Hankel Determinants of Geometric Moments and Rational Lambert Values

Will Cook

Short paper

Hankel Determinants of Geometric Moments and Rational Lambert Values

Erdős #1049 By Will Cook September 2026 Authorship, AI use and citation

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Summary. Positive geometric moments and a uniform partition-sum bound give the fixed-base asymptotic of Zudilin’s normalised Hankel determinants, including its constant and power correction. A separate calculation of degrees after cancellation in his 2004 forms extends the rational-base irrationality region. It includes every positive integral power of $31 / 4$ and excludes $3 / 2$ .

This paper covers the fixed-base Hankel asymptotic and rational-base irrationality theorem; the long paper develops exact orders, content and endpoint obstructions, and the remaining problem at $3 / 2$ .

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #1049, which remains open.

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Positive weights

For $r \geq 1$ and $k \geq 0$ , set

R_{k}^{(r)} (q) = \sum_{n_{1} + \dots + n_{r} = k} \frac{(q; q)_{k}}{\prod_{j = 1}^{r} (q; q)_{n_{j}}},

over $r$ nonnegative parts. These are the multivariate Rogers–Szegő polynomials at unit arguments, written as sums of Gaussian multinomial polynomials [19]. At $q = 0$ , every Gaussian multinomial equals $1$ , so they count weak compositions (zero parts are allowed), and

c_{k} := R_{k}^{(2)} (0) R_{k}^{(3)} (0) = \frac{(k + 1)^{2} (k + 2)}{2}, \prod_{k < N} c_{k} = C_{N} .

Proposition 3.1 (the positive moment weights). For every $k \geq 0$ ,

a_{k} = P^{4} \frac{R_{k}^{(2)} R_{k}^{(3)}}{(q; q)_{k}} .

At fixed $0 < q < 1$ these weights satisfy

P^{5} c_{k} \leq a_{k} \leq P^{- 1} c_{k}, \frac{a_{k + h}}{a_{k}} \leq P^{- 6} (1 + h)^{3} (k, h \geq 0) .

Proof. We use the Gaussian binomial coefficients ${[\binom{n}{j}]}_{q} = (q; q)_{n} / ((q; q)_{j} (q; q)_{n - j})$ for $0 \leq j \leq n$ , and put $E (z) = (z; q)_{\infty}^{- 1}$ . Euler’s expansion gives $E (z)^{r} = \sum_{n \geq 0} R_{n}^{(r)} z^{n} / (q; q)_{n}$ . The desired factorisation multiplies two coefficients at the same index $k$ , rather than taking a coefficient of $E^{5}$ . We obtain it by splitting $k = n + j$ : the factor $R_{k}^{(3)}$ stays fixed, and summing the two factorial denominators supplies $R_{k}^{(2)}$ . To produce this split, use the unnormalised $q$ -difference operator $\partial_{q} f (z) = (f (z) - f (q z)) / z$ . Since $\partial_{q} z^{j} = (1 - q^{j}) z^{j - 1}$ , it cancels the last factor of $(q; q)_{j}$ and shifts the numerator index:

\partial_{q} E = E, \partial_{q}^{n} E (z)^{3} = \sum_{j \geq 0} \frac{R_{n + j}^{(3)}}{(q; q)_{j}} z^{j} .

The $q$ -Leibniz rule, obtained by induction from the product rule, is

\partial_{q}^{n} (f g) (z) = \sum_{j = 0}^{n} {[\binom{n}{j}]}_{q} (\partial_{q}^{j} f) (q^{n - j} z) (\partial_{q}^{n - j} g) (z) .

