n\\item[Coefficient crossing]\nThe strict inequality $2^{S_r}>3^r$. Equivalently the homogeneous transfer coefficient is less than one. This is a necessary coefficient condition for descent of that prefix.\n\\item[Actual descent]\nAn endpoint below the same original source. It requires $(2^{S_r}-3^r)n>C_r$, so it includes the offset budget and the","\\item[Depth $S_r$ and offset $C_r$]\nThe depth sums the removed dyadic powers. The offset sums the transported additions. The first determines the multiplier, while both determine the endpoint.\n\\item[Cylinder]\nThe exact odd residue class modulo $2^{S+1}$ that realizes a word. It is infinite as an arithmetic progression, even when a chosen finite test interval misses it.\n\\item[Coefficient crossing]\nThe strict inequality $2^{S_r}>3^r$. Equivalently the homogeneous transfer coefficient is less than one. This is a necessary coefficient condition for descent of that prefix.\n\\item[Actual descent]\nAn endpoint below the same original source. It requires $(2^{S_r}-3^r)n>C_r$, so it includes the offset budget and the source value.\n\\it","\\item[Cylinder]\nThe exact odd residue class modulo $2^{S+1}$ that realizes a word. It is infinite as an arithmetic progression, even when a chosen finite test interval misses it.\n\\item[Coefficient crossing]\nThe strict inequality $2^{S_r}>3^r$. Equivalently the homogeneous transfer coefficient is less than one. This is a necessary coefficient condition for descent of that prefix.\n\\item[Actual descent]\nAn endpoint below the same original source. It requires $(2^{S_r}-3^r)n>C_r$, so it includes the offset budget and the source value.\n\\item[Source floor]\nThe condition that all relevant states remain at least the original source. It differs from staying above a later local minimum or falling below a local peak.\n\\item[Permanent coefficient noncrossing]\nTh","\\item[Coefficient crossing]\nThe strict inequality $2^{S_r}>3^r$. Equivalently the homogeneous transfer coefficient is less than one. This is a necessary coefficient condition for descent of that prefix.\n\\item[Actual descent]\nAn endpoint below the same original source. It requires $(2^{S_r}-3^r)n>C_r$, so it includes the offset budget and the source value.\n\\item[Source floor]\nThe condition that all relevant states remain at least the original source. It differs from staying above a later local minimum or falling below a local peak.\n\\item[Permanent coefficient noncrossing]\nThe condition $\\rho_r\\le1$ for every time. It implies every homogeneous prefix coefficient is at least one. It is stronger than permanent absence of actual descent.\n\\item[Deficit $d_r$","\\item[Actual descent]\nAn endpoint below the same original source. It requires $(2^{S_r}-3^r)n>C_r$, so it includes the offset budget and the source value.\n\\item[Source floor]\nThe condition that all relevant states remain at least the original source. It differs from staying above a later local minimum or falling below a local peak.\n\\item[Permanent coefficient noncrossing]\nThe condition $\\rho_r\\le1$ for every time. It implies every homogeneous prefix coefficient is at least one. It is stronger than permanent absence of actual descent.\n\\item[Deficit $d_r$]\nThe integral gap between the critical dyadic depth and the actual depth in the noncrossing domain. A bounded deficit strip yields a linear height bound. An unbounded deficit does not supply that bound.\n\\item[Sound certificate and comple","\\item[Permanent coefficient noncrossing]\nThe condition $\\rho_r\\le1$ for every time. It implies every homogeneous prefix coefficient is at least one. It is stronger than permanent absence of actual descent.\n\\item[Deficit $d_r$]\nThe integral gap between the critical dyadic depth and the actual depth in the noncrossing domain. A bounded deficit strip yields a linear height bound. An unbounded deficit does not supply that bound.\n\\item[Sound certificate and complete certificate system]\nA sound certificate proves its claimed conclusion whenever its checks pass. A complete system supplies a successful certificate for every source in its stated domain. The latter quantifier is the open global obligation here.\n\\end{description}","\\item[Deficit $d_r$]\nThe integral gap between the critical dyadic depth and the actual depth in the noncrossing domain. A bounded deficit strip yields a linear height bound. An unbounded deficit does not supply that bound.\n\\item[Sound certificate and complete certificate system]\nA sound certificate proves its claimed conclusion whenever its checks pass. A complete system supplies a successful certificate for every source in its stated domain. The latter quantifier is the open global obligation here.\n\\end{description}","The long run results use a direct orbit identity for $2^{j+1}q-1$. The terminating seed family adds an elementary divisibility induction. The progression lifts add a base descent and a slope comparison. None of those steps assumes a global density theorem or an eventual probabilistic law for valuations.","For a complex number \\(s\\) with real part greater than one, the zeta function is\n\\begin{equation}\\label{rh:zeta}\n\\zeta(s)=\\sum_{n=1}^{\\infty}n^{-s}\n=\\prod_{p}(1-p^{-s})^{-1},\n\\end{equation}\nwhere \\(n\\) ranges over positive integers and \\(p\\) ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at \\(s=1\\). The completed function\n\\begin{equation}\\label{rh:xi}\n\\xi(s)=\\frac12s(s-1)\\pi^{-s/2}\\Gamma(s/2)\\zeta(s)\n\\end{equation}\nis entire and satisfies \\(\\xi(s)=\\xi(1-s)\\). Here \\(\\Gamma\\) is the gamma function. These are classical facts, with the normalization in \\eqref{rh:xi} fixed throughout this article \\cite{DLMFZeta}.","x number \\(s\\) with real part greater than one, the zeta function is\n\\begin{equation}\\label{rh:zeta}\n\\zeta(s)=\\sum_{n=1}^{\\infty}n^{-s}\n=\\prod_{p}(1-p^{-s})^{-1},\n\\end{equation}\nwhere \\(n\\) ranges over positive integers and \\(p\\) ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at \\(s=1\\). The completed function\n\\begin{equation}\\label{rh:xi}\n\\xi(s)=\\frac12s(s-1)\\pi^{-s/2}\\Gamma(s/2)\\zeta(s)\n\\end{equation}\nis entire and satisfies \\(\\xi(s)=\\xi(1-s)\\). Here \\(\\Gamma\\) is the gamma function. These are classical facts, with the normalization in \\eqref{rh:xi} fixed throughout this article \\cite{DLMFZeta}.","a(s)=\\sum_{n=1}^{\\infty}n^{-s}\n=\\prod_{p}(1-p^{-s})^{-1},\n\\end{equation}\nwhere \\(n\\) ranges over positive integers and \\(p\\) ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at \\(s=1\\). The completed function\n\\begin{equation}\\label{rh:xi}\n\\xi(s)=\\frac12s(s-1)\\pi^{-s/2}\\Gamma(s/2)\\zeta(s)\n\\end{equation}\nis entire and satisfies \\(\\xi(s)=\\xi(1-s)\\). Here \\(\\Gamma\\) is the gamma function. These are classical facts, with the normalization in \\eqref{rh:xi} fixed throughout this article \\cite{DLMFZeta}.","{-s}\n=\\prod_{p}(1-p^{-s})^{-1},\n\\end{equation}\nwhere \\(n\\) ranges over positive integers and \\(p\\) ranges over primes. Analytic continuation extends it to a meromorphic function on the complex plane with a simple pole at \\(s=1\\). The completed function\n\\begin{equation}\\label{rh:xi}\n\\xi(s)=\\frac12s(s-1)\\pi^{-s/2}\\Gamma(s/2)\\zeta(s)\n\\end{equation}\nis entire and satisfies \\(\\xi(s)=\\xi(1-s)\\). Here \\(\\Gamma\\) is the gamma function. These are classical facts, with the normalization in \\eqref{rh:xi} fixed throughout this article \\cite{DLMFZeta}.","Let \\(\\mathcal Z\\) be the multiset of nontrivial zeros, with every zero repeated according to its multiplicity. They lie in the strip \\(0<\\Re s<1\\). Conjugation and the functional equation imply that a zero \\(\\rho\\) is accompanied by \\(\\overline\\rho\\), \\(1-\\rho\\), and \\(1-\\overline\\rho\\), allowing coincidences on the critical line \\cite{DLMFZeros}.","\\begin{conjecture}[Riemann hypothesis]\\label{rh:conjecture}\nEvery \\(\\rho\\in\\mathcal Z\\) satisfies \\(\\Re\\rho=1/2\\).\n\\end{conjecture}","Write \\(\\rho=\\beta+i\\gamma\\), with real numbers \\(\\beta\\) and \\(\\gamma\\), and define its centered coordinate\n\\begin{equation}\\label{rh:centered}\nz_\\rho=\\rho-\\frac12.\n\\end{equation}\nThe symmetries become \\(z\\mapsto-z\\) and \\(z\\mapsto\\overline z\\). Under the conjecture, every centered coordinate is purely imaginary. Away from the line, the real part of \\(z\\) becomes an exponential growth parameter in the packet below.","Continuation and completion do different jobs. Continuation lets the same analytic function be studied outside its first convergence region. Completion packages the factors needed for reflection about the critical line. The factor \\(s(s-1)\\) removes the apparent exceptional behavior at zero and one, while the gamma factor balances the trivial zeros at the negative even integers. The resulting entire function has the nontrivial zeros as its zero set. This is why a plot or packet built from \\(\\xi\\) can focus on the conjecture without also carrying the pole of \\(\\zeta\\).","tudied outside its first convergence region. Completion packages the factors needed for reflection about the critical line. The factor \\(s(s-1)\\) removes the apparent exceptional behavior at zero and one, while the gamma factor balances the trivial zeros at the negative even integers. The resulting entire function has the nontrivial zeros as its zero set. This is why a plot or packet built from \\(\\xi\\) can focus on the conjecture without also carrying the pole of \\(\\zeta\\).","actors needed for reflection about the critical line. The factor \\(s(s-1)\\) removes the apparent exceptional behavior at zero and one, while the gamma factor balances the trivial zeros at the negative even integers. The resulting entire function has the nontrivial zeros as its zero set. This is why a plot or packet built from \\(\\xi\\) can focus on the conjecture without also carrying the pole of \\(\\zeta\\).","The reflection is \\(s\\mapsto1-s\\), not ordinary complex conjugation. These two operations commute, and together they generate the symmetric zero pattern. For example, a hypothetical zero\n\\[\n\\rho=\\frac12+\\delta+i\\gamma,\\qquad \\delta>0,\n\\]\nwould bring centered coordinates\n\\[\n\\delta+i\\gamma,\\quad \\delta-i\\gamma,\\quad\n-\\delta+i\\gamma,\\quad-\\delta-i\\gamma.\n\\]\nThe quartet may collapse when \\(\\delta=0\\) or \\(\\gamma=0\\). We sum the actual multiset, so coincidences are handled by multiplicity rather than by pretending that four distinct zeros are always present.","eflection is \\(s\\mapsto1-s\\), not ordinary complex conjugation. These two operations commute, and together they generate the symmetric zero pattern. For example, a hypothetical zero\n\\[\n\\rho=\\frac12+\\delta+i\\gamma,\\qquad \\delta>0,\n\\]\nwould bring centered coordinates\n\\[\n\\delta+i\\gamma,\\quad \\delta-i\\gamma,\\quad\n-\\delta+i\\gamma,\\quad-\\delta-i\\gamma.\n\\]\nThe quartet may collapse when \\(\\delta=0\\) or \\(\\gamma=0\\). We sum the actual multiset, so coincidences are handled by multiplicity rather than by pretending that four distinct zeros are always present.","On the critical line, conjugation and reflection identify the same point:\n\\(\\overline{1/2+i\\gamma}=1-(1/2+i\\gamma)\\). Therefore \\(\\xi(1/2+i\\gamma)\\) is real. Real values on that line explain why one can track sign changes in a one dimensional plot. They do not explain why every zero must lie there. A real analytic function restricted to one line can have additional zeros elsewhere in the complex plane.","The centered coordinate separates this geometry from the notation of the completed function. Its imaginary part is an oscillation frequency. Its real part is a growth or decay rate. In the zero packet, increasing the translation \\(u\\) tests precisely that rate. The heat parameter suppresses large frequencies, which makes the sum convergent, while preserving every mode with a nonzero coefficient.","\\figplaceholder{rh_xi}{A finite evaluation of the completed function on the critical line. The lower panel multiplies the normalized value by \\(e^{\\pi t/4}\\), where \\(t\\) is the plotted height. The symmetry and normalization are exact classical facts. The plotted interval is a finite numerical illustration and supplies no conclusion about all zeros.}","\\begin{definition}[Zero heat packet]\\label{rh:heatdef}\nFor a real heat parameter \\(t>0\\) and a complex translation \\(u\\), define\n\\begin{equation}\\label{rh:theta}\n\\Theta_t(u)=\\sum_{\\rho\\in\\mathcal Z}\n\\exp\\bigl(tz_\\rho^2+uz_\\rho\\bigr).\n\\end{equation}\n\\end{definition}","The series converges normally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real \\(u\\),\n\\begin{equation}\\label{rh:maj}\n\\left|e^{tz_\\rho^2+uz_\\rho}\\right|\n\\le e^{t/4+|u|/2}e^{-t\\gamma^2}.\n\\end{equation}\nFor complex \\(u\\), there is an additional exponential factor linear in \\(|\\gamma|\\) on each compact set. Gaussian decay still dominates. The classical zero counting estimate \\(N(Y)=O(Y\\log Y)\\), where \\(N(Y)\\) counts zeros with \\(0<\\Im\\rho\\le Y\\), gives summability \\cite[Section 2.4, equation (2.11)]{BondarenkoRadchenkoSeip}. The same domination permits differentiation. Consequently\n\\begin{equation}\\label{rh:heatflow}\n\\partial_t\\Theta_t=\\partial_u^2\\Theta_t.\n\\end{equation}\nThe zero symmetries make \\(\\Theta_t(u)\\) real and even when \\(u\\) is real.","ally on compact sets of translations and on compact intervals of positive heat times. Indeed, for real \\(u\\),\n\\begin{equation}\\label{rh:maj}\n\\left|e^{tz_\\rho^2+uz_\\rho}\\right|\n\\le e^{t/4+|u|/2}e^{-t\\gamma^2}.\n\\end{equation}\nFor complex \\(u\\), there is an additional exponential factor linear in \\(|\\gamma|\\) on each compact set. Gaussian decay still dominates. The classical zero counting estimate \\(N(Y)=O(Y\\log Y)\\), where \\(N(Y)\\) counts zeros with \\(0<\\Im\\rho\\le Y\\), gives summability \\cite[Section 2.4, equation (2.11)]{BondarenkoRadchenkoSeip}. The same domination permits differentiation. Consequently\n\\begin{equation}\\label{rh:heatflow}\n\\partial_t\\Theta_t=\\partial_u^2\\Theta_t.\n\\end{equation}\nThe zero symmetries make \\(\\Theta_t(u)\\) real and even when \\(u\\) is real.","nslations and on compact intervals of positive heat times. Indeed, for real \\(u\\),\n\\begin{equation}\\label{rh:maj}\n\\left|e^{tz_\\rho^2+uz_\\rho}\\right|\n\\le e^{t/4+|u|/2}e^{-t\\gamma^2}.\n\\end{equation}\nFor complex \\(u\\), there is an additional exponential factor linear in \\(|\\gamma|\\) on each compact set. Gaussian decay still dominates. The classical zero counting estimate \\(N(Y)=O(Y\\log Y)\\), where \\(N(Y)\\) counts zeros with \\(0<\\Im\\rho\\le Y\\), gives summability \\cite[Section 2.4, equation (2.11)]{BondarenkoRadchenkoSeip}. The same domination permits differentiation. Consequently\n\\begin{equation}\\label{rh:heatflow}\n\\partial_t\\Theta_t=\\partial_u^2\\Theta_t.\n\\end{equation}\nThe zero symmetries make \\(\\Theta_t(u)\\) real and even when \\(u\\) is real.","tervals of positive heat times. Indeed, for real \\(u\\),\n\\begin{equation}\\label{rh:maj}\n\\left|e^{tz_\\rho^2+uz_\\rho}\\right|\n\\le e^{t/4+|u|/2}e^{-t\\gamma^2}.\n\\end{equation}\nFor complex \\(u\\), there is an additional exponential factor linear in \\(|\\gamma|\\) on each compact set. Gaussian decay still dominates. The classical zero counting estimate \\(N(Y)=O(Y\\log Y)\\), where \\(N(Y)\\) counts zeros with \\(0<\\Im\\rho\\le Y\\), gives summability \\cite[Section 2.4, equation (2.11)]{BondarenkoRadchenkoSeip}. The same domination permits differentiation. Consequently\n\\begin{equation}\\label{rh:heatflow}\n\\partial_t\\Theta_t=\\partial_u^2\\Theta_t.\n\\end{equation}\nThe zero symmetries make \\(\\Theta_t(u)\\) real and even when \\(u\\) is real.","r complex \\(u\\), there is an additional exponential factor linear in \\(|\\gamma|\\) on each compact set. Gaussian decay still dominates. The classical zero counting estimate \\(N(Y)=O(Y\\log Y)\\), where \\(N(Y)\\) counts zeros with \\(0<\\Im\\rho\\le Y\\), gives summability \\cite[Section 2.4, equation (2.11)]{BondarenkoRadchenkoSeip}. The same domination permits differentiation. Consequently\n\\begin{equation}\\label{rh:heatflow}\n\\partial_t\\Theta_t=\\partial_u^2\\Theta_t.\n\\end{equation}\nThe zero symmetries make \\(\\Theta_t(u)\\) real and even when \\(u\\) is real.","Gaussian decay still dominates. The classical zero counting estimate \\(N(Y)=O(Y\\log Y)\\), where \\(N(Y)\\) counts zeros with \\(0<\\Im\\rho\\le Y\\), gives summability \\cite[Section 2.4, equation (2.11)]{BondarenkoRadchenkoSeip}. The same domination permits differentiation. Consequently\n\\begin{equation}\\label{rh:heatflow}\n\\partial_t\\Theta_t=\\partial_u^2\\Theta_t.