# Public copy of an independent AI analytic review

**Verdict: PASS, conditional on the published Bui–Hall moment representation.** The frozen candidate's equations (1)–(8), including real-exponent sharpness, all three magnitude bounds, and the stated boundary/equality cases, are analytically correct. This is an independent agent review, not outside expert review, Lean verification, an originality finding, or an RH result.

Reviewed target: `source/research/millennium/iterations/0029/sharp-factorization.md`.

- First-read SHA256: `bac6ab41c3eaff2fff9e75b7ccaddabc141d121f05c46389d25145005c3287cc`.
- Last-read SHA256: `bac6ab41c3eaff2fff9e75b7ccaddabc141d121f05c46389d25145005c3287cc`. Hashes immediately before and after the final complete read agree; the candidate is unchanged.

## Evidence and weakest step

The weakest step in the standalone candidate is equation (6): it imports the integral normalization and parity transform instead of deriving them. A sign, Jacobian, or ordinary-versus-exponential coefficient error there would invalidate the Hardy bounds while leaving the formal factorization correct. I therefore reconstructed that link explicitly below before selecting a verdict. No gap remained.

I read cycle 24 only for the integral, random-variable and generating-function definitions and normalization being imported; I did not use its positivity proof or any prior PASS/audit/acceptance file as evidence. I directly checked the definition of HARDY and Theorem 3 in [Bui–Hall, arXiv:2304.05178v1](https://arxiv.org/html/2304.05178v1#Thmtheorem3). That source supplies the published integral representation, which remains an external input rather than a theorem reproved here. HARDY is the leading coefficient multiplied by pi squared, not the unscaled moment coefficient. The external formula has exactly the factor \(3(-1)^{m+n}i^S\) and the four linear factors used below.

## 1. Character calculation and coefficientwise positivity

Work in the total-degree completion of the real polynomial ring in four variables. Every denominator has constant term one, and every logarithm argument is one plus a positive-degree series. The formal logarithm, exponential, binomial expansion, differentiation and all coefficient extractions below are well-defined with finitely many contributions per coefficient.

Let \(H=\{\epsilon\in\{\pm1\}^4:\prod_i\epsilon_i=1\}\). The eight forms \(\epsilon\cdot t\) are precisely \(p_0,p_1,p_2,p_3\) and their negatives. Consequently

\[
\log R=\sum_{\epsilon\in H}\sum_{s\ge1}\frac{(\epsilon\cdot t)^s}{s}.
\]

For a multi-index \(\beta\), parametrize \(\epsilon=(a,b,c,abc)\). Its character sum factors as

\[
\sum_{\epsilon\in H}\epsilon^\beta
=\prod_{i=1}^3\bigl(1+(-1)^{\beta_i+\beta_4}\bigr).
\]

This equals eight exactly when all four entries of \(\beta\) have the same parity, and zero otherwise. This independently establishes the character assertion, including mixed parity and odd total degree. With \(s=|\beta|>0\), the contribution to \(s[t^\beta]\log R\) is therefore \(8\binom{s}{\beta}\) in the two allowed parity classes. The contribution of \(4\log(1-Q)\) is zero unless \(\beta=2a\), in which case it is

\[
-4\frac{s}{s/2}\binom{s/2}{\beta/2}
=-8\binom{s/2}{\beta/2}.
\]

This proves exactly candidate (1), with no division by zero because the constant index is excluded. In the all-even case, replacing each letter of a word with two adjacent copies is injective: recover the original word by retaining positions 1,3,5,... . Thus the multinomial difference is nonnegative, also when some entries vanish and on an axis, where the two counts are both one. All-odd weights are strictly positive. The other weights are zero.

It follows that \(L=\log R+4\log(1-Q)\) has nonnegative coefficients and zero constant coefficient. Every coefficient of \(\exp L\) is a finite sum of products of nonnegative numbers, with positive factorial denominators. Since formal logarithms add on commuting units, \(\exp L=(1-Q)^4R=C\). This proves Theorem 1's first assertion.

For any real \(\lambda\le4\), put \(a=4-\lambda\ge0\). The coefficient of \(Q^j\) in \((1-Q)^{-a}\) is \(a(a+1)\cdots(a+j-1)/j!\) for \(j\ge1\), and its constant is one. These numbers are nonnegative; when \(a=0\), every nonconstant coefficient is zero. Multiplication by \(C\) proves sufficiency, including every nonintegral real exponent and the endpoint \(\lambda=4\). Conversely, \([t_i^2]R=4\) and \([t_i^2](1-Q)^\lambda=-\lambda\), so \([t_i^2](1-Q)^\lambda R=4-\lambda\). Every \(\lambda>4\) fails at degree two. The claimed iff and its exact endpoint are proved. This is optimality only within the stated family.

