Euler numbers through packets, polynomials, and residuals

Read side by side

An extended hand-worked Komorebi example · Komorebi P. · 29 October 2026 · Navigation and FIR example updated 30 October 2026

We will reconstruct the Euler secant numbers through a different sequence, turn the calculation into polynomial identities, and then use the same Euler data in a refinement calculation. At each change of coordinates we will write the transfer explicitly. When a shortcut fails, we will calculate the residual that explains the failure.

The two main numerical destinations are

for the coefficient of \(t^{12}/12!\) in \(\sec t\), and

for a refinement of the signed Euler sequence. The derivations connect them to reciprocal series, a two-packet matrix, finite-difference coordinates, and an exact correction law. We will also reconstruct \(E_8=1385\) from alternating odd powers and from a separate dyadic source. Those calculations will give exact partial sums and precision certificates.

The Euler numbers and the quartic sequence used below are classical. This example develops them through Komorebi's packet, transfer, and residual language. Its purpose is to make a substantial calculation inspectable from beginning to end. The source notes identify the classical references and the project sources.

You can follow the algebra without running code. The accompanying Python program uses exact integers and fractions and needs only Python's standard library.

Part 1 / Packets and exact reconstruction

Conventions and coordinate systems

Euler number conventions vary. We will use the positive secant numbers \(s_n\), indexed by half the ordinary degree:

The first values are

These are OEIS A000364, with offset zero. The signed Euler numbers \(E_j\) use

Thus

In the sequence-refinement part we will write \(a(n)=E_{2n}\). In particular, \(a(1)=-1\), even though the full-index number \(E_1\) is zero. The DLMF definitions and secant series fix this sign convention.

We will use several coordinate systems. Their operations have different meanings.

The series calculations take place over the rational numbers. A formal series is a sequence of coefficients with a multiplication rule. Every coefficient of a product is a finite sum, so the calculations do not require an analytic convergence argument. A series with a nonzero constant coefficient has a unique formal reciprocal.

For a prime \(p\), \(v_p(x)\) denotes the exponent of \(p\) in the reduced numerator of a nonzero rational number \(x\), minus that in its denominator. We set \(v_p(0)=\infty\). A rational number can be reduced modulo \(p^h\) when its reduced denominator is coprime to \(p\), using the modular inverse of that denominator.

Build the secant numbers from the reciprocal equation

Start with

Write an unknown reciprocal as

The equation \(\cos t\,S(t)=1\) gives \(s_0=1\). For \(n\geq1\), its coefficient at ordinary degree \(2n\) is

Multiplying by \((2n)!\) gives

The binomial coefficient comes from converting an ordinary product back to exponential normalization. It is part of the transfer law.

Now perform the first six steps. Each row uses only previously calculated values.

The series begins

For example, the ordinary coefficient of \(t^4\) is \(5/24\). The compressed index is two, but the denominator is \(4!\).

Split the series into two packets

Every even series can be written uniquely as

The series \(A\) collects ordinary degrees \(0,4,8,\ldots\). The series \(t^2B\) collects degrees \(2,6,10,\ldots\). These two packets retain every coefficient of the even series.

For the secant series, the packets we want are

We will now derive those packets without using their already known values.

Split the denominator in the same way:

where

Multiply the two packet expressions:

Because \(t^4=u\), the product has packets

The product of two degree-two packets returns to the degree-zero packet, with one factor of \(u\). This is the carry that a residue calculation must retain.

Multiplication by \(\cos t\) is therefore represented over \(\mathbb Q[[u]]\) by

The reciprocal problem becomes

This is an exact change of representation. We have replaced multiplication of even series with a two-packet transfer matrix.

Try a coarser representation and calculate its failure

Suppose we retain only the first packet and replace the full equation by

Then

The degree-four numerator would be \(-1\). The correct secant numerator is \(5\).

We can locate the error before calculating any more coefficients. Insert the proposed pair \((C^{-1},0)\) into the full equation. Its residual, with the convention actual output minus required output, is

The second residual has constant coefficient \(-1/2\). The discarded channel already contains a nonzero obligation.

There is also a structural obstruction. Let

be the coarsening map, and let the proposed coarse operator be multiplication by \(C\). Its transport residual is

For the vector \(v=(0,1)\),

The coarse description identifies \(v\) with zero, but the next operation distinguishes them. There is no operator on the first packet alone that reproduces \(fL\) for every input pair. Any such operator would have to send the same coarse input to both zero and \(uD\).

At constant order in \(u\), the carry is invisible. It appears at the next order. This explains how a coarse calculation can pass an initial check and still fail when the requested resolution increases.

Repair the transfer by inverting the packet matrix

The determinant is

Its constant coefficient is one, so it is a formal unit. The inverse matrix is

Consequently,

We can verify both equations without an appeal to the matrix formula:

and

Calculate the first terms of the determinant:

Writing \(H=G^{-1}\), coefficient inversion gives

Now multiply:

The missing first-packet coefficient was

In the correct product, the order-\(u\) term from \(CA\) is cancelled by the term from \(uDB\), whose leading value is \(-u/4\). The residual identifies the missing interaction exactly.

Eliminate a packet while retaining its effect

There is a useful second way to solve the same system. Its second equation gives

because \(C(0)=1\). Substitute this expression into the first equation:

The effective first-packet operator is therefore

This is the Schur complement of the second diagonal block. It retains the trip from the first packet to the second and back. Solving the reduced equation gives

and the eliminated packet is recovered by \(B=-C^{-1}DA\).

The first two correction coefficients make the effect visible. Since

we obtain

Its inverse begins

These are the correct first-packet secant coefficients.

The reduction uses the known second equation of the reciprocal problem. It is an elimination with a reconstruction rule. The earlier coarsening obstruction concerned arbitrary input pairs, for which the first coordinate alone does not determine the next output. A reduced representation becomes useful when the operation and its retained constraints are transformed along with it.

Identify the quartic reciprocal family

The classical quartic sequence is OEIS A211212, an entry by Peter Luschny. Its recurrence is credited to Ilya Gutkovskiy (27 January 2020), and its cosine/cosh generating function to Michael Wallner (17 November 2020). See sources and contributions for the conventions and comparison pattern.

The determinant has a useful scalar interpretation. Since

we have

Its full series is

Here is a proof of that coefficient formula. Work temporarily over \(\mathbb Q(i)\):

For an even cosine index \(m=2j\), both powers \((1+i)^{4j}\) and \((1-i)^{4j}\) equal \((-4)^j\). For an odd index \(m=2j+1\), the powers of degree \(4j+2\) cancel. Only degrees divisible by four remain. All resulting coefficients are rational, giving the displayed identity in \(\mathbb Q[[t]]\).

Define the Paper IV denominator and reciprocal coefficients by

We have \(D_2(t)=\cos t\) and \(b_2(j)=s_j\). For \(N=4\), put \(b_j=b_4(j)\). The inverse of the determinant is

This follows by replacing \(t^4\) with \(4u\) in the reciprocal relation for \(D_4\). It is a formal substitution in the reduced variable, with no square root needed in the coefficient field.

The quartic recurrence is

Work out four steps:

The resulting sequence

is OEIS A211212. Its published generating function is equivalent to \(D_4(t)^{-1}\). The OEIS also identifies it as counting four-alternating permutations. We will use that interpretation as a separate computational control later.

Reconstruct the Euler numbers through the quartic sequence

From the matrix calculation,

Take the coefficient of \(t^{2n}\). A term of degree \(4k\) from \(H\) leaves degree \(2n-4k\) for \(\cosh t\). Therefore

After restoring exponential normalization, the bridge is

Equivalently, the two packets obey

The packet matrix has supplied one quartic carrier for both Euler lanes. The fixed numerator \(\cosh t\) determines how that carrier reconstructs them.

Check several cases by hand:

For the degree-twelve target,

This agrees with the direct secant recurrence, but its intermediate objects are different. One route uses all preceding secant coefficients. The other constructs the quartic carrier and transfers it to the two packets. We have proved why the routes agree, so a disagreement in an implementation would expose a convention or arithmetic error.

Recover the quartic carrier from an Euler packet

The bridge can also be read backwards. In the first packet, the coefficient of \(b_m\) is \(4^m\), so

The case \(m=0\) gives \(b_0=s_0=1\), with an empty sum. The next steps give

Once that carrier is known, the second packet is determined by the same reconstruction rule. For example,

Thus the Euler coefficients at degrees \(0,4,8,12\) recover the carrier through \(b_3\), which also predicts the coefficient at degree fourteen. This uses the reciprocal equation and its fixed numerator. An arbitrary even series would have two independent packets. The reconstruction constraint is what connects them in this example.

Count the first quartic values without series inversion

There is a concrete permutation interpretation of the same integers. Arrange \(4n\) distinct letters in descending blocks of four. Require an ascent at each boundary between blocks. This is the four-alternating pattern used by the cited OEIS entries.

At length four, only \(4321\) occurs, so \(b_1=1\). At length eight, choose which four letters occupy the first block. Each block then has a unique descending order. There are \(\binom84=70\) choices. Exactly one choice gives a descent at the boundary as well, namely the globally descending permutation. Removing it leaves

At length twelve, three descending blocks can be filled in

ways. A forbidden descent at the first boundary merges its two blocks into a descending block of eight. That leaves \(\binom{12}4=495\) choices. The second forbidden boundary gives another 495. Their intersection is the unique globally descending permutation. Inclusion-exclusion gives

At length sixteen there are three internal boundaries. The same calculation gives

The double-boundary intersections have different shapes. Two adjacent merges give blocks of sizes twelve and four. Two disjoint merges give blocks of sizes eight and eight. Keeping those cases separate explains the two different binomial terms.

For general block length \(N\geq1\) and \(n>0\), a selection of forbidden boundaries merges the blocks into an ordered composition \(q_1+\cdots+q_k=n\). There are \(n-k\) merged boundaries and

ways to fill the resulting descending blocks. Hence the count is

Expanding \(D_N^{-1}=(1+(D_N-1))^{-1}\) gives precisely this expression for \(b_N(n)\). This proves the enumeration identity. Direct permutation enumeration and rank-based dynamic programming provide separate finite checks of its implementation.