Let $h_{k} (z) = \sum_{j = 0}^{k} {[\binom{k}{j}]}_{q} (z; q)_{j}$ . Applying the rule to $E \cdot E$ and using $E (q^{m} z) = (z; q)_{m} E (z)$ , we obtain $\partial_{q}^{k} E (z)^{2} = E (z)^{2} h_{k} (z)$ (for $k = 1$ , both sides equal $E (z)^{2} (2 - z)$ ). A second application, this time to $E \cdot E^{2}$ , gives

\partial_{q}^{n} E (z)^{3} = E (z)^{3} \sum_{k = 0}^{n} {[\binom{n}{k}]}_{q} (z; q)_{k} h_{k} (z),

and hence

\sum_{n \geq 0} \frac{w^{n}}{(q; q)_{n}} \partial_{q}^{n} E (z)^{3} = E (w) E (z)^{3} \sum_{k \geq 0} \frac{w^{k} (z; q)_{k}}{(q; q)_{k}} h_{k} (z) .

The left-hand side is $\sum_{n, j \geq 0} R_{n + j}^{(3)} w^{n} z^{j} / ((q; q)_{n} (q; q)_{j})$ . Putting $z = w$ collects all pairs with the same sum $n + j$ . The coefficient of $w^{k}$ is consequently $R_{k}^{(3)} \sum_{j \leq k} ((q; q)_{j} (q; q)_{k - j})^{- 1} = R_{k}^{(2)} R_{k}^{(3)} / (q; q)_{k}$ . On the right-hand side, put $z = w$ , expand $h_{k}$ , and write $k = j + t$ . The identity $(w; q)_{j + t} = (w; q)_{j} (w q^{j}; q)_{t}$ gives

E (w)^{4} \sum_{j \geq 0} \frac{w^{j} (w; q)_{j}^{2}}{(q; q)_{j}} \sum_{t \geq 0} \frac{(w q^{j}; q)_{t}}{(q; q)_{t}} w^{t} .

The inner sum is evaluated by the $q$ -binomial theorem:

\sum_{t \geq 0} \frac{(w q^{j}; q)_{t}}{(q; q)_{t}} w^{t} = \frac{(w^{2} q^{j}; q)_{\infty}}{(w; q)_{\infty}} .

Finally, $(w; q)_{j} = (w; q)_{\infty} / (w q^{j}; q)_{\infty}$ cancels two of the five factors $E (w)$ , leaving

E (w)^{3} \sum_{j \geq 0} \frac{w^{j}}{(q; q)_{j}} \frac{(w^{2} q^{j}; q)_{\infty}}{(w q^{j}; q)_{\infty}^{2}} = G_{q} (w) .

We may read each identity coefficientwise, or as an absolutely convergent expansion when $| z |$ and $| w |$ are sufficiently small.

For the bounds, we write $b_{k}^{(r)} = [z^{k}] (z; q)_{\infty}^{- r}$ , so that $R_{k}^{(r)} = (q; q)_{k} b_{k}^{(r)}$ and $a_{k} = P^{4} (q; q)_{k} b_{k}^{(2)} b_{k}^{(3)}$ . By Euler’s expansion, $b_{k}^{(r)}$ is an $r$ -fold convolution of $1 / (q; q)_{n}$ , whose terms lie in $[1, P^{- 1}]$ . There are $(\binom{k + r - 1}{r - 1})$ compositions of $k$ into $r$ nonnegative parts, and we deduce

(\binom{k + r - 1}{r - 1}) \leq b_{k}^{(r)} \leq P^{- r} (\binom{k + r - 1}{r - 1}) .

Combining these at $r = 2, 3$ with $P \leq (q; q)_{k} \leq 1$ and $c_{k} = (k + 1) (\binom{k + 2}{2})$ gives $P^{5} c_{k} \leq a_{k} \leq P^{- 1} c_{k}$ . The shift bound follows from $c_{k + h} / c_{k} \leq (1 + h)^{3}$ . ◻

Positivity holds at each fixed real $q \in (0, 1)$ , not coefficientwise in the formal $q$ -series: $a_{0} (q) = P^{4} = 1 - 4 q + O (q^{2})$ . The bounds control shifted ratios, but do not determine the weight product. A relative error of order $1 / k$ accumulates logarithmically; one of order $1 / k^{2}$ is summable. Computing $a_{k} / c_{k}$ to first order will therefore determine the power of $N$ and give the fixed-shift limit.

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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