\n\\end{equation}\nThe zero symmetries make \\(\\Theta_t(u)\\) real and even when \\(u\\) is real.","classical zero counting estimate \\(N(Y)=O(Y\\log Y)\\), where \\(N(Y)\\) counts zeros with \\(0<\\Im\\rho\\le Y\\), gives summability \\cite[Section 2.4, equation (2.11)]{BondarenkoRadchenkoSeip}. The same domination permits differentiation. Consequently\n\\begin{equation}\\label{rh:heatflow}\n\\partial_t\\Theta_t=\\partial_u^2\\Theta_t.\n\\end{equation}\nThe zero symmetries make \\(\\Theta_t(u)\\) real and even when \\(u\\) is real.","It helps to make the normal convergence argument quantitative enough to see which operations it permits. Write \\(z=x+i\\gamma\\), where \\(|x|<1/2\\), and \\(u=a+ib\\). Then\n\\[\n\\Re(tz^2+uz)=t(x^2-\\gamma^2)+ax-b\\gamma.\n\\]\nIf \\(t\\in[t_0,T]\\) and \\(|a|,|b|\\le B\\), this is at most\n\\[\nT/4+B/2-t_0\\gamma^2+B|\\gamma|\n\\le T/4+B/2+\\frac{B^2}{2t_0}-\\frac{t_0}2\\gamma^2.\n\\]\nThe last step is the elementary inequality\n\\(B|\\gamma|\\le(t_0/2)\\gamma^2+B^2/(2t_0)\\).\nThus a compact set of heat times and translations has one Gaussian majorant. A finite number of derivatives only inserts a polynomial in \\(z\\). Such a polynomial is still summable against this Gaussian.","helps to make the normal convergence argument quantitative enough to see which operations it permits. Write \\(z=x+i\\gamma\\), where \\(|x|<1/2\\), and \\(u=a+ib\\). Then\n\\[\n\\Re(tz^2+uz)=t(x^2-\\gamma^2)+ax-b\\gamma.\n\\]\nIf \\(t\\in[t_0,T]\\) and \\(|a|,|b|\\le B\\), this is at most\n\\[\nT/4+B/2-t_0\\gamma^2+B|\\gamma|\n\\le T/4+B/2+\\frac{B^2}{2t_0}-\\frac{t_0}2\\gamma^2.\n\\]\nThe last step is the elementary inequality\n\\(B|\\gamma|\\le(t_0/2)\\gamma^2+B^2/(2t_0)\\).\nThus a compact set of heat times and translations has one Gaussian majorant. A finite number of derivatives only inserts a polynomial in \\(z\\). Such a polynomial is still summable against this Gaussian.","t(x^2-\\gamma^2)+ax-b\\gamma.\n\\]\nIf \\(t\\in[t_0,T]\\) and \\(|a|,|b|\\le B\\), this is at most\n\\[\nT/4+B/2-t_0\\gamma^2+B|\\gamma|\n\\le T/4+B/2+\\frac{B^2}{2t_0}-\\frac{t_0}2\\gamma^2.\n\\]\nThe last step is the elementary inequality\n\\(B|\\gamma|\\le(t_0/2)\\gamma^2+B^2/(2t_0)\\).\nThus a compact set of heat times and translations has one Gaussian majorant. A finite number of derivatives only inserts a polynomial in \\(z\\). Such a polynomial is still summable against this Gaussian.","For completeness, divide the positive ordinates into intervals \\([j,j+1)\\). The number in an interval is no greater than \\(N(j+1)\\), which is \\(O((j+1)\\log(j+2))\\). Even this deliberately loose bound gives\n\\[\n\\sum_{\\rho\\in\\mathcal Z}(1+|\\gamma|)^A e^{-c\\gamma^2}<\\infty\n\\qquad(c>0,\\ A\\ge0).\n\\]\nThe finitely many small ordinates cause no difficulty, and the negative ones follow by conjugation. This establishes uniform convergence on the compact sets just described. It permits differentiation in \\(u\\), differentiation in a positive heat interval, and multiplication by the polynomial zero weight introduced later. It does not permit taking \\(t=0\\), where the Gaussian suppression disappears.","al is no greater than \\(N(j+1)\\), which is \\(O((j+1)\\log(j+2))\\). Even this deliberately loose bound gives\n\\[\n\\sum_{\\rho\\in\\mathcal Z}(1+|\\gamma|)^A e^{-c\\gamma^2}<\\infty\n\\qquad(c>0,\\ A\\ge0).\n\\]\nThe finitely many small ordinates cause no difficulty, and the negative ones follow by conjugation. This establishes uniform convergence on the compact sets just described. It permits differentiation in \\(u\\), differentiation in a positive heat interval, and multiplication by the polynomial zero weight introduced later. It does not permit taking \\(t=0\\), where the Gaussian suppression disappears.",")^A e^{-c\\gamma^2}<\\infty\n\\qquad(c>0,\\ A\\ge0).\n\\]\nThe finitely many small ordinates cause no difficulty, and the negative ones follow by conjugation. This establishes uniform convergence on the compact sets just described. It permits differentiation in \\(u\\), differentiation in a positive heat interval, and multiplication by the polynomial zero weight introduced later. It does not permit taking \\(t=0\\), where the Gaussian suppression disappears.","For a single centered mode, \\(\\partial_t e^{tz^2+uz}=z^2e^{tz^2+uz}\\) and \\(\\partial_u^2e^{tz^2+uz}=z^2e^{tz^2+uz}\\). Normal convergence transfers this equality to the entire packet. The heat equation here is an exact identity for the chosen series, rather than an empirical fit to a finite list of zeros.","Under the Riemann hypothesis the packet is\n\\begin{equation}\\label{rh:criticalpacket}\n\\Theta_t(u)=\\sum_{\\rho\\in\\mathcal Z}e^{-t\\gamma^2}e^{iu\\gamma}.\n\\end{equation}\nIt is then the Fourier transform of a finite positive discrete measure. In particular, its magnitude is bounded by its value at the origin. The converse can be proved without estimating cancellation among different zeros.","\\begin{theorem}[One heat time criterion]\\label{rh:bounded}\nFor any fixed \\(t>0\\), the Riemann hypothesis is equivalent to boundedness of \\(\\Theta_t\\) on \\([0,\\infty)\\). It is also equivalent to\n\\begin{equation}\\label{rh:det}\n\\Theta_t(0)^2-\\Theta_t(u)^2\\ge0\n\\qquad\\text{for every real }u.\n\\end{equation}\n\\end{theorem}","\\begin{proof}\nUnder the hypothesis, \\eqref{rh:criticalpacket} gives the asserted bound and determinant inequality. Conversely, suppose the packet is bounded on the positive half line. Its Laplace transform\n\\[\nF_t(w)=\\int_0^{\\infty}\\Theta_t(u)e^{-wu}\\,du\n\\]\nis holomorphic for \\(\\Re w>0\\). For \\(\\Re w>1/2\\), absolute convergence gives\n\\begin{equation}\\label{rh:laplace}\nF_t(w)=\\sum_{\\rho\\in\\mathcal Z}\n\\frac{e^{tz_\\rho^2}}{w-z_\\rho}.\n\\end{equation}\nThe series on the right is meromorphic, with normal convergence away from its poles. At each distinct centered zero its residue is its positive integer multiplicity times \\(e^{tz_\\rho^2}\\), which is nonzero. Analytic continuation therefore excludes a centered zero with positive real part, since it would give a pole in the domain where the left side is holomorphic. Symmetry excludes negative real parts as well. Finally, \\eqref{rh:det} itself bounds the packet by \\(|\\Theta_t(0)|\\), so it also gives the","\\begin{proof}\nUnder the hypothesis, \\eqref{rh:criticalpacket} gives the asserted bound and determinant inequality. Conversely, suppose the packet is bounded on the positive half line. Its Laplace transform\n\\[\nF_t(w)=\\int_0^{\\infty}\\Theta_t(u)e^{-wu}\\,du\n\\]\nis holomorphic for \\(\\Re w>0\\). For \\(\\Re w>1/2\\), absolute convergence gives\n\\begin{equation}\\label{rh:laplace}\nF_t(w)=\\sum_{\\rho\\in\\mathcal Z}\n\\frac{e^{tz_\\rho^2}}{w-z_\\rho}.\n\\end{equation}\nThe series on the right is meromorphic, with normal convergence away from its poles. At each distinct centered zero its residue is its positive integer multiplicity times \\(e^{tz_\\rho^2}\\), which is nonzero. Analytic continuation therefore excludes a centered zero with positive real part, since it would give a pole in the domain where the left side is holomorphic. Symmetry excludes negative real parts as well. Finally, \\eqref{rh:det} itself bounds the packet by \\(|\\Theta_t(0)|\\), so it also gives the converse.\n\\end{proof}","^{\\infty}\\Theta_t(u)e^{-wu}\\,du\n\\]\nis holomorphic for \\(\\Re w>0\\). For \\(\\Re w>1/2\\), absolute convergence gives\n\\begin{equation}\\label{rh:laplace}\nF_t(w)=\\sum_{\\rho\\in\\mathcal Z}\n\\frac{e^{tz_\\rho^2}}{w-z_\\rho}.\n\\end{equation}\nThe series on the right is meromorphic, with normal convergence away from its poles. At each distinct centered zero its residue is its positive integer multiplicity times \\(e^{tz_\\rho^2}\\), which is nonzero. Analytic continuation therefore excludes a centered zero with positive real part, since it would give a pole in the domain where the left side is holomorphic. Symmetry excludes negative real parts as well. Finally, \\eqref{rh:det} itself bounds the packet by \\(|\\Theta_t(0)|\\), so it also gives the converse.\n\\end{proof}","from its poles. At each distinct centered zero its residue is its positive integer multiplicity times \\(e^{tz_\\rho^2}\\), which is nonzero. Analytic continuation therefore excludes a centered zero with positive real part, since it would give a pole in the domain where the left side is holomorphic. Symmetry excludes negative real parts as well. Finally, \\eqref{rh:det} itself bounds the packet by \\(|\\Theta_t(0)|\\), so it also gives the converse.\n\\end{proof}","If \\(\\Theta_t\\) is bounded, then on a compact subset of \\(\\Re w>0\\) the integrand of its Laplace transform is bounded by a constant times \\(e^{-au}\\), for some \\(a>0\\). Differentiation with respect to \\(w\\) inserts powers of \\(u\\), all of which remain integrable. This proves holomorphy on the claimed half plane.","In the smaller initial domain \\(\\Re w>1/2\\), absolute summation justifies integration of each zero term:\n\\[\n\\int_0^\\infty e^{tz^2+uz-wu}\\,du\n=\\frac{e^{tz^2}}{w-z}.\n\\]\nFor a compact set of \\(w\\) avoiding all centered zeros, large ordinates make \\(|w-z|\\) large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero \\(z_0\\), isolate all copies of that zero. If its multiplicity is \\(m_0\\), their sum is\n\\[\n\\frac{m_0e^{tz_0^2}}{w-z_0}.\n\\]\nEvery other term is holomorphic in a sufficiently small disk. The residue cannot vanish because \\(m_0\\ge1\\) and an exponential is never zero. The reflected zero lies at another pole location unless it coincides with \\(z_0\\), in which case multip","In the smaller initial domain \\(\\Re w>1/2\\), absolute summation justifies integration of each zero term:\n\\[\n\\int_0^\\infty e^{tz^2+uz-wu}\\,du\n=\\frac{e^{tz^2}}{w-z}.\n\\]\nFor a compact set of \\(w\\) avoiding all centered zeros, large ordinates make \\(|w-z|\\) large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero \\(z_0\\), isolate all copies of that zero. If its multiplicity is \\(m_0\\), their sum is\n\\[\n\\frac{m_0e^{tz_0^2}}{w-z_0}.\n\\]\nEvery other term is holomorphic in a sufficiently small disk. The residue cannot vanish because \\(m_0\\ge1\\) and an exponential is never zero. The reflected zero lies at another pole location unless it coincides with \\(z_0\\), in which case multiplicity adds rather than cancels.","\\), absolute summation justifies integration of each zero term:\n\\[\n\\int_0^\\infty e^{tz^2+uz-wu}\\,du\n=\\frac{e^{tz^2}}{w-z}.\n\\]\nFor a compact set of \\(w\\) avoiding all centered zeros, large ordinates make \\(|w-z|\\) large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero \\(z_0\\), isolate all copies of that zero. If its multiplicity is \\(m_0\\), their sum is\n\\[\n\\frac{m_0e^{tz_0^2}}{w-z_0}.\n\\]\nEvery other term is holomorphic in a sufficiently small disk. The residue cannot vanish because \\(m_0\\ge1\\) and an exponential is never zero. The reflected zero lies at another pole location unless it coincides with \\(z_0\\), in which case multiplicity adds rather than cancels.","justifies integration of each zero term:\n\\[\n\\int_0^\\infty e^{tz^2+uz-wu}\\,du\n=\\frac{e^{tz^2}}{w-z}.\n\\]\nFor a compact set of \\(w\\) avoiding all centered zeros, large ordinates make \\(|w-z|\\) large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero \\(z_0\\), isolate all copies of that zero. If its multiplicity is \\(m_0\\), their sum is\n\\[\n\\frac{m_0e^{tz_0^2}}{w-z_0}.\n\\]\nEvery other term is holomorphic in a sufficiently small disk. The residue cannot vanish because \\(m_0\\ge1\\) and an exponential is never zero. The reflected zero lies at another pole location unless it coincides with \\(z_0\\), in which case multiplicity adds rather than cancels.","avoiding all centered zeros, large ordinates make \\(|w-z|\\) large. The Gaussian coefficient then gives a summable bound for the meromorphic series. Near one distinct centered zero \\(z_0\\), isolate all copies of that zero. If its multiplicity is \\(m_0\\), their sum is\n\\[\n\\frac{m_0e^{tz_0^2}}{w-z_0}.\n\\]\nEvery other term is holomorphic in a sufficiently small disk. The residue cannot vanish because \\(m_0\\ge1\\) and an exponential is never zero. The reflected zero lies at another pole location unless it coincides with \\(z_0\\), in which case multiplicity adds rather than cancels.","for the meromorphic series. Near one distinct centered zero \\(z_0\\), isolate all copies of that zero. If its multiplicity is \\(m_0\\), their sum is\n\\[\n\\frac{m_0e^{tz_0^2}}{w-z_0}.\n\\]\nEvery other term is holomorphic in a sufficiently small disk. The residue cannot vanish because \\(m_0\\ge1\\) and an exponential is never zero. The reflected zero lies at another pole location unless it coincides with \\(z_0\\), in which case multiplicity adds rather than cancels.","The two representations agree in their initial half plane. Uniqueness of analytic continuation makes them agree wherever both can be continued. A positive real part of \\(z_0\\) would place its nonremovable pole inside the domain of the bounded packet's holomorphic transform. This is the contradiction. The argument is qualitative: even a coefficient too small to be visible in a practical plot still has a nonzero residue.","The determinant formulation has the same content. Under the conjecture, the packet is real and \\(|\\Theta_t(u)|\\le\\Theta_t(0)\\), with \\(\\Theta_t(0)>0\\). Conversely, the displayed determinant inequality forces \\(\\Theta_t(u)^2\\le\\Theta_t(0)^2\\). No new positivity theorem for an arbitrary zero packet is being assumed.","The packet also makes the obstruction quantitative. Define\n\\begin{equation}\\label{rh:delta}\n\\delta_\\zeta=\\sup_{\\rho\\in\\mathcal Z}(\\Re\\rho-1/2),\n\\qquad\nM_t(U)=\\max\\left(1,\\sup_{0\\le u\\le U}|\\Theta_t(u)|\\right).\n\\end{equation}\nThen \\(0\\le\\delta_\\zeta\\le1/2\\). The same Laplace argument proves\n\\begin{equation}\\label{rh:front}\n\\limsup_{U\\to\\infty}\\frac{\\log M_t(U)}{U}=\\delta_\\zeta.\n\\end{equation}\nThe upper bound follows from summing absolute values. If a smaller exponential rate held, the Laplace transform would be holomorphic beyond that rate, contradicting a pole at a zero further to the right. An off line zero may have a very large ordinate and a very small Gaussian coefficient. Its coefficient remains nonzero, so it eventually determines the possible exponential growth.","To spell out the last implication, suppose the limit superior in \\eqref{rh:front} were below some centered zero's real part. Choose a number \\(a\\) strictly between them. Beyond a finite initial interval the packet would then satisfy \\(|\\Theta_t(u)|\\le e^{au}\\), after slightly enlarging \\(a\\) if necessary. Its Laplace transform would be holomorphic for \\(\\Re w>a\\), but the chosen zero would still place a pole there. This proves the lower bound without requiring that the supremum \\(\\delta_\\zeta\\) be attained. When \\(\\delta_\\zeta=0\\), the definition \\(M_t(U)\\ge1\\) supplies the matching nonnegative lower bound.","e below some centered zero's real part. Choose a number \\(a\\) strictly between them. Beyond a finite initial interval the packet would then satisfy \\(|\\Theta_t(u)|\\le e^{au}\\), after slightly enlarging \\(a\\) if necessary. Its Laplace transform would be holomorphic for \\(\\Re w>a\\), but the chosen zero would still place a pole there. This proves the lower bound without requiring that the supremum \\(\\delta_\\zeta\\) be attained. When \\(\\delta_\\zeta=0\\), the definition \\(M_t(U)\\ge1\\) supplies the matching nonnegative lower bound.","art. Choose a number \\(a\\) strictly between them. Beyond a finite initial interval the packet would then satisfy \\(|\\Theta_t(u)|\\le e^{au}\\), after slightly enlarging \\(a\\) if necessary. Its Laplace transform would be holomorphic for \\(\\Re w>a\\), but the chosen zero would still place a pole there. This proves the lower bound without requiring that the supremum \\(\\delta_\\zeta\\) be attained. When \\(\\delta_\\zeta=0\\), the definition \\(M_t(U)\\ge1\\) supplies the matching nonnegative lower bound.","between them. Beyond a finite initial interval the packet would then satisfy \\(|\\Theta_t(u)|\\le e^{au}\\), after slightly enlarging \\(a\\) if necessary. Its Laplace transform would be holomorphic for \\(\\Re w>a\\), but the chosen zero would still place a pole there. This proves the lower bound without requiring that the supremum \\(\\delta_\\zeta\\) be attained. When \\(\\delta_\\zeta=0\\), the definition \\(M_t(U)\\ge1\\) supplies the matching nonnegative lower bound.","\\figplaceholder{rh_heatfront}{Amplitude envelopes of hypothetical centered modes. A displacement \\(\\delta\\) gives amplitude growth \\(e^{\\delta u}\\), with a fixed nonzero heat multiplier omitted in the illustration. These model curves illustrate the theorem's exponent and are not sampled off line zeta zeros or a complete symmetric zero packet.