Part 2 / Polynomial and local coordinates

Turn the reciprocal calculation into parameter polynomials

The same construction has a polynomial parameter. Define

Here \(r\) may be an indeterminate. The definition takes place in \(\mathbb Q[r][[t]]\). Write \(D_N=1+U\), where \(U\) has zero constant coefficient. The formal binomial expansion is

where

Only \(k\leq n\) can contribute at degree \(Nn\). Thus \(P_{N,n}\) is a polynomial of degree at most \(n\). More explicitly, for \(n>0\),

The inner sum records ordered compositions of the compressed index. Its factorials still use ordinary degrees. The degree is exactly \(n\), because the leading coefficient is

For a convenient recurrence, differentiate the defining reciprocal power. If \(F=D_N^{-r}\), then

At degree \(Nn-1\), this gives

We use \(P_{N,0}=1\) and set a binomial coefficient to zero when its lower index lies outside its range.

For \(N=2\), the first steps are

Continuing gives

Setting \(r=1\) recovers the secant numbers. Setting \(r=2\) gives the coefficients of \(\sec^2t\), beginning

At \(N=4\), the same recurrence gives

The values at one are \(1,69,33661\). The scalar quartic sequence is one specialization of this polynomial family.

There are useful boundary checks. For \(n>0\), \(P_{N,n}(0)=0\). Also

because reciprocal parameter \(-1\) returns the original denominator. For \(N=1\), \(D_1(t)=e^{-t}\), so \(P_{1,n}(r)=r^n\). The program checks these boundaries as well as positive parameters.

Derive the parameter-addition law

Formal powers of the same unit satisfy

One way to verify this for formal parameters is to differentiate the product. It satisfies the same equation \(D_NF'+(r+\rho)D_N'F=0\) and has the same constant coefficient as \(D_N^{-(r+\rho)}\). The recurrence above determines its coefficients uniquely.

Comparing the coefficient of \(t^{Nn}\) gives

For \(N=2,n=2\), this becomes

The left side expands to

which is exactly the right side. With \(r=\rho=1\), it gives

This is a reusable transfer identity for reciprocal powers. Its weights depend on the ordinary degree \(Nn\). Replacing them by \(\binom nk\) changes the law.

Recover the classical Euler polynomials

There is a second polynomial family that should be kept distinct from \(P_{N,n}(r)\). The classical Euler polynomials \(\mathcal E_m(x)\) are defined by

The parameter \(x\) is a translation variable. The earlier parameter \(r\) was a reciprocal power. Their roles differ even when some coefficients come from the same Euler numbers.

Factor the generating function about the midpoint:

Using the signed Euler numbers already reconstructed, coefficient multiplication gives

The first values follow directly:

For degree four,

The constants cancel: \(1/16-3/8+5/16=0\). The quadratic terms cancel as well. These cancellations provide simple controls on the signs and scaling.

Three whole-family identities follow from the generating function:

For the first, add the generating functions at \(x+1\) and \(x\). The factor \(e^t+1\) cancels, leaving \(2e^{xt}\). For the derivative identity, differentiation in \(x\) multiplies the generating function by \(t\). For reflection, the generating function at \(1-x\) is the original one with \(t\) replaced by \(-t\). Each conclusion follows by coefficient comparison.

We have now used one Euler sequence to construct both its residue-packet representation and a classical polynomial basis. The parameter-addition law and the Euler translation identities remain separate, explicitly typed operations.

Use Euler polynomials to sum alternating powers

The translation law also turns a finite sum into two boundary values. Multiply

by \((-1)^j\), and sum over \(0\leq j<L\). Every interior polynomial value cancels with its neighbor. For every integer \(L\geq0\),

The empty sum at \(L=0\) is zero. No limiting process is involved.

For example,

The polynomial calculation gives the same answer:

The alternating odd-power window needed next is obtained by taking \(x=1/2\) and multiplying by \(2^m\). For an odd positive length \(p\), define

Then

This includes \(m=0\): the factor is \(1/2\), and the bracket is two. For \(p=3,m=4\), the direct value is \(1-81+625=545\). Since

the polynomial route gives \(8(5+1085)/16=545\) as well.

Reconstruct Euler numbers from an odd-length window

We now use the local Euler coordinates developed in Paper IX. The source numbers \(A_{p,2n}\) are finite alternating sums. The first question is what operator reconstructs \(E_{2n}\) from them.

Start with their full exponential generating series. The finite geometric sum gives, for positive odd \(p\),

Average the expression at \(t\) and \(-t\) to retain only even degrees:

The reconstruction kernel is therefore \(\operatorname{sech}^2(pt)\). Write

The earlier reciprocal-power polynomials give

These coefficients are integers. One can also see their integrality directly by multiplying the two signed Euler series. Multiplying \(\mathcal A_p\) by the reconstruction kernel and comparing EGF coefficients proves

The inverse transfer is equally explicit. Since

its EGF coefficients are

Consequently,

For \(p=3,n=2\), this inverse calculation is already visible:

The forward and inverse transfers are triangular and have diagonal one. Oddness of \(p\) is needed for this source identity. Primality is needed only when we interpret the resulting precision in terms of a prime.

Work the odd-prime reconstruction through to 1385

Choose \(p=3\). The source and kernel values needed for ordinary degree eight are

Insert them into the degree-eight reconstruction:

Let \(T_L^{(p)}(n)\) retain terms \(0\leq r\leq\min(L,n)\). Every omitted term contains \(p^{2r}\), and all its remaining factors are integers. Thus, for every odd positive \(p\),

If \(L\geq n\), the omitted sum is empty and the error is zero. For an odd prime this gives a prime-adic precision certificate. It follows from the explicit finite sum for every \(n\), rather than from the size of a tested prefix.

For this particular row, the complete partial sums and errors are

For instance, \(1385\equiv8\pmod9\), while \(-7124023\equiv1385\equiv8\pmod{81}\). The partial sums fluctuate in ordinary size, but their differences from the target acquire the stated powers of three. These are two different ways of measuring the same rational or integer error.

Build the separate dyadic source and kernel

The odd-length source cannot simply be reused with \(p=2\). Its constant coefficient would be \(1-1=0\). Multiplication by any formal series would still have constant coefficient zero, so it could not reconstruct \(E_0=1\).

Paper IX uses a different source for the dyadic chart. In ordinary even degree \(m\), set

Its generating series is

The numerator is divisible by \(t^2\), so this is a well-defined formal series with constant coefficient one. The identity \(\cosh(3t)-\cosh t=4\cosh t\sinh^2t\) gives

The quotient \(\sinh(2t)/(2t)\) also has constant coefficient one. Define

Then \(K(t)\mathcal R(t)=\operatorname{sech}t\). If

the dyadic reconstruction is

We can obtain the kernel entirely by rational inversion. The inverse kernel has coefficients

The constant coefficient is one. For \(r>0\), the product equation \(JK=1\) gives

For example,

The relevant source and kernel values are

The dyadic source, kernel and factorial normalization have all changed together. Merely substituting two into the odd-prime formula would miss that change of coordinates.

Reconstruct 1385 in the dyadic chart

At ordinary degree eight, calculate the five terms separately:

Define \(T_L^{(2)}(n)\) by retaining the first \(L+1\) terms, or all terms when \(L\geq n\). The error table for this row is

All denominators in this table are odd, so the congruences are meaningful in the ring \(\mathbb Z_{(2)}\) of rationals with odd denominator.

The first term alone already gives the correct answer modulo 32. Indeed,

so

This modular division uses the inverse of an odd denominator. It is a different operation from attempting to reduce \(5/24\) modulo a power of two, which is undefined because 24 is not a unit.

Prove the dyadic precision bound and locate its sharp edge

The exact kernel explains the precision pattern. Define the Bernoulli numbers by

The regular even series \(Q(z)=z\coth z\) has expansion

This follows from \(Q(z)=z+2z/(e^{2z}-1)\). Differentiation gives

Substitute \(z=2t\) and compare coefficients:

The von Staudt–Clausen denominator theorem states that the reduced denominator of \(B_{2r}\), for \(r\geq1\), contains each prime \(q\) with \((q-1)\mid2r\) exactly once. In particular it contains two exactly once. Thus \(v_2(B_{2r})=-1\), and

We also need the source coefficients to be dyadic units. Write \(n+1=2^du\), with \(u\) odd. In

the first factor has valuation three, since \((9^u-1)/(9-1)\) is odd. Each factor in the product has valuation one. Hence \(v_2(9^{n+1}-1)=3+d\). The denominator of \(R_{2n}\) has the same valuation, so \(v_2(R_{2n})=0\).

It follows that the \(r\)-th reconstruction term, for \(1\leq r\leq n\), has exact weight

Every term omitted after depth \(L\) has weight at least \(4L+3\). Their finite sum therefore proves

The bound includes the empty tail, where the error is zero. It is uniformly sharp: at \(n=L+1\), exactly one term is omitted, and that term is \(\kappa_{L+1}\), because the binomial coefficient and \(R_0\) are both one. Its valuation is exactly \(4L+3\).

The first sharpness examples are

For the fixed row \(E_8\), the four nonconstant term weights are

They are distinct and increase strictly. In each nonempty tail there is a unique term of least valuation, so it cannot be cancelled by terms divisible by a higher power of two. This proves the exact error valuations \(5,8,13,15\) in the table. The uniform bound and the stronger precision of a particular row answer different questions.

Factor the omitted dyadic tail exactly

The uniform boundary is already optimal. There is nevertheless more information about the error at an individual row. Paper IX's classical-tail appendix gives an exact factorization.

Write \(r=L+1\) for the first omitted index, and let

Here \(\mathcal R\) denotes the error, while \(T\) remains the retained partial sum. Define

and retain the earlier source \(A(n)=R_{2n}\). With \((z/\sinh z)^2=\sum_{m\geq0}\beta_m z^{2m}\), put

At these nonnegative integer arguments the expression is a finite rational sum. The exact tail is

To verify it, substitute \(m=r+q\) in the omitted terms. The coefficient ratio simplifies to

The last equality separates each consecutive even and odd factor. This proves the factorization without an infinite-sum argument. At \(x=0\), \(H_r(0)=A(0)=1\), recovering the sharp first-omitted-row error \(\kappa_r\).

Legendre's binary formula gives \(v_2((2m)!)=2m-s_2(m)\), where \(s_2(m)\) counts binary ones. Combining this with the already proved \(v_2(\kappa_m)=4m-1\) yields

Also \(v_2\binom{2N}{2r}=v_2\binom Nr\). Consequently the factorization gives the exact valuation identity

The equality includes zero errors by taking their valuation to be infinity. The binomial factor accounts for one source of extra precision. The remaining factor records the cancellations that the uniform bound cannot resolve.

