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A coefficient proof of the Bui–Hall sign conjecture

2026-09-13. Candidate full proof; independent review and later-literature comparison pending. This concerns a specific conjecture about the leading constants in mixed fourth moments of Hardy's Z-function. It proves neither RH nor the cycle19 defect estimate G. No originality or formal-verification claim is made here.

Exact external statement

Let alpha=(k,l,m,n) be any four nonnegative integers, S=k+l+m+n, and r the number of odd entries. Bui–Hall, On the derivatives of Hardy's function Z(t), BLMS 55 (2023), 2304–2323, Theorem 3, gives

$$\operatorname{HARDY}(k,l,m,n)=3(-1)^{m+n}i^S I_\alpha, \tag{1}$$

where

$$I_\alpha=\int_{[0,1]^4}(u_1-u_2)^2 A^k B^l C^m D^n\,d\mathbf u,$$ $$A=\tfrac12-u_1+(u_1-u_2)u_3,\quad B=\tfrac12-u_2-(u_1-u_2)u_3,$$ $$C=\tfrac12-u_1+(u_1-u_2)u_4,\quad D=\tfrac12-u_2-(u_1-u_2)u_4.$$

The same formula and Conjecture 1 were directly inspected in arXiv:2304.05178v1. The source defines HARDY as pi squared times the coefficient of T(log T)^(S+4) in the mixed fourth moment. We use its moment theorem as an external published input, without re-proving it.

Theorem. For every alpha of even total degree,

$$(-1)^{\lfloor k/2\rfloor+\lfloor l/2\rfloor+\lfloor m/2\rfloor+\lfloor n/2\rfloor} \operatorname{HARDY}(k,l,m,n)>0. \tag{2}$$

Thus HARDY is nonzero, and is negative exactly when |i^k+i^l+i^m+i^n|=2, as in Conjecture 1. Odd total degree gives zero, consistently with the source's Corollary 1.

1. Exact reduction to three independent Laplace variables

Put x=1/2-u1 and y=1/2-u2. For any a,b define the oriented integral

$$J_{a,b}(x,y)=\int_x^y t^a(x+y-t)^b\,dt.$$

The changes in u3 and u4 each divide by y-x, and cancel the original (y-x)^2. The diagonal x=y has measure zero. Hence

$$I_\alpha=\int_{[-1/2,1/2]^2}J_{k,l}(x,y)J_{m,n}(x,y)\,dx\,dy. \tag{3}$$

The product is unchanged by exchanging x,y. Restrict to x<=y and multiply by two. Write s=x+y, d=y-x. The Jacobian is 1/2 and the range is |s|<=1, 0<=d<=1-|s|. In the two inner integrals put t=(s+v)/2 and t'=(s+w)/2, so their combined Jacobian is 1/4 and |v|,|w|<=d. Integrating d first gives

$$I_\alpha=\frac14\int_{|s|+\max(|v|,|w|)\le1} (1-|s|-\max(|v|,|w|)) \left(\frac{s+v}{2}\right)^k\left(\frac{s-v}{2}\right)^l \left(\frac{s+w}{2}\right)^m\left(\frac{s-w}{2}\right)^n,ds\,dv\,dw. \tag{4}$$

Set u=s, z=(v-w)/2, h=(v+w)/2. Then dv dw=2 dz dh and max(|v|,|w|)=|z|+|h|. Define the four unscaled forms

$$L_1=u+z+h,\quad L_2=u-z-h,\quad L_3=-u+z-h,\quad L_4=-u-z+h,$$

and P_alpha=L1^k L2^l L3^m L4^n. With rho=|u|+|z|+|h|, (4) becomes

$$I_\alpha=(-1)^{m+n}2^{-S-1} \int_{\rho\le1}(1-\rho)P_\alpha(u,z,h)\,du\,dz\,dh. \tag{5}$$

For any homogeneous polynomial P of degree S in three variables, integration separately on the eight orthants and then in the radial variable rho gives

$$\int_{\rho\le1}(1-\rho)P =\frac1{(S+4)!}\int_{\mathbb R^3}e^{-\rho}P. \tag{6}$$

Indeed the radial factors are respectively 1/((S+3)(S+4)) and (S+2)!; signed angular integrals cause no difficulty. All integrals are absolutely convergent. Let U,Z,H be independent real variables of density e^(-|t|)/2, and let X_i=L_i(U,Z,H). Equations (1),(5),(6) yield, for even S,

$$\boxed{\operatorname{HARDY}(\alpha) =\frac{12(-1)^{S/2}}{2^S(S+4)!} \mathbb E[X_1^kX_2^lX_3^mX_4^n].} \tag{7}$$

The constant gives HARDY(0,0,0,0)=1/2. For odd S the expectation vanishes by simultaneous reflection.