}","Let \\(\\Lambda(n)\\) denote the von Mangoldt function: it is \\(\\log p\\) when \\(n=p^k\\) for a prime \\(p\\) and an integer \\(k\\ge1\\), and it is zero otherwise. Let \\(\\psi=\\Gamma'/\\Gamma\\) be the digamma function, and set\n\\begin{equation}\\label{rh:psiinf}\n\\Psi_\\infty(v)=-\\log\\pi+\n\\Re\\psi\\left(\\frac14+\\frac{iv}{2}\\right)\n\\qquad(v\\in\\mathbb R).\n\\end{equation}\nDefine the archimedean term and prime packet by\n\\begin{align}\n\\mathfrak A_t(u)&=\\frac1{2\\pi}\\int_{\\mathbb R}\ne^{-tv^2}\\cos(uv)\\Psi_\\infty(v)\\,dv,\n\\label{rh:arch}",")\\) denote the von Mangoldt function: it is \\(\\log p\\) when \\(n=p^k\\) for a prime \\(p\\) and an integer \\(k\\ge1\\), and it is zero otherwise. Let \\(\\psi=\\Gamma'/\\Gamma\\) be the digamma function, and set\n\\begin{equation}\\label{rh:psiinf}\n\\Psi_\\infty(v)=-\\log\\pi+\n\\Re\\psi\\left(\\frac14+\\frac{iv}{2}\\right)\n\\qquad(v\\in\\mathbb R).\n\\end{equation}\nDefine the archimedean term and prime packet by\n\\begin{align}\n\\mathfrak A_t(u)&=\\frac1{2\\pi}\\int_{\\mathbb R}\ne^{-tv^2}\\cos(uv)\\Psi_\\infty(v)\\,dv,\n\\label{rh:arch}\\\\\n\\mathfrak P_t(u)&=\\frac1{2\\sqrt{\\pi t}}\n\\sum_{n\\ge2}\\frac{\\Lambda(n)}{\\sqrt n}\n\\left(e^{-(u-\\log n)^2/(4t)}+\ne^{-(u+\\log n)^2/(4t)}\\right).\n\\label{rh:primepacket}\n\\end{align}\nThe Gaussian makes the prime series absolutely convergent. The completed explicit formula gives\n\\begin{equation}\\label{rh:explicit}\n\\Theta_t(u)=2e^{t/4}\\cosh(u/2)+\\mathfrak A_t(u)-\\mathfrak P_t(u).\n\\end{equation}\nWe use the formula in the recovered normalization \\cite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has tr…","\\mathfrak P_t(u)&=\\frac1{2\\sqrt{\\pi t}}\n\\sum_{n\\ge2}\\frac{\\Lambda(n)}{\\sqrt n}\n\\left(e^{-(u-\\log n)^2/(4t)}+\ne^{-(u+\\log n)^2/(4t)}\\right).\n\\label{rh:primepacket}\n\\end{align}\nThe Gaussian makes the prime series absolutely convergent. The completed explicit formula gives\n\\begin{equation}\\label{rh:explicit}\n\\Theta_t(u)=2e^{t/4}\\cosh(u/2)+\\mathfrak A_t(u)-\\mathfrak P_t(u).\n\\end{equation}\nWe use the formula in the recovered normalization \\cite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has transform\n\\[\ng(x)=\\frac1{4\\sqrt{\\pi t}}\n\\left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\\right).\n\\]\nThe zero symmetry identifies its spectral sum with \\(\\Theta_t(u)\\).\nIts two pole evaluations give \\(2e^{t/4}\\cosh(u/2)\\). The prime contribution uses \\(g(\\log n)+g(-\\log n)\\), which fixes the constants in \\eqref{rh:primepacket}. The even archimedean integral reduces to its real digamma part, giving \\eqref{rh:explicit} with the stated signs.","e^{-(u+\\log n)^2/(4t)}\\right).\n\\label{rh:primepacket}\n\\end{align}\nThe Gaussian makes the prime series absolutely convergent. The completed explicit formula gives\n\\begin{equation}\\label{rh:explicit}\n\\Theta_t(u)=2e^{t/4}\\cosh(u/2)+\\mathfrak A_t(u)-\\mathfrak P_t(u).\n\\end{equation}\nWe use the formula in the recovered normalization \\cite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has transform\n\\[\ng(x)=\\frac1{4\\sqrt{\\pi t}}\n\\left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\\right).\n\\]\nThe zero symmetry identifies its spectral sum with \\(\\Theta_t(u)\\).\nIts two pole evaluations give \\(2e^{t/4}\\cosh(u/2)\\). The prime contribution uses \\(g(\\log n)+g(-\\log n)\\), which fixes the constants in \\eqref{rh:primepacket}. The even archimedean integral reduces to its real digamma part, giving \\eqref{rh:explicit} with the stated signs.","d{align}\nThe Gaussian makes the prime series absolutely convergent. The completed explicit formula gives\n\\begin{equation}\\label{rh:explicit}\n\\Theta_t(u)=2e^{t/4}\\cosh(u/2)+\\mathfrak A_t(u)-\\mathfrak P_t(u).\n\\end{equation}\nWe use the formula in the recovered normalization \\cite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has transform\n\\[\ng(x)=\\frac1{4\\sqrt{\\pi t}}\n\\left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\\right).\n\\]\nThe zero symmetry identifies its spectral sum with \\(\\Theta_t(u)\\).\nIts two pole evaluations give \\(2e^{t/4}\\cosh(u/2)\\). The prime contribution uses \\(g(\\log n)+g(-\\log n)\\), which fixes the constants in \\eqref{rh:primepacket}. The even archimedean integral reduces to its real digamma part, giving \\eqref{rh:explicit} with the stated signs.","ies absolutely convergent. The completed explicit formula gives\n\\begin{equation}\\label{rh:explicit}\n\\Theta_t(u)=2e^{t/4}\\cosh(u/2)+\\mathfrak A_t(u)-\\mathfrak P_t(u).\n\\end{equation}\nWe use the formula in the recovered normalization \\cite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has transform\n\\[\ng(x)=\\frac1{4\\sqrt{\\pi t}}\n\\left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\\right).\n\\]\nThe zero symmetry identifies its spectral sum with \\(\\Theta_t(u)\\).\nIts two pole evaluations give \\(2e^{t/4}\\cosh(u/2)\\). The prime contribution uses \\(g(\\log n)+g(-\\log n)\\), which fixes the constants in \\eqref{rh:primepacket}. The even archimedean integral reduces to its real digamma part, giving \\eqref{rh:explicit} with the stated signs.","thfrak A_t(u)-\\mathfrak P_t(u).\n\\end{equation}\nWe use the formula in the recovered normalization \\cite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has transform\n\\[\ng(x)=\\frac1{4\\sqrt{\\pi t}}\n\\left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\\right).\n\\]\nThe zero symmetry identifies its spectral sum with \\(\\Theta_t(u)\\).\nIts two pole evaluations give \\(2e^{t/4}\\cosh(u/2)\\). The prime contribution uses \\(g(\\log n)+g(-\\log n)\\), which fixes the constants in \\eqref{rh:primepacket}. The even archimedean integral reduces to its real digamma part, giving \\eqref{rh:explicit} with the stated signs.","We use the formula in the recovered normalization \\cite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has transform\n\\[\ng(x)=\\frac1{4\\sqrt{\\pi t}}\n\\left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\\right).\n\\]\nThe zero symmetry identifies its spectral sum with \\(\\Theta_t(u)\\).\nIts two pole evaluations give \\(2e^{t/4}\\cosh(u/2)\\). The prime contribution uses \\(g(\\log n)+g(-\\log n)\\), which fixes the constants in \\eqref{rh:primepacket}. The even archimedean integral reduces to its real digamma part, giving \\eqref{rh:explicit} with the stated signs.","ite{KomorebiHeat,BondarenkoRadchenkoSeip}. With Fourier convention\n\\(g(x)=(2\\pi)^{-1}\\int h(v)e^{-ivx}\\,dv\\), the even test\n\\(h(v)=e^{-tv^2}\\cos(uv)\\) has transform\n\\[\ng(x)=\\frac1{4\\sqrt{\\pi t}}\n\\left(e^{-(x-u)^2/(4t)}+e^{-(x+u)^2/(4t)}\\right).\n\\]\nThe zero symmetry identifies its spectral sum with \\(\\Theta_t(u)\\).\nIts two pole evaluations give \\(2e^{t/4}\\cosh(u/2)\\). The prime contribution uses \\(g(\\log n)+g(-\\log n)\\), which fixes the constants in \\eqref{rh:primepacket}. The even archimedean integral reduces to its real digamma part, giving \\eqref{rh:explicit} with the stated signs.","The normalization can be checked without any arithmetic approximation. For real \\(x\\), the elementary Gaussian Fourier integral is\n\\[\n\\int_{\\mathbb R}e^{-tv^2}e^{-ivx}\\,dv\n=\\sqrt{\\frac\\pi t}\\,e^{-x^2/(4t)}.\n\\]\nWrite \\(\\cos(uv)=(e^{iuv}+e^{-iuv})/2\\). The factor one half from this cosine and the factor \\((2\\pi)^{-1}\\) in the Fourier convention give the coefficient \\(1/(4\\sqrt{\\pi t})\\) in \\(g\\). The function \\(g\\) is even, so the sum \\(g(\\log n)+g(-\\log n)\\) doubles it. This produces the coefficient \\(1/(2\\sqrt{\\pi t})\\) in the prime packet, not twice that coefficient.","tion can be checked without any arithmetic approximation. For real \\(x\\), the elementary Gaussian Fourier integral is\n\\[\n\\int_{\\mathbb R}e^{-tv^2}e^{-ivx}\\,dv\n=\\sqrt{\\frac\\pi t}\\,e^{-x^2/(4t)}.\n\\]\nWrite \\(\\cos(uv)=(e^{iuv}+e^{-iuv})/2\\). The factor one half from this cosine and the factor \\((2\\pi)^{-1}\\) in the Fourier convention give the coefficient \\(1/(4\\sqrt{\\pi t})\\) in \\(g\\). The function \\(g\\) is even, so the sum \\(g(\\log n)+g(-\\log n)\\) doubles it. This produces the coefficient \\(1/(2\\sqrt{\\pi t})\\) in the prime packet, not twice that coefficient.","t any arithmetic approximation. For real \\(x\\), the elementary Gaussian Fourier integral is\n\\[\n\\int_{\\mathbb R}e^{-tv^2}e^{-ivx}\\,dv\n=\\sqrt{\\frac\\pi t}\\,e^{-x^2/(4t)}.\n\\]\nWrite \\(\\cos(uv)=(e^{iuv}+e^{-iuv})/2\\). The factor one half from this cosine and the factor \\((2\\pi)^{-1}\\) in the Fourier convention give the coefficient \\(1/(4\\sqrt{\\pi t})\\) in \\(g\\). The function \\(g\\) is even, so the sum \\(g(\\log n)+g(-\\log n)\\) doubles it. This produces the coefficient \\(1/(2\\sqrt{\\pi t})\\) in the prime packet, not twice that coefficient.","sian Fourier integral is\n\\[\n\\int_{\\mathbb R}e^{-tv^2}e^{-ivx}\\,dv\n=\\sqrt{\\frac\\pi t}\\,e^{-x^2/(4t)}.\n\\]\nWrite \\(\\cos(uv)=(e^{iuv}+e^{-iuv})/2\\). The factor one half from this cosine and the factor \\((2\\pi)^{-1}\\) in the Fourier convention give the coefficient \\(1/(4\\sqrt{\\pi t})\\) in \\(g\\). The function \\(g\\) is even, so the sum \\(g(\\log n)+g(-\\log n)\\) doubles it. This produces the coefficient \\(1/(2\\sqrt{\\pi t})\\) in the prime packet, not twice that coefficient.","The two pole evaluations can be checked just as directly. In the explicit formula the corresponding spectral arguments are \\(i/2\\) and \\(-i/2\\). Both Gaussian values contribute \\(e^{t/4}\\), and the cosine contributes \\(\\cosh(u/2)\\). Their sum is the explicit term in \\eqref{rh:explicit}. For a zero, the spectral argument is \\((\\rho-1/2)/i\\). The Gaussian value becomes \\(e^{tz_\\rho^2}\\), while the cosine becomes \\(\\cosh(uz_\\rho)\\). Summing over the symmetric multiset identifies this sum with \\(\\sum e^{tz_\\rho^2+uz_\\rho}\\).","luations can be checked just as directly. In the explicit formula the corresponding spectral arguments are \\(i/2\\) and \\(-i/2\\). Both Gaussian values contribute \\(e^{t/4}\\), and the cosine contributes \\(\\cosh(u/2)\\). Their sum is the explicit term in \\eqref{rh:explicit}. For a zero, the spectral argument is \\((\\rho-1/2)/i\\). The Gaussian value becomes \\(e^{tz_\\rho^2}\\), while the cosine becomes \\(\\cosh(uz_\\rho)\\). Summing over the symmetric multiset identifies this sum with \\(\\sum e^{tz_\\rho^2+uz_\\rho}\\).","ng spectral arguments are \\(i/2\\) and \\(-i/2\\). Both Gaussian values contribute \\(e^{t/4}\\), and the cosine contributes \\(\\cosh(u/2)\\). Their sum is the explicit term in \\eqref{rh:explicit}. For a zero, the spectral argument is \\((\\rho-1/2)/i\\). The Gaussian value becomes \\(e^{tz_\\rho^2}\\), while the cosine becomes \\(\\cosh(uz_\\rho)\\). Summing over the symmetric multiset identifies this sum with \\(\\sum e^{tz_\\rho^2+uz_\\rho}\\).","At \\(u=0\\), for example, the prime packet is\n\\[\n\\mathfrak P_t(0)=\\frac1{\\sqrt{\\pi t}}\n\\sum_{n\\ge2}\\frac{\\Lambda(n)}{\\sqrt n}\ne^{-(\\log n)^2/(4t)}.\n\\]\nEvery summand is nonnegative. The complete formula is nonetheless a difference involving this positive quantity. At large positive \\(u\\), the Gaussian centered at \\(\\log n\\) selects prime powers with logarithms near \\(u\\), so the active arithmetic scale is around \\(n=e^u\\). A fixed heat time gives a fixed width in logarithmic coordinates, which corresponds to a multiplicative window in the integers.","example, the prime packet is\n\\[\n\\mathfrak P_t(0)=\\frac1{\\sqrt{\\pi t}}\n\\sum_{n\\ge2}\\frac{\\Lambda(n)}{\\sqrt n}\ne^{-(\\log n)^2/(4t)}.\n\\]\nEvery summand is nonnegative. The complete formula is nonetheless a difference involving this positive quantity. At large positive \\(u\\), the Gaussian centered at \\(\\log n\\) selects prime powers with logarithms near \\(u\\), so the active arithmetic scale is around \\(n=e^u\\). A fixed heat time gives a fixed width in logarithmic coordinates, which corresponds to a multiplicative window in the integers.","Absolute convergence of the prime series requires no prime number theorem. Use the elementary bound \\(\\Lambda(n)\\le\\log n\\) and partition the integers by \\(j\\le\\log n<j+1\\). There are at most \\(e^{j+1}\\) terms in such a group. The Gaussian in \\(j\\) dominates the resulting exponential factor. The same reasoning covers differentiated packets because their derivatives insert only polynomials in \\(u-\\log n\\). This supplies a direct analytic justification for the operations applied to the prime side.","The integral over \\(\\Psi_\\infty\\), the explicit pole term, and the prime powers have distinct origins. Keeping all three visible prevents a mistaken inference from one individually large or small channel. The exact formula is useful because it states which difference must be controlled.","\\begin{definition}[Pole annihilated packet]\\label{rh:polefree}\nLet \\(L=1/4-\\partial_u^2\\), and define\n\\begin{equation}\\label{rh:tilde}\n\\widetilde\\Theta_t=L\\Theta_t\n=\\sum_{\\rho\\in\\mathcal Z}\\rho(1-\\rho)\ne^{tz_\\rho^2+uz_\\rho}.\n\\end{equation}\n\\end{definition}","The identity \\(1/4-z_\\rho^2=\\rho(1-\\rho)\\) explains the weight. It is nonzero for every nontrivial zero. Thus the operator kills the explicit pole contribution in \\eqref{rh:explicit} while retaining every zero mode. Differentiated normal convergence gives\n\\begin{equation}\\label{rh:tildeexplicit}\n\\widetilde\\Theta_t=\\widetilde{\\mathfrak A}_t-\n\\widetilde{\\mathfrak P}_t,\n\\end{equation}\nwhere tildes on the right mean application of \\(L\\).","1-\\rho)\\) explains the weight. It is nonzero for every nontrivial zero. Thus the operator kills the explicit pole contribution in \\eqref{rh:explicit} while retaining every zero mode. Differentiated normal convergence gives\n\\begin{equation}\\label{rh:tildeexplicit}\n\\widetilde\\Theta_t=\\widetilde{\\mathfrak A}_t-\n\\widetilde{\\mathfrak P}_t,\n\\end{equation}\nwhere tildes on the right mean application of \\(L\\).","For a translated Gaussian \\(G_{t,a}(u)=e^{-(u-a)^2/(4t)}\\), direct differentiation gives\n\\[\nLG_{t,a}(u)=\n\\left(\\frac14+\\frac1{2t}-\\frac{(u-a)^2}{4t^2}\\right)\nG_{t,a}(u).\n\\]\nThe annihilated prime packet therefore carries a polynomial factor with both signs. Positivity of the original Gaussian sum is no longer positivity of every transformed term. On the zero side the same operator is simpler because each exponential is an eigenfunction of \\(\\partial_u^2\\). On the explicit pole channel, \\(\\partial_u^2\\cosh(u/2)=\\frac14\\cosh(u/2)\\), so cancellation is exact.","u-a)^2/(4t)}\\), direct differentiation gives\n\\[\nLG_{t,a}(u)=\n\\left(\\frac14+\\frac1{2t}-\\frac{(u-a)^2}{4t^2}\\right)\nG_{t,a}(u).\n\\]\nThe annihilated prime packet therefore carries a polynomial factor with both signs. Positivity of the original Gaussian sum is no longer positivity of every transformed term. On the zero side the same operator is simpler because each exponential is an eigenfunction of \\(\\partial_u^2\\). On the explicit pole channel, \\(\\partial_u^2\\cosh(u/2)=\\frac14\\cosh(u/2)\\), so cancellation is exact.","LG_{t,a}(u)=\n\\left(\\frac14+\\frac1{2t}-\\frac{(u-a)^2}{4t^2}\\right)\nG_{t,a}(u).\n\\]\nThe annihilated prime packet therefore carries a polynomial factor with both signs. Positivity of the original Gaussian sum is no longer positivity of every transformed term. On the zero side the same operator is simpler because each exponential is an eigenfunction of \\(\\partial_u^2\\). On the explicit pole channel, \\(\\partial_u^2\\cosh(u/2)=\\frac14\\cosh(u/2)\\), so cancellation is exact.","The choice of operator is tied to a declared removal target. Its two scalar roots are the centered coordinates \\(\\pm1/2\\) of the exceptional points \\(s=0,1\\). Actual nontrivial zeros are strictly inside the critical strip, so none is removed. A different operator which vanished at an unknown nontrivial zero would no longer preserve the obstruction required by the criterion.","The appearance of \\(\\Lambda\\) can be traced directly to the Euler product in its first convergence half plane. There,\n\\[\n\\log\\zeta(s)=\\sum_p\\sum_{k\\ge1}\\frac{p^{-ks}}k.\n\\]\nDifferentiating gives\n\\[\n-\\frac{\\zeta'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).","appearance of \\(\\Lambda\\) can be traced directly to the Euler product in its first convergence half plane. There,\n\\[\n\\log\\zeta(s)=\\sum_p\\sum_{k\\ge1}\\frac{p^{-ks}}k.