Find exact row precision from binary carries

The scalar coefficient at offset \(q\) in \(H_r\) has valuation

Since \(r+q>0\), this is at least \(3q+1-s_2(r)\). Thus a minimum can occur only at

This is a proved finite cutoff, not an empirical search limit. In particular, if \(s_2(r)\leq3\), every \(q\geq1\) has positive weight. At an integer \(x\geq0\), the \(q=0\) term is the odd unit \(A(x)\), and every later term is even. Hence \(H_r(x)\) is a unit, giving the exact all-row law

Every first omitted index from one through fourteen belongs to this family. It explains the earlier \(E_8\) error table directly: for \(r=1,2,3,4\),

This strengthens particular rows while retaining the same sharp uniform boundary.

For general \(r\), define \(\mu_r=\min_q\lambda_r(q)\) and let \(J_r\) be the set of minimizing offsets. Some representative carry calculations are

For example, at \(r=15\), the two weights are \(0\) and \(3+1-4=0\). At \(r=31\), the second weight becomes \(3+1-5=-1\). The binary carry into the next power of two changes which offset controls the normalized factor.

Check a cancellation beyond the uniform boundary

The first omitted index with four binary ones is \(r=15\), corresponding to retained depth fourteen. At the next row, \(N=16\), precisely two terms remain. Both have valuation 63:

Their sum has greater precision. Since \(P_1(1)=1\), the remaining factor is

Both integers in the final fraction are odd, so \(v_2(H_{15}(1))=5\). The exact tail valuation is therefore

The uniform guarantee at depth fourteen is only 59. The row's binomial weight raises the termwise lower bound to 63, and cancellation adds five more digits. None of this permits a stronger uniform exponent: at \(N=15\), the error is exactly \(\kappa_{15}\), with valuation 59.

State the remaining boundary precisely

The appendix goes further than the finite integer-row identity. It constructs a restricted 2-adic series extending the source, then normalizes \(G_r=2^{-\mu_r}H_r\). Its coefficient reduction is the finite polynomial

It proves an exact preparation \(G_r=W_rU_r\), where \(W_r\) is monic of degree \(\max J_r\), and \(U_r\) is an invertible restricted series congruent to one modulo two. Thus

For the carry rows above, the reductions are respectively \(1\), \(1+X\), \(X\), \(X\), and \(1+X+X^2\). These are reductions modulo two, not complete formulas for the 2-adic preparation polynomials.

The sharp uniform boundary, finite carry cutoff, exact factorization, and finite-polynomial reduction are proved results. A universal explicit classification of those polynomials and a proof that every dyadic root is simple remain open in the checked source. The appendix resolves particular families, but those results do not settle the universal statement. This remaining cancellation problem is separate from the already proved optimal uniform bound.

Part 3 / Matrices, congruences, and residuals

State the supercongruences and check an Euler index shift

The proved local truncation laws can be written as explicit supercongruences. For an odd prime \(p\), every \(n,L\geq0\) satisfies

For the separate dyadic source,

Both statements include exact equality at full depth. The word supercongruence here refers to the proved powers of the prime in the reconstruction error. It does not remove the source, depth, or coefficient-domain hypotheses.

For \(E_8\), the guaranteed moduli and resulting residues are

Each residue can be obtained from the corresponding partial sum, using a modular inverse for the odd dyadic denominator. The actual error table gives stronger row-specific dyadic precision at the first three depths. The uniform bound remains sharp across all Euler rows.

There is also a useful supercongruence between Euler indices. For an odd prime \(p\) and an even integer \(m\geq2\),

Here is a proof from the depth-zero source. We already know \(E_m\equiv A_{p,m}\pmod{p^2}\). If \(u\) is not divisible by \(p\), multiplication by \(u\) permutes the nonzero residues modulo \(p\). Multiplying them and cancelling their invertible product proves \(u^{p-1}\equiv1\pmod p\). Write \(u^{p-1}=1+pb\). Raising to the \(p\)-th power shows \(u^{p(p-1)}\equiv1\pmod{p^2}\).

Apply this to each odd integer \(2a+1\) in the source window. The one divisible by \(p\) contributes zero modulo \(p^2\), because both exponents are at least two. Every other term is unchanged when the exponent increases by \(p(p-1)\). The two source sums, and therefore the two Euler numbers, are congruent modulo \(p^2\).

At \(p=3,m=2\),

Its three-adic valuation is exactly two. The starting-index restriction matters: \(E_6-E_0=-62\) is not divisible by three. Deeper index-shift claims must also control the binomial weights in the retained correction terms. The depth-zero argument alone does not supply them.

Write the coefficient transform as an infinite matrix

The local reconstruction formulas are rows of one infinite coefficient transform. Let

and let \(A(t)\) have even EGF coefficients \(A_0,A_1,\ldots\), so that its coefficient at \(t^{2j}/(2j)!\) is \(A_j\). Multiplication by \(K_q\) transforms these coefficients through the matrix

Its opening rows are

The row index is the output's compressed degree. The column index is the source's compressed degree. Both refer to ordinary degrees twice as large. If \(\boldsymbol A=(A_0,A_1,\ldots)^T\), then the transformed vector is \(\mathsf C(q)\boldsymbol A\).

This infinite matrix acts on every rational coefficient sequence. Row \(n\) contains only \(n+1\) possible nonzero entries. Each output is therefore a finite sum. No convergence condition or summability prescription is being imposed on the infinite column.

For the odd chart at three, put \(q_r=\gamma_r3^{2r}\), and call the resulting matrix \(\mathsf C_3\). Its block with rows and columns zero through four is

The entire five-degree calculation is

For instance, the degree-four row gives \(1296-108\cdot17+545=5\). The final row is the earlier five-term reconstruction of \(E_8\), written in ascending source degree. Its column index \(j\) corresponds to correction index \(r=4-j\).

For the dyadic chart, use \(q_r=\kappa_r\). The corresponding block is

It acts on the different source vector:

The same Euler vector has two different coefficient descriptions. Their source vectors and transform matrices must be kept together.

Compose, invert, and truncate the coefficient matrices

Multiplying two kernels corresponds exactly to composing their matrices. For another normalized kernel sequence \(r\), define the EGF convolution

Then

To verify the entry in row \(n\), column \(k\), only intermediate indices \(k\leq j\leq n\) occur. Use

The product entry becomes

which is the claimed entry of \(\mathsf C(q\star r)\). All sums used in the proof are finite.

If the series with coefficients \(r\) is the reciprocal of \(K_q\), their convolution is \((1,0,0,\ldots)\). Therefore

For the odd chart at three, the inverse kernel has coefficients \(1,18,648,\ldots\). The first three rows give an explicit check:

The lower-left entry is \(1296-1944+648=0\). The next entry in that row is \(-108+108=0\). Thus the inverse restores the source coefficients, rather than merely reproducing one chosen Euler vector.

Finite leading blocks are exact here. Let \(\pi_N\) retain coordinates zero through \(N\), and let \(\mathsf C(q)^{[N]}\) be the corresponding square block. Lower triangularity gives

Every product entry in these rows uses intermediate indices at most \(N\), so composition and inversion also commute with taking this block. Each block has diagonal one and determinant one. The infinitely many blocks are compatible finite coefficient calculations. No input coefficient beyond \(N\) can change an output coefficient at or below \(N\).

Change the local coefficient basis and track congruence precision

Let \(\boldsymbol A^{(p)}\) denote the local source vector, with entries \(A_{p,2n}\) for odd \(p\), and \(R_{2n}\) for \(p=2\). Both reconstruct the same Euler vector through \(\boldsymbol e=\mathsf C_p\boldsymbol A^{(p)}\). The exact change from source chart \(p\) to source chart \(q\) is

These lower-triangular matrices satisfy

The intervening factors \(\mathsf C_q\mathsf C_q^{-1}\) cancel, and every matrix entry uses finitely many terms. Changing coordinates directly or through an intermediate chart gives the same result.

For odd source lengths three and five, the scalar kernel of \(\mathsf B_{5\leftarrow3}\) is \(\cosh^2(5t)\operatorname{sech}^2(3t)\). Its first coefficients are \(1,32,896,\ldots\). Thus

The last source value also follows directly from \(1-3^4+5^4-7^4+9^4=4705\). Reducing the input coefficients modulo 81 gives \((1,17,59)\). The final transformed row is

which agrees with \(4705\equiv7\pmod{81}\). Odd-chart base changes have integer entries and integer inverses, so they preserve coefficientwise congruences for every positive integer modulus. Choosing a new chart does not by itself choose a new modulus.

More generally, consider a finite block \(B\) at a fixed prime \(\ell\), with integers \(d\geq0\) and \(h\). If every entry has valuation at least \(-d\), and an input error vector has every coordinate of valuation at least \(h\), then

Each summand has that bound, so its finite sum does too. Integral matrices have \(d=0\). A denominator divisible by the working prime can cause a loss of precision.

For a concrete loss, change from the three-chart to the dyadic source. The first kernel coefficients of \(\mathsf B_{2\leftarrow3}\) are

The entry taking source degree four to output degree eight is

If only that input coordinate is perturbed by nine, the output perturbation is

The input error had valuation two. This contribution loses one ternary digit. Some dyadic-source values themselves have denominators divisible by three, so this statement is about the valuation of their difference, not an ordinary residue of each value modulo nine.

At the prime two, the dyadic and odd-chart matrices and their inverses all have odd denominators. Their base changes therefore preserve coefficientwise dyadic precision. An exact rational change of basis and its behavior at a specified prime are related facts that must both be checked.

Separate factorial normalization from ordinary Toeplitz multiplication

The binomial weights arise from a diagonal change of coordinates. Set \(x=t^2\), and define ordinary coefficients

Multiplication in ordinary \(x\)-coefficients is the lower-triangular Toeplitz matrix

with zero entries above the diagonal. Its diagonals are constant. Introduce

The EGF matrix is its factorial conjugate:

Indeed, its \((n,j)\) entry is

For the Euler output, write \(\widehat e_n=E_{2n}/(2n)!\). At the first nonconstant coefficient of the three-chart example, ordinary multiplication gives

Multiplying by \(2!\) recovers \(E_2=-1\). Using the ordinary Toeplitz row on the unnormalized EGF numerators would instead give \(-9+17=8\). The kernel has not changed, but the coordinate convention has been mixed.

The diagonal conjugation is valid over \(\mathbb Q\), where all factorials are invertible. It does not justify cancelling factorials modulo a prime that divides them. The integral odd-chart matrices and the dyadic matrices over \(\mathbb Z_{(2)}\) retain the coefficient domains established earlier.

