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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Geometric moment determinants

For constant weights, the moments are geometric series: $M_{m} = \sum_{k \geq 0} q^{(m + 1) k} = (1 - q^{m + 1})^{- 1}$ . Cauchy’s determinant formula evaluates their determinant exactly. Writing

P = (q; q)_{\infty}, M (q) = \prod_{d \geq 1} (1 - q^{d})^{- d}, B_{N} = \sum_{j < N} j^{2},

we obtain

\begin{matrix} (2) & D_{N}^{(0)} := det {(\frac{1}{1 - q^{i + j + 1}})}_{0 \leq i, j < N} = \frac{q^{B_{N}} Δ_{N}}{\prod_{0 \leq i, j < N} (1 - q^{i + j + 1})}, Δ_{N} = \prod_{d = 1}^{N - 1} (1 - q^{d})^{2 (N - d)} . \end{matrix}

For each fixed $d$ , the multiplicity of $1 - q^{d}$ in the denominator increases to $d$ , giving the factor $M (q)$ in the limit. The numerator supplies two more copies:

\frac{Δ_{N}}{P^{2 N}} = \prod_{d < N} (1 - q^{d})^{- 2 d} \prod_{d \geq N} (1 - q^{d})^{- 2 N} ⟶ M (q)^{2} .

Indeed, $\sum d | \log (1 - q^{d}) | < \infty$ , and the logarithm of the second product is $O_{q} (N q^{N})$ . Thus

\begin{matrix} (3) & D_{N}^{(0)} \sim M (q)^{3} q^{B_{N}} P^{2 N} . \end{matrix}

Replacing the constant weights by $a_{k}$ will contribute $\prod_{k < N} a_{k}$ . The two hypotheses below do different jobs: the ratio limit handles each fixed displacement; the polynomial bound permits summation over all displacements.

Theorem 2.1 (geometric moment determinants). Let $0 < q < 1$ and $a_{k} > 0$ for $k \geq 0$ . Suppose that, for fixed constants $C > 0$ and $κ \geq 0$ ,

\begin{matrix} (4) & \frac{a_{k + h}}{a_{k}} ⟶ 1 (k \to \infty) for each fixed h \geq 0, \frac{a_{k + h}}{a_{k}} \leq C (1 + h)^{κ} (k, h \geq 0) . \end{matrix}

For $M_{m} = \sum_{k \geq 0} a_{k} q^{(m + 1) k}$ and $D_{N} = det (M_{i + j})_{0 \leq i, j < N}$ , we have

\begin{matrix} (5) & D_{N} \sim M (q)^{3} q^{B_{N}} P^{2 N} \prod_{k = 0}^{N - 1} a_{k} (N \to \infty) . \end{matrix}

For every real $s$ , the weights $a_{k} = (k + 1)^{s}$ satisfy (4) with $C = 1$ and $κ = max (s, 0)$ , and their first $N$ terms have product $(N!)^{s}$ . Taking $s = 1$ and summing the moments gives

\begin{matrix} (6) & det {(\frac{1}{(1 - q^{i + j + 1})^{2}})}_{0 \leq i, j < N} \sim M (q)^{3} N! q^{B_{N}} P^{2 N} . \end{matrix}

The polynomial bound excludes $a_{k} = e^{\sqrt{k}}$ , although its adjacent ratio tends to $1$ . The companion’s ordinary ratio-limit argument allows $a_{k + 1} / a_{k} \to ρ \in (0, q^{- 1})$ and, at $ρ = 1$ , gives (5) without that bound. This broader claim remains unformalised and is not used here: the theorem and its applications retain both hypotheses in (4).

Proof of Theorem 2.1. The bound $a_{h} \leq C a_{0} (1 + h)^{κ}$ makes $\sum_{k \geq 0} a_{k} q^{k} δ_{q^{k}}$ a finite positive measure with moments $M_{m}$ . Apply Cauchy–Binet to its finite truncations. The entries, hence the determinants, converge; the nonnegative tuple sums increase. Passing to the limit gives

\begin{matrix} (7) & D_{N} = \sum_{0 \leq k_{0} < \dots < k_{N - 1}} \prod_{i < N} a_{k_{i}} q^{k_{i}} \prod_{i < j < N} (q^{k_{i}} - q^{k_{j}})^{2} . \end{matrix}

The tuple $k_{i} = i$ contributes $q^{B_{N}} Δ_{N} \prod_{i < N} a_{i}$ . It does not account for the whole asymptotic. For example, replace $(0, 1, \dots, N - 2, N - 1)$ by $(0, 1, \dots, N - 2, N)$ . The ratio of its contribution to that of the original tuple is

\begin{matrix} (8) & q \frac{a_{N}}{a_{N - 1}} \prod_{d = 1}^{N - 1} {(\frac{1 - q^{d + 1}}{1 - q^{d}})}^{2} = q \frac{a_{N}}{a_{N - 1}} {(\frac{1 - q^{N}}{1 - q})}^{2} ⟶ \frac{q}{(1 - q)^{2}} . \end{matrix}

Only pairs involving the moved node change, and their product telescopes. At $q = 1 / 2$ the displaced tuple is asymptotically twice as large as the normalising tuple.