\n\\]\nDifferentiating gives\n\\[\n-\\frac{\\zeta'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).","of \\(\\Lambda\\) can be traced directly to the Euler product in its first convergence half plane. There,\n\\[\n\\log\\zeta(s)=\\sum_p\\sum_{k\\ge1}\\frac{p^{-ks}}k.\n\\]\nDifferentiating gives\n\\[\n-\\frac{\\zeta'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).","raced directly to the Euler product in its first convergence half plane. There,\n\\[\n\\log\\zeta(s)=\\sum_p\\sum_{k\\ge1}\\frac{p^{-ks}}k.\n\\]\nDifferentiating gives\n\\[\n-\\frac{\\zeta'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).","There,\n\\[\n\\log\\zeta(s)=\\sum_p\\sum_{k\\ge1}\\frac{p^{-ks}}k.\n\\]\nDifferentiating gives\n\\[\n-\\frac{\\zeta'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).","_p\\sum_{k\\ge1}\\frac{p^{-ks}}k.\n\\]\nDifferentiating gives\n\\[\n-\\frac{\\zeta'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).",".\n\\]\nDifferentiating gives\n\\[\n-\\frac{\\zeta'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).","'(s)}{\\zeta(s)}\n=\\sum_p\\sum_{k\\ge1}(\\log p)p^{-ks}\n=\\sum_{n\\ge2}\\frac{\\Lambda(n)}{n^s}.\n\\]\nThe factor \\(k\\) from differentiation cancels the denominator \\(k\\), which is why the coefficient of \\(p^k\\) is \\(\\log p\\), rather than \\(\\log(p^k)\\). Absolute convergence of these derivative series holds for \\(\\Re s>1\\). For instance, \\(\\Lambda(n)\\le\\log n\\), and \\(\\sum(\\log n)n^{-\\sigma}\\) converges for \\(\\sigma>1\\).","Thus \\(\\Lambda(8)=\\log2\\), \\(\\Lambda(9)=\\log3\\), and \\(\\Lambda(12)=0\\). The exact prime packet includes these prime powers. Omitting them would change the explicit formula, even if primes supply its largest familiar contribution. The completed formula then carries this arithmetic logarithmic derivative into a Gaussian test against zeros, with the gamma and pole terms accounting for completion.","The archimedean term has a uniform bound on every positive heat strip. A direct proof uses the classical digamma series \\cite{DLMFDigamma}:\n\\[\n\\psi(a+iy)-\\psi(a)=\n\\sum_{k\\ge0}\\frac{iy}{(k+a)(k+a+iy)},\n\\qquad a>0.\n\\]\nIts absolute value is at most \\(|y|\\sum_{k\\ge0}(k+a)^{-2}\\). Therefore \\(|\\Psi_\\infty(v)|\\le C(1+|v|)\\) for a fixed constant \\(C\\). Since\n\\[\n\\widetilde{\\mathfrak A}_t(u)=\\frac1{2\\pi}\n\\int_{\\mathbb R}(1/4+v^2)e^{-tv^2}\\cos(uv)\n\\Psi_\\infty(v)\\,dv,\n\\]\nits magnitude is bounded independently of \\(u\\) and \\(t\\in[t_0,T]\\), whenever \\(0<t_0<T<\\infty\\). Its energy on a unit translation band is consequently \\(O(1)\\). Removing this explicit channel is an analytic estimate, rather than the missing arithmetic cancellation.","mma series \\cite{DLMFDigamma}:\n\\[\n\\psi(a+iy)-\\psi(a)=\n\\sum_{k\\ge0}\\frac{iy}{(k+a)(k+a+iy)},\n\\qquad a>0.\n\\]\nIts absolute value is at most \\(|y|\\sum_{k\\ge0}(k+a)^{-2}\\). Therefore \\(|\\Psi_\\infty(v)|\\le C(1+|v|)\\) for a fixed constant \\(C\\). Since\n\\[\n\\widetilde{\\mathfrak A}_t(u)=\\frac1{2\\pi}\n\\int_{\\mathbb R}(1/4+v^2)e^{-tv^2}\\cos(uv)\n\\Psi_\\infty(v)\\,dv,\n\\]\nits magnitude is bounded independently of \\(u\\) and \\(t\\in[t_0,T]\\), whenever \\(0<t_0<T<\\infty\\). Its energy on a unit translation band is consequently \\(O(1)\\). Removing this explicit channel is an analytic estimate, rather than the missing arithmetic cancellation.","\\cite{DLMFDigamma}:\n\\[\n\\psi(a+iy)-\\psi(a)=\n\\sum_{k\\ge0}\\frac{iy}{(k+a)(k+a+iy)},\n\\qquad a>0.\n\\]\nIts absolute value is at most \\(|y|\\sum_{k\\ge0}(k+a)^{-2}\\). Therefore \\(|\\Psi_\\infty(v)|\\le C(1+|v|)\\) for a fixed constant \\(C\\). Since\n\\[\n\\widetilde{\\mathfrak A}_t(u)=\\frac1{2\\pi}\n\\int_{\\mathbb R}(1/4+v^2)e^{-tv^2}\\cos(uv)\n\\Psi_\\infty(v)\\,dv,\n\\]\nits magnitude is bounded independently of \\(u\\) and \\(t\\in[t_0,T]\\), whenever \\(0<t_0<T<\\infty\\). Its energy on a unit translation band is consequently \\(O(1)\\). Removing this explicit channel is an analytic estimate, rather than the missing arithmetic cancellation.","si(a+iy)-\\psi(a)=\n\\sum_{k\\ge0}\\frac{iy}{(k+a)(k+a+iy)},\n\\qquad a>0.\n\\]\nIts absolute value is at most \\(|y|\\sum_{k\\ge0}(k+a)^{-2}\\). Therefore \\(|\\Psi_\\infty(v)|\\le C(1+|v|)\\) for a fixed constant \\(C\\). Since\n\\[\n\\widetilde{\\mathfrak A}_t(u)=\\frac1{2\\pi}\n\\int_{\\mathbb R}(1/4+v^2)e^{-tv^2}\\cos(uv)\n\\Psi_\\infty(v)\\,dv,\n\\]\nits magnitude is bounded independently of \\(u\\) and \\(t\\in[t_0,T]\\), whenever \\(0<t_0<T<\\infty\\). Its energy on a unit translation band is consequently \\(O(1)\\). Removing this explicit channel is an analytic estimate, rather than the missing arithmetic cancellation.","absolute value is at most \\(|y|\\sum_{k\\ge0}(k+a)^{-2}\\). Therefore \\(|\\Psi_\\infty(v)|\\le C(1+|v|)\\) for a fixed constant \\(C\\). Since\n\\[\n\\widetilde{\\mathfrak A}_t(u)=\\frac1{2\\pi}\n\\int_{\\mathbb R}(1/4+v^2)e^{-tv^2}\\cos(uv)\n\\Psi_\\infty(v)\\,dv,\n\\]\nits magnitude is bounded independently of \\(u\\) and \\(t\\in[t_0,T]\\), whenever \\(0<t_0<T<\\infty\\). Its energy on a unit translation band is consequently \\(O(1)\\). Removing this explicit channel is an analytic estimate, rather than the missing arithmetic cancellation.","For a nonnegative quantity \\(E_J\\), call its growth subexponential when\n\\begin{equation}\\label{rh:subexp}\n\\lim_{J\\to\\infty}\\frac{\\log(1+E_J)}J=0.\n\\end{equation}\nEquivalently, for every \\(\\varepsilon>0\\), there is a constant\n\\(C_\\varepsilon\\) such that \\(E_J\\le C_\\varepsilon e^{\\varepsilon J}\\)\nfor all positive integers \\(J\\). This is an upper growth condition. It does\nnot assert a positive lower bound.","\\begin{theorem}[Local energy criterion]\\label{rh:energy}\nFix \\(0<t_0<T<\\infty\\). For positive integers \\(J\\), set\n\\begin{equation}\\label{rh:energydef}\n\\widetilde E_J=\\int_{t_0}^T\\int_J^{J+1}\n|\\widetilde\\Theta_t(u)|^2\\,du\\,dt.\n\\end{equation}\nThe Riemann hypothesis is equivalent to subexponential growth of \\(\\widetilde E_J\\).\n\\end{theorem}","\\begin{proof}\nUnder the hypothesis, \\(\\rho(1-\\rho)=1/4+\\gamma^2\\). Absolute summation in \\eqref{rh:tilde} gives a uniform bound on the strip, and hence \\(\\widetilde E_J=O(1)\\).","Conversely, regard \\(u\\mapsto\\widetilde\\Theta_\\cdot(u)\\) as a function taking values in \\(H=L^2([t_0,T];\\mathbb C)\\). Cauchy Schwarz on each unit interval and the assumed energy bound imply\n\\[\n\\int_0^\\infty\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du<\\infty\n\\qquad(a>0).\n\\]\nIts \\(H\\)-valued Laplace transform is holomorphic for \\(\\Re w>0\\). Initially for \\(\\Re w>1/2\\), it also equals the normally convergent meromorphic series with terms\n\\[\n\\frac{\\rho(1-\\rho)e^{t z_\\rho^2}}{w-z_\\rho}.\n\\]\nThe residue at a distinct centered zero is a nonzero function of \\(t\\) in \\(H\\). Indeed, the multiplicity and \\(\\rho(1-\\rho)\\) are nonzero, and the exponential never vanishes. A zero with positive real part would therefore give a nonremovable pole in the holomorphi","Conversely, regard \\(u\\mapsto\\widetilde\\Theta_\\cdot(u)\\) as a function taking values in \\(H=L^2([t_0,T];\\mathbb C)\\). Cauchy Schwarz on each unit interval and the assumed energy bound imply\n\\[\n\\int_0^\\infty\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du<\\infty\n\\qquad(a>0).\n\\]\nIts \\(H\\)-valued Laplace transform is holomorphic for \\(\\Re w>0\\). Initially for \\(\\Re w>1/2\\), it also equals the normally convergent meromorphic series with terms\n\\[\n\\frac{\\rho(1-\\rho)e^{t z_\\rho^2}}{w-z_\\rho}.\n\\]\nThe residue at a distinct centered zero is a nonzero function of \\(t\\) in \\(H\\). Indeed, the multiplicity and \\(\\rho(1-\\rho)\\) are nonzero, and the exponential never vanishes. A zero with positive real part would therefore give a nonremovable pole in the holomorphic domain. Symmetry now gives the hypothesis.\n\\end{proof}","Theta_\\cdot(u)\\) as a function taking values in \\(H=L^2([t_0,T];\\mathbb C)\\). Cauchy Schwarz on each unit interval and the assumed energy bound imply\n\\[\n\\int_0^\\infty\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du<\\infty\n\\qquad(a>0).\n\\]\nIts \\(H\\)-valued Laplace transform is holomorphic for \\(\\Re w>0\\). Initially for \\(\\Re w>1/2\\), it also equals the normally convergent meromorphic series with terms\n\\[\n\\frac{\\rho(1-\\rho)e^{t z_\\rho^2}}{w-z_\\rho}.\n\\]\nThe residue at a distinct centered zero is a nonzero function of \\(t\\) in \\(H\\). Indeed, the multiplicity and \\(\\rho(1-\\rho)\\) are nonzero, and the exponential never vanishes. A zero with positive real part would therefore give a nonremovable pole in the holomorphic domain. Symmetry now gives the hypothesis.\n\\end{proof}","nd the assumed energy bound imply\n\\[\n\\int_0^\\infty\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du<\\infty\n\\qquad(a>0).\n\\]\nIts \\(H\\)-valued Laplace transform is holomorphic for \\(\\Re w>0\\). Initially for \\(\\Re w>1/2\\), it also equals the normally convergent meromorphic series with terms\n\\[\n\\frac{\\rho(1-\\rho)e^{t z_\\rho^2}}{w-z_\\rho}.\n\\]\nThe residue at a distinct centered zero is a nonzero function of \\(t\\) in \\(H\\). Indeed, the multiplicity and \\(\\rho(1-\\rho)\\) are nonzero, and the exponential never vanishes. A zero with positive real part would therefore give a nonremovable pole in the holomorphic domain. Symmetry now gives the hypothesis.\n\\end{proof}","sumed energy bound imply\n\\[\n\\int_0^\\infty\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du<\\infty\n\\qquad(a>0).\n\\]\nIts \\(H\\)-valued Laplace transform is holomorphic for \\(\\Re w>0\\). Initially for \\(\\Re w>1/2\\), it also equals the normally convergent meromorphic series with terms\n\\[\n\\frac{\\rho(1-\\rho)e^{t z_\\rho^2}}{w-z_\\rho}.\n\\]\nThe residue at a distinct centered zero is a nonzero function of \\(t\\) in \\(H\\). Indeed, the multiplicity and \\(\\rho(1-\\rho)\\) are nonzero, and the exponential never vanishes. A zero with positive real part would therefore give a nonremovable pole in the holomorphic domain. Symmetry now gives the hypothesis.\n\\end{proof}","infty\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du<\\infty\n\\qquad(a>0).\n\\]\nIts \\(H\\)-valued Laplace transform is holomorphic for \\(\\Re w>0\\). Initially for \\(\\Re w>1/2\\), it also equals the normally convergent meromorphic series with terms\n\\[\n\\frac{\\rho(1-\\rho)e^{t z_\\rho^2}}{w-z_\\rho}.\n\\]\nThe residue at a distinct centered zero is a nonzero function of \\(t\\) in \\(H\\). Indeed, the multiplicity and \\(\\rho(1-\\rho)\\) are nonzero, and the exponential never vanishes. A zero with positive real part would therefore give a nonremovable pole in the holomorphic domain. Symmetry now gives the hypothesis.\n\\end{proof}","The energy statement is deliberately local in \\(u\\). It does not assume a global square integrable packet, which would already rule out many possible growth behaviors. The Hilbert space \\(H\\) collects only the fixed positive heat interval. Tonelli's theorem identifies\n\\[\n\\widetilde E_J=\\int_J^{J+1}\n\\|\\widetilde\\Theta_\\cdot(u)\\|_H^2\\,du.\n\\]\nLet \\(a>0\\). On a unit band, Cauchy Schwarz gives\n\\[\n\\int_J^{J+1}\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du\n\\le e^{-aJ}\\widetilde E_J^{1/2}.\n\\]\nChoose the energy exponent to be \\(\\varepsilon=a\\). Subexponential growth then bounds the right side by a constant times \\(e^{-aJ/2}\\). The geometric sum converges. The compact initial band \\([0,1]\\) is harmless by normal convergence. This proves the weighted in","The energy statement is deliberately local in \\(u\\). It does not assume a global square integrable packet, which would already rule out many possible growth behaviors. The Hilbert space \\(H\\) collects only the fixed positive heat interval. Tonelli's theorem identifies\n\\[\n\\widetilde E_J=\\int_J^{J+1}\n\\|\\widetilde\\Theta_\\cdot(u)\\|_H^2\\,du.\n\\]\nLet \\(a>0\\). On a unit band, Cauchy Schwarz gives\n\\[\n\\int_J^{J+1}\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du\n\\le e^{-aJ}\\widetilde E_J^{1/2}.\n\\]\nChoose the energy exponent to be \\(\\varepsilon=a\\). Subexponential growth then bounds the right side by a constant times \\(e^{-aJ/2}\\). The geometric sum converges. The compact initial band \\([0,1]\\) is harmless by normal convergence. This proves the weighted integrability used in the theorem for every \\(a>0\\).","eady rule out many possible growth behaviors. The Hilbert space \\(H\\) collects only the fixed positive heat interval. Tonelli's theorem identifies\n\\[\n\\widetilde E_J=\\int_J^{J+1}\n\\|\\widetilde\\Theta_\\cdot(u)\\|_H^2\\,du.\n\\]\nLet \\(a>0\\). On a unit band, Cauchy Schwarz gives\n\\[\n\\int_J^{J+1}\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du\n\\le e^{-aJ}\\widetilde E_J^{1/2}.\n\\]\nChoose the energy exponent to be \\(\\varepsilon=a\\). Subexponential growth then bounds the right side by a constant times \\(e^{-aJ/2}\\). The geometric sum converges. The compact initial band \\([0,1]\\) is harmless by normal convergence. This proves the weighted integrability used in the theorem for every \\(a>0\\).","fixed positive heat interval. Tonelli's theorem identifies\n\\[\n\\widetilde E_J=\\int_J^{J+1}\n\\|\\widetilde\\Theta_\\cdot(u)\\|_H^2\\,du.\n\\]\nLet \\(a>0\\). On a unit band, Cauchy Schwarz gives\n\\[\n\\int_J^{J+1}\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du\n\\le e^{-aJ}\\widetilde E_J^{1/2}.\n\\]\nChoose the energy exponent to be \\(\\varepsilon=a\\). Subexponential growth then bounds the right side by a constant times \\(e^{-aJ/2}\\). The geometric sum converges. The compact initial band \\([0,1]\\) is harmless by normal convergence. This proves the weighted integrability used in the theorem for every \\(a>0\\).","lde E_J=\\int_J^{J+1}\n\\|\\widetilde\\Theta_\\cdot(u)\\|_H^2\\,du.\n\\]\nLet \\(a>0\\). On a unit band, Cauchy Schwarz gives\n\\[\n\\int_J^{J+1}\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du\n\\le e^{-aJ}\\widetilde E_J^{1/2}.\n\\]\nChoose the energy exponent to be \\(\\varepsilon=a\\). Subexponential growth then bounds the right side by a constant times \\(e^{-aJ/2}\\). The geometric sum converges. The compact initial band \\([0,1]\\) is harmless by normal convergence. This proves the weighted integrability used in the theorem for every \\(a>0\\).","[\n\\int_J^{J+1}\\|\\widetilde\\Theta_\\cdot(u)\\|_H e^{-au}\\,du\n\\le e^{-aJ}\\widetilde E_J^{1/2}.\n\\]\nChoose the energy exponent to be \\(\\varepsilon=a\\). Subexponential growth then bounds the right side by a constant times \\(e^{-aJ/2}\\). The geometric sum converges. The compact initial band \\([0,1]\\) is harmless by normal convergence. This proves the weighted integrability used in the theorem for every \\(a>0\\).","On compact subsets of \\(\\Re w>0\\), the same estimate absorbs any fixed power of \\(u\\). Thus the Bochner integral defining the \\(H\\)-valued Laplace transform can be differentiated with respect to \\(w\\), and the transform is holomorphic there. In its initial absolute convergence half plane, one can instead integrate the zero expansion. The compact positive heat interval provides one Gaussian majorant for all the vector coefficients.","Near a distinct centered zero \\(z_0\\) of multiplicity \\(m_0\\), the residue vector is\n\\[\nr_{z_0}(t)=m_0\\rho_0(1-\\rho_0)e^{tz_0^2}.\n\\]\nIt is continuous and never zero on \\([t_0,T]\\), so its \\(H\\) norm is positive. To reduce the contradiction to ordinary complex analysis, choose the continuous linear functional given by pairing with this residue vector. The scalar transform then has nonzero residue \\(\\|r_{z_0}\\|_H^2\\), whereas the Laplace integral remains holomorphic. This makes explicit why integrating over heat times cannot average the residue away.","ear a distinct centered zero \\(z_0\\) of multiplicity \\(m_0\\), the residue vector is\n\\[\nr_{z_0}(t)=m_0\\rho_0(1-\\rho_0)e^{tz_0^2}.