Read the packet matrix as an infinite block operator

The earlier two-by-two matrix has formal series as entries. It also represents an infinite matrix of scalar coefficients. To see the relation, write

Group the coefficients of any \(f(x)=\sum f_nx^n\) into two packets:

The scalar multiplication operator becomes

Here each block is an ordinary coefficient Toeplitz operator in its indicated variable. The formula follows by multiplying \((C+xD)(A+xB)\) and assigning its even and odd powers of \(x\) to the two packets. Each scalar row in each block has finite support. The top-right block starts one degree later because of the factor \(u\).

For the first two coefficients of each packet, use the vector order \((f_0,f_2,f_1,f_3)^T\). The block matrix is

The entry \(-1/2\) in row two, column three is the first returning carry. The corresponding entry in row one is zero. For the secant input, \([u^0]B=1/2\), so this carry contributes \(-1/4\) to the order-\(u\) first output. It is exactly the interaction missing in the earlier coarse calculation.

For \(h\geq1\), retaining \(h\) coefficients in each packet is exactly the quotient calculation modulo \(u^h\), or modulo \(x^{2h}\) before regrouping. Thus the two-by-two formal matrix and its infinite scalar block operator describe the same operation at their stated resolutions.

Keep the congruential family and its address visible

There is another kind of base change in the Euler work: changing the family index or its address. Paper IX defines normalized congruential coefficients by

The normalization makes \(C_{N,j}(0)=1\). Removing it gives the original coefficients \(U_{N,j}(n)=j!C_{N,j}(n)\), whose constant is \(j!\).

Clearing factorials gives an infinite triangular system:

Its diagonal entry is \(\binom{Nn+j}{Nn}\). Unlike the unit diagonals of the local coefficient matrices, this need not be one. Its divisibility determines where denominators can enter.

For the first nonconstant coefficient,

At \(N=4\), the four addresses give \(-1,-1/5,-1/15,-1/35\). In particular, the address-one coefficient can be reduced modulo powers of two using its odd denominator, but cannot be treated as an integral residue modulo five.

A general denominator bound follows directly from the triangular recurrence. Put

Then \(D_{N,j}(n)C_{N,j}(n)\) is an integer. For the induction step, \(D_{N,j}(n-1)\) clears every earlier coefficient's denominator, and the next diagonal factor clears the division in the recurrence. The empty product at \(n=0\) is one.

Two specializations explain how this family meets the preceding example:

The second identity follows because the denominator is the earlier \(C(u)=\sum u^n/(4n)!\), while the quartic carrier denominator uses alternating signs. Consequently,

The coarse inverse \(C(u)^{-1}\) was a valid member of this different family. It failed when used as the classical Euler first packet, whose numerators are \(1,5,1385,2702765,\ldots\). Changing the family index does not automatically preserve the original sequence.

There is also a Bernoulli boundary address. At \(N=2,j=1\), the inverse is \(t/\sinh t\). If \(B(t)=t/(e^t-1)\), then a common-denominator calculation gives \(2B(t)-B(2t)=t/\sinh t\). Thus

This identifies another exact coefficient family while keeping its normalization and modular denominator obligations visible.

Combine supercongruences with coprime moduli

A coordinate change can preserve a known modulus. Replacing that modulus by an unrelated one requires additional information. For example, 1 and 82 agree modulo 81 but differ modulo 25. There is no well-defined conversion of that residue class alone into a class modulo 25.

The odd and dyadic reconstructions supply the additional information here. The depth-one odd calculation gives \(E_8\equiv8\pmod{81}\). The row-specific dyadic error bound gives \(E_8\equiv9\pmod{32}\). Solve both conditions by writing

Since \(32\cdot38-81\cdot15=1\), the inverse of 32 modulo 81 is 38. Therefore

Any two solutions differ by a multiple of both 81 and 32, hence by their product 2592. The unique combined residue is 1385 modulo 2592. It agrees with the exact reconstruction already obtained. A residue alone would still need a separate size bound to identify an arbitrary integer uniquely.

We have now distinguished an exact change of source coordinates, a change of congruential family and address, and the combination of independent modular information. Each uses a different map and retains its own hypotheses.

Use a residual to certify and improve a finite reciprocal

Return to a formal unit \(D\). For a candidate inverse \(B\), define the reciprocal residual by

This sign convention is chosen for the correction rule below. It is the negative of an output-minus-target residual for the multiplication equation.

Set

In a commutative coefficient ring,

If \(R\) is divisible by \(t^q\), then the new residual is divisible by \(t^{2q}\). Also

Since \(D^{-1}\) is a unit, the residual certifies exactly which initial coefficients of \(B\) agree with the true inverse. This is an algebraic precision statement.

For \(D=\cos t\), start with \(B_0=1\). Put

The corrections have a particularly simple form:

Inductively,

The residual orders are \(2,4,8,16,\ldots\). Thus \(B_2\) determines the secant series through degree six, and \(B_3\) determines it through degree fourteen.

Work out the degree-six calculation:

Here \(O(t^8)\) means divisibility by \(t^8\) in a formal series ring. It is not a numerical error estimate at a chosen real argument.

Separate a correct jet from a complete reciprocal

Let

It is a polynomial of degree six. It agrees with \(\sec t\) through that degree. Its first uncancelled reciprocal residual is

To see the coefficient directly, the degree-eight term in the product is

The residual supplies the next missing ordinary coefficient. A polynomial can be a correct finite jet while still failing the full reciprocal equation.

The program carries the correction calculation out in

In that quotient, the fourth correction has zero residual. The correct conclusion is that the residual has order at least 32 in the untruncated series. The quotient cannot reveal whether its later coefficients vanish.

No runtime comparison follows from the doubling of formal precision alone. Polynomial multiplication, coefficient sizes, and representation costs would also have to be counted in a performance claim.

Part 4 / Polyphase and filter taps

Change from series coefficients to sequence differences

We now use the signed sequence

The independent variable is the integer sequence index \(n\). Define the forward shift and difference by

Write

These are Newton difference coordinates. For every nonnegative integer \(n\),

Indeed, \(S=I+\Delta\), so the ordinary finite binomial theorem gives

Apply both sides to \(a\) and evaluate at zero. This proof works for any sequence. There is no assumption that the sequence is the restriction of an analytic function.

Construct the difference triangle from the first five Euler values:

Taking the first entry of each row gives

Reconstruct \(a(4)\):

This is another exact coordinate representation of the Euler data.

Derive the polyphase transfer kernel

For spacing \(M\geq1\) and offset \(s\geq0\), define the sampled lane

For the usual polyphase decomposition, \(0\leq s<M\). The polynomial identity below also allows any nonnegative offset.

Advancing one position in the lane advances \(M\) positions in the original sequence. Therefore

and hence

The lane's Newton coefficient is

Substitute \(S=I+\Delta\). Define the finite polynomial

Then

For \(k=0\), the lower bound is zero. For \(k>0\), the factor raised to the \(k\)-th power has order \(k\) and degree \(Mk\), which proves the stated support. The identity transfers the difference coordinates directly. It does not require reconstructing every intermediate sequence value.

At \(M=2,s=0\),

The first three kernel rows are therefore

They give

Check directly from the even lane:

Its differences start with \(5-1=4\) and

For the odd lane, the first-difference kernel is

Thus

which agrees with \(a(3)-a(1)=-61-(-1)=-60\). Offset and spacing are both part of the transfer.

Conjugate sampling by the Pascal coefficient transform

The change from sequence values to Newton coefficients is another infinite matrix pair. Define lower-triangular matrices

with zero entries for \(j>n\). For the value vector \(\boldsymbol a\) and difference vector \(\boldsymbol c\),

To verify the inverse, use \(\binom nj\binom jk=\binom nk\binom{n-k}{j-k}\). The \((n,k)\) entry of \(\mathsf P\mathsf Q\) becomes \(\binom nk\) times an alternating binomial sum. It vanishes for \(n>k\) and is one for \(n=k\). The same argument in the reverse order gives \(\mathsf Q\mathsf P=I\).

Let \(\mathsf R_{M,s}\) be the selector matrix whose row \(q\) has a single one in column \(Mq+s\). This is sampling in value coordinates. Sampling in difference coordinates is its Pascal conjugate:

The rightmost matrix reconstructs values. The middle one selects a lane. The leftmost one forms the lane's differences. Row \(k\) of the product is finite because \(\mathsf Q\) uses only rows \(0,\ldots,k\), the selector reaches at most \(Mk+s\), and \(\mathsf P\) uses only the preceding input columns.

Its entries are

Thus the infinite matrix product recovers exactly the previously derived kernel, including its offset and finite support.

For \(M=2,s=0\), the opening rows are

This matrix is row-finite but is not lower triangular. Its third row needs five input difference coefficients. For the Euler data,

Cutting the matrix down to three columns instead would give \((1,4,32)^T\). The missing contribution to the third output is

In general, outputs through difference order \(K\) require input coefficients through index \(MK+s\). The correct finite calculation therefore uses \(K+1\) rows and \(MK+s+1\) columns. A square cutoff is valid for the earlier lower-triangular coefficient transforms. It cannot be applied to a sampling matrix without checking this larger support.

There is a further infinite-matrix distinction. Every square leading block of \(\mathsf T_{2,0}\) is upper triangular with diagonal \(1,2,4,8,\ldots\), so it is invertible over \(\mathbb Q\). Nevertheless, the infinite operator is not injective on arbitrary coefficient sequences. Consider

Its first output is zero because \(v_0=0\). For every row \(k\geq1\), the constant kernel coefficient is zero, so the remaining finite sum can be evaluated as a polynomial:

Thus a nonzero infinite vector lies in the kernel, despite all those nonsingular square blocks. Its later coordinates supply cancellations that a square cutoff discards. This vector is the difference-coordinate form of the bounded signal \((-1)^n-1\), whose even samples are all zero. We will return to that signal below.

Compose two refinements and recover the same answer

Sample the lane \(R_{M,s}a\) again, now with spacing \(L\) and offset \(r\). Direct substitution gives

Therefore

The offsets compose as well as the spacings. When \(s<M\) and \(r<L\), the new offset lies between zero and \(ML-1\).

The same equality holds in difference coordinates. The Pascal conjugation makes it

All matrices here are row-finite, so their products can be associated using finite sums. To verify the same identity directly from the kernels, substitute \((1+z)^M-1\) for the difference variable of the second transfer and multiply its offset factor by \((1+z)^s\). The product is

which is the direct kernel for spacing \(ML\). Each row has finite support, so the row composition is well defined without an infinite summation argument.

For two even refinements, start from the lane difference coordinates

Refining that lane by two gives first difference

The direct spacing-four kernel gives

Finally, the original values give

We have three descriptions of the same result: direct sampling, two successive coordinate transfers, and one composed kernel.