## 2. Euler recurrence, support, and integrality

For \(\mathcal D=\sum_i t_i\partial_{t_i}\), the formal chain rule gives \(\mathcal DC=(\mathcal DL)C\). Extracting \(t^\alpha\) gives candidate (2), since \([t^\beta]\mathcal DL=w_\beta\). The restriction \(0<\beta\le\alpha\) is finite and guarantees strictly smaller predecessor degree. Initial value \(c_0=1\) therefore determines a unique solution and proves its nonnegativity inductively, independently of numerical sampling.

Modulo two, the allowed indices form the additive subgroup \(\{0000,1111\}\). Thus all coefficients outside these parity classes vanish. On the \(i\)-th axis all \(p_j^2=t_i^2\), so \(C=(1-t_i^2)^4(1-t_i^2)^{-4}=1\); every positive-degree axis coefficient vanishes. At \(\alpha=(1,1,1,1)\), the only nonzero weight with \(0<\beta\le\alpha\) is \(\beta=\alpha\), giving \(c_\alpha=8\cdot24/4=48\). In particular, no unmentioned lower-degree terms contribute to candidate (3).

The numerator and denominator of \(C\) have integer coefficients and the denominator constant is one. Inverting such a denominator recursively uses only integer arithmetic. Thus the coefficients are integers even though recurrence (2) is written with division by \(S\).

## 3. The actual rational identity and coefficients of B

Use the imported random variables only as definitions: independent \(U,Z,H\) have density \(e^{-|x|}/2\), and

\[
X=(U+Z+H,\ U-Z-H,\ -U+Z-H,\ -U-Z+H).
\]

Their MGF is \(M=1/(A_1A_2A_3)\), where \(A_j=1-p_j^2\), because the Laplace MGF is \(1/(1-t^2)\). For an even-total index, the coefficient multiplier in the signed combination

\[
\tfrac12\{M(t_1,t_2,-t_3,-t_4)+M(t_1,-t_2,t_3,-t_4)
+M(t_1,-t_2,-t_3,t_4)-M(t)\}
\]

is \(\{(-1)^{m+n}+(-1)^{l+n}+(-1)^{l+m}-1\}/2\). It is \(+1\) when the last three entries share parity, forcing all four to share parity, and \(-1\) otherwise. This is exactly \((-1)^{r/2}\), for \(r=0,2,4\) odd entries. Odd-total coefficients of \(M\) vanish by simultaneous reflection. Hence this combination is the desired \(G\), with no parity assumption omitted.

The three substitutions permute the squared Hadamard forms, and replace the missing denominator \(A_0\) by \(A_1,A_2,A_3\), respectively. Over the common denominator \(\prod_{j=0}^3 A_j\), their combination has numerator

\[
\tfrac12(A_1+A_2+A_3-A_0)
=1+p_0^2-2Q=1-Q+2E_2.
\]

Here \(\sum_jp_j^2=4Q\) follows by expanding the four squares. Thus \(G=(1-Q+2E_2)R=BC\) as actual rational functions and their formal expansions, proving candidate (4).

Expand \((1-Q)^{-d}=\sum_{j\ge0}\binom{j+d-1}{d-1}Q^j\). The first summand of \(B\) contributes only at \(2a\), with coefficient \(\binom{|a|+2}{2}\binom{|a|}{a}\). The second contributes only at \(2a+e_i+e_j\), with coefficient \(2\binom{|a|+3}{3}\binom{|a|}{a}\). A monomial with two odd entries has a unique odd pair, so there is no hidden multiplicity. The supports of the two summands are disjoint. This proves (5), including zero entries and the constant coefficient.

## 4. Independent normalization check for equation (6)

Begin with the published integral \(I_\alpha\) and factor \(3(-1)^{m+n}i^S\). Put \(x=1/2-u_1\), \(y=1/2-u_2\). For fixed \(x,y\), substitute \(t=x+(y-x)u_3\) and \(t'=x+(y-x)u_4\). The weight \((u_1-u_2)^2\) cancels the two substitution denominators. Reversing \(x,y\) reverses both oriented inner integrals, leaving their product unchanged. Restrict to \(x\le y\) and double; the diagonal is measure zero and introduces no singularity.

Set \(s=x+y\), \(d=y-x\). The Jacobian \(1/2\) cancels that doubling, and \(0\le d\le1-|s|\). Next \(t=(s+v)/2\), \(t'=(s+w)/2\), with \(|v|,|w|\le d\), contributes Jacobian \(1/4\). Integrating \(d\) over \([\max(|v|,|w|),1-|s|]\) gives

\[
I_\alpha=2^{-S-2}\int_{|s|+\max(|v|,|w|)\le1}
(1-|s|-\max(|v|,|w|))
(s+v)^k(s-v)^l(s+w)^m(s-w)^n\,ds\,dv\,dw.
\]