\n\\]\nIt is continuous and never zero on \\([t_0,T]\\), so its \\(H\\) norm is positive. To reduce the contradiction to ordinary complex analysis, choose the continuous linear functional given by pairing with this residue vector. The scalar transform then has nonzero residue \\(\\|r_{z_0}\\|_H^2\\), whereas the Laplace integral remains holomorphic. This makes explicit why integrating over heat times cannot average the residue away.","The criterion survives the archimedean removal in the form needed for a future estimate. Since that channel has bounded unit band energy, the inequalities\n\\(\\|f-g\\|^2\\le2\\|f\\|^2+2\\|g\\|^2\\) and\n\\(\\|f\\|^2\\le2\\|f-g\\|^2+2\\|g\\|^2\\)\nshow that subtracting it preserves the property of subexponential energy. This uses bounded energy, not pointwise arithmetic cancellation. It leaves the full transformed prime expression as the place where an estimate must still be proved.","The prime packet is naturally expressed as a sum of translated Gaussians. Treating every integer translate as an independent coordinate produces an unstable frame at large integers. The centers \\(\\log n\\) and \\(\\log(n+1)\\) differ by about \\(1/n\\), much less than the fixed heat width. Their near coincidence is a coordinate problem which has an exact repair.","For \\(t>0\\) and a real center \\(a\\), define the normalized Gaussian\n\\begin{equation}\\label{rh:gaussian}\ng_{t,a}(u)=(2\\pi t)^{-1/4}e^{-(u-a)^2/(4t)}.\n\\end{equation}\nIt has norm one in \\(L^2(\\mathbb R,du)\\). For a chosen center \\(c\\), let \\(\\psi_{r,t,c}\\), \\(r=0,1,2,\\ldots\\), be the orthonormal scaled Hermite functions\n\\begin{equation}\\label{rh:hermite}\n\\psi_{r,t,c}(u)=\n\\frac{(2\\pi t)^{-1/4}}{\\sqrt{2^r r!}}\nH_r\\left(\\frac{u-c}{\\sqrt{2t}}\\right)e^{-(u-c)^2/(4t)}.\n\\end{equation}\nHere \\(H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}\\) is the physicists' Hermite polynomial. If \\(P_R\\) is orthogonal projection onto indices \\(0\\le r\\le R\\), the coherent expansion gives\n\\begin{align}\ng_{t,a}&=e^{-\\lambda/2}\n\\sum_{r\\","For \\(t>0\\) and a real center \\(a\\), define the normalized Gaussian\n\\begin{equation}\\label{rh:gaussian}\ng_{t,a}(u)=(2\\pi t)^{-1/4}e^{-(u-a)^2/(4t)}.\n\\end{equation}\nIt has norm one in \\(L^2(\\mathbb R,du)\\). For a chosen center \\(c\\), let \\(\\psi_{r,t,c}\\), \\(r=0,1,2,\\ldots\\), be the orthonormal scaled Hermite functions\n\\begin{equation}\\label{rh:hermite}\n\\psi_{r,t,c}(u)=\n\\frac{(2\\pi t)^{-1/4}}{\\sqrt{2^r r!}}\nH_r\\left(\\frac{u-c}{\\sqrt{2t}}\\right)e^{-(u-c)^2/(4t)}.\n\\end{equation}\nHere \\(H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}\\) is the physicists' Hermite polynomial. If \\(P_R\\) is orthogonal projection onto indices \\(0\\le r\\le R\\), the coherent expansion gives\n\\begin{align}\ng_{t,a}&=e^{-\\lambda/2}\n\\sum_{r\\ge0}\\frac{\\sigma^r}{\\sqr","For \\(t>0\\) and a real center \\(a\\), define the normalized Gaussian\n\\begin{equation}\\label{rh:gaussian}\ng_{t,a}(u)=(2\\pi t)^{-1/4}e^{-(u-a)^2/(4t)}.\n\\end{equation}\nIt has norm one in \\(L^2(\\mathbb R,du)\\). For a chosen center \\(c\\), let \\(\\psi_{r,t,c}\\), \\(r=0,1,2,\\ldots\\), be the orthonormal scaled Hermite functions\n\\begin{equation}\\label{rh:hermite}\n\\psi_{r,t,c}(u)=\n\\frac{(2\\pi t)^{-1/4}}{\\sqrt{2^r r!}}\nH_r\\left(\\frac{u-c}{\\sqrt{2t}}\\right)e^{-(u-c)^2/(4t)}.\n\\end{equation}\nHere \\(H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}\\) is the physicists' Hermite polynomial. If \\(P_R\\) is orthogonal projection onto indices \\(0\\le r\\le R\\), the coherent expansion gives\n\\begin{align}\ng_{t,a}&=e^{-\\lambda/2}\n\\sum_{r\\ge0}\\frac{\\sigma^r}{\\sqrt{r!}}\\psi_{r,t,c},\n\\label{rh:coherent}","bel{rh:gaussian}\ng_{t,a}(u)=(2\\pi t)^{-1/4}e^{-(u-a)^2/(4t)}.\n\\end{equation}\nIt has norm one in \\(L^2(\\mathbb R,du)\\). For a chosen center \\(c\\), let \\(\\psi_{r,t,c}\\), \\(r=0,1,2,\\ldots\\), be the orthonormal scaled Hermite functions\n\\begin{equation}\\label{rh:hermite}\n\\psi_{r,t,c}(u)=\n\\frac{(2\\pi t)^{-1/4}}{\\sqrt{2^r r!}}\nH_r\\left(\\frac{u-c}{\\sqrt{2t}}\\right)e^{-(u-c)^2/(4t)}.\n\\end{equation}\nHere \\(H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}\\) is the physicists' Hermite polynomial. If \\(P_R\\) is orthogonal projection onto indices \\(0\\le r\\le R\\), the coherent expansion gives\n\\begin{align}\ng_{t,a}&=e^{-\\lambda/2}\n\\sum_{r\\ge0}\\frac{\\sigma^r}{\\sqrt{r!}}\\psi_{r,t,c},\n\\label{rh:coherent}","rm one in \\(L^2(\\mathbb R,du)\\). For a chosen center \\(c\\), let \\(\\psi_{r,t,c}\\), \\(r=0,1,2,\\ldots\\), be the orthonormal scaled Hermite functions\n\\begin{equation}\\label{rh:hermite}\n\\psi_{r,t,c}(u)=\n\\frac{(2\\pi t)^{-1/4}}{\\sqrt{2^r r!}}\nH_r\\left(\\frac{u-c}{\\sqrt{2t}}\\right)e^{-(u-c)^2/(4t)}.\n\\end{equation}\nHere \\(H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}\\) is the physicists' Hermite polynomial. If \\(P_R\\) is orthogonal projection onto indices \\(0\\le r\\le R\\), the coherent expansion gives\n\\begin{align}\ng_{t,a}&=e^{-\\lambda/2}\n\\sum_{r\\ge0}\\frac{\\sigma^r}{\\sqrt{r!}}\\psi_{r,t,c},\n\\label{rh:coherent}","center \\(c\\), let \\(\\psi_{r,t,c}\\), \\(r=0,1,2,\\ldots\\), be the orthonormal scaled Hermite functions\n\\begin{equation}\\label{rh:hermite}\n\\psi_{r,t,c}(u)=\n\\frac{(2\\pi t)^{-1/4}}{\\sqrt{2^r r!}}\nH_r\\left(\\frac{u-c}{\\sqrt{2t}}\\right)e^{-(u-c)^2/(4t)}.\n\\end{equation}\nHere \\(H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}\\) is the physicists' Hermite polynomial. If \\(P_R\\) is orthogonal projection onto indices \\(0\\le r\\le R\\), the coherent expansion gives\n\\begin{align}\ng_{t,a}&=e^{-\\lambda/2}\n\\sum_{r\\ge0}\\frac{\\sigma^r}{\\sqrt{r!}}\\psi_{r,t,c},\n\\label{rh:coherent}","e the orthonormal scaled Hermite functions\n\\begin{equation}\\label{rh:hermite}\n\\psi_{r,t,c}(u)=\n\\frac{(2\\pi t)^{-1/4}}{\\sqrt{2^r r!}}\nH_r\\left(\\frac{u-c}{\\sqrt{2t}}\\right)e^{-(u-c)^2/(4t)}.\n\\end{equation}\nHere \\(H_r(x)=(-1)^r e^{x^2}(d^r/dx^r)e^{-x^2}\\) is the physicists' Hermite polynomial. If \\(P_R\\) is orthogonal projection onto indices \\(0\\le r\\le R\\), the coherent expansion gives\n\\begin{align}\ng_{t,a}&=e^{-\\lambda/2}\n\\sum_{r\\ge0}\\frac{\\sigma^r}{\\sqrt{r!}}\\psi_{r,t,c},\n\\label{rh:coherent}\\\\\n\\sigma&=\\frac{a-c}{2\\sqrt t},\\qquad \\lambda=\\sigma^2,\n\\notag\\\\\n\\|(I-P_R)g_{t,a}\\|_2^2\n&=e^{-\\lambda}\\sum_{r>R}\\frac{\\lambda^r}{r!}.\n\\label{rh:poisson}\n\\end{align}\nThe last expression is a Poisson upper tail. These formulas and the compression estimates are recovered manuscript results \\cite{KomorebiHermite}.","The word physical refers to the measure \\(du\\) in which the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap\n\\begin{equation}\\label{rh:overlap}\n\\langle g_{t,a},g_{t,b}\\rangle\n=e^{-(a-b)^2/(8t)}.\n\\end{equation}\nFor two translates, the Gram matrix has diagonal entries one and off diagonal entry \\(\\kappa=e^{-(a-b)^2/(8t)}\\). Its eigenvalues are \\(1+\\kappa\\) and \\(1-\\kappa\\). When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients are individually large.","e word physical refers to the measure \\(du\\) in which the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap\n\\begin{equation}\\label{rh:overlap}\n\\langle g_{t,a},g_{t,b}\\rangle\n=e^{-(a-b)^2/(8t)}.\n\\end{equation}\nFor two translates, the Gram matrix has diagonal entries one and off diagonal entry \\(\\kappa=e^{-(a-b)^2/(8t)}\\). Its eigenvalues are \\(1+\\kappa\\) and \\(1-\\kappa\\). When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients are individually large.","ich the packet is actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap\n\\begin{equation}\\label{rh:overlap}\n\\langle g_{t,a},g_{t,b}\\rangle\n=e^{-(a-b)^2/(8t)}.\n\\end{equation}\nFor two translates, the Gram matrix has diagonal entries one and off diagonal entry \\(\\kappa=e^{-(a-b)^2/(8t)}\\). Its eigenvalues are \\(1+\\kappa\\) and \\(1-\\kappa\\). When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients are individually large.","actually evaluated. It does not mean that a separately chosen coefficient norm can be substituted. Completing the square gives the exact overlap\n\\begin{equation}\\label{rh:overlap}\n\\langle g_{t,a},g_{t,b}\\rangle\n=e^{-(a-b)^2/(8t)}.\n\\end{equation}\nFor two translates, the Gram matrix has diagonal entries one and off diagonal entry \\(\\kappa=e^{-(a-b)^2/(8t)}\\). Its eigenvalues are \\(1+\\kappa\\) and \\(1-\\kappa\\). When the centers are close, their difference is almost invisible in the physical norm even if the two raw coefficients are individually large.","For example, take \\(t=1\\), \\(a=\\log1000\\), and \\(b=\\log1001\\). The separation is approximately \\(0.00099950\\), the overlap is approximately \\(0.999999875\\), and the Gram condition number \\((1+\\kappa)/(1-\\kappa)\\) is about \\(1.60\\times10^7\\). These values diagnose a coordinate system, not the size of a zeta coefficient. The asymptotic explanation is exact:\n\\[\n1-e^{-(a-b)^2/(8t)}\\sim\\frac{(a-b)^2}{8t},\n\\qquad \\log(n+1)-\\log n\\sim\\frac1n.\n\\]\nConsecutive integer centers therefore produce a Gram condition number of order \\(n^2\\) at fixed heat width.","s approximately \\(0.999999875\\), and the Gram condition number \\((1+\\kappa)/(1-\\kappa)\\) is about \\(1.60\\times10^7\\). These values diagnose a coordinate system, not the size of a zeta coefficient. The asymptotic explanation is exact:\n\\[\n1-e^{-(a-b)^2/(8t)}\\sim\\frac{(a-b)^2}{8t},\n\\qquad \\log(n+1)-\\log n\\sim\\frac1n.\n\\]\nConsecutive integer centers therefore produce a Gram condition number of order \\(n^2\\) at fixed heat width.","An orthonormal Hermite basis replaces this almost singular coordinate system. If a finite packet is \\(f=\\sum_r a_r\\psi_{r,t,c}\\), Parseval's identity gives \\(\\|f\\|_2^2=\\sum_r|a_r|^2\\) with no unknown Gram loss. This exact equality is the repair. It remains possible for the arithmetic coefficients \\(a_r\\) to be large, or to contain a conjecture level obstruction in one of the first modes.","Put \\(x=(u-c)/\\sqrt{2t}\\). The ratio of a translated Gaussian to the centered one is\n\\[\n\\frac{g_{t,a}(u)}{g_{t,c}(u)}\n=e^{\\sqrt2\\sigma x-\\sigma^2}.\n\\]\nThe Hermite generating identity\n\\(e^{2xv-v^2}=\\sum_{r\\ge0}H_r(x)v^r/r!\\), with\n\\(v=\\sigma/\\sqrt2\\), turns this ratio into\n\\[\ne^{-\\sigma^2/2}\\sum_{r\\ge0}\n\\frac{H_r(x)\\sigma^r}{2^{r/2}r!}.\n\\]\nSubstitution of the normalized Hermite functions gives \\eqref{rh:coherent}. Squaring its orthogonal coefficients gives\n\\(e^{-\\lambda}\\lambda^r/r!\\). Their total is one because the exponential series sums to \\(e^\\lambda\\). The discarded squared norm is therefore exactly a Poisson probability, rather than a bound inferred from a plot.","to the centered one is\n\\[\n\\frac{g_{t,a}(u)}{g_{t,c}(u)}\n=e^{\\sqrt2\\sigma x-\\sigma^2}.\n\\]\nThe Hermite generating identity\n\\(e^{2xv-v^2}=\\sum_{r\\ge0}H_r(x)v^r/r!\\), with\n\\(v=\\sigma/\\sqrt2\\), turns this ratio into\n\\[\ne^{-\\sigma^2/2}\\sum_{r\\ge0}\n\\frac{H_r(x)\\sigma^r}{2^{r/2}r!}.\n\\]\nSubstitution of the normalized Hermite functions gives \\eqref{rh:coherent}. Squaring its orthogonal coefficients gives\n\\(e^{-\\lambda}\\lambda^r/r!\\). Their total is one because the exponential series sums to \\(e^\\lambda\\). The discarded squared norm is therefore exactly a Poisson probability, rather than a bound inferred from a plot.","The Hermite generating identity\n\\(e^{2xv-v^2}=\\sum_{r\\ge0}H_r(x)v^r/r!\\), with\n\\(v=\\sigma/\\sqrt2\\), turns this ratio into\n\\[\ne^{-\\sigma^2/2}\\sum_{r\\ge0}\n\\frac{H_r(x)\\sigma^r}{2^{r/2}r!}.\n\\]\nSubstitution of the normalized Hermite functions gives \\eqref{rh:coherent}. Squaring its orthogonal coefficients gives\n\\(e^{-\\lambda}\\lambda^r/r!\\). Their total is one because the exponential series sums to \\(e^\\lambda\\). The discarded squared norm is therefore exactly a Poisson probability, rather than a bound inferred from a plot.","As finite examples, the following entries are rounded values of that exact tail. The cutoff includes all indices from zero through \\(R\\).\n\\begin{center}\n\\begin{tabular}{rrrr}\n\\toprule\n\\(\\lambda\\)&\\(R=4\\)&\\(R=8\\)&\\(R=12\\)","1&\\(3.66\\times10^{-3}\\)&\\(1.13\\times10^{-6}\\)&\\(6.36\\times10^{-11}\\)","4&\\(3.71\\times10^{-1}\\)&\\(2.14\\times10^{-2}\\)&\\(2.74\\times10^{-4}\\)","Suppose a fixed positive width constant \\(C>0\\) is chosen and the active centers in a shell satisfy\n\\(|a-c_J|\\le C\\sqrt J+O(1)\\) and \\(t\\ge t_0>0\\). Put\n\\(\\alpha=C^2/(4t_0)\\). Then \\(\\lambda\\le\\alpha J+O(\\sqrt J)\\). For a fixed constant \\(q>\\alpha\\), projection through \\(R_J=\\lceil qJ\\rceil\\) has norm error bounded by\n\\begin{equation}\\label{rh:rate}\n\\exp\\left(-\\frac12 I(q\\mid\\alpha)J+o(J)\\right),\n\\qquad I(q\\mid\\alpha)=q\\log(q/\\alpha)-q+\\alpha.\n\\end{equation}\nThe coefficient mass of the positive prime packet through\n\\(X_J=\\exp(J+O(\\sqrt J))\\) is at most\n\\[\n\\sum_{n\\le X_J}\\frac{\\Lambda(n)}{\\sqrt n}\n\\le2\\sqrt{X_J}\\log X_J=\\exp(J/2+o(J)).\n\\]\nThus \\(I(q\\mid\\alpha)>1\\) makes the aggregate compression error exponentially small. The pole annih","Suppose a fixed positive width constant \\(C>0\\) is chosen and the active centers in a shell satisfy\n\\(|a-c_J|\\le C\\sqrt J+O(1)\\) and \\(t\\ge t_0>0\\). Put\n\\(\\alpha=C^2/(4t_0)\\). Then \\(\\lambda\\le\\alpha J+O(\\sqrt J)\\). For a fixed constant \\(q>\\alpha\\), projection through \\(R_J=\\lceil qJ\\rceil\\) has norm error bounded by\n\\begin{equation}\\label{rh:rate}\n\\exp\\left(-\\frac12 I(q\\mid\\alpha)J+o(J)\\right),\n\\qquad I(q\\mid\\alpha)=q\\log(q/\\alpha)-q+\\alpha.\n\\end{equation}\nThe coefficient mass of the positive prime packet through\n\\(X_J=\\exp(J+O(\\sqrt J))\\) is at most\n\\[\n\\sum_{n\\le X_J}\\frac{\\Lambda(n)}{\\sqrt n}\n\\le2\\sqrt{X_J}\\log X_J=\\exp(J/2+o(J)).\n\\]\nThus \\(I(q\\mid\\alpha)>1\\) makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leaves this exponen","Suppose a fixed positive width constant \\(C>0\\) is chosen and the active centers in a shell satisfy\n\\(|a-c_J|\\le C\\sqrt J+O(1)\\) and \\(t\\ge t_0>0\\). Put\n\\(\\alpha=C^2/(4t_0)\\). Then \\(\\lambda\\le\\alpha J+O(\\sqrt J)\\). For a fixed constant \\(q>\\alpha\\), projection through \\(R_J=\\lceil qJ\\rceil\\) has norm error bounded by\n\\begin{equation}\\label{rh:rate}\n\\exp\\left(-\\frac12 I(q\\mid\\alpha)J+o(J)\\right),\n\\qquad I(q\\mid\\alpha)=q\\log(q/\\alpha)-q+\\alpha.\n\\end{equation}\nThe coefficient mass of the positive prime packet through\n\\(X_J=\\exp(J+O(\\sqrt J))\\) is at most\n\\[\n\\sum_{n\\le X_J}\\frac{\\Lambda(n)}{\\sqrt n}\n\\le2\\sqrt{X_J}\\log X_J=\\exp(J/2+o(J)).\n\\]\nThus \\(I(q\\mid\\alpha)>1\\) makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leaves this exponential conclusion unc","Suppose a fixed positive width constant \\(C>0\\) is chosen and the active centers in a shell satisfy\n\\(|a-c_J|\\le C\\sqrt J+O(1)\\) and \\(t\\ge t_0>0\\). Put\n\\(\\alpha=C^2/(4t_0)\\). Then \\(\\lambda\\le\\alpha J+O(\\sqrt J)\\). For a fixed constant \\(q>\\alpha\\), projection through \\(R_J=\\lceil qJ\\rceil\\) has norm error bounded by\n\\begin{equation}\\label{rh:rate}\n\\exp\\left(-\\frac12 I(q\\mid\\alpha)J+o(J)\\right),\n\\qquad I(q\\mid\\alpha)=q\\log(q/\\alpha)-q+\\alpha.\n\\end{equation}\nThe coefficient mass of the positive prime packet through\n\\(X_J=\\exp(J+O(\\sqrt J))\\) is at most\n\\[\n\\sum_{n\\le X_J}\\frac{\\Lambda(n)}{\\sqrt n}\n\\le2\\sqrt{X_J}\\log X_J=\\exp(J/2+o(J)).\n\\]\nThus \\(I(q\\mid\\alpha)>1\\) makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leaves this exponential conclusion unchanged.","(1)\\) and \\(t\\ge t_0>0\\). Put\n\\(\\alpha=C^2/(4t_0)\\). Then \\(\\lambda\\le\\alpha J+O(\\sqrt J)\\). For a fixed constant \\(q>\\alpha\\), projection through \\(R_J=\\lceil qJ\\rceil\\) has norm error bounded by\n\\begin{equation}\\label{rh:rate}\n\\exp\\left(-\\frac12 I(q\\mid\\alpha)J+o(J)\\right),\n\\qquad I(q\\mid\\alpha)=q\\log(q/\\alpha)-q+\\alpha.\n\\end{equation}\nThe coefficient mass of the positive prime packet through\n\\(X_J=\\exp(J+O(\\sqrt J))\\) is at most\n\\[\n\\sum_{n\\le X_J}\\frac{\\Lambda(n)}{\\sqrt n}\n\\le2\\sqrt{X_J}\\log X_J=\\exp(J/2+o(J)).\n\\]\nThus \\(I(q\\mid\\alpha)>1\\) makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leaves this exponential conclusion unchanged.","=C^2/(4t_0)\\). Then \\(\\lambda\\le\\alpha J+O(\\sqrt J)\\). For a fixed constant \\(q>\\alpha\\), projection through \\(R_J=\\lceil qJ\\rceil\\) has norm error bounded by\n\\begin{equation}\\label{rh:rate}\n\\exp\\left(-\\frac12 I(q\\mid\\alpha)J+o(J)\\right),\n\\qquad I(q\\mid\\alpha)=q\\log(q/\\alpha)-q+\\alpha.\n\\end{equation}\nThe coefficient mass of the positive prime packet through\n\\(X_J=\\exp(J+O(\\sqrt J))\\) is at most\n\\[\n\\sum_{n\\le X_J}\\frac{\\Lambda(n)}{\\sqrt n}\n\\le2\\sqrt{X_J}\\log X_J=\\exp(J/2+o(J)).\n\\]\nThus \\(I(q\\mid\\alpha)>1\\) makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leaves this exponential conclusion unchanged.","ojection through \\(R_J=\\lceil qJ\\rceil\\) has norm error bounded by\n\\begin{equation}\\label{rh:rate}\n\\exp\\left(-\\frac12 I(q\\mid\\alpha)J+o(J)\\right),\n\\qquad I(q\\mid\\alpha)=q\\log(q/\\alpha)-q+\\alpha.