Factorial precision at section order four

Write \(C_j=C_{4,j}\) and \(F_j=F_{4,j}\). The all-row theorem removes the threshold, but at \(N=4\) it leaves a further factorial gain unused. The exact algebra below proves that gain for all four addresses and identifies several infinite families where it is sharp.

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The proof of the first two addresses uses one classical arithmetic input: Frobenius's exact congruence for the ordinary Euler numbers in the \(\operatorname{sech}\) convention, as stated and derived by Gessel [gessel], Section 9. The address-two and address-three bounds below are finite Bernoulli calculations and do not require that input. A divided-power argument is necessary: ordinary integral moments alone do not justify division by \(k!\).

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Let \(\mathcal L\) be the ordinary Euler evaluation functional on a variable \(\mathcal E\), defined by

The functional is linear; it is not a multiplicative evaluation. Independent variables mean that moments of polynomials in separate variables factor. Odd moments vanish. Frobenius's congruence is

For \(m=0\), the assertion is that every even Euler number is odd. With \(P=(\mathcal E^2-1)/2\) and \(\gamma_m(P)=P^m/m!\), the case \(n=0\) says

Introduce an independent Euler variable \(\mathcal F\) and \(Q=(\mathcal F^2-1)/2\).

Put \(K=\mathbb Q_2(i)\), \(R=\mathbb Z_2[i]\), \(a=(1+i)/2\), and \(b=(1-i)/2\). The exact identity

gives, for \(U=a\mathcal E+b\mathcal F\), \(\mathcal L(U^{4n})=C_0(n)\). Although \(a\) and \(b\) have negative valuation, their denominators disappear from the following exact polynomial identities:

Let \(\Gamma_R(P,Q)\) be the \(R\)-subalgebra generated by all \(\gamma_m(P)\) and \(\gamma_r(Q)\). Adjoin \(\mathcal E,\mathcal F\) with

and call the resulting algebra \(\mathcal A_0\). Its elements are finite expressions; no completion is being presumed. It is generated as a \(\Gamma_R(P,Q)\)-module by \(1,\mathcal E,\mathcal F,\mathcal E\mathcal F\). The functional maps this module to \(R\). For an even basis term this is (5.4) and independence. For a basis term with an odd Euler factor the moment is zero, since \(P\) and \(Q\) are even polynomials in their respective variables.

Every displayed component of \(H_0\) has all its divided powers in \(\mathcal A_0\). Besides the generators, the needed identities are

The first scalar counts partitions into pairs and is an integer. For the last term of \(H_0\), multiply the third identity by \((i\mathcal E\mathcal F)^m\) and reduce the Euler powers using the integral quadratic relations. Finally,

proves \(\gamma_k(H_0)\in\mathcal A_0\). Applying the functional and putting \(W=U^4-1=2H_0\) gives

These moments are rational. The identity \(R\cap\mathbb Q_2=\mathbb Z_2\), obtained from the linear independence of \(1,i\) over \(\mathbb Q_2\), descends every such divisibility statement to \(\mathbb Z_2\). No field property of \(R/2R\) is needed.

For every starting row,

Each summand lies in \(2^k k!\mathbb Z_2\), proving the factorial bound for address zero from row zero onward.

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The conversion from unit steps to arbitrary mixed steps is used several times, so it is useful to make it explicit. Write \(\Delta=\Delta_1\). For \(H>0\),

A mixed term with local orders \(m_1,\ldots,m_k\geq1\) and \(M=\sum_\nu m_\nu\) therefore has valuation at least

if the unit-order estimate has offset \(\beta\) for every positive order. The inequalities

make this at least \(k+v_2(k!)+\beta+\sum_\nu v_2(H_\nu)\). Every expansion is finite. This proves the desired mixed-step transfer without an exchange of limits.

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Let \(B_m\) denote the Bernoulli numbers with \(B_1=-1/2\). The identity obtained by substituting \(u=e^t-1\) into \(\log(1+u)/u\) is

where the braces are Stirling numbers of the second kind. Each \(2r!/(r+1)\) is dyadically integral. Indeed, if \(s=v_2(r+1)>0\), then \(r\geq2^s-1\) and

The last inequality is \(2^s\geq2s\); the case \(s=0\) is immediate. Consequently

Thus the transfer factors below require no appeal to the von Staudt–Clausen theorem.

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The address-one source is \(G_{4,1}(t)=(\sinh t+\sin t)/(2t)\). The identities \(a+b=1\), \(a-b=i\), and \(ab=1/2\) give

For example, its numerator follows from

while \(\cosh t+\cos t=2\cosh(at)\cosh(bt)\). Only degrees divisible by four survive. Its actual Hurwitz packets are

The constant is one. Since \(v_2(4i+2)=1\), the elementary Bernoulli bound shows \(v_2(d_i)\geq2i\). Hurwitz inversion therefore gives

This follows by induction from the inverse recurrence: for packet index \(i\), every product \(d_rh_{i-r}\) has valuation at least \(2i\), and its binomial multiplier is integral. Because \(F_1=F_0D_1^{-1}\),

At a natural \(n\) this is a finite sum, with \(i\leq n\).

