Geometric Moments and Rational Lambert Values: Proofs and Further Results

Finite certificates and an all-rank degree check.

Put $D_{N,h}(p)=det(s_{i+j+h}(p){)}_{0\le i,j<N}$, with $D_{0,h}=1$. Exact polynomial calculations give strictly positive coefficients in $t=p-1$ for $D_{N,h}(1+t)$ when $h=0,1$ and $1\le N\le 8$. Thus sixteen determinant polynomials are positive for every real $p\ge 1$. Their degrees, in increasing rank, are

The calculation uses exact arithmetic in $\mathbb{Z}[p]$. Starting from $D_{0,h}=1$ and $D_{1,h}=s_h$, the Desnanot–Jacobi identity gives

The degree argument below shows that each denominator is nonzero. The computation checks that every division has zero remainder, then substitutes $p=1+t$ and tests every coefficient. All $8824$ coefficients in the sixteen polynomials are strictly positive. A program computes the full lists. An independent checker evaluates each stored polynomial at more integer bases than its degree and compares the values with fraction-free determinants of the integer Hankel matrices, which identifies the stored list with $D_{N,h}(1+t)$ in $\mathbb{Z}[t]$. As a separate check of the normalisation, direct integer determinants at $p=1,2,3,5,11$ agree with evaluations of all sixteen polynomials. Those evaluations are checks, not the proof of polynomial positivity. This is a finite computer-algebra calculation and gives no all-rank result.

For the eight unshifted determinants, positivity is also a Lean theorem, proved by a different certificate. Put $T=2^{2929}$. An explicit bound shows that every coefficient of $D_{N,0}(1+t)$, $N\le 8$, has absolute value less than $T/2$. An integer polynomial with this property is determined by its value at $t=T$, and its coefficients are the base-$T$ digits of that value whenever every digit is less than $T/2$. The Lean kernel computes the eight integers $D_{N,0}(1+T)$ exactly and checks this digit condition and that the lowest digit is nonzero. Every coefficient of $D_{N,0}(1+t)$ is therefore nonnegative and the constant term is positive, so $D_{N,0}(p)>0$ for every real $p\ge 1$. This is the positivity used in Proposition 3.9. The strict positivity of every coefficient, and the shifted determinants $D_{N,1}$, rest on the computation above. Neither the finite computation nor the Lean proof yields an all-rank coefficient-positivity theorem.

The degrees admit a separate all-rank proof. The Gaussian binomial ${[\genfrac{}{}{0pt}{}{u}{v}]}_p$ is monic of degree $v(u-v)$. In the signed sum for $R_m$, the degree increases by $2m-k$ from the $k$th to the $(k+1)$st term. The unique maximal term is $k=m$, with positive leading coefficient. Hence $R_m$ is monic of degree $(3m^2+m)/2$, and $s_m$ is monic of degree $3m^2-m$. In the determinant, the only permutation-dependent part of the degree is $6\sum _{i}i\sigma (i)$; strict rearrangement makes the identity its unique maximum. Therefore

More generally the same argument works for a fixed minor with distinct increasing row and column indices. Each such minor is positive for all sufficiently large $p$, with a threshold that may depend on the minor. At $p=1$, both leading Hankel families are positive definite by the infinite-support product measure, so each fixed rank is also positive in some neighbourhood of $1$. Neither argument provides a neighbourhood or large-base threshold uniform over every rank.

A Stieltjes moment representation requires positive semidefiniteness of both Hankel families at all ranks. The positive-definite formulation for nondegenerate sequences is stated in [33]; finite support requires allowing semidefinite matrices. The necessity of both conditions is visible from their quadratic forms: if a positive measure $\mu$ on $[0,\mathrm{\infty })$ satisfies $s_m={\int }_0^{\mathrm{\infty }}{x}^{m}d\mu (x)$, and $P(x)=\sum _{i=0}^{N-1}{c}_i{x}^i$ has real coefficients, then

The shifted test records nonnegative support, not merely positivity of the measure. For example, the positive point mass at $-1$ has moments $(-1{)}^m$. Its unshifted quadratic form is $P(-1{)}^2$, whereas its shifted form is $-P(-1{)}^2$. Thus a Hamburger moment sequence, which permits support anywhere on $\mathbb{R}$, need not be a Stieltjes moment sequence. The two tests above do not assert a measure for the entire coefficient sequence considered here. Coefficientwise total positivity would be stronger than the sixteen certificates above. Positive production matrices and path constructions can supply such a mechanism in other families [38][39]; no such matrix for this moving-degree sequence has been established here.