\n\\end{equation}\nThe coefficient mass of the positive prime packet through\n\\(X_J=\\exp(J+O(\\sqrt J))\\) is at most\n\\[\n\\sum_{n\\le X_J}\\frac{\\Lambda(n)}{\\sqrt n}\n\\le2\\sqrt{X_J}\\log X_J=\\exp(J/2+o(J)).\n\\]\nThus \\(I(q\\mid\\alpha)>1\\) makes the aggregate compression error exponentially small. The pole annihilator adds only polynomial factors to the Hermite ladder and leaves this exponential conclusion unchanged.","There are only \\(O(J)\\) retained coordinates, and they are orthogonal in the physical metric. This closes a conditioning problem. It does not make the arithmetic coefficients small.","The exponent in \\eqref{rh:rate} comes from a standard elementary exponential bound for a Poisson variable \\(X\\) of mean \\(\\lambda\\). For \\(h>0\\),\n\\[\n\\Pr(X\\ge R)\\le e^{-hR}\\mathbb E(e^{hX})\n=\\exp\\bigl(-hR+\\lambda(e^h-1)\\bigr).\n\\]\nWhen \\(R>\\lambda>0\\), choose \\(e^h=R/\\lambda\\). For \\(\\lambda=0\\), every positive tail is zero. The positive mean case gives the rate\n\\(R\\log(R/\\lambda)-R+\\lambda\\). Passing to the squared norm tail and then taking a square root supplies the factor one half in \\eqref{rh:rate}. Changing \\(R\\) by one has no effect on its leading shell exponent.","n \\(\\lambda\\). For \\(h>0\\),\n\\[\n\\Pr(X\\ge R)\\le e^{-hR}\\mathbb E(e^{hX})\n=\\exp\\bigl(-hR+\\lambda(e^h-1)\\bigr).\n\\]\nWhen \\(R>\\lambda>0\\), choose \\(e^h=R/\\lambda\\). For \\(\\lambda=0\\), every positive tail is zero. The positive mean case gives the rate\n\\(R\\log(R/\\lambda)-R+\\lambda\\). Passing to the squared norm tail and then taking a square root supplies the factor one half in \\eqref{rh:rate}. Changing \\(R\\) by one has no effect on its leading shell exponent.","For a concrete choice, let \\(C=2\\sqrt{t_0}\\), so \\(\\alpha=1\\), and take \\(q=4\\). Then\n\\[\nI(4\\mid1)=4\\log4-3\\approx2.54518>1.\n\\]\nThe individual norm error decays with exponent approximately \\(-1.27259J\\). Multiplication by a coefficient mass with exponent at most \\(J/2+o(J)\\) still leaves decay with exponent approximately \\(-0.77259J\\). This example verifies the scale balance in the compression argument. It does not give a bound on the retained coordinates. For different shell widths or heat minima, one must recompute \\(\\alpha\\) and choose \\(q\\) accordingly.","\\]\nThe individual norm error decays with exponent approximately \\(-1.27259J\\). Multiplication by a coefficient mass with exponent at most \\(J/2+o(J)\\) still leaves decay with exponent approximately \\(-0.77259J\\). This example verifies the scale balance in the compression argument. It does not give a bound on the retained coordinates. For different shell widths or heat minima, one must recompute \\(\\alpha\\) and choose \\(q\\) accordingly.","error decays with exponent approximately \\(-1.27259J\\). Multiplication by a coefficient mass with exponent at most \\(J/2+o(J)\\) still leaves decay with exponent approximately \\(-0.77259J\\). This example verifies the scale balance in the compression argument. It does not give a bound on the retained coordinates. For different shell widths or heat minima, one must recompute \\(\\alpha\\) and choose \\(q\\) accordingly.","The simple coefficient mass bound also has an elementary proof. Since \\(\\Lambda(n)\\le\\log X\\) for \\(n\\le X\\), and \\(\\sum_{n\\le X}n^{-1/2}\\le2\\sqrt X\\), their product gives the stated estimate. Compression may therefore be arranged without first proving any delicate prime cancellation theorem. The remaining arithmetic estimate lies in the compressed packet, not in the omitted Gaussian tail.","The ground channel shows why the distinction matters. Direct integration gives\n\\begin{equation}\\label{rh:ground}\n\\int_{\\mathbb R}g_{s,c}(u)\\widetilde\\Theta_t(u)\\,du\n=K_s\\widetilde\\Theta_{t+s}(c),\n\\qquad K_s=2\\sqrt{\\pi s}(2\\pi s)^{-1/4}.\n\\end{equation}\nThe Gaussian test makes this integral absolutely convergent even without the conjecture. It is a Gaussian pairing, with no assertion that the unweighted zero packet belongs to \\(L^2(\\mathbb R)\\). Completing the square proves the identity term by term. Normal convergence and the Gaussian majorant justify summation \\cite{KomorebiGround}.","he distinction matters. Direct integration gives\n\\begin{equation}\\label{rh:ground}\n\\int_{\\mathbb R}g_{s,c}(u)\\widetilde\\Theta_t(u)\\,du\n=K_s\\widetilde\\Theta_{t+s}(c),\n\\qquad K_s=2\\sqrt{\\pi s}(2\\pi s)^{-1/4}.\n\\end{equation}\nThe Gaussian test makes this integral absolutely convergent even without the conjecture. It is a Gaussian pairing, with no assertion that the unweighted zero packet belongs to \\(L^2(\\mathbb R)\\). Completing the square proves the identity term by term. Normal convergence and the Gaussian majorant justify summation \\cite{KomorebiGround}.","To verify \\eqref{rh:ground}, the scalar Gaussian integral is\n\\[\n\\int_{\\mathbb R}(2\\pi s)^{-1/4}\ne^{-(u-c)^2/(4s)}e^{uz}\\,du\n=2\\sqrt{\\pi s}(2\\pi s)^{-1/4}e^{cz+sz^2}.\n\\]\nIt holds for every complex \\(z\\), either by completing the square first for real \\(z\\) and continuing analytically, or by the entire Gaussian transform. Multiplying by the mode coefficient at heat time \\(t\\) replaces \\(t\\) by \\(t+s\\). The annihilator weight is unchanged. Absolute values before integration are bounded using \\(|\\Re z|<1/2\\) and Gaussian decay in the ordinate, so the sum and integral can be interchanged without the Riemann hypothesis.","e^{-(u-c)^2/(4s)}e^{uz}\\,du\n=2\\sqrt{\\pi s}(2\\pi s)^{-1/4}e^{cz+sz^2}.\n\\]\nIt holds for every complex \\(z\\), either by completing the square first for real \\(z\\) and continuing analytically, or by the entire Gaussian transform. Multiplying by the mode coefficient at heat time \\(t\\) replaces \\(t\\) by \\(t+s\\). The annihilator weight is unchanged. Absolute values before integration are bounded using \\(|\\Re z|<1/2\\) and Gaussian decay in the ordinate, so the sum and integral can be interchanged without the Riemann hypothesis.","The parameter \\(s\\) in this display is a positive test width, not the complex variable of \\(\\zeta\\). At center \\(c=0\\), the pairing reads the packet at its origin after additional smoothing. At large \\(c\\), it reads precisely the translations where an off line zero could create growth. Smoothing delays and rescales that mode but cannot make its coefficient vanish. The ground channel is therefore a particularly direct test of whether a proposed coordinate argument has preserved the original question.","The Möbius function \\(\\mu(n)\\) is zero when a square of a prime divides \\(n\\). Otherwise it equals \\((-1)^k\\), where \\(k\\) is the number of distinct prime divisors of \\(n\\), and \\(\\mu(1)=1\\). For an integer \\(K\\ge1\\), put\n\\begin{equation}\\label{rh:w}\nw_K(n)=\\frac1{n^2}\\left(1-\\frac1{n^2}\\right)^K,\n\\qquad c_K=\\sum_{n\\ge1}\\mu(n)w_K(n).\n\\end{equation}\nThe series is absolutely convergent. Its natural active scale is \\(n\\) of order \\(M=\\sqrt K\\). The coefficient \\(c_K\\) is classical. Báez Duarte proved the following exact criterion \\cite{BaezDuarte}.","n \\(\\mu(n)\\) is zero when a square of a prime divides \\(n\\). Otherwise it equals \\((-1)^k\\), where \\(k\\) is the number of distinct prime divisors of \\(n\\), and \\(\\mu(1)=1\\). For an integer \\(K\\ge1\\), put\n\\begin{equation}\\label{rh:w}\nw_K(n)=\\frac1{n^2}\\left(1-\\frac1{n^2}\\right)^K,\n\\qquad c_K=\\sum_{n\\ge1}\\mu(n)w_K(n).\n\\end{equation}\nThe series is absolutely convergent. Its natural active scale is \\(n\\) of order \\(M=\\sqrt K\\). The coefficient \\(c_K\\) is classical. Báez Duarte proved the following exact criterion \\cite{BaezDuarte}.","zero when a square of a prime divides \\(n\\). Otherwise it equals \\((-1)^k\\), where \\(k\\) is the number of distinct prime divisors of \\(n\\), and \\(\\mu(1)=1\\). For an integer \\(K\\ge1\\), put\n\\begin{equation}\\label{rh:w}\nw_K(n)=\\frac1{n^2}\\left(1-\\frac1{n^2}\\right)^K,\n\\qquad c_K=\\sum_{n\\ge1}\\mu(n)w_K(n).\n\\end{equation}\nThe series is absolutely convergent. Its natural active scale is \\(n\\) of order \\(M=\\sqrt K\\). The coefficient \\(c_K\\) is classical. Báez Duarte proved the following exact criterion \\cite{BaezDuarte}.","divides \\(n\\). Otherwise it equals \\((-1)^k\\), where \\(k\\) is the number of distinct prime divisors of \\(n\\), and \\(\\mu(1)=1\\). For an integer \\(K\\ge1\\), put\n\\begin{equation}\\label{rh:w}\nw_K(n)=\\frac1{n^2}\\left(1-\\frac1{n^2}\\right)^K,\n\\qquad c_K=\\sum_{n\\ge1}\\mu(n)w_K(n).\n\\end{equation}\nThe series is absolutely convergent. Its natural active scale is \\(n\\) of order \\(M=\\sqrt K\\). The coefficient \\(c_K\\) is classical. Báez Duarte proved the following exact criterion \\cite{BaezDuarte}.","\\begin{theorem}[Báez Duarte criterion]\\label{rh:bd}\nThe Riemann hypothesis holds if and only if, for every \\(\\varepsilon>0\\),\n\\begin{equation}\\label{rh:bdcriterion}\n|c_K|=O_\\varepsilon(K^{-3/4+\\varepsilon}).\n\\end{equation}\nEquivalently, for every \\(\\varepsilon>0\\),\n\\(|c_{M^2}|=O_\\varepsilon(M^{-3/2+\\varepsilon})\\), with \\(M\\) restricted to square roots of positive integers.\n\\end{theorem}","The change in \\(\\varepsilon\\) between these equivalent formulations is harmless because each is quantified over every positive \\(\\varepsilon\\). We do not replace this with the stronger estimate having \\(\\varepsilon=0\\).","Although \\(c_K\\) is written as an infinite arithmetic series, each fixed \\(K\\) has an exact finite expression in even zeta values. Expand its power by the binomial theorem:\n\\begin{equation}\\label{rh:binomial}\nc_K=\\sum_{j=0}^K(-1)^j\\binom Kj\\frac1{\\zeta(2j+2)}.\n\\end{equation}\nTo justify this identity, note first that the expansion has only \\(K+1\\) terms. Each inner series \\(\\sum\\mu(n)n^{-2j-2}\\) is absolutely convergent. Its Euler product is \\(\\prod_p(1-p^{-2j-2})\\), which equals the reciprocal zeta value in the same half plane of absolute convergence. Thus no limiting interchange near the critical line is involved.","hmetic series, each fixed \\(K\\) has an exact finite expression in even zeta values. Expand its power by the binomial theorem:\n\\begin{equation}\\label{rh:binomial}\nc_K=\\sum_{j=0}^K(-1)^j\\binom Kj\\frac1{\\zeta(2j+2)}.\n\\end{equation}\nTo justify this identity, note first that the expansion has only \\(K+1\\) terms. Each inner series \\(\\sum\\mu(n)n^{-2j-2}\\) is absolutely convergent. Its Euler product is \\(\\prod_p(1-p^{-2j-2})\\), which equals the reciprocal zeta value in the same half plane of absolute convergence. Thus no limiting interchange near the critical line is involved.","For example,\n\\[\nc_1=\\frac1{\\zeta(2)}-\\frac1{\\zeta(4)},\\qquad\nc_2=\\frac1{\\zeta(2)}-\\frac2{\\zeta(4)}+\\frac1{\\zeta(6)}.\n\\]\nThese formulas are exact. At large \\(K\\), their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at \\(n=N\\), the elementary certified remainder bound is instead\n\\[\n\\left|c_K-\\sum_{n\\le N}\\mu(n)w_K(n)\\right|\n\\le\\sum_{n>N}n^{-2}\\le\\frac1N.\n\\]\nThis bound is useful for checking a finite computation, but it is much too coarse to prove the asymptotic criterion by itself.","zeta(2)}-\\frac1{\\zeta(4)},\\qquad\nc_2=\\frac1{\\zeta(2)}-\\frac2{\\zeta(4)}+\\frac1{\\zeta(6)}.\n\\]\nThese formulas are exact. At large \\(K\\), their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at \\(n=N\\), the elementary certified remainder bound is instead\n\\[\n\\left|c_K-\\sum_{n\\le N}\\mu(n)w_K(n)\\right|\n\\le\\sum_{n>N}n^{-2}\\le\\frac1N.\n\\]\nThis bound is useful for checking a finite computation, but it is much too coarse to prove the asymptotic criterion by itself.","\\zeta(4)}+\\frac1{\\zeta(6)}.\n\\]\nThese formulas are exact. At large \\(K\\), their alternating binomial terms can be individually much larger than their sum, so straightforward floating point evaluation may lose accuracy. A finite expression is not automatically a stable evaluation scheme. For a direct truncation of the original series at \\(n=N\\), the elementary certified remainder bound is instead\n\\[\n\\left|c_K-\\sum_{n\\le N}\\mu(n)w_K(n)\\right|\n\\le\\sum_{n>N}n^{-2}\\le\\frac1N.\n\\]\nThis bound is useful for checking a finite computation, but it is much too coarse to prove the asymptotic criterion by itself.","The active scale follows from the weight. For a real variable \\(x>1\\), write \\(y=x^{-2}\\). Then \\(w_K(x)=y(1-y)^K\\), and differentiation in \\(y\\) puts its maximum at \\(y=1/(K+1)\\). The maximum of the continuously extended weight is therefore at \\(x=\\sqrt{K+1}\\), close to \\(M=\\sqrt K\\). For fixed \\(r>0\\), as \\(M\\) tends to infinity through square roots of integers,\n\\[\nM^2w_{M^2}(rM)\n=r^{-2}\\left(1-\\frac1{r^2M^2}\\right)^{M^2}\n\\longrightarrow r^{-2}e^{-1/r^2}.\n\\]\nThis is a scaling description of the positive weight. It says nothing about the sign distribution of \\(\\mu(n)\\) across its active window.","y=1/(K+1)\\). The maximum of the continuously extended weight is therefore at \\(x=\\sqrt{K+1}\\), close to \\(M=\\sqrt K\\). For fixed \\(r>0\\), as \\(M\\) tends to infinity through square roots of integers,\n\\[\nM^2w_{M^2}(rM)\n=r^{-2}\\left(1-\\frac1{r^2M^2}\\right)^{M^2}\n\\longrightarrow r^{-2}e^{-1/r^2}.\n\\]\nThis is a scaling description of the positive weight. It says nothing about the sign distribution of \\(\\mu(n)\\) across its active window.","The size difference behind the criterion is now visible. There are on the order of \\(M\\) integers in a window of length proportional to \\(M\\), and a typical weight there has order \\(M^{-2}\\). An unsigned mass therefore naturally has order \\(M^{-1}\\). The desired signed amplitude has the smaller scale \\(M^{-3/2+\\varepsilon}\\). This count is only orientation. The barrier proof below supplies a rigorous lower bound for the actual full absolute majorant, with its squarefree support and cube dependence included.","Let \\(Q\\) be a squarefree positive integer, and let \\(d=\\omega(Q)\\) be its number of distinct prime divisors. Its alternating prime cube acts on the weight by\n\\begin{equation}\\label{rh:deltaq}\n\\Delta_Qw_K(m)=\\sum_{b\\mid Q}\\mu(b)w_K(bm).\n\\end{equation}\nHere \\(b\\) ranges over all positive divisors of \\(Q\\). The result is a sum of \\(2^d\\) terms, one for each subset of the prime coordinates.","\\begin{proposition}[Exact cube decomposition]\\label{rh:cube}\nFor every \\(K\\ge1\\) and squarefree \\(Q\\),\n\\begin{equation}\\label{rh:cubeidentity}\nc_K=\\sum_{\\substack{m\\ge1","\\begin{proposition}[Exact cube decomposition]\\label{rh:cube}\nFor every \\(K\\ge1\\) and squarefree \\(Q\\),\n\\begin{equation}\\label{rh:cubeidentity}\nc_K=\\sum_{\\substack{m\\ge1\\\\(m,Q)=1}}\n\\mu(m)\\Delta_Qw_K(m).\n\\end{equation}\nThe condition \\((m,Q)=1\\) means that \\(m\\) and \\(Q\\) have greatest common divisor one.\n\\end{proposition}","(m,Q)=1}}\n\\mu(m)\\Delta_Qw_K(m).\n\\end{equation}\nThe condition \\((m,Q)=1\\) means that \\(m\\) and \\(Q\\) have greatest common divisor one.\n\\end{proposition}","\\begin{proof}\nEvery squarefree integer \\(n\\) has a unique expression \\(n=bm\\), where \\(b\\mid Q\\) and \\((m,Q)=1\\). For these factors, \\(\\mu(n)=\\mu(b)\\mu(m)\\). Integers which are not squarefree contribute zero. Splitting the absolutely convergent series according to this expression gives \\eqref{rh:cubeidentity}.\n\\end{proof}","The cube creates an exact alternating difference, but the remaining bases still carry \\(\\mu(m)\\). The next section measures what happens if those signs are removed.","For \\(Q=6\\), the divisors correspond to the subsets of \\(\\{2,3\\}\\), and the exact packet is\n\\[\n\\Delta_6w_K(m)=w_K(m)-w_K(2m)-w_K(3m)+w_K(6m).\n\\]\nFor \\(Q=30\\), one adds a prime coordinate:\n\\begin{align*}\n\\Delta_{30}w_K(m)={}&w_K(m)-w_K(2m)-w_K(3m)-w_K(5m)","For \\(Q=6\\), the divisors correspond to the subsets of \\(\\{2,3\\}\\), and the exact packet is\n\\[\n\\Delta_6w_K(m)=w_K(m)-w_K(2m)-w_K(3m)+w_K(6m).\n\\]\nFor \\(Q=30\\), one adds a prime coordinate:\n\\begin{align*}\n\\Delta_{30}w_K(m)={}&w_K(m)-w_K(2m)-w_K(3m)-w_K(5m)\\\\\n&+w_K(6m)+w_K(10m)+w_K(15m)-w_K(30m).\n\\end{align*}\nThe complete divisor signs are shown here to make the grouping unambiguous.\n\\begin{center}\n\\begin{tabular}{rrrrrrrrr}\n\\toprule\n\\(b\\)&1&2&3&5&6&10&15&30","&+w_K(6m)+w_K(10m)+w_K(15m)-w_K(30m).\n\\end{align*}\nThe complete divisor signs are shown here to make the grouping unambiguous.