2. Parity transform of the generating function

All power series below may be read formally over the rationals, or analytically near zero. Write

$$p_0=t_1+t_2+t_3+t_4,\quad p_1=t_1+t_2-t_3-t_4,$$ $$p_2=t_1-t_2+t_3-t_4,\quad p_3=t_1-t_2-t_3+t_4, \quad A_j=1-p_j^2.$$

The Laplace moment generating function is

$$M(\mathbf t)=\mathbb E e^{\sum t_iX_i}=\frac1{A_1A_2A_3}. \tag{8}$$

For a monomial of exponent alpha, multiply its coefficient in M by i^r, where r is the number of odd entries of alpha. Since odd total degrees in M vanish, the transformed series G is real. Averaging over coordinate sign changes gives

$$G=\tfrac12\{M(t_1,t_2,-t_3,-t_4)+M(t_1,-t_2,t_3,-t_4) +M(t_1,-t_2,-t_3,t_4)-M(\mathbf t)\}. \tag{9}$$

For example this follows from 2^(-4) sum_epsilon product_j(1+i epsilon_j) M(epsilon_1 t1,...,epsilon_4 t4); terms with an odd number of negative signs have purely imaginary weights and cancel in antipodal pairs. Multiplying out the four denominators in (9) gives

$$\boxed{G=\frac{1-\sum_jt_j^2+2\sum_{i<j}t_it_j} {\prod_{j=0}^3(1-p_j^2)}=(1-Q)R+2\sum_{i<j}t_it_jR,} \tag{10}$$

where Q=sum t_j^2 and R=1/product A_j. This uses sum_(j=0)^3 p_j^2=4Q.

3. Strict positivity of every even-degree coefficient

Let E be the eight sign vectors epsilon in {+1,-1}^4 with product epsilon_j=1. Factoring each 1-p_j^2 shows

$$\log R=\sum_{d\ge1}\frac1d\sum_{\epsilon\in E}(\epsilon\cdot\mathbf t)^d. \tag{11}$$

The character sum sum_E epsilon^alpha is eight if the four entries of alpha are all even or all odd, and zero otherwise. (These are exactly the characters trivial on E.) Thus log R has nonnegative coefficients and zero constant term. For |alpha|=2N>0,

$$[\mathbf t^\alpha]\log R= \begin{cases}\dfrac4N\binom{2N}{\alpha_1,\alpha_2,\alpha_3,\alpha_4},&\text{all entries have the same parity},\\ 0,&\text{otherwise}. \end{cases} \tag{12}$$

Set F=log R+log(1-Q). The second summand subtracts a coefficient only at all-even alpha=2a with |a|=N, namely (1/N) binomial(N;a). Consequently

$$[\mathbf t^{2a}]F=\frac1N\left(4\binom{2N}{2a}-\binom N a\right)>0. \tag{13}$$

The inequality follows from binomial(2N;2a)>=binomial(N;a): map each word with a_j copies of letter j to the word with every letter duplicated. At every all-odd exponent vector the coefficient of F is the strictly positive coefficient in (12). All other coefficients of F are zero. In particular F is coefficientwise nonnegative, with zero constant term. Therefore

$$(1-Q)R=\exp F \tag{14}$$

is coefficientwise nonnegative, and has strictly positive coefficients at every all-even nonzero or all-odd exponent vector; its constant is one. Similarly R=exp(log R) is coefficientwise nonnegative and positive at every all-even vector, including zero.

Every even-total alpha has r=0,2,or4. For r=0 or4, its coefficient in (14) is strictly positive. For r=2, choose the two odd positions i,j; then alpha-e_i-e_j is nonnegative and all-even, and the term 2t_i t_j R in (10) supplies a strictly positive coefficient. All remaining contributions in (10) are nonnegative. Hence

$$[\mathbf t^\alpha]G>0\qquad (|\alpha|\text{ even}). \tag{15}$$

By the definition of G and the exponential coefficient convention, (15) says (-1)^(r/2) E[X^alpha]>0. Equation (7) proves (2), since sum floor(alpha_j/2)=(S-r)/2 and (S+r)/2 has the same parity.