Keep the prime residual when the precision increases

For a prime \(p\), expand

where

For \(1\leq j<p\), the numerator of \(\binom pj\) contains the factor \(p\), while \(j!\) does not. This proves that each displayed quotient is an integer.

Modulo \(p\), the correction vanishes. At higher precision it remains part of the exact operator. For \(p=2\),

Apply this to the Euler data at zero:

Dropping the residual \(2\Delta\) produces eight. The correct result is four.

Both values are zero modulo four. Modulo eight, they differ:

The coarse agreement does not authorize the higher-precision replacement. In this particular calculation the two-adic valuations are also different:

These congruences are calculated in integer sequence coordinates. The ordinary Taylor coefficient \(5/24\) from the series calculation cannot be reduced directly modulo two, because its denominator is not invertible there. The integer EGF numerator \(s_2=5\) can be reduced. Moving from rational series to modular arithmetic therefore requires the coefficient and normalization choices we made at the start.

The spacing-four output is \(1384=8\cdot173\), with valuation three. These are values for the displayed inputs. A congruence for an entire branch of Euler indices would require a separate argument controlling all terms.

Primality matters in the integer-polynomial statement. For the composite value four, \(\binom42=6\) is not divisible by four. One cannot replace the prime hypothesis by an arbitrary positive integer while keeping \(B_p\) integral in this form.

Compare the residuals at three and five

The next difference coordinate of the Euler sequence is

At spacing three, the operator is

Its Euler output is

The residual 18 has valuation \(v_3(18)=2\). Thus omitting it happens to preserve this output modulo nine. It fails modulo 27. The extra factor of three comes from the particular input values, since \(c_1+c_2=6\). The operator coefficients alone guarantee only the explicit factor three.

At spacing five,

The correction is

Therefore

which agrees with \(-50521-1\). Here \(v_5(7590)=1\), so the discarded residual is already visible modulo 25. Its residue there is 15.

These examples distinguish a divisibility statement about the transfer polynomial from an additional cancellation in one Euler row. Both can be calculated exactly, but they are separate pieces of information. The local reconstruction ladders proved earlier concern different operators and retain their own hypotheses.

Read the same shift polynomial as a signal filter

The operator \((S^M-I)^k\) also acts on a complex exponential sequence

Since \(Sz=e^{i\omega}z\), its response is

For \(k\geq1\), it vanishes at \(\omega=2\pi j/M\). For example,

has response \((e^{4i\omega}-1)^2\), with zeros at the fourth roots of unity on the frequency circle. On two-sided signals the causal convention is

It differs by a pure shift and has the same gain. The earlier one-sided Euler calculation uses only forward shifts and does not need an inverse shift.

This frequency response belongs to the finite filter. We have not taken a convergent Fourier transform of the rapidly growing Euler sequence. The arithmetic reading of the polynomial measures divisibility of integer outputs. The signal reading evaluates it on exponential modes. The common shift polynomial connects the two calculations while preserving what each one means.

Split an FIR filter into polyphase taps

The signal and the filter can both be split into phases. A tap is one coefficient of the finite filter. For \(L\geq1\) and \(n\geq0\), we keep the forward-shift convention already used above:

Take the four-tap filter

Its even and odd tap polynomials are

For a general finite tap list, the same split is \(H_0(z)=\sum_jh_{2j}z^j\) and \(H_1(z)=\sum_jh_{2j+1}z^j\), with only the available tap indices included. Split the input and output into

and let \((S_qb)(q)=b(q+1)\) advance a phase sequence by one place, for \(q\geq0\). Separating the even and odd tap indices in the finite sum gives

Thus the complete polyphase transfer is

The extra \(S_q\) in the lower-left entry is the lane-boundary carry: an odd output index plus an odd tap index lands in the next even input position. This location follows from the forward shift. The earlier formal-series multiplication matrix uses its own coefficient convention.

Use the signed Euler sequence \(a(n)=E_{2n}\). Its two phases begin

The first even and odd outputs are therefore

One more coarse index gives the following complete windows:

Interleaving the two output phases recovers the direct four-tap output \((-43,1191,-46239,2548371,\ldots)\). For a failure control, omit the advance in the lower-left entry. The first odd output would then be \((-2\cdot1+5)+(-1+3(-61))=-181\), which disagrees with \(1191\).

The tap-split program shows both implementations. Download the standalone checker, then run:

python -B euler_polyphase_taps.py

Even taps: 1, 3
Odd taps: -2, 1
Direct outputs: -43, 1191, -46239, 2548371
Polyphase outputs: -43, 1191, -46239, 2548371
Missing-advance control: -181 (correct: 1191)
Finite comparisons: 600 signal cases and 468 impulse cases
PASS: exact polyphase/tap split

The exact result record contains the phase contributions and finite test ranges. These are exact integer calculations with complete input windows; no zero padding, periodic extension, or Fourier transform of the Euler sequence is used. The finite comparisons support the implementation in their stated ranges. The general phase law follows from the finite-sum reindexing above.

See what a single sampled lane forgets

A small bounded signal makes the sampling issue visible without the growth of the Euler numbers. Compare

Their even lanes agree:

Their odd lanes are different: \(R_{2,1}x=(-1,-1,\ldots)\), while \(R_{2,1}y=(1,1,\ldots)\). Consequently, knowing the even lane alone does not determine the even lane after a one-step shift of the original signal.

More explicitly, set \(v=x-y\) and \(f=R_{2,0}\). Then

This is the same kind of kernel obstruction that appeared when we discarded the second formal-series packet. An operator on the retained data alone could not send the same zero input to both of these outputs.

Retaining both phases repairs the representation. Write \(b_0=R_{2,0}x\) and \(b_1=R_{2,1}x\). A unit shift of the original signal sends the pair to

Two such operations give \((S_qb_0,S_qb_1)\), as required for an original shift by two. The advance of the lane index is part of this two-phase transfer.

Finally, \((S^2-I)x=0\), because \(x(n+2)=x(n)\). The filter cannot detect this alternating mode at that spacing. This is its zero at frequency \(\omega=\pi\), evaluated on an actual bounded signal. The statement concerns the filter and the retained phases, with no Fourier transform assigned to the Euler sequence.

Part 5 / Euler and Fibonacci coordinates

Change Euler polynomials to Fibonacci coordinates

The Fibonacci connection can be made at two levels: as a polynomial basis change, and as a recurrence model for a sequence. We will calculate both before constructing a matrix that sends the even Euler sequence to Fibonacci values.

Define the Fibonacci polynomials by

For a basis indexed by degree, put \(B_k(x)=F_{k+1}(x)\). Its first terms are

Each \(B_k\) is monic of degree \(k\), so these polynomials form a basis of \(\mathbb Q[x]\). Descending elimination already gives

There is an explicit formula for every monomial:

Here a binomial coefficient outside its usual range is zero. To prove the formula, start with \(m=0\), then multiply by \(x\). The Fibonacci recurrence gives \(xB_k=B_{k+1}-B_{k-1}\) for \(k\geq1\), while \(xB_0=B_1\). Thus the new coefficient at \(B_k\), for \(k\geq1\), is the old coefficient at \(B_{k-1}\) minus the old coefficient at \(B_{k+1}\). Pascal's identity gives exactly the displayed difference of binomial coefficients for degree \(m+1\). At \(k=0\), the new coefficient is minus the old coefficient at \(B_1\). When this boundary is present, the same identity follows from \(\binom{2r+1}{r}=\binom{2r+1}{r+1}\). This proves the induction, including the constant term.

Write the classical Euler polynomials in this basis as

Combining the monomial formula with the Appell expansion obtained from the defining generating function \(\mathcal E_n(x)=\sum_{j=0}^n\binom nj\mathcal E_{n-j}(0)x^j\) gives

for \(0\leq k\leq n\), and \(C_{k,n}=0\) when \(k>n\). No sequence limit is needed. This is a finite substitution for one polynomial.

Rows of \(\mathsf C\) are Fibonacci degrees and columns are Euler degrees. The degree-four block is

If \(p(x)=\sum_n\alpha_n\mathcal E_n(x)=\sum_k\beta_kB_k(x)\), then \(\beta=\mathsf C\alpha\). In particular, its fifth column says

This also recovers an Euler number. At \(x=1/2\), the first five basis values are \(1,1/2,5/4,9/8,29/16\). Therefore

The infinite matrix here has a different domain from the earlier lower-triangular coefficient matrices. It is upper triangular and each column is finite. It acts on finite-support columns, the coefficient space of polynomials. Its inverse has the same property, and every degree-bounded square block is invertible with diagonal one. An arbitrary infinite Euler expansion would require an additional meaning for its sums. We do not extend this polynomial basis change to all infinite coefficient columns by assumption.

Write the Fibonacci polynomial transfer matrix

The polynomial recurrence has a two-dimensional transfer:

For \(n\geq1\), induction gives

At \(n=0\) the matrix is the identity. Multiplication by \(\mathsf T(x)\) advances each column by the defining recurrence, proving the next step. For example,

At \(x=1\), this becomes \(\bigl(\begin{smallmatrix}5&3\\3&2\end{smallmatrix}\bigr)\), and the usual Fibonacci numbers appear. Its determinant is \((-1)^n\), since \(\det\mathsf T(x)=-1\). In particular, the step is reversible over \(\mathbb Z[x]\):

This matrix advances a Fibonacci polynomial state. The connection matrix \(\mathsf C\) changes the coordinates of an entire polynomial. Their sizes, inputs, and purposes are different, even though both come from the same basis recurrence.

Keep the defect when Euler values enter a Fibonacci recurrence

Return to the compressed even Euler stream

This is distinct from the full-index stream \(E_n=1,0,-1,0,5,0,-61,\ldots\). Define the Fibonacci defect of the chosen stream by

The first four calculations are

The coordinate column is now \((a_0,a_1,f_0,f_1,f_2,\ldots)^{\mathsf T}\). Its five-term analysis and synthesis matrices are

In particular,

The last row of the inverse recovers \(2(1)+3(-1)+2(5)-65+1441=1385\). The matrices are lower triangular with diagonal one. At every prefix length at least two, solving \(a_{n+2}=a_{n+1}+a_n+f_n\) gives the inverse uniquely. The compatible infinite matrices act on arbitrary rational streams, with a finite sum in every row.