Set \(s=u\), \(v=z+h\), \(w=h-z\). This has absolute Jacobian two, and \(\max(|z+h|,|h-z|)=|z|+|h|\). The last two linear factors are \(-X_3\) and \(-X_4\) evaluated at \(u,z,h\); the first two are \(X_1,X_2\). Therefore, for the homogeneous degree-\(S\) polynomial \(P=X_1^kX_2^lX_3^mX_4^n\) and \(\rho=|u|+|z|+|h|\),

\[
I_\alpha=(-1)^{m+n}2^{-S-1}\int_{\rho\le1}(1-\rho)P.
\]

On each orthant the radial Jacobian is proportional to \(\rho^2\). The ratio of radial integrals is

\[
\frac{\int_0^1(1-r)r^{S+2}\,dr}{\int_0^\infty e^{-r}r^{S+2}\,dr}
=\frac{1}{(S+3)(S+4)(S+2)!}=\frac1{(S+4)!}.
\]

This argument is linear in \(P\) and does not require a positive angular integrand. Polynomial integrability under the exponential density justifies the integrals. The joint Laplace density is \(e^{-\rho}/8\), so

\[
\operatorname{HARDY}(\alpha)
=\frac{12(-1)^{S/2}}{2^S(S+4)!}\mathbb E[X^\alpha]
\quad(S\text{ even}).
\]

Finally \([t^\alpha]M=\mathbb E[X^\alpha]/\alpha!\), and \([t^\alpha]G=(-1)^{r/2}[t^\alpha]M\). Since \(\sum_i\lfloor\alpha_i/2\rfloor=(S-r)/2\), the resulting sign equals the candidate's sign (their exponents differ by the even number \(S-r\)). This proves (6) with exactly \(K_\alpha=12\alpha!/(2^S(S+4)!)>0\). For odd total degree simultaneous reflection makes the expectation zero; the positive-bound assertion does not extend to those indices.

## 5. All three bounds and boundary/equality checks

Every even-total four-index has exactly zero, two, or four odd entries. Writing it uniquely in the corresponding form, the sizes of \(a\) are \(N,N-1,N-2\), respectively.

- **All even:** retain \(c_0=1\). This gives \(\binom{N+2}{2}\binom Na\), including \(N=0\), \(a=0\), value one. The resulting HARDY bound there is \(K_0=12/4!=1/2\), attained.
- **Two odd:** retain \(c_0=1\). Formula (5) gives \(2\binom{N+2}{3}\binom{N-1}{a}\). Necessarily \(N\ge1\). At \(N=1\), \(a=0\), the coefficient bound is two. There is no nonconstant coefficient of \(C\) that can fit beneath \(e_i+e_j\), so equality holds. Here \(K=12/(4\cdot6!)=1/240\), hence HARDY is \(1/120\).
- **Four odd:** retain \(c_{1111}=48\) and the all-even \(B\) coefficient at \(2a\). This gives \(48\binom N2\binom{N-2}{a}\), since \(|a|=N-2\). Necessarily \(N\ge2\). At \(N=2\), the bound is 48. \(B\) has no four-odd monomial and the only contributing nonconstant \(C\) index below \(1111\) is \(1111\) itself. Thus equality holds. Here \(K=12/(16\cdot8!)=1/53760\), hence HARDY is \(48/53760=1/1120\).

All binomial and multinomial factors in these cases are strictly positive at every allowed index, including zeros among the entries of \(a\); no negative upper argument occurs. Every discarded convolution term is nonnegative. Since the signed HARDY value is positive, these are indeed bounds on its absolute magnitude, not merely one-sided bounds with an uncontrolled sign. The two minimal odd cases have positive parity sign, as asserted. Permutations of their coordinates give the same checks.

On every coordinate axis \(C=1\) and \(E_2=0\), giving \(G=(1-t^2)^{-3}\) exactly. Hence the all-even bound is attained for every \(N\ge0\) on any axis. Direct simplification yields

\[
\frac{12(2N)!}{4^N(2N+4)!}\binom{N+2}{2}
=\frac{3}{2\cdot4^N(2N+1)(2N+3)},
\]

and multiplication by \((-1)^N\) proves (8). In particular \(N=0\) gives \(1/2\) and \(N=1\) gives \(-1/40\). All denominators remain nonzero throughout the claimed domain.

The candidate does not claim an exhaustive classification of equality at other indices, and this verdict does not add one. For clarity the bounds need not be exact generally: at \(\alpha=2e_i+2e_j\), the weight is \(8(6-2)=32\), so \(c_\alpha=8>0\); its contribution with \(B_0=1\) makes the all-even bound strict. This is an exact symbolic check, not numerical sampling.

## Scope and disposition

There is no identified analytic gap in the frozen candidate. Its conclusions hold for all four nonnegative integer derivative orders of even total degree, with the published moment representation as input. The coefficient theorem additionally covers every real \(\lambda\), precisely with cutoff four. No test program, finite table, Lean lemma, prior acceptance, or novelty inference supports this verdict. The Bui–Hall theorem itself, literature priority, full formal verification, RH, zero-free regions, and zero proportions remain outside the reviewed claim.