Geometric Moments and Rational Lambert Values: Proofs and Further Results

Will Cook

Long paper

Geometric Moments and Rational Lambert Values: Proofs and Further Results

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

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Summary. The record proves the fixed-base Hankel asymptotic for Zudilin's normalised moments and the rational-base irrationality criterion log b/log a<0.4056830213840605…. The determinant proof separates geometric node configurations from positive weights; the irrationality proof instead specialises Zudilin's 2004 polynomial forms after cancellation. Coefficient moments, evaluated lattices and congruence conditions distinguish integer divisibility from nonzero remainder decay. A broader ratio-limit argument and a rational-value model remain unformalised. None of these arguments settles $F (3 / 2)$ .

This paper covers the sharp fixed-base Hankel asymptotic and rational-base irrationality proofs, coefficient moments and arithmetic obstructions, with ordinary ratio-limit and rational-value models distinguished from checked results.

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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The row identity.

We choose row operations whose first surviving coefficients can be computed explicitly. The auxiliary lemma applies to more general products than the moments above; its use here is to identify the coefficient that remains after each row operation.

Work over $A = Z [[q]]$ . Let $H (X) = 1 + \sum_{s \geq 1} a_{s} (q) X^{s}$ . For nonnegative integers $m, t$ , put

W_{m} (t) = q^{(m + 1) t} \prod_{r = 1}^{m} H (q^{r}), D_{j} = \prod_{r = 0}^{j - 1} (I - q^{r} N), E (m, j) = m j - \frac{j (j - 1)}{2},

where $N$ is the backward shift $(N f)_{m} = f_{m - 1}$ in the index $m$ . Thus $D_{j}$ is Zudilin’s backward-difference operator in [9]. Write $\bar{H} = H mod q$ and $h_{r} = [X^{r}] \bar{H} (X)^{- 1}$ , with $h_{r} = 0$ for $r < 0$ and $h_{0} = 1$ .

The lemma allows arbitrary coefficients $a_{s} (q) \in Z [[q]]$ ; the only normalisation imposed on $H$ is its constant term $1$ . It therefore covers the product series used below, but not an unnormalised series with a different constant term. All statements here are coefficientwise identities of formal series, with no analytic convergence assumption. For a general $H$ , the coefficient in the lemma can vanish. The application below computes it for the chosen products and proves the required nonvanishing.

Lemma 3.5 (a coefficient of each transformed row). For $m \geq j \geq 0$ one has $D_{j} W_{m} (t) \in q^{E (m, j)} A$ and

[q^{E (m, j)}] D_{j} W_{m} (t) = (- 1)^{j} h_{j - t} .

Proof. Let $e_{u}$ denote a formal basis vector indexed by $u \geq 0$ , and suppress the argument $q$ in $a_{s} (q)$ . We use the transition rule

Φ e_{0} = \sum_{s \geq 1} a_{s} e_{s - 1}, Φ e_{u} = (q^{u} - 1) e_{u - 1} + q^{u} \sum_{s \geq 1} a_{s} e_{u + s - 1} (u > 0) .

In $Φ^{j} e_{t}$ , a coefficient means the sum of the weights of length- $j$ paths from $t$ to the specified endpoint. Each step lowers its index by at most one. A path ending at $u$ therefore visits only indices at most $max (t, u + j)$ , so each coefficient is a finite sum. No action on the algebraic direct sum is assumed.

For $m \geq 1$ , a direct expansion gives

W_{m} (u) - W_{m - 1} (u) = q^{m} \sum_{v} (Φ e_{u})_{v} W_{m - 1} (v),

since both sides equal

q^{m u} \prod_{r = 1}^{m - 1} H (q^{r}) (q^{u} H (q^{m}) - 1) .