\n\\begin{center}\n\\begin{tabular}{rrrrrrrrr}\n\\toprule\n\\(b\\)&1&2&3&5&6&10&15&30","\\midrule\n\\(\\mu(b)\\)&1&\\(-1\\)&\\(-1\\)&\\(-1\\)&1&1&1&\\(-1\\)","\\bottomrule\n\\end{tabular}\n\\end{center}\nThe \\(Q=6\\) row uses only columns \\(1,2,3,6\\). The condition on the remaining base matters: \\(m\\) must be coprime to the chosen \\(Q\\). For example, the squarefree integer \\(42\\) becomes \\(6\\cdot7\\) for both cubes. The integer \\(35\\) becomes \\(1\\cdot35\\) for \\(Q=6\\), but \\(5\\cdot7\\) for \\(Q=30\\). In each case the coefficient factors correctly. A nonsquarefree integer such as \\(49\\) contributes zero in the original sum and remains zero in the grouped sum.","abular}\n\\end{center}\nThe \\(Q=6\\) row uses only columns \\(1,2,3,6\\). The condition on the remaining base matters: \\(m\\) must be coprime to the chosen \\(Q\\). For example, the squarefree integer \\(42\\) becomes \\(6\\cdot7\\) for both cubes. The integer \\(35\\) becomes \\(1\\cdot35\\) for \\(Q=6\\), but \\(5\\cdot7\\) for \\(Q=30\\). In each case the coefficient factors correctly. A nonsquarefree integer such as \\(49\\) contributes zero in the original sum and remains zero in the grouped sum.","At \\(K=1\\), the weights are rational, since \\(w_1(n)=n^{-2}-n^{-4}\\). Factoring the divisor sums gives\n\\begin{equation}\\label{rh:smallcubes}\n\\Delta_6w_1(m)=\\frac2{3m^2}-\\frac{25}{27m^4},\n\\qquad\n\\Delta_{30}w_1(m)=\\frac{16}{25m^2}-\\frac{208}{225m^4}.\n\\end{equation}\nThe following entries are exact. Every listed base is coprime to the \\(Q\\) of its row. The final column multiplies the cube value by the remaining Möbius sign.\n\\begin{center}\n\\begin{tabular}{rrrrr}\n\\toprule\n\\(Q\\)&\\(m\\)&\\(\\mu(m)\\)&\\(\\Delta_Qw_1(m)\\)&Contribution","toring the divisor sums gives\n\\begin{equation}\\label{rh:smallcubes}\n\\Delta_6w_1(m)=\\frac2{3m^2}-\\frac{25}{27m^4},\n\\qquad\n\\Delta_{30}w_1(m)=\\frac{16}{25m^2}-\\frac{208}{225m^4}.\n\\end{equation}\nThe following entries are exact. Every listed base is coprime to the \\(Q\\) of its row. The final column multiplies the cube value by the remaining Möbius sign.\n\\begin{center}\n\\begin{tabular}{rrrrr}\n\\toprule\n\\(Q\\)&\\(m\\)&\\(\\mu(m)\\)&\\(\\Delta_Qw_1(m)\\)&Contribution","the divisor sums gives\n\\begin{equation}\\label{rh:smallcubes}\n\\Delta_6w_1(m)=\\frac2{3m^2}-\\frac{25}{27m^4},\n\\qquad\n\\Delta_{30}w_1(m)=\\frac{16}{25m^2}-\\frac{208}{225m^4}.\n\\end{equation}\nThe following entries are exact. Every listed base is coprime to the \\(Q\\) of its row. The final column multiplies the cube value by the remaining Möbius sign.\n\\begin{center}\n\\begin{tabular}{rrrrr}\n\\toprule\n\\(Q\\)&\\(m\\)&\\(\\mu(m)\\)&\\(\\Delta_Qw_1(m)\\)&Contribution","ivisor sums gives\n\\begin{equation}\\label{rh:smallcubes}\n\\Delta_6w_1(m)=\\frac2{3m^2}-\\frac{25}{27m^4},\n\\qquad\n\\Delta_{30}w_1(m)=\\frac{16}{25m^2}-\\frac{208}{225m^4}.\n\\end{equation}\nThe following entries are exact. Every listed base is coprime to the \\(Q\\) of its row. The final column multiplies the cube value by the remaining Möbius sign.\n\\begin{center}\n\\begin{tabular}{rrrrr}\n\\toprule\n\\(Q\\)&\\(m\\)&\\(\\mu(m)\\)&\\(\\Delta_Qw_1(m)\\)&Contribution","gives\n\\begin{equation}\\label{rh:smallcubes}\n\\Delta_6w_1(m)=\\frac2{3m^2}-\\frac{25}{27m^4},\n\\qquad\n\\Delta_{30}w_1(m)=\\frac{16}{25m^2}-\\frac{208}{225m^4}.\n\\end{equation}\nThe following entries are exact. Every listed base is coprime to the \\(Q\\) of its row. The final column multiplies the cube value by the remaining Möbius sign.\n\\begin{center}\n\\begin{tabular}{rrrrr}\n\\toprule\n\\(Q\\)&\\(m\\)&\\(\\mu(m)\\)&\\(\\Delta_Qw_1(m)\\)&Contribution","\\midrule\n6&1&1&\\(-7/27\\)&\\(-7/27\\)","6&5&\\(-1\\)&\\(17/675\\)&\\(-17/675\\)","6&7&\\(-1\\)&\\(857/64827\\)&\\(-857/64827\\)","6&25&0&\\(449/421875\\)&0","6&35&1&\\(881/1620675\\)&\\(881/1620675\\)","\\midrule\n30&1&1&\\(-64/225\\)&\\(-64/225\\)","30&7&\\(-1\\)&\\(6848/540225\\)&\\(-6848/540225\\)","30&11&\\(-1\\)&\\(17216/3294225\\)&\\(-17216/3294225\\)","30&49&0&\\(345536/1297080225\\)&0","The cube's shape is also scale dependent. At fixed \\(K\\), large \\(m\\) makes every factor \\((1-1/(bm)^2)^K\\) close to one, so its leading coefficient is \\(\\prod_{p\\mid Q}(1-p^{-2})\\). At the active scale \\(m\\asymp\\sqrt K\\), the different divisors sample different parts of the weight profile. This difference can create local cancellation within a cube. It does not remove the remaining sign \\(\\mu(m)\\), which is the quantity the signed criterion still asks us to control.","Define the full triangle majorant\n\\begin{equation}\\label{rh:majorant}\n\\mathcal A_Q(M)=\\sum_{\\substack{m\\ge1\\\\(m,Q)=1}}\n|\\mu(m)|\\,|\\Delta_Qw_{M^2}(m)|.\n\\end{equation}\nIt is finite and satisfies \\(|c_{M^2}|\\le\\mathcal A_Q(M)\\). We retain the squarefree support through \\(|\\mu(m)|\\). A lower bound on this majorant proves that this particular upper bound is insufficient. It gives no lower bound on \\(|c_{M^2}|\\).","(m,Q)=1}}\n|\\mu(m)|\\,|\\Delta_Qw_{M^2}(m)|.\n\\end{equation}\nIt is finite and satisfies \\(|c_{M^2}|\\le\\mathcal A_Q(M)\\). We retain the squarefree support through \\(|\\mu(m)|\\). A lower bound on this majorant proves that this particular upper bound is insufficient. It gives no lower bound on \\(|c_{M^2}|\\).","For any finite interval \\(I\\) of remaining bases, distinguish its signed contribution from its absolute mass:\n\\[\nS_I=\\sum_{\\substack{m\\in I","For any finite interval \\(I\\) of remaining bases, distinguish its signed contribution from its absolute mass:\n\\[\nS_I=\\sum_{\\substack{m\\in I\\\\(m,Q)=1}}\\mu(m)\\Delta_Qw_{M^2}(m),\n\\qquad\nV_I=\\sum_{\\substack{m\\in I\\\\(m,Q)=1}}\n|\\mu(m)|\\,|\\Delta_Qw_{M^2}(m)|.\n\\]\nThe valid inequalities are \\(|S_I|\\le V_I\\le\\mathcal A_Q(M)\\) and \\(|c_{M^2}|\\le\\mathcal A_Q(M)\\). There is no general inequality \\(|c_{M^2}|\\le V_I\\). Terms outside the chosen interval remain in the full signed sum. Even a two term toy sum with values \\(3\\) inside the interval and \\(10\\) outside it shows why its interval mass cannot bound the full sum.","(m,Q)=1}}\n|\\mu(m)|\\,|\\Delta_Qw_{M^2}(m)|.\n\\]\nThe valid inequalities are \\(|S_I|\\le V_I\\le\\mathcal A_Q(M)\\) and \\(|c_{M^2}|\\le\\mathcal A_Q(M)\\). There is no general inequality \\(|c_{M^2}|\\le V_I\\). Terms outside the chosen interval remain in the full signed sum. Even a two term toy sum with values \\(3\\) inside the interval and \\(10\\) outside it shows why its interval mass cannot bound the full sum.","The opposite confusion is equally serious. A large absolute mass does not force a large signed value. The finite sum \\(3-2-1\\) is zero while its absolute mass is six. The barrier will establish a lower bound on an absolute quantity which is available as an upper bound for the desired signed quantity. Such a lower bound prevents this upper bound from proving the desired estimate. It does not contradict the estimate itself.","Finally, dropping \\(|\\mu(m)|\\) creates a different mass. Its terms include nonsquarefree bases which the signed sum never sees. The exact cube table at \\(m=25\\) or \\(m=49\\) gives explicit examples: the cube value is positive, but the contribution to both the signed sum and its true triangle majorant is zero. This is why the support count below is part of the proof rather than an optional refinement.","\\begin{lemma}[Uniform positive cube tail]\\label{rh:positive}\nFor a fixed \\(R>0\\), put\n\\begin{equation}\\label{rh:cr}\n\\kappa_R=\\frac1{\\zeta(2)}-\\frac{\\zeta(4)}{R^2}.\n\\end{equation}\nFor every squarefree \\(Q\\) and every integer \\(m\\ge RM\\),\n\\begin{equation}\\label{rh:positivebound}\n\\Delta_Qw_{M^2}(m)\\ge\\frac{\\kappa_R}{m^2}.\n\\end{equation}\nIn particular, \\(R=3\\) gives \\(\\kappa_R>0\\).\n\\end{lemma}","\\begin{proof}\nThe limiting divisor sum is\n\\[\n\\sum_{b\\mid Q}\\frac{\\mu(b)}{b^2}\n=\\prod_{p\\mid Q}(1-p^{-2})\\ge\\zeta(2)^{-1}.\n\\]\nFor \\(0\\le x\\le1\\), Bernoulli's inequality gives\n\\(0\\le1-(1-x)^K\\le Kx\\). With \\(K=M^2\\), the error in replacing each power by one is at most\n\\[\n\\frac{M^2}{m^2}\\sum_{b\\mid Q}b^{-4}\n\\le\\frac{\\zeta(4)}{R^2}.\n\\]\nMultiplying the resulting lower bound by \\(m^{-2}\\) proves the lemma.\n\\end{proof}","The bound is uniform in the chosen primes. The finite product cannot fall below its product over all primes, while the error sum cannot exceed its sum over all positive integers. Taking \\(R=3\\) gives\n\\[\n\\kappa_3=\\frac6{\\pi^2}-\\frac{\\pi^4}{810}>0.\n\\]\nThe first term is approximately \\(0.6079\\) and the subtracted term approximately \\(0.1203\\). Thus the selected tail lies beyond the central region where the cube's own signs may fluctuate. Its positivity is a useful rigorous fact, but it still leaves the external signs \\(\\mu(m)\\).","or sum cannot exceed its sum over all positive integers. Taking \\(R=3\\) gives\n\\[\n\\kappa_3=\\frac6{\\pi^2}-\\frac{\\pi^4}{810}>0.\n\\]\nThe first term is approximately \\(0.6079\\) and the subtracted term approximately \\(0.1203\\). Thus the selected tail lies beyond the central region where the cube's own signs may fluctuate. Its positivity is a useful rigorous fact, but it still leaves the external signs \\(\\mu(m)\\).","\\begin{lemma}[Squarefree coprime bases]\\label{rh:count}\nFor fixed \\(R>0\\), the number \\(N_{Q,R}(M)\\) of squarefree integers \\(m\\) in \\([RM,2RM]\\) with \\((m,Q)=1\\) satisfies, uniformly in squarefree \\(Q\\),\n\\begin{equation}\\label{rh:countformula}\nN_{Q,R}(M)=RM\\frac{\\varphi(Q)}Q\n\\prod_{p\\nmid Q}(1-p^{-2})+O_R(2^d\\sqrt M).\n\\end{equation}\nHere \\(\\varphi(Q)\\) is Euler's totient function, which counts the positive residues coprime to \\(Q\\).\n\\end{lemma}","bases]\\label{rh:count}\nFor fixed \\(R>0\\), the number \\(N_{Q,R}(M)\\) of squarefree integers \\(m\\) in \\([RM,2RM]\\) with \\((m,Q)=1\\) satisfies, uniformly in squarefree \\(Q\\),\n\\begin{equation}\\label{rh:countformula}\nN_{Q,R}(M)=RM\\frac{\\varphi(Q)}Q\n\\prod_{p\\nmid Q}(1-p^{-2})+O_R(2^d\\sqrt M).\n\\end{equation}\nHere \\(\\varphi(Q)\\) is Euler's totient function, which counts the positive residues coprime to \\(Q\\).\n\\end{lemma}","\\begin{proof}\nUse the exact identity \\(\\mu(m)^2=\\sum_{\\ell^2\\mid m}\\mu(\\ell)\\). In the coprime sector only \\((\\ell,Q)=1\\) contributes. Inclusion and exclusion over the prime factors of \\(Q\\) counts the possible multiples of \\(\\ell^2\\) in the interval with main term\n\\(RM\\ell^{-2}\\varphi(Q)/Q\\) and error \\(O(2^d)\\).\nThere are \\(O_R(\\sqrt M)\\) values of \\(\\ell\\), so the summed error is \\(O_R(2^d\\sqrt M)\\). Extending the main sum from \\(\\ell\\le\\sqrt{2RM}\\) to all positive integers contributes only \\(O_R(\\sqrt M)\\). Absolute convergence identifies the extended sum as \\(\\prod_{p\\nmid Q}(1-p^{-2})\\), giving \\eqref{rh:countformula}.\n\\end{proof}","\\(\\mu(m)^2=\\sum_{\\ell^2\\mid m}\\mu(\\ell)\\). In the coprime sector only \\((\\ell,Q)=1\\) contributes. Inclusion and exclusion over the prime factors of \\(Q\\) counts the possible multiples of \\(\\ell^2\\) in the interval with main term\n\\(RM\\ell^{-2}\\varphi(Q)/Q\\) and error \\(O(2^d)\\).\nThere are \\(O_R(\\sqrt M)\\) values of \\(\\ell\\), so the summed error is \\(O_R(2^d\\sqrt M)\\). Extending the main sum from \\(\\ell\\le\\sqrt{2RM}\\) to all positive integers contributes only \\(O_R(\\sqrt M)\\). Absolute convergence identifies the extended sum as \\(\\prod_{p\\nmid Q}(1-p^{-2})\\), giving \\eqref{rh:countformula}.\n\\end{proof}","only \\((\\ell,Q)=1\\) contributes. Inclusion and exclusion over the prime factors of \\(Q\\) counts the possible multiples of \\(\\ell^2\\) in the interval with main term\n\\(RM\\ell^{-2}\\varphi(Q)/Q\\) and error \\(O(2^d)\\).\nThere are \\(O_R(\\sqrt M)\\) values of \\(\\ell\\), so the summed error is \\(O_R(2^d\\sqrt M)\\). Extending the main sum from \\(\\ell\\le\\sqrt{2RM}\\) to all positive integers contributes only \\(O_R(\\sqrt M)\\). Absolute convergence identifies the extended sum as \\(\\prod_{p\\nmid Q}(1-p^{-2})\\), giving \\eqref{rh:countformula}.\n\\end{proof}","me factors of \\(Q\\) counts the possible multiples of \\(\\ell^2\\) in the interval with main term\n\\(RM\\ell^{-2}\\varphi(Q)/Q\\) and error \\(O(2^d)\\).\nThere are \\(O_R(\\sqrt M)\\) values of \\(\\ell\\), so the summed error is \\(O_R(2^d\\sqrt M)\\). Extending the main sum from \\(\\ell\\le\\sqrt{2RM}\\) to all positive integers contributes only \\(O_R(\\sqrt M)\\). Absolute convergence identifies the extended sum as \\(\\prod_{p\\nmid Q}(1-p^{-2})\\), giving \\eqref{rh:countformula}.\n\\end{proof}","The squarefree identity in the proof can be understood prime by prime. If no prime square divides \\(m\\), the only possible square divisor in its Möbius sum is built from no primes, and the sum is one. If some primes do have squares dividing \\(m\\), summing over their squarefree subsets gives \\((1-1)^r=0\\), where \\(r\\ge1\\) is their number. Thus the identity is exactly the indicator of squarefree support.","For each permitted \\(\\ell\\), write \\(m=\\ell^2v\\). Coprimality of \\(m\\) with \\(Q\\) forces both \\((\\ell,Q)=1\\) and \\((v,Q)=1\\). Inclusion and exclusion gives\n\\[\n\\#\\{v:A<v\\le B,\\ (v,Q)=1\\}\n=\\sum_{b\\mid Q}\\mu(b)\n\\left(\\left\\lfloor\\frac B b\\right\\rfloor-\n\\left\\lfloor\\frac A b\\right\\rfloor\\right).\n\\]\nReplacing each floor by its real argument creates an error bounded by a constant times the number of divisors, namely \\(2^d\\). This explains exactly where cube size enters the density proof. Summing over \\(\\ell\\) up to order \\(\\sqrt M\\) creates the error in Lemma~\\ref{rh:count}. No distribution theorem for primes or Möbius signs is used.","ed \\(\\ell\\), write \\(m=\\ell^2v\\). Coprimality of \\(m\\) with \\(Q\\) forces both \\((\\ell,Q)=1\\) and \\((v,Q)=1\\). Inclusion and exclusion gives\n\\[\n\\#\\{v:A<v\\le B,\\ (v,Q)=1\\}\n=\\sum_{b\\mid Q}\\mu(b)\n\\left(\\left\\lfloor\\frac B b\\right\\rfloor-\n\\left\\lfloor\\frac A b\\right\\rfloor\\right).\n\\]\nReplacing each floor by its real argument creates an error bounded by a constant times the number of divisors, namely \\(2^d\\). This explains exactly where cube size enters the density proof. Summing over \\(\\ell\\) up to order \\(\\sqrt M\\) creates the error in Lemma~\\ref{rh:count}. No distribution theorem for primes or Möbius signs is used.","Q)=1\\) and \\((v,Q)=1\\). Inclusion and exclusion gives\n\\[\n\\#\\{v:A<v\\le B,\\ (v,Q)=1\\}\n=\\sum_{b\\mid Q}\\mu(b)\n\\left(\\left\\lfloor\\frac B b\\right\\rfloor-\n\\left\\lfloor\\frac A b\\right\\rfloor\\right).\n\\]\nReplacing each floor by its real argument creates an error bounded by a constant times the number of divisors, namely \\(2^d\\). This explains exactly where cube size enters the density proof. Summing over \\(\\ell\\) up to order \\(\\sqrt M\\) creates the error in Lemma~\\ref{rh:count}. No distribution theorem for primes or Möbius signs is used.","). Inclusion and exclusion gives\n\\[\n\\#\\{v:A<v\\le B,\\ (v,Q)=1\\}\n=\\sum_{b\\mid Q}\\mu(b)\n\\left(\\left\\lfloor\\frac B b\\right\\rfloor-\n\\left\\lfloor\\frac A b\\right\\rfloor\\right).\n\\]\nReplacing each floor by its real argument creates an error bounded by a constant times the number of divisors, namely \\(2^d\\). This explains exactly where cube size enters the density proof. Summing over \\(\\ell\\) up to order \\(\\sqrt M\\) creates the error in Lemma~\\ref{rh:count}. No distribution theorem for primes or Möbius signs is used.","For \\(Q=6\\), the density constant simplifies to\n\\[\n\\frac{\\varphi(6)}6\\prod_{p\\nmid6}(1-p^{-2})\n=\\frac1{2\\zeta(2)}.\n\\]\nFor \\(Q=30\\), it is \\(5/(12\\zeta(2))\\). As a finite illustration, take \\(M=10\\) and \\(R=3\\). The interval is \\([30,60]\\). Its \\(Q=6\\) squarefree coprime bases are\n\\[\n31,35,37,41,43,47,53,55,59.\n\\]\nFor \\(Q=30\\), remove \\(35\\) and \\(55\\), leaving seven bases. In both coprime sectors, \\(49\\) would be counted if squarefree support were forgotten. The main term predicts approximately \\(9.12\\) bases for \\(Q=6\\) and \\(7.60\\) for \\(Q=30\\). These small examples illustrate the meaning of the density constant. The asymptotic lemma includes its error and does not claim that a main term rounded to","For \\(Q=6\\), the density constant simplifies to\n\\[\n\\frac{\\varphi(6)}6\\prod_{p\\nmid6}(1-p^{-2})\n=\\frac1{2\\zeta(2)}.\n\\]\nFor \\(Q=30\\), it is \\(5/(12\\zeta(2))\\). As a finite illustration, take \\(M=10\\) and \\(R=3\\). The interval is \\([30,60]\\). Its \\(Q=6\\) squarefree coprime bases are\n\\[\n31,35,37,41,43,47,53,55,59.\n\\]\nFor \\(Q=30\\), remove \\(35\\) and \\(55\\), leaving seven bases. In both coprime sectors, \\(49\\) would be counted if squarefree support were forgotten. The main term predicts approximately \\(9.12\\) bases for \\(Q=6\\) and \\(7.60\\) for \\(Q=30\\). These small examples illustrate the meaning of the density constant. The asymptotic lemma includes its error and does not claim that a main term rounded to the nearest integer always gives the exact finite count.","\\(Q=6\\), the density constant simplifies to\n\\[\n\\frac{\\varphi(6)}6\\prod_{p\\nmid6}(1-p^{-2})\n=\\frac1{2\\zeta(2)}.