Subtracting $(0, 1, \dots, N - 1)$ gives the shifts $λ_{i} = k_{i} - i$ . Since $k_{i + 1} \geq k_{i} + 1$ , these are nonnegative and weakly increasing; reversing their order gives a partition. For $N = 4$ , for example,

\underset{(k_{i})}{\underset{⏟}{(0, 1, 3, 5)}} ⟼ \underset{(λ_{i})}{\underset{⏟}{(0, 0, 1, 2)}} ⟼ \underset{(μ_{j})}{\underset{⏟}{(2, 1, 0, 0)}} .

In general, put $μ_{j} = λ_{N - j}$ for $1 \leq j \leq N$ , so that $μ_{1} \geq \dots \geq μ_{N} \geq 0$ . Division by the contribution of $k_{i} = i$ turns the summand for this tuple into

\begin{matrix} (9) & \begin{aligned} W_{N} (μ) = & q^{\sum_{j \leq N} (2 j - 1) μ_{j}} \prod_{j \leq N} \frac{a_{N - j + μ_{j}}}{a_{N - j}} \\ \cdot \prod_{1 \leq j < k \leq N} {(\frac{1 - q^{k - j + μ_{j} - μ_{k}}}{1 - q^{k - j}})}^{2} . \end{aligned} \end{matrix}

The $j$ th index from the right enters once through the mass $q^{k_{i}}$ and twice through each of its $j - 1$ pairs with later indices. This gives the exponent $1 + 2 (j - 1) = 2 j - 1$ . “Right” refers to increasing indices, not to the nodes $q^{k_{i}}$ , which decrease towards zero.

Set $W_{N} (μ) = 0$ for partitions with more than $N$ positive parts, giving every sum the same index set. A fixed partition with $ℓ$ positive parts affects only the last $ℓ$ indices of $(0, 1, \dots, N - 1)$ . These indices tend to infinity while their displacements stay fixed, so the weight ratios in (9) tend to $1$ .

The Vandermonde quotient still contains a growing number of factors, but its dependence on $N$ telescopes. The factors with $j < k \leq ℓ$ are fixed, and those with $j > ℓ$ equal $1$ . For $N \geq ℓ$ and each $j \leq ℓ$ , the remaining factors satisfy

\begin{matrix} (10) & \prod_{k = ℓ + 1}^{N} \frac{1 - q^{k - j + μ_{j}}}{1 - q^{k - j}} = \frac{(q^{N + 1 - j}; q)_{μ_{j}}}{(q^{ℓ + 1 - j}; q)_{μ_{j}}} ⟶ \frac{1}{(q^{ℓ + 1 - j}; q)_{μ_{j}}} . \end{matrix}

Thus each $W_{N} (μ)$ has the same limit as for constant weights.

The fixed-partition limit does not yet control partitions whose length or parts grow with $N$ . For a uniform bound, group the Vandermonde factors by $j$ . Only $j \leq ℓ$ contribute; each group’s denominator product is at least $P$ and its numerator product at most $1$ . The polynomial shift bound therefore gives

W_{N} (μ) \leq (C P^{- 2})^{ℓ} \prod_{j = 1}^{ℓ} (1 + μ_{j})^{κ} q^{(2 j - 1) μ_{j}} .

Let $A = \sum_{u \geq 1} (1 + u)^{κ} q^{u - 1} < \infty$ . Dropping the ordering of the positive parts, we obtain

\begin{aligned} \sum_{ℓ (μ) = ℓ} sup_{N} W_{N} (μ) & \leq (C P^{- 2})^{ℓ} \prod_{j = 1}^{ℓ} \sum_{u \geq 1} (1 + u)^{κ} q^{(2 j - 1) u} \\ \leq (C P^{- 2} A)^{ℓ} q^{ℓ^{2}} . \end{aligned}

Here $q^{(2 j - 1) u} \leq q^{2 j - 1} q^{u - 1}$ and $1 + 3 + \dots + (2 ℓ - 1) = ℓ^{2}$ . The size of each part is absorbed in $A$ ; their number is controlled by $q^{ℓ^{2}}$ , which dominates $(C P^{- 2} A)^{ℓ}$ . Thus the bound is summable over $ℓ$ and independent of $N$ .

Dominated convergence gives the same limit for $\sum_{μ} W_{N} (μ)$ as for $a_{k} = 1$ . Equation (2) evaluates that limit as $M (q)$ , without a separate partition identity. Restoring the normalising tuple supplies the weight product and, through $Δ_{N} / P^{2 N}$ , the other two copies of $M (q)$ . Hence

\frac{D_{N}}{D_{N}^{(0)} \prod_{k < N} a_{k}} ⟶ 1.

Combining this limit with (3) proves the theorem. ◻

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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