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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In this paper

The coefficient asymptotic

Proof of Theorem 1.1. Put $L = F (1 / q)$ , $e_{k} = (q; q)_{k}^{- 1}$ and $E = P^{- 1}$ . Replacing every $e_{k}$ by $E$ gives $E^{2} (k + 1)$ and $E^{3} (\binom{k + 2}{2})$ , counting nonnegative pairs and triples with sum $k$ . Terms with one deficit $E - e_{i}$ give the first correction. That coordinate has two possible positions in a pair and three in a triple; fixing it at $i$ leaves one pair or $k - i + 1$ triples. Thus the same total deficit produces a constant correction in the first convolution and a linear correction in the second. We calculate it and control its first moment.

Writing $d_{k} = E - e_{k}$ , the identity $e_{k + 1} - e_{k} = q^{k + 1} e_{k + 1}$ gives

d_{k} = \sum_{j > k} q^{j} e_{j}, 0 \leq d_{k} \leq \frac{E q^{k + 1}}{1 - q} .

Thus $\sum d_{k}$ and $\sum k d_{k}$ converge. To evaluate the former, use Euler’s identity $E (z) = \sum_{k \geq 0} e_{k} z^{k} = (z; q)_{\infty}^{- 1}$ for $| z | < 1$ . Tonelli’s theorem and logarithmic differentiation at $z = q$ give

S := \sum_{k \geq 0} d_{k} = \sum_{j \geq 1} j q^{j} e_{j} = q E^{'} (q) = E \sum_{r \geq 1} \frac{q^{r}}{1 - q^{r}} = E L .

In the double sum, $q^{j} e_{j}$ occurs once for each $k = 0, \dots, j - 1$ , which explains the factor $j$ . Logarithmic differentiation is justified by locally uniform convergence on $| z | < 1$ .

Let $b_{k}^{(r)} = [z^{k}] E (z)^{r}$ . The finite-sum formula in Proposition 3.1 becomes

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

We expand each finite convolution using $e_{k} = E - d_{k}$ . For two factors,

b_{k}^{(2)} = E^{2} (k + 1) - 2 E \sum_{i = 0}^{k} d_{i} + \sum_{i = 0}^{k} d_{i} d_{k - i} = E^{2} (k + 1 - 2 L) + O_{q} ((k + 1) q^{k}) .

For three factors, the terms with one $d_{i}$ contribute $- 3 E^{2} \sum_{i = 0}^{k} (k - i + 1) d_{i}$ . Here the sum equals $(k + 1) S + O_{q} (1)$ because $\sum i d_{i} < \infty$ . Terms with two $d_{i}$ are bounded by $3 E S^{2}$ , and the term with three is $O_{q} ((k + 1)^{2} q^{k})$ . It follows that

b_{k}^{(3)} = E^{3} ((\binom{k + 2}{2}) - 3 L (k + 1)) + O_{q} (1) .

Division by $(\binom{k + 2}{2})$ turns $- 3 L (k + 1)$ into the relative correction $- 6 L / (k + 2)$ ; the two-fold sum gives $- 2 L / (k + 1)$ . Since $(q; q)_{k} = P (1 + O_{q} (q^{k}))$ , the leading constants in $a_{k}$ cancel: $P^{5} E^{5} = 1$ . Replacing $(k + 2)^{- 1}$ by $(k + 1)^{- 1}$ costs only $O_{q} ((k + 1)^{- 2})$ . The two corrections therefore add, and with $c_{k} = (k + 1)^{2} (k + 2) / 2$ we obtain

\begin{aligned} \frac{a_{k}}{c_{k}} & = (1 - \frac{2 L}{k + 1} + O_{q} (q^{k})) (1 - \frac{6 L}{k + 2} + O_{q} ((k + 1)^{- 2})) (1 + O_{q} (q^{k})) \\ = 1 - \frac{8 L}{k + 1} + O_{q} ((k + 1)^{- 2}) . \end{aligned}

In particular $a_{k + 1} / a_{k} \to 1$ , and multiplication of finitely many adjacent ratios gives $a_{k + h} / a_{k} \to 1$ for each fixed $h$ . The uniform shift bound follows from Proposition 3.1, so Theorem 2.1 applies.

Let $γ_{E}$ denote Euler’s constant and put

\begin{matrix} (13) & A (q) = e^{- 8 γ_{E} L} \prod_{k \geq 0} (\frac{a_{k}}{c_{k}} e^{8 L / (k + 1)}), K (q) = A (q) M (q)^{3} . \end{matrix}

For sufficiently large $k$ , $\log (1 + u) = u + O (u^{2})$ gives $\log (a_{k} / c_{k}) + 8 L / (k + 1) = O_{q} ((k + 1)^{- 2})$ . This tail is summable, and the finitely many initial factors are positive by Proposition 3.1. Thus $0 < A (q) < \infty$ . Using $\prod_{k < N} c_{k} = C_{N}$ , separate the harmonic sum from this convergent remainder:

\log \frac{\prod_{k < N} a_{k}}{C_{N}} = - 8 L \sum_{k < N} \frac{1}{k + 1} + \sum_{k < N} (\log \frac{a_{k}}{c_{k}} + \frac{8 L}{k + 1}) .

The second sum tends to $\log A (q) + 8 γ_{E} L$ . Using $\sum_{k < N} (k + 1)^{- 1} = \log N + γ_{E} + o (1)$ , the Euler-constant terms cancel, giving $\prod_{k < N} a_{k} \sim A (q) C_{N} N^{- 8 L}$ . Substitution in (5) proves the theorem, or equivalently

\log V_{N}^{*} = B_{N} \log q + \log C_{N} + 2 N \log P - 8 L \log N + \log K (q) + o (1) .

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The scales are now separate: the node geometry gives the cubic term $B_{N} \log q$ ; the cubic growth of the weights gives $\log C_{N}$ of order $N \log N$ ; their first relative correction gives $- 8 F (1 / q) \log N$ . The other terms are linear and constant. In particular, the power correction does not change the cubic rate.

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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