To see the Fibonacci weights in that inverse, use the state order \(v_n=(a_n,a_{n+1})^{\mathsf T}\) and set

This order reverses the polynomial state used above. With \(\mathsf J=\bigl(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\bigr)\), we have \(\mathsf A=\mathsf J\mathsf T(1)\mathsf J\). The forced recurrence is exactly

Writing \(F_n=F_n(1)\), repeated substitution gives the finite variation-of-constants formula

The empty sum at \(n=0\) is zero. Induction proves the formula by applying \(\mathsf A\) and adding the next defect. For \(n\geq1\),

For example, the state containing \(-61\) and \(1385\) is

Discarding the defects would leave the homogeneous sequence \(1,-1,0,-1,-1,-2,\ldots\). It already disagrees with the Euler stream at \(a_2=5\). The Fibonacci model is exact when its forcing coordinates remain attached.

Move with the Fibonacci transfer

Since \(\det\mathsf A=-1\), the companion matrix has an integer inverse:

Define moving coordinates \(z_n=\mathsf A^{-n}v_n\). Multiplying the forced recurrence by \(\mathsf A^{-(n+1)}\) gives

The coordinate change is invertible at every step, with \(v_n=\mathsf A^nz_n\). For the Euler stream, the first values are

The increments can be checked without a matrix power routine:

For a homogeneous Fibonacci solution, every defect is zero and these moving coordinates are constant. Conversely, constant moving coordinates imply zero defect, because \(\mathsf A^{-(n+1)}e\) is nonzero. The Euler example has nonzero increments. The change of coordinates exposes that difference rather than removing it.

Transfer the Euler sequence to Fibonacci values

We can also ask for a matrix whose output values actually are Fibonacci numbers. For this calculation, use an ordinary formal series in a new variable \(z\):

There are no factorials here. This series differs from the earlier exponential generating function \(\operatorname{sech}t\). All operations take place in \(\mathbb Z[[z]]\), so analytic convergence is unnecessary.

Choose the shifted Fibonacci stream

The shift makes both constant coefficients one. A unit lower-triangular convolution cannot send \(a_0=1\) to the unshifted \(F_0=0\).

First apply the recurrence polynomial to the source:

Its coefficients after degree one are exactly the defects: \(r_0=1\), \(r_1=a_1-a_0=-2\), and \(r_n=f_{n-2}\) for \(n\geq2\). Because \(r_0=1\), there is a unique integral formal inverse \(H=R^{-1}\), obtained from

Here are the first five steps:

Consequently,

Indeed, multiplying the left side by \(1-z-z^2\) gives \(HR=1\). The formal unit \(1-z-z^2\) then gives the claimed identity uniquely.

For ordinary coefficients, let \(\mathsf L(H)_{n,j}=h_{n-j}\) when \(j\leq n\), and zero otherwise. The first six rows give the requested sequence transfer matrix:

For the last three displayed rows, the arithmetic is

This is an invertible infinite matrix on all integer streams. Every row is finite, its diagonal is one, and its inverse is \(\mathsf L(R)\). For example, the inverse recovers

The transfer was constructed from the known Euler source and the chosen Fibonacci target. It demonstrates an exact reversible encoding. It does not establish a faster way to discover or compute either sequence.

Transport the operation along with the Fibonacci coordinates

A sequence transformation should carry its operation as well as its values. Let \(S\) be the forward shift, \((Sc)_n=c_{n+1}\), and let \(\varepsilon_0(c)=c_0\). For any ordinary convolution kernel \(H\), direct cancellation of the finite sums gives

Their difference is the boundary term \(h_{n+1}c_0\). Set \(u_n=h_{n+1}\). Then

The second identity uses \(\varepsilon_0\mathsf L(R)=\varepsilon_0\), since \(r_0=1\). Thus a shift of the source is represented by a shift plus an explicit boundary correction in the new coordinates. For the Euler input, the first commutator component is \(g_1-a_1=1-(-1)=2=h_1a_0\). At component one, it is \(g_2-(2a_1+a_2)=2-3=-1=h_2a_0\).

Finally, the target obeys \((S^2-S-I)g=0\). Pulling that operation back to the Euler coordinates gives

This is the transported recurrence. Applying \(S^2-S-I\) directly to \(a\) produces the nonzero defect stream already calculated. The coordinate matrix, inverse, recurrence, and boundary term together specify what the transfer preserves.

Sample Fibonacci packets and derive their Lucas recurrence

Paper IX develops companion jumps and the Fibonacci specialization. Its companion-jump theorem and Fibonacci consequence let us continue the transfer calculation at a coarser sampling step.

Use the state order of the polynomial companion at \(x=1\):

Fix a step \(N\geq1\) and an address \(j\geq0\). The packet state is

Its transition is \(\mathsf C^N\). For example,

To derive the scalar recurrence, define Lucas numbers independently by \(L_0=2\), \(L_1=1\), and \(L_{n+2}=L_{n+1}+L_n\). Since \(\mathsf C^2=\mathsf C+I\), the traces \(\operatorname{tr}(\mathsf C^n)\) satisfy this recurrence and these initial values. Hence

Every two-by-two matrix \(J\) satisfies \(J^2-(\operatorname{tr}J)J+(\det J)I=0\), as direct multiplication of its four entries verifies. Apply this to \(J=\mathsf C^N\), then take the second component of the packet state:

For \(N=3\), \(L_3=4\), so every address obeys \(u_{k+2}=4u_{k+1}+u_k\):

For the last row, \(21=4(5)+1\), \(89=4(21)+5\), and \(377=4(89)+21\). The first two values and this recurrence determine every subsequent packet value by induction.

The parity sign matters. At step two, the recurrence is \(u_{k+2}=3u_{k+1}-u_k\). The packet \(F_{2k+1}=1,2,5,13,34,\ldots\) gives \(5=3(2)-1\). Replacing the minus with a plus gives the wrong value seven. The formula also has a consistent zero-step boundary: \(L_0=2\), and a constant packet satisfies \(u=2u-u\).

We can compose this sampling with the earlier Euler transfer. If \(\mathsf R_{2,0}\) selects even entries, then

The first three rows of the composed map are

Each retained row is complete. The target now follows the step-two Lucas recurrence. This is the composition of a reversible sequence encoding with a sampling operation. Recovering the full Fibonacci stream also requires the omitted lane or equivalent extra data.

Refine a Fibonacci packet and increase its congruence precision

Paper IX's matrix precision amplification and dyadic Fibonacci example start from an explicit integral jump:

Set \(H_a=12\cdot2^a\). We claim that

The displayed jump proves the base case. Squaring proves the next step and gives the factor itself:

For example,

and indeed \(\mathsf C^{24}=I+16\mathsf B_1\). Every entry remains integral. More generally, if \(X=I+p^LB\) with integral \(B\) and \(L\geq1\), the binomial expansion of \(X^p\) shows \(X^p=I+p^{L+1}B'\) for integral \(B'\). Iteration gives the analogous refinement at any prime. The positive exponent and integral coefficient domain are part of this statement.

Define the step difference by \(\Delta_Hf_n=f_{n+H}-f_n\). Repeated differences of a matrix orbit give

This follows by applying \(\mathsf C^{H_a}-I\) once for each difference. All factors commute because \(\mathsf C^{H_a}-I\) is a polynomial in \(\mathsf C\). Thus

Here \(a,n,r\geq0\), and all states and observations are integral. No assertion about arbitrary rational observation vectors is needed.

This bound is sharp at \(n=1\). The base factor satisfies \(\mathsf B_0\equiv I\pmod2\), and its refinement recurrence preserves that property. Therefore \(e_2^{\mathsf T}\mathsf C\mathsf B_a^rw_0\equiv F_1=1\pmod2\). We have the exact valuation

The refinement bound is a source statement. This sharpness calculation is a direct consequence of the displayed factor's parity. At the first step,

The three remaining factors are odd, giving valuations three, six, and nine exactly.

Reconstruct a large Fibonacci value through the precision ladder

The same jump gives a finite binomial expansion. For integers \(x,j\geq0\),

Indeed, expand \((I+2^{a+3}\mathsf B_a)^x\), then apply the indicated state and observation. If we retain only \(r<L\), every omitted term is divisible by \(2^{(a+3)L}\). When \(L>x\), no term is omitted and the answer is exact.

Take \(a=0\), \(j=1\), and \(x=3\). This reconstructs \(F_{37}\). The normalized difference coefficients just calculated are \(1,29,1165,46745\), so

The intermediate answers carry a precise meaning:

For example, the second row means \(F_{37}\equiv697\equiv57\pmod{64}\). The third means \(F_{37}\equiv224377\equiv121\pmod{512}\). Doubling the packet step would increase the guaranteed exponent per omitted depth from three to four. The state factor records why that extra precision is available.

This is an exact finite expansion at nonnegative integer indices. Extending it to other arguments would require a stated domain and convergence argument.

Obtain simultaneous congruences from one integral jump

Paper IX also gives a simultaneous Fibonacci jump:

This proves \(\Delta_{24}^{\,r}F_n\equiv0\pmod{24^r}\) by the same matrix-difference calculation. In this particular matrix, the remaining entries share a factor six. Factoring it out yields the stronger explicit statement

and therefore

This improvement follows from the displayed entries. The two prime consequences are divisibility by \(2^{4r}\) and by \(3^{2r}\). They hold simultaneously, since \(144^r=2^{4r}3^{2r}\).

For a hand check at \(n=1\),

For the second equality, \(\mathsf P^2\) has upper-left entry \(521^2+322^2=375125\), which is the scalar observed by \(e_2^{\mathsf T}\mathsf C\mathsf P^2w_0\). The ordinary recurrence, integer matrix calculation, and congruence agree on the same quantity.

Carry differentiation across the Euler and Fibonacci bases

The typed-expression example in Komorebi Math supplies a second continuation. It carries a polynomial through actual operations in both bases. We can write its matrices in the conventions already used here.

Let \(\mathcal P_d^F\) and \(\mathcal P_d^E\) denote polynomials of degree at most \(d\), expressed in the Fibonacci and Euler bases. They have dimension \(d+1\). Write \(\mathsf C_d:\mathcal P_d^E\to\mathcal P_d^F\) for the connection matrix derived above.

Euler polynomials form an Appell family, so differentiation from degree four to degree three is

To apply this physical operation in Fibonacci coordinates, change the input to Euler coordinates and the output back:

This is a rectangular transport: the input and output degree bounds differ. Each column can be checked by differentiating its basis polynomial. For instance,

Using the Euler derivative matrix directly on Fibonacci coefficients would instead turn \(B_3\) into \(3B_2=3x^2+3\), missing the constant correction. The connection must accompany the operation.