Finally the equivalence to the source's absolute-value condition is elementary. If r=0 or4, four signs on one coordinate axis sum to a vector of length two exactly when an odd number of their signs are negative. If r=2, the two real and two imaginary signs give length two exactly when one pair has opposite signs and the other equal signs. In both cases this is exactly (-1)^(sum floor(alpha_j/2))=-1. This covers zero indices and all mod-four patterns.

Trust and remaining decisions

The argument is an elementary integral and formal-series proof for every quadruple, not a finite verification. Its application to zeta uses the external Bui–Hall Theorem 3. An independent source/parity/geometry review and an exact arithmetic cross-check are required before repository acceptance. A later-literature audit is required before an originality claim. No RH, new zero-free region, simple-zero proportion, G, or Lean result follows.

A sharp positive factorization and explicit Hardy-moment bounds

September 14, 2026. Candidate strengthening of the cycle-24 proof, pending independent review. This note makes no originality claim. It concerns an auxiliary mixed-moment theorem, not RH. The original proof and its published moment input are retained unchanged.

1. The sharp factorization

Use four variables and the Hadamard forms

$$p_0=t_1+t_2+t_3+t_4,\quad p_1=t_1+t_2-t_3-t_4, \quad p_2=t_1-t_2+t_3-t_4,\quad p_3=t_1-t_2-t_3+t_4.$$

Put $Q=\sum_i t_i^2$, $E_2=\sum_{i<j}t_it_j$, and

$$R=\prod_{j=0}^3(1-p_j^2)^{-1},\qquad C=(1-Q)^4R.$$

Theorem 1. Every coefficient of $C$ is nonnegative. For any real $\lambda$, the formal series $(1-Q)^\lambda R$ is coefficientwise nonnegative if and only if $\lambda\le4$.

All formal real powers mean the binomial expansion with constant term one. Formal logarithms and exponentials are valid in the total-degree completion over the reals; for any fixed coefficient only finitely many terms contribute.

Proof. Let $L=\log R+4\log(1-Q)$, so $C=\exp L$. The eight even-product sign vectors give, for a nonzero multi-index $\beta$ with $s=|\beta|$,

$$s[\mathbf t^\beta]L=w_\beta, \qquad w_\beta=\begin{cases} 8\left\{\binom{s}{\beta}-\binom{s/2}{\beta/2}\right\},&\beta\text{ all even},\\ 8\binom{s}{\beta},&\beta\text{ all odd},\\ 0,&\text{otherwise}. \end{cases} \tag{1}$$

Here a multinomial has four lower entries. The character calculation is precisely the one used in cycle 24; subtracting four logarithms instead of one changes its all-even coefficient to the displayed difference.

Duplicating every letter in a word injects words counted by $\binom{s/2}{\beta/2}$ into those counted by $\binom{s}{\beta}$. Thus $w_\beta\ge0$. Since $L$ has zero constant term, every coefficient of $\exp L$ is nonnegative. For $\lambda\le4$, write

$$(1-Q)^\lambda R=(1-Q)^{-(4-\lambda)}C.$$

The first factor has nonnegative coefficients by the binomial expansion, including the case $4-\lambda=0$. Conversely its coefficient at $t_1^2$ before this factorization is $4-\lambda$, since $[t_1^2]R=4$. Nonnegativity therefore requires $\lambda\le4$. This proves necessity and sufficiency. $\square$

This is an optimal exponent within this specific family of coefficientwise factorizations, not an optimal zeta bound or a novelty certificate.

2. A recurrence with nonnegative weights

Write $c_\alpha=[\mathbf t^\alpha]C$, with $c_0=1$. The Euler derivation $\mathcal D=\sum_i t_i\partial_{t_i}$ gives $\mathcal DC=(\mathcal DL)C$. Consequently, for $S=|\alpha|>0$,

$$\boxed{c_\alpha=\frac1S\sum_{0<\beta\le\alpha}w_\beta c_{\alpha-\beta}.} \tag{2}$$

The componentwise inequality makes this a finite sum and every predecessor has smaller total degree. Thus (2) uniquely determines the coefficients, uses nonnegative weights, and proves their nonnegativity by induction without an appeal to numerical sampling. In fact they are integers, because $(1-Q)^4R$ is a rational series over the integers with denominator constant one.