Here is the polynomial estimate that supplies one additional bit. For \(i\geq1\), put \(P_i(x)=4^i\binom{4x}{4i}\). Its constant is zero, and its degree-\(d\) coefficient is

where \(s(m,d)\) is a signed Stirling number of the first kind. In each product contributing to \(s(4i,d)\), at most \(d-1\) of the \(2i-1\) even integers in \(1,\ldots,4i-1\) can be omitted. Thus

Writing \(s_2(i)\) for the number of ones in the binary expansion of \(i\), and using \(v_2((4i)!)=4i-s_2(i)\), gives

For \(k\geq1\), \(\Delta_1^k x^d\) vanishes when \(d<k\) and is an integer polynomial with values divisible by \(k!\). To see the latter, expand \(x^d=\sum_r\left\{\begin{smallmatrix}d\\r\end{smallmatrix}\right\} r!\binom xr\) and apply \(\Delta_1^k\binom xr=\binom{x}{r-k}\). It follows that

For \(k=0\) the bound follows directly from the coefficient estimate.

To apply discrete products in (5.15) without a moving truncation, first extend \(C_0\) to \(\mathbb Z_2\) by its uniformly convergent Newton series. The all-row bound \(v_2(\Delta_1^m C_0(0))\geq m\) proves this convergence. Its factorial bounds extend to every dyadic starting point by continuity and approximation with natural rows. The sum in (5.15) is then uniformly convergent on \(\mathbb Z_2\): both the binomial polynomial and \(C_0(x-i)\) are integral-valued and \(v_2(h_i)\geq2i\).

For the pointwise product \(P_i(x)C_0(x-i)\), the discrete Leibniz formula has scalar divisibility

Multiplication by \(h_i/4^i\in\mathbb Z_2\) and uniform summation now give

The same extra bit holds for positive mixed steps by (5.10). Combining this with the address-zero bound proves Theorem 5.1 for address one.

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The variable \(\mathcal B=2Z+1\) has transform

Its odd moments vanish, and \(P=(\mathcal B^2-1)/4=Z(Z+1)\) has the divided-power moments of Lemma 5.2. Introduce an independent copy \(\mathcal D\) and \(Q=(\mathcal D^2-1)/4\). The exact factorization

shows that \(U=a\mathcal B+b\mathcal D\) satisfies \(\mathcal L(U^{4n})=C_2(n)\). Direct expansion gives

Let \(\mathcal A_2\) be the divided-power algebra over \(R\) generated by \(P,Q\), adjoining \(\mathcal B,\mathcal D\) with \(\mathcal B^2=1+4P\), \(\mathcal D^2=1+4Q\). The functional is integral on its module basis \(1,\mathcal B,\mathcal D,\mathcal B\mathcal D\) by the Bernoulli lemma, independence, and vanishing of odd moments. Every component in (5.26) has integral divided powers, by the same finite identities as before; \(P^2\) is even easier than \(P^2/2\). Consequently

The starting-row conclusion follows from the finite base expansion (5.9), now with \(W=4H_2\).

The low Newton coefficients of the actual reciprocal are

where \(a_m=\Delta_1^m C_2(0)\). Thus \(v_2(a_1)=4\) and \(v_2(a_2)=6\). For \(m\geq3\), (5.27) bounds \(a_m\) by \(2m+v_2(m!)\). The finite shifted Newton formula

therefore gives \(v_2(\Delta_1^kC_2(n))\geq k+v_2(k!)+3\) for every \(k\geq1\). For \(k\geq3\) this is \(2k\geq k+3\); for \(k=1,2\) it uses the exact initial coefficients and the higher tail bound. The finite step expansion proves the offset three in (5.1).

There is a still stronger filtration. Put

Then the exact identities are

Every divided power of \(G\) is integral component by component. The finite addition identity gives

because the \(r\)th term belongs to \(2^{r+\lfloor(k-r)/2\rfloor}\mathcal A_2\). Applying the functional, rational descent, and the base-index expansion proves the unit-step version of (5.2). For a mixed term of total order \(M\), its lower bound exceeds the target at order \(k\) by at least

Indeed \(\sum_\nu v_2(m_\nu)\leq M-k\). This proves the stronger mixed bound. The earlier offset-three bound is retained, since it is stronger at the lowest orders.

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The address-three source is \(G_{4,3}(t)=3(\sinh t-\sin t)/t^3\). Addition identities give

Consequently

This expression is a power series: its two polar terms cancel because \(a^{-2}+b^{-2}=0\). Its constant is one, only degrees divisible by four survive, and the Bernoulli expansion gives

Again \(v_2(d_i)\geq2i\), and the inverse coefficients \(h_i\) have the same bound. Since \(F_3=F_2D_3^{-1}\),

The scaled-binomial argument (5.18) applies unchanged. The basic bound for \(C_2\) includes order zero; its continuous extension is available from the threshold-free theorem. Thus the shifted sums converge uniformly on \(\mathbb Z_2\), and the pointwise product rule proves

At every positive order \(C_2\) has three extra bits, so this gives the one-bit offset for \(C_3\). Finite step expansion finishes the proof of Theorem 5.1.

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