Complete an integration and differentiation round trip

Let \(J_0p(x)=\int_0^x p(t)\,dt\). Its matrix from \(\mathcal P_3^F\) to \(\mathcal P_4^F\) is

Every column is obtained by integrating one \(B_k\), expressing the result in the degree-four Fibonacci basis, and imposing value zero at the basepoint. For example, the last column reconstructs \(-B_0/2+B_2/4+B_4/4=x^4/4+x^2\), which vanishes at zero.

Start with \(p=B_3=x^3+2x\), represented by \(v_0=(0,0,0,1)^{\mathsf T}\). Integration gives

The inverse degree-four connection is

Changing to Euler coordinates, differentiating there, and returning to Fibonacci coordinates gives

For a direct check of the middle vector, \(\tfrac32\mathcal E_0+\tfrac72\mathcal E_1+\tfrac32\mathcal E_2+\mathcal E_3=x^3+2x\). The whole operator is exactly

The identity holds for every polynomial of degree at most three. It follows from \((J_0p)'=p\), and can also be verified by multiplying these four matrices.

The reverse order retains a boundary defect. Since \(J_0p'=p-p(0)\), its degree-four Fibonacci matrix is

The row \((1,0,1,0,1)\) evaluates the Fibonacci basis at zero. It is not merely the first coefficient of a Fibonacci coordinate column. For example, \(B_2(0)=1\), so the reverse operation sends \(B_2\) to \(B_2-B_0=x^2\). On the constant polynomial it gives zero. These checks expose the boundary term that a false reverse-identity claim would lose.

Finally, integrate the recovered polynomial on \([0,1]\). The integration covectors in degree three are

Their entries are the integrals of the corresponding basis polynomials. They satisfy the exact dual-coordinate law \(\ell_F\mathsf C_3=\ell_E\). Thus both paths give

in agreement with \(\int_0^1(x^3+2x)\,dx=1/4+1\). The coefficient columns, operator matrices, degree bounds, and scalar observation have all moved together.

Part 6 / Quartic and logarithmic coordinates

Work with all four quartic addresses

The four-address results in Paper IX concern \(C_{4,j}\). These differ from the classical signed Euler numbers, which belong to \(C_{2,0}\). Keep the modulus and address attached to the coefficients:

The defining triangular system becomes

For these four addresses the diagonal is odd. More explicitly,

Induction therefore shows that every \(C_j(n)\) is a rational number with odd denominator. Reduction modulo \(2^r\) means inverting that denominator modulo \(2^r\). The first rows are

For instance, at address two and row two,

At address zero, \(C_0(n)=(-1)^n b_n\), where \(b_n\) is the quartic carrier already reconstructed from the secant sequence. The other three addresses now give further arithmetic coordinates for the same reciprocal-section construction.

Use the all-row bound to isolate a dominant Newton coefficient

Paper IX proves the all-row dyadic difference bound: for \(n,k\geq0\) and positive integer \(u\),

Zero differences have valuation infinity. We use this theorem as an input. Its source proof controls the triangular inverse at every row, including row zero. The following finite calculation shows how that bound leads to an exact displacement law.

Set \(a_{j,k}=\Delta_1^kC_j(0)\). The first Newton coefficients are

Their valuations are \((c_0,c_1,c_2,c_3)=(1,1,4,2)\). Already at order two we can calculate

Continuing the same rational recurrence gives the finite initial data used in Paper IX's Newton-dominance argument:

These values and the universal bound \(v_2(a_{j,k})\geq k\) establish

Only the displayed initial orders require separate checking. Beyond them, \(c_j+1+\lfloor\log_2k\rfloor\leq k\). Thus a finite initial computation and a proved uniform estimate cover every order.

Finite Newton translation gives

For \(k\geq2\), the same logarithmic valuation envelope survives translation. For \(k=1\), every term after \(a_{j,1}\) has strictly higher valuation. Consequently \(v_2(\Delta_1C_j(n))=c_j\) at every row.

Now expand a positive displacement \(T\):

The identity \(k\binom Tk=T\binom{T-1}{k-1}\) implies \(v_2\binom Tk\geq v_2(T)-v_2(k)\). Each term in the remaining sum therefore has valuation at least \(v_2(T)+c_j+1\). The leading term has exact valuation \(v_2(T)+c_j\), so it cannot be canceled by that sum.

Determine the least periods of the quartic packets

We have recovered Paper IX's exact displacement law:

The difference is nonzero for every \(n\geq0\) and \(T>0\). Multiplication by \(j!\) gives the corresponding law for the original coefficients:

For example,

Their valuations are five and six. This is the predicted one-digit gain when the displacement changes from one to two.

A positive integer \(T\) is a period modulo \(2^r\) precisely when

Hence the least positive period is exactly \(2^{\max(0,r-d_j)}\). At precision \(2^{10}\), the four periods are

Each proposed half-period fails modulo \(1024\) at every row. The exact law classifies all positive periods, including non-powers of two. It also gives \(U_2(n)\equiv2\pmod{32}\) and \(U_3(n)\equiv6\pmod8\) for all \(n\).

These are periods in the compressed index \(n\). Their corresponding supported degree displacements are \(4T\). Changing the address is a different operation.

Invert a quartic packet one binary digit at a time

Normalize the exact gain by setting

The displacement law gives \(\Phi_j(n)\in\mathbb Z_2\), \(\Phi_j(0)=0\), and

This is the isometric coordinate map from Paper IX. It has a constructive inverse. Modulo \(2^s\), its value depends only on the input residue, and two distinct input residues have distinct output residues. Since there are \(2^s\) of each, it permutes them.

At the next precision, an input \(n_s\) has two possible children: \(n_s\) and \(n_s+2^s\). Their images agree modulo \(2^s\) and differ by exactly the next binary digit. Exactly one child matches any prescribed target modulo \(2^{s+1}\).

Take address two and solve \(\Phi_2(n)\equiv13\pmod{64}\). The complete lift is

Every image in a row is reduced modulo that row's modulus, as is the target thirteen. Thus \(n\equiv21\pmod{64}\) is the unique answer. The first nonzero coordinate can be checked directly: \(\Phi_2(1)=(-2/15-2)/32=-1/15\), which is one modulo two and modulo four.

We can also inspect the final coefficient calculation. At row 21, the triangular diagonal is \(\binom{86}{84}=3655\), whose inverse modulo 2048 is 887. The 21 terms \(\binom{86}{4q}C_2(q)\), reduced modulo 2048, are

They sum to \(22537\equiv9\pmod{2048}\). Therefore

Dividing the last difference by 32 gives \(\Phi_2(21)\equiv13\pmod{64}\), exactly as the lifting table predicts.

For a full 2-adic target, repeat the lift indefinitely. The compatible input residues define one element of \(\mathbb Z_2\), and the isometry passes to the limit. This proves the continuous extension is a bijective isometry. Consequently, the image of the continuous extension \(U_j:\mathbb Z_2\to\mathbb Z_2\) is exactly

with one preimage in \(\mathbb Z_2\) for each point of that image. The statement concerns 2-adic coordinates; no complex-analytic continuation is being inferred.

Invert the Euler source in a finite Hurwitz mode algebra

Another part of Paper IX keeps coefficient multiplication meaningful in finite characteristic. For sequences over a ring, define

This is Hurwitz multiplication. Over \(\mathbb Q\) it corresponds to multiplying exponential generating functions. Over \(\mathbb F_p\) it is defined by the displayed integer binomial coefficients, without dividing by factorials.

For \(u\in\mathbb F_p\), let \(\epsilon_u=(u^n)_{n\geq0}\), with \(0^0=1\). The binomial theorem gives

Thus finite mode weights multiply by cyclic convolution on the additive group \(\mathbb F_p\). Their representation is faithful: the first \(p\) sequence entries form a Vandermonde system on the distinct field elements, with nonzero determinant.

If a mode combination \(x\) has total weight one, then \(x^p=\delta\). Indeed, Frobenius removes mixed terms, each field coefficient satisfies \(c^p=c\), and \(\epsilon_u^p=\epsilon_{pu}=\epsilon_0=\delta\). Therefore \(N=x-\delta\) satisfies \(N^p=0\), giving a finite inverse:

The normalized cosh source at an odd prime is

Its total weight is one. The signed full-index Euler sequence is its Hurwitz inverse, since its defining binomial recurrence says \(q\star E=\delta\).

At \(p=5\), the source weights at \(u=0,1,2,3,4\) are \((0,3,0,0,3)\). The finite inverse gives

The cyclic convolution can be checked in five entries. At zero it is \(3(1+1)=1\pmod5\). At every other address it is \(3(1-1)=0\). Hence, at every full index \(n\geq0\),

For example, at \(n=6\), \(1-64-729+4096=3304\equiv4\equiv-61\pmod5\). This finite five-mode expression reproduces every coefficient modulo five. It is a formula in the Hurwitz ring, not a truncated exponential series in which \(5!\) has been inverted.

Keep the degree-zero mode when stating a period

The mode \(\epsilon_0\) in the preceding formula is concentrated at index zero. It is essential:

For positive indices, every remaining base has fourth power one modulo five. The full-index sequence therefore has period four on its positive tail. Its first values are

Deleting the degree-zero mode would change the initial value. Extending the tail period to row zero would make a false statement. In the compressed even stream, the same distinction is a period of two for \(E_{2n}\) at \(n\geq1\), with the exceptional initial value still retained.

This also explains why the finite-characteristic calculation used Hurwitz coordinates. Factorial conjugation is available over \(\mathbb Q\), but \(5!\) is zero modulo five. A normalization valid over the rationals cannot simply be inverted in this field.

Change to logarithmic Euler coordinates with Stirling matrices

Paper IX's formal Amice pullback gives another exact coordinate system. Here the coefficient domain is \(\mathbb Q\), and all substitutions are formal. Put

Define signed Stirling numbers by \(x(x-1)\cdots(x-n+1)=\sum_k s(n,k)x^k\). They satisfy \(s(n+1,k)=s(n,k-1)-ns(n,k)\). The formal identity

follows from the same recurrence: differentiate the left side and multiply by \(1+T\). Therefore, if \(F(t)=\sum a_nt^n/n!\) and \(\widehat F(T)=\sum b_nT^n/n!\), then

The inverse substitution is \(T=e^t-1\). Differentiating \((e^t-1)^k/k!\) gives the recurrence \(S(n+1,k)=S(n,k-1)+kS(n,k)\), with \(S(0,0)=1\) and zero outside the triangular range. These are the Stirling numbers of the second kind, giving

These are inverse lower-triangular infinite matrices, with a finite sum in every row. The first transformation begins

Use the full-index Euler coefficients here, including their odd zeros. Substituting the compressed even stream would give a different object.