\n\\]\nFor \\(Q=30\\), it is \\(5/(12\\zeta(2))\\). As a finite illustration, take \\(M=10\\) and \\(R=3\\). The interval is \\([30,60]\\). Its \\(Q=6\\) squarefree coprime bases are\n\\[\n31,35,37,41,43,47,53,55,59.\n\\]\nFor \\(Q=30\\), remove \\(35\\) and \\(55\\), leaving seven bases. In both coprime sectors, \\(49\\) would be counted if squarefree support were forgotten. The main term predicts approximately \\(9.12\\) bases for \\(Q=6\\) and \\(7.60\\) for \\(Q=30\\). These small examples illustrate the meaning of the density constant. The asymptotic lemma includes its error and does not claim that a main term rounded to the nearest integer always gives the exact finite count.","{2\\zeta(2)}.\n\\]\nFor \\(Q=30\\), it is \\(5/(12\\zeta(2))\\). As a finite illustration, take \\(M=10\\) and \\(R=3\\). The interval is \\([30,60]\\). Its \\(Q=6\\) squarefree coprime bases are\n\\[\n31,35,37,41,43,47,53,55,59.\n\\]\nFor \\(Q=30\\), remove \\(35\\) and \\(55\\), leaving seven bases. In both coprime sectors, \\(49\\) would be counted if squarefree support were forgotten. The main term predicts approximately \\(9.12\\) bases for \\(Q=6\\) and \\(7.60\\) for \\(Q=30\\). These small examples illustrate the meaning of the density constant. The asymptotic lemma includes its error and does not claim that a main term rounded to the nearest integer always gives the exact finite count.","\\(Q=30\\), it is \\(5/(12\\zeta(2))\\). As a finite illustration, take \\(M=10\\) and \\(R=3\\). The interval is \\([30,60]\\). Its \\(Q=6\\) squarefree coprime bases are\n\\[\n31,35,37,41,43,47,53,55,59.\n\\]\nFor \\(Q=30\\), remove \\(35\\) and \\(55\\), leaving seven bases. In both coprime sectors, \\(49\\) would be counted if squarefree support were forgotten. The main term predicts approximately \\(9.12\\) bases for \\(Q=6\\) and \\(7.60\\) for \\(Q=30\\). These small examples illustrate the meaning of the density constant. The asymptotic lemma includes its error and does not claim that a main term rounded to the nearest integer always gives the exact finite count.","t is \\(5/(12\\zeta(2))\\). As a finite illustration, take \\(M=10\\) and \\(R=3\\). The interval is \\([30,60]\\). Its \\(Q=6\\) squarefree coprime bases are\n\\[\n31,35,37,41,43,47,53,55,59.\n\\]\nFor \\(Q=30\\), remove \\(35\\) and \\(55\\), leaving seven bases. In both coprime sectors, \\(49\\) would be counted if squarefree support were forgotten. The main term predicts approximately \\(9.12\\) bases for \\(Q=6\\) and \\(7.60\\) for \\(Q=30\\). These small examples illustrate the meaning of the density constant. The asymptotic lemma includes its error and does not claim that a main term rounded to the nearest integer always gives the exact finite count.","zeta(2))\\). As a finite illustration, take \\(M=10\\) and \\(R=3\\). The interval is \\([30,60]\\). Its \\(Q=6\\) squarefree coprime bases are\n\\[\n31,35,37,41,43,47,53,55,59.\n\\]\nFor \\(Q=30\\), remove \\(35\\) and \\(55\\), leaving seven bases. In both coprime sectors, \\(49\\) would be counted if squarefree support were forgotten. The main term predicts approximately \\(9.12\\) bases for \\(Q=6\\) and \\(7.60\\) for \\(Q=30\\). These small examples illustrate the meaning of the density constant. The asymptotic lemma includes its error and does not claim that a main term rounded to the nearest integer always gives the exact finite count.","\\begin{theorem}[Absolute cube barrier]\\label{rh:barrier}\nFix \\(R\\) with \\(\\kappa_R>0\\). Suppose squarefree \\(Q=Q(M)\\) has \\(d(M)=o(\\log M)\\) distinct prime divisors. For all sufficiently large \\(M\\),\n\\begin{equation}\\label{rh:barrierbound}\n\\mathcal A_Q(M)\\ge\n\\frac{\\kappa_R}{8R\\zeta(2)}\n\\frac{\\varphi(Q)}{QM}\n\\ge\\frac{\\kappa_R}{8R\\zeta(2)M(d+1)}.\n\\end{equation}\nConsequently \\(\\mathcal A_Q(M)\\ge M^{-1-o(1)}\\). It cannot itself satisfy the target bound in Theorem~\\ref{rh:bd}.\n\\end{theorem}","\\begin{proof}\nOrder the prime divisors as \\(p_1<\\cdots<p_d\\). Since \\(p_i\\ge i+1\\),\n\\[\n\\frac{\\varphi(Q)}Q=\\prod_{i=1}^d(1-p_i^{-1})\n\\ge\\prod_{i=1}^d\\frac i{i+1}=\\frac1{d+1}.\n\\]\nAlso \\(\\prod_{p\\nmid Q}(1-p^{-2})\\ge1/\\zeta(2)\\).\nThe error in \\eqref{rh:countformula} is \\(M^{1/2+o(1)}\\), while its main term is at least \\(RM/[\\zeta(2)(d+1)]\\). Therefore eventually\n\\[\nN_{Q,R}(M)\\ge\\frac{RM}{2\\zeta(2)}\\frac{\\varphi(Q)}Q.\n\\]\nOn every counted base, \\(|\\mu(m)|=1\\), and Lemma~\\ref{rh:positive} gives\n\\(|\\Delta_Qw_{M^2}(m)|\\ge\\kappa_R/(4R^2M^2)\\).\nTheir contribution is a lower bound for the full majorant. Multiplication gives \\eqref{rh:barrierbound}.\n\\end{proof}","1<\\cdots<p_d\\). Since \\(p_i\\ge i+1\\),\n\\[\n\\frac{\\varphi(Q)}Q=\\prod_{i=1}^d(1-p_i^{-1})\n\\ge\\prod_{i=1}^d\\frac i{i+1}=\\frac1{d+1}.\n\\]\nAlso \\(\\prod_{p\\nmid Q}(1-p^{-2})\\ge1/\\zeta(2)\\).\nThe error in \\eqref{rh:countformula} is \\(M^{1/2+o(1)}\\), while its main term is at least \\(RM/[\\zeta(2)(d+1)]\\). Therefore eventually\n\\[\nN_{Q,R}(M)\\ge\\frac{RM}{2\\zeta(2)}\\frac{\\varphi(Q)}Q.\n\\]\nOn every counted base, \\(|\\mu(m)|=1\\), and Lemma~\\ref{rh:positive} gives\n\\(|\\Delta_Qw_{M^2}(m)|\\ge\\kappa_R/(4R^2M^2)\\).\nTheir contribution is a lower bound for the full majorant. Multiplication gives \\eqref{rh:barrierbound}.\n\\end{proof}","rphi(Q)}Q=\\prod_{i=1}^d(1-p_i^{-1})\n\\ge\\prod_{i=1}^d\\frac i{i+1}=\\frac1{d+1}.\n\\]\nAlso \\(\\prod_{p\\nmid Q}(1-p^{-2})\\ge1/\\zeta(2)\\).\nThe error in \\eqref{rh:countformula} is \\(M^{1/2+o(1)}\\), while its main term is at least \\(RM/[\\zeta(2)(d+1)]\\). Therefore eventually\n\\[\nN_{Q,R}(M)\\ge\\frac{RM}{2\\zeta(2)}\\frac{\\varphi(Q)}Q.\n\\]\nOn every counted base, \\(|\\mu(m)|=1\\), and Lemma~\\ref{rh:positive} gives\n\\(|\\Delta_Qw_{M^2}(m)|\\ge\\kappa_R/(4R^2M^2)\\).\nTheir contribution is a lower bound for the full majorant. Multiplication gives \\eqref{rh:barrierbound}.\n\\end{proof}","The dimension condition means that the cube size \\(2^d\\) is \\(M^{o(1)}\\). In this regime the target amplitude is \\(M^{-3/2+o(1)}\\), whereas the triangle majorant is at least \\(M^{-1-o(1)}\\). The missing factor is \\(M^{-1/2+o(1)}\\). It must arise from cancellation or another operation which retains and controls the signs.","There are three separate estimates in the theorem. First, the totient factor loses at most \\(d+1\\). This follows from ordering the primes, not from assuming that the chosen primes are the first \\(d\\) primes. Larger prime divisors make the totient density larger, so the lower bound is uniform over every squarefree \\(Q\\) with the stated dimension.","Second, the support error is negligible compared with its main term. The ratio is bounded by a constant times\n\\[\n\\frac{2^d(d+1)}{\\sqrt M}.\n\\]\nIf \\(d=o(\\log M)\\), then \\(2^d=M^{o(1)}\\) and \\(d+1=M^{o(1)}\\). The ratio therefore tends to zero. This is why the statement can retain an increasing cube, while still concluding that a fixed positive fraction of its predicted support is present. If the dimension condition is removed, this particular proof need not give a negligible error. We do not silently extend the theorem beyond that range.","Third, each selected base lies below \\(2RM\\), so its positive cube tail is at least \\(\\kappa_R/(4R^2M^2)\\). Multiplying this by the proven support count gives the constant in \\eqref{rh:barrierbound}. The argument uses only one interval as a lower bound on the full absolute sum, which is the direction an interval restriction actually supports.","To see the incompatibility with the target, choose any fixed amplitude exponent loss \\(\\varepsilon<1/2\\) in the \\(M\\) formulation of the criterion. A quantity bounded below by \\(M^{-1-o(1)}\\) cannot also be bounded above by a constant times \\(M^{-3/2+\\varepsilon}\\). Their ratio grows as \\(M^{1/2-\\varepsilon-o(1)}\\). The criterion requires every positive \\(\\varepsilon\\), so a failure for one such exponent already shows that this majorant cannot supply it. The signed coefficient can still be much smaller than the majorant.","\\figplaceholder{rh_barrier}{The two exponents in the repaired cube barrier. The curves show theorem scales, rather than measured Möbius amplitudes. The absolute majorant has a lower bound of order \\(M^{-1-o(1)}\\), while the Báez Duarte criterion asks for the signed amplitude to be at most \\(M^{-3/2+o(1)}\\).}","First, an exact prime high pass followed by exact coprime reconstruction preserves a power mode. If a mode has local multiplier \\(1-p^{-\\rho}\\), reconstruction contributes the reciprocal multiplier whenever it is defined. Their product is one. The high pass alone can look smaller, but exact reconstruction restores the removed factor. The historical Lean module \\texttt{HighpassReconstructionNoGo.lean} certifies this algebraic cancellation. It does not estimate a prime sum.","Third, global Hardy norms on a prime torus are blind to the squarefree Möbius sign. For a finite prime set, a squarefree monomial is \\(\\prod_{p\\in S}z_p\\). Replacing every coordinate by \\(-z_p\\) multiplies it by \\((-1)^{|S|}\\), exactly its Möbius coefficient. The substitution preserves Haar measure on the torus. Every global \\(L^q\\) norm formed with that measure is consequently unchanged by this sign operation. A bound in that invariant norm cannot distinguish the signed packet from the corresponding unsigned squarefree polynomial. This statement does not say that local restrictions or correlations with a separate observable are invariant.","Fourth, a raw positive fourth moment cannot lie below the square of its quadratic energy in a normalized probability space. If \\(\\nu\\) is its probability measure and \\(F\\in L^4(\\nu)\\) is the packet, Cauchy Schwarz gives\n\\[\n\\int |F|^4\\,d\\nu\\ge\\left(\\int |F|^2\\,d\\nu\\right)^2.\n\\]\nThis is the precise obstruction to a strict saving below that scale for the raw moment. It does not assert positivity of every separately grouped collision term for arbitrary signed coefficients. A viable replacement would require a signed expression, a justified quotient, or an independently proved correlation estimate.","\\begin{conjecture}[Signed rough base bound]\\label{rh:signed}\nFor every \\(\\varepsilon>0\\), there is a constant \\(C_\\varepsilon\\) such that, for all integers \\(K\\ge1\\) and every squarefree \\(Q\\),\n\\begin{equation}\\label{rh:signedbound}\n\\left|\\sum_{\\substack{m\\ge1","\\begin{conjecture}[Signed rough base bound]\\label{rh:signed}\nFor every \\(\\varepsilon>0\\), there is a constant \\(C_\\varepsilon\\) such that, for all integers \\(K\\ge1\\) and every squarefree \\(Q\\),\n\\begin{equation}\\label{rh:signedbound}\n\\left|\\sum_{\\substack{m\\ge1\\\\(m,Q)=1}}\n\\mu(m)\\Delta_Qw_K(m)\\right|\n\\le C_\\varepsilon K^{-3/4+\\varepsilon}.\n\\end{equation}\n\\end{conjecture}","\\begin{proposition}[Exact conditional implication]\\label{rh:conditional}\nConjecture~\\ref{rh:signed} is equivalent to the Riemann hypothesis. The same is true if \\eqref{rh:signedbound} is required only along one declared choice \\(Q=Q(K)\\), including an admissible growing prime cube.\n\\end{proposition}","\\begin{proof}\nBy Proposition~\\ref{rh:cube}, the sum on the left is exactly \\(c_K\\), regardless of \\(Q\\). Theorem~\\ref{rh:bd} supplies both implications. Its bound is independent of \\(Q\\) because its value is independent of \\(Q\\).\n\\end{proof}","This equivalence does not make the estimate easier by itself. It states precisely what must be proved in this coordinate system. A choice of \\(Q\\) is useful only if it exposes a provable signed estimate. The barrier theorem shows that an admissible cube followed by its absolute majorant does not do so.","The heat formulation is equally precise: prove subexponential growth of \\(\\widetilde E_J\\) in \\eqref{rh:energydef}. The Hermite formulation places that task in stable physical coordinates. The saved program proposes a lower bound on prime detail energy, combined with a historical upper bound and a record argument, as one possible route. That lower bound is currently unproved. We do not promote it to an established estimate by writing it as a spectral gap.","The heat criterion accepts a very specific type of conclusion. For every \\(\\varepsilon>0\\), one needs a constant \\(C_\\varepsilon\\), depending on the fixed heat strip but not on \\(J\\), such that\n\\[\n\\int_{t_0}^T\\int_J^{J+1}\n|\\widetilde{\\mathfrak A}_t(u)-\n\\widetilde{\\mathfrak P}_t(u)|^2\\,du\\,dt\n\\le C_\\varepsilon e^{\\varepsilon J}.\n\\]\nThe explicit archimedean estimate allows this to be replaced by the corresponding subexponential estimate for the transformed prime packet. An estimate after removing its ground component would require an independent argument recovering that component. An estimate averaged over an additional variable would require an argument bringing it back to the actual packet. These are mathematical obligations of those proposed routes, rather than changes to the","The heat criterion accepts a very specific type of conclusion. For every \\(\\varepsilon>0\\), one needs a constant \\(C_\\varepsilon\\), depending on the fixed heat strip but not on \\(J\\), such that\n\\[\n\\int_{t_0}^T\\int_J^{J+1}\n|\\widetilde{\\mathfrak A}_t(u)-\n\\widetilde{\\mathfrak P}_t(u)|^2\\,du\\,dt\n\\le C_\\varepsilon e^{\\varepsilon J}.\n\\]\nThe explicit archimedean estimate allows this to be replaced by the corresponding subexponential estimate for the transformed prime packet. An estimate after removing its ground component would require an independent argument recovering that component. An estimate averaged over an additional variable would require an argument bringing it back to the actual packet. These are mathematical obligations of those proposed routes, rather than changes to the criterion.","The arithmetic criterion is equally demanding about quantifiers. It asks for a bound at all positive integers \\(K\\), with a constant uniform in that scale. A computed range, a favorable subsequence, or a choice of cube which changes at each scale does not replace the required bound unless a theorem covers the missing scales. Since every exact cube gives the same \\(c_K\\), uniformity over \\(Q\\) in Conjecture~\\ref{rh:signed} is automatic once the true signed bound is known. It cannot be supplied by numerical optimization of an unsigned cube profile.","The strongest completed boundary in the present arithmetic route is therefore constructive in a limited sense. It identifies which operation cannot provide the closing saving: the declared full absolute majorant for a cube with \\(d=o(\\log M)\\). A new attempt can keep the cube while seeking signed correlations, change the observable while proving an exact recovery estimate, or work directly with the heat packet. In each case the final output must reach one of the established criteria with its constants and quantifiers intact.","\\midrule\n\\(\\rho=\\beta+i\\gamma\\)&A nontrivial zero, counted with its multiplicity.","\\(z_\\rho\\)&The centered coordinate \\(\\rho-1/2\\).","\\(t,u\\)&Positive heat time and translation, respectively.","\\(\\Theta_t\\)&The Gaussian sum over centered zero modes.","\\(L\\)&The operator \\(1/4-\\partial_u^2\\) removing the explicit pole channel.","\\(\\widetilde E_J\\)&Squared packet energy on one unit translation band and one fixed heat strip.","\\(g_{t,a}\\)&A Gaussian of norm one in the physical measure \\(du\\).","\\(\\lambda,R\\)&Squared coherent displacement and Hermite cutoff. The cutoff \\(R\\) in this section is separate from the fixed tail multiplier in the barrier.","\\(K,M\\)&Integer sequence index and its natural scale \\(M=\\sqrt K\\).","\\(Q,d\\)&A squarefree product of primes and its number of prime coordinates.","\\(\\Delta_Qw_K\\)&The alternating divisor cube acting on the positive weight.","\\(c_K,\\mathcal A_Q\\)&The signed coefficient and its full absolute majorant.","The recovered v0.13 report records four focused Lean modules checked in a pinned offline Lean \\(4.34.0\\) release candidate environment: physical metric bookkeeping, diagonal preservation, exact high pass reconstruction, and common normalized modes. It also records ordinary Mathlib logical foundations in the axiom audit and the rejection of two false controls. One rejected control exposed a missing nonnegativity hypothesis in a squared energy estimate, which was repaired in that checkpoint \\cite{KomorebiFinalGap}."]}