Every nonzero weight has all-even or all-odd index. This parity class is closed under addition. Therefore $C$ is supported on all-even and all-odd indices. On a single coordinate axis, $C=1$, so its nonconstant axis coefficients vanish. Also

$$c_{(1,1,1,1)}=\frac{8\cdot4!}{4}=48. \tag{3}$$

The parity-corrected generating function from cycle 24 is now

$$\boxed{G=\big\{(1-Q)^{-3}+2E_2(1-Q)^{-4}\big\}C.} \tag{4}$$

This is an identity of the actual rational functions, since the right side equals $(1-Q+2E_2)R$. It provides the same coefficients as the signed three-factor MGF after the previously proved parity transform.

For use in (4), write $B=(1-Q)^{-3}+2E_2(1-Q)^{-4}$. Its coefficients are explicit. For $a\in\mathbb N^4$,

$$[\mathbf t^{2a}]B=\binom{|a|+2}{2}\binom{|a|}{a},\qquad [\mathbf t^{2a+e_i+e_j}]B=2\binom{|a|+3}{3}\binom{|a|}{a}\quad(i<j). \tag{5}$$

All other coefficients vanish. There is a unique odd pair in the second case. Convolving these nonnegative coefficients with (2) computes every coefficient of $G$ using sums of nonnegative terms. The multinomial differences defining the weights are exactly nonnegative by the injection above; the method does not rely on cancellation in a signed moment expansion.

3. Explicit lower bounds for every allowed derivative order

For an even-total index $\alpha=(k,l,m,n)$, put $S=|\alpha|=2N$, $\alpha!=k!l!m!n!$, and

$$K_\alpha=\frac{12\alpha!}{2^S(S+4)!}.$$

The original integral reduction and parity transform give

$$(-1)^{\sum_i\lfloor\alpha_i/2\rfloor}\operatorname{HARDY}(\alpha) =K_\alpha[\mathbf t^\alpha]G. \tag{6}$$

Theorem 2. The right side is at least $K_\alpha b_\alpha>0$, where

$$b_\alpha=\begin{cases} \displaystyle\binom{N+2}{2}\binom N a,&\alpha=2a,\\[4pt] \displaystyle2\binom{N+2}{3}\binom{N-1}{a},&\alpha=2a+e_i+e_j,\ i<j,\\[4pt] \displaystyle48\binom N2\binom{N-2}{a},&\alpha=2a+(1,1,1,1). \end{cases} \tag{7}$$

Each case automatically has a nonnegative upper multinomial argument. These three cases exhaust even total degree.

Proof. In the first two cases retain just the constant coefficient $c_0=1$ in (4) and apply (5). In the last case retain $c_{(1,1,1,1)}=48$ and the all-even coefficient of $B$ at $2a$. Every omitted convolution term is nonnegative. Every displayed factor is strictly positive, including $N=0$ in the first case, $N=1$ in the second, and $N=2$ in the third. $\square$

The bound is exact on every coordinate axis. Indeed, on such an axis $C=1$, $E_2=0$, and $G=(1-t^2)^{-3}$. Simplifying (6) gives

$$\operatorname{HARDY}(2N,0,0,0) =(-1)^N\frac{3}{2\cdot4^N(2N+1)(2N+3)}\qquad(N\ge0). \tag{8}$$

The minimal two-odd and four-odd cases also attain the bound: $\operatorname{HARDY}(1,1,0,0)=1/120$ and $\operatorname{HARDY}(1,1,1,1)=1/1120$. These equality statements do not assert optimality for every other index.

4. Scope and remaining proof obligations

The all-orders statements rest on the analytic/formal-series proof above and the original published moment representation, not on finite computations. The recurrence and lower bounds strengthen the project’s constructive presentation of the auxiliary sign theorem; they do not improve RH, a zero-free region, or a zero proportion.

The exact even-sign character identity is being addressed separately in Lean. Even if that lemma compiles, this note’s series factorization, recurrence-to-rational-function identification, integral reduction, and complete Hardy conclusion are not thereby formally verified.

A separate reviewer must check (1)–(8), boundary cases, sharpness, and the link to the original normalization before acceptance. Whether this strengthening or the underlying sign theorem is new in the literature remains unresolved.