For \(F(t)=\operatorname{sech}t\), the transformed series is rational:

This follows by writing \(2\cosh t=e^t+e^{-t}\) and substituting \(e^t=1+T\). Comparing coefficients of the last rational identity gives an integer recurrence for its Hurwitz numerators:

The first coefficients are

For example, \(b_6=90\), \(b_7=-7(90)=-630\), and \(b_8=-8(-630)-28(90)=2520\). The inverse Stirling row then reconstructs the familiar value:

The omitted terms have zero \(b_k\). Thus the same Euler coefficient has now been recovered from a rational logarithmic series and two inverse integer coefficient transforms.

Transport the derivative through the logarithmic coordinate

The coordinate change also carries differentiation. Formal chain differentiation gives

Write \(\mathsf S\) for the signed Stirling matrix and \(\mathsf R\) for its inverse. Let \(\mathsf D\) shift Hurwitz coefficients forward, \((\mathsf D a)_n=a_{n+1}\), and let \(\mathsf N\) be diagonal with \((\mathsf Nb)_n=nb_n\). Then

Indeed, differentiating \(\sum b_nT^n/n!\) gives \(b_{n+1}\), and multiplying its derivative by \(T\) contributes \(nb_n\). Every resulting coefficient uses finitely many input entries.

For the Euler example, the first transported derivative coefficients are

For instance, at row three the value is \(-6+3(3)=3\). Directly transforming the derivative's coefficients \((E_1,E_2,E_3,E_4,\ldots)\) gives \(s(3,1)E_2+s(3,2)E_3+s(3,3)E_4=2(-1)-3(0)+5=3\).

The rational form, Stirling matrices, and conjugated derivative describe the same formal object. No convergence claim is needed for these coefficient identities. Evaluating the infinite series at a 2-adic or complex argument would require the appropriate convergence conditions separately.

Part 7 / Checks, downloads, and sources

Run the exact checks

Download the standalone checker, or the complete source and results bundle. From the download directory, run:

python -B euler_packets.py

Python 3.10 or later is sufficient. On Windows, py -3 can replace python when that is how Python is installed. The program makes no network requests. With no options, it writes only this report:

Secant numbers: 1, 1, 5, 61, 1385, 50521, 2702765
Quartic reciprocal: 1, 1, 69, 33661, 60376809
Odd-length and dyadic reconstructions: E_8 = 1385
Dyadic E_8 tail valuations: 5, 8, 13, 15, exact
Matrix checks: inverse, composition, factorial normalization, packets, Pascal sampling
Sampling cutoff control: 3 columns give 32; 5 columns give 1376
Base changes: exact cocycle; prime-dependent precision checked
Supercongruence combination: 8 mod 81 and 9 mod 32 give 1385 mod 2592
Fibonacci polynomial coordinates: E_4(x) = 2 B_0 + 5 B_1 - 3 B_2 - 2 B_3 + B_4
Fibonacci recurrence defects: 5, -65, 1441, -51845
Reversible sequence transfer: 1, -1, 5, -61, 1385 -> 1, 1, 2, 3, 5
Paper IX Fibonacci packets: step 3 gives u[k+2] = 4*u[k+1] + u[k]
Fibonacci precision ladder: F_37 = 24157817; tail valuations 3, 6, 9, exact
Euler/Fibonacci calculus: full round trip is I_4; integral = 5/4
Quartic address gains: normalized 1, 1, 4, 2; original 1, 1, 5, 3
Dyadic inverse: address 2, target 13 mod 64 -> index 21
Finite mode inverse mod 5: weights 1, 1, -1, -1, 1
Logarithmic Euler coordinates: 1, 0, -1, 3, -6, 0, 90, -630, 2520
Dyadic edge: exact tail factorization and sparse-binary row valuations checked
Signed Euler difference coordinates: 1, -2, 8, -80, 1664
Spacing-two lane: 1, 5, 1385; differences: 1, 4, 1376
Refinement: 2*4 + 1376 = 1384
Prime residual: 2*(-2) + 8 = 4
Newton residual orders: 2, 4, 8, 16, at least 32
PASS: Euler packets, polynomials and residuals

To save exact JSON and CSV files, choose a new results directory:

python -B euler_packets.py --output my-euler-results

The program refuses to replace existing result files. The CSV values retain integers and rational numbers, including the ordinary coefficients and their factorial normalization.

You can also inspect the published coefficient table, parameter-polynomial table, local reconstruction and error table, matrix-entry table, supercongruence table, Fibonacci transfer matrices, Fibonacci precision ladders, polynomial calculus matrices, quartic address coefficients, binary inversion trace, Euler coordinate table, exact dyadic tails, exact results, and source identities.

The matrix table contains twelve-by-twelve blocks of each forward and inverse local transform for charts 2, 3, 5 and 7, the Pascal pair, the displayed three-by-five sampling block, and five-by-five base changes from chart three to charts five and two. The Fibonacci table adds the degree-twelve polynomial connection and its inverse, and sixteen-by-sixteen recurrence analysis, synthesis, ordinary sequence transfer and inverse matrices. These finite tables can be inspected entry by entry. The formulas and proofs above define the infinite operators.

The checks use several different routes:

The permutation control uses a different definition of the numbers. At length \(m\), keep the number of admissible permutations ending at each possible rank. If the next comparison is an ascent, sum the preceding counts below the new rank. For a descent, sum those at or above it. Repeating the prescribed pattern gives the total. The program also enumerates the small permutation sets directly to check this dynamic program. It does not infer a universal enumeration theorem from those samples.

The bounded run manifest records the command, Python version, source hashes before and after execution, time limit, observed completion, and result hashes. The dyadic-tail self-review links the preceding proof reviews and checks the added arguments separately. There is no new Lean run for this example.

A separate check compares 605 reciprocal coefficient values with the pinned Komorebi Math source, using both its integer sequence API and its logarithmic reciprocal-power implementation. It also makes 1,012 comparisons in the local Euler charts: 220 source, kernel and reconstruction values, 528 truncation values and errors, and 264 row weights, across prime charts 2, 3, 5 and 7 through ordinary degree twenty. Its comparison record retains the source identity and declared ranges.

That supplementary checker also verifies twelve complete rows of the infinite sampling-kernel witness and 18 Euler index-shift supercongruences, for primes 3, 5 and 7 with even starting degree 2–12. It retains the failing degree-zero boundary as a negative control. It also makes 921 Fibonacci comparisons with the pinned library: 169 polynomial connection entries, 52 polynomial companion entries, 13 centered Euler values, 512 recurrence matrix entries, 150 recurrence checks, and 25 full-index records. The full and even-index streams are checked separately at every prefix length from two through twenty-six. A further 3,506 comparisons check the new literature examples against pinned APIs: 780 packet values, 1,560 packet-state coordinates, 68 jump-matrix entries, and 1,098 polynomial-calculus entries. The source check records the exact source statements and distinguishes their written consequences. The coordinate atlas adds 1,164 comparisons: 96 quartic coefficients, 150 modular Hurwitz inverse/product values, 578 Stirling entries, 51 formal-coordinate values, and 289 substitution entries. Its source record identifies the all-row theorem used in the universal displacement proof. The universal statements have the written proofs or checked source inputs above.

The tail continuation adds 234 library comparisons: 46 partial sums, 46 exact errors, 132 omitted-term weights, and ten kernel values. The tail source check records the finite-polynomial result and its remaining simplicity boundary.

What the example establishes

The written calculation gives exact reciprocal recurrences, the two-packet multiplication law and its coarsening obstruction, the quartic reconstruction and enumeration identities, reciprocal parameter polynomials, Euler-polynomial power sums, odd-length and dyadic local supercongruences, infinite coefficient matrices and inverse transforms, prime-dependent base changes, congruential families with explicit addresses, coprime modular reconstruction, formal residual correction, sequence-refinement kernels with their Pascal conjugation, composition and prime correction, and explicit Euler-to-Fibonacci polynomial, recurrence and sequence transfers with their inverses and transported operation. The continuation derives Lucas packet recurrences, dyadic and simultaneous-prime Fibonacci precision ladders, and a complete Euler/Fibonacci polynomial calculus round trip. The quartic continuation derives exact displacement and period laws, binary inverse coordinates, finite-characteristic Euler mode inversion, and logarithmic/Stirling coordinates with a transported derivative.

The finite program checks implementations of these laws in its declared ranges. Its deliberately wrong controls expose missing carry factors, wrong factorial normalization, discarded residuals, and an invalid composite-number generalization.

The role of Komorebi is visible in the intermediate objects. A packet that appears dispensable at one resolution is needed by the next operation. A residual identifies the lost interaction. A transfer matrix reconstructs the original coefficients through a different carrier. A polynomial kernel moves the resulting sequence into sampled coordinates, and a second residual records what a coarse congruence discards.

The method keeps the representation, operation, preserved data, and remaining defect together throughout the calculation. That is the concrete connection to machine-native mathematics. Establishing an advantage for an automated discovery system would additionally require a specified system and a controlled comparison.

Sources and further reading

• Polyphase/tap example review and finite scope.

• Machine-native mathematics essay and sequence experiments.

• Quartic and coefficient-atlas source check, for the exact four-address, isometry, Hurwitz-mode and formal-logarithmic source statements and their uses here.

• Paper IX: Euler Congruences, for companion jumps, Fibonacci packets, and integral matrix precision amplification. Exact source identities and the distinction between source statements and new worked consequences appear in the source check.
• Komorebi Math typed-expression example, for the polynomial integration, basis changes, derivative, boundary defect, and integral covector.
• NIST DLMF, Euler numbers and polynomials, including the signed convention and polynomial generating function, and the secant expansion.
• NIST DLMF, von Staudt–Clausen theorem, for the precise Bernoulli denominator input in the dyadic kernel valuation.
• OEIS A000364 and its coefficient table, for the positive secant sequence and offset.
• OEIS A211212, for the quartic reciprocal sequence, and A181985, for generalized alternating permutations.
• Paper IV, for cyclic denominators and reciprocal-power polynomials. The sequence experiments use the same exponential normalization.
• Euler Signal Bridge, for the existing finite-difference/FIR interface developed in greater detail here.
• Paper III, for the broader typed transfer and residual language, and Paper IX, for the odd-length and dyadic Euler reconstruction, sharp precision bounds, arithmetic packets, and local polynomial deformations.
• Source and overlap notes, target and route record, and proof self-review.

Back to the essay · Explore the sequence graphs · All demonstrations