For the induction step take $m \geq j + 1$ and note that $E (m, j) - E (m - 1, j) = j$ . Applying $I - q^{j} N$ to $q^{E (m, j)} W_{m - j} (u)$ therefore gives $q^{E (m, j)} (W_{m - j} (u) - W_{m - j - 1} (u))$ . The preceding identity contributes one further factor $q^{m - j}$ , and $E (m, j) + m - j = E (m, j + 1)$ . Induction, starting with $D_{0} = I$ , gives

\begin{matrix} (13) & D_{j} W_{m} (t) = q^{E (m, j)} \sum_{u} (Φ^{j} e_{t})_{u} W_{m - j} (u) . \end{matrix}

The endpoint sum is also locally finite: $ord W_{m - j} (u) = (m - j + 1) u$ , so only finitely many endpoints contribute to any fixed power of $q$ .

Modulo $q$ the operator simplifies: $\bar{Φ} e_{u} = - e_{u - 1}$ for $u > 0$ , and $\bar{Φ} e_{0} = \sum_{s \geq 1} a_{s} (0) e_{s - 1}$ . Put $c_{j, t} = ({\bar{Φ}}^{j} e_{t})_{0}$ . Then $c_{j + 1, t} = - c_{j, t - 1}$ for $t > 0$ and $c_{j + 1, 0} = \sum_{s = 1}^{j + 1} a_{s} (0) c_{j, s - 1}$ , the sum terminating because a state above $j$ cannot reach $0$ in $j$ steps. Now $c_{0, t} = δ_{t, 0} = h_{- t}$ , and if $c_{j, t} = (- 1)^{j} h_{j - t}$ for all $t$ then $c_{j + 1, t} = (- 1)^{j + 1} h_{j + 1 - t}$ for $t > 0$ at once, while for $t = 0$ the identity $\bar{H} {\bar{H}}^{- 1} = 1$ gives $h_{j + 1} = - \sum_{s \geq 1} a_{s} (0) h_{j + 1 - s}$ and hence $c_{j + 1, 0} = (- 1)^{j} \sum_{s \geq 1} a_{s} (0) h_{j + 1 - s} = (- 1)^{j + 1} h_{j + 1}$ . So $c_{j, t} = (- 1)^{j} h_{j - t}$ for all $j, t$ . Since $ord W_{m - j} (u) = (m - j + 1) u$ and $m \geq j$ , in (13) only the state $u = 0$ contributes at degree $E (m, j)$ , which gives both assertions. ◻

Apply the lemma to a summand of the normalised remainder:

T_{m, t} = q^{(m + 1) t} (q; q)_{m}^{3} \frac{(q^{t + 1}; q)_{m}}{(q^{m + t + 1}; q)_{m + 1}} .

Set

H_{t} (X) = \frac{(1 - X)^{3} (1 - q^{t} X)^{2}}{(1 - q^{t} X^{2}) (1 - q^{t + 1} X^{2})} .

Then $T_{m, t} = (1 - q^{t + 1})^{- 1} W_{m}^{H_{t}} (t)$ , by the telescoping identity

\prod_{r = 1}^{m} H_{t} (q^{r}) = \frac{(q; q)_{m}^{3} (1 - q^{t + 1}) (q^{t + 1}; q)_{m}}{(q^{m + t + 1}; q)_{m + 1}} .

The two denominator products contribute the consecutive factors $1 - q^{t + 2}, \dots, 1 - q^{t + 2 m + 1}$ . Reducing modulo $q$ gives ${\bar{H}}_{t} = (1 - X)^{3}$ for $t > 0$ , so $h_{r}^{(t)} = (\binom{r + 2}{2})$ , and ${\bar{H}}_{0} = (1 - X)^{4} / (1 + X)$ , so $h_{r}^{(0)} = (r + 1) (r + 2) (2 r + 3) / 6$ . The scalar factor $(1 - q^{t + 1})^{- 1}$ has constant term $1$ . The rows are $v_{m}^{*} = \sum_{t \geq 0} T_{m, t}$ . In $D_{j}$ only the shifts $m, m - 1, \dots, m - j$ occur. Since $m \geq j$ and $ord T_{m - r, t} = (m - r + 1) t \geq (m - j + 1) t$ , only finitely many $t$ can affect any fixed coefficient after any of these shifts. This justifies interchanging the sum and $D_{j}$ , and then extracting its coefficient at degree $E (m, j)$ . Terms with $t > j$ contribute $h_{j - t}^{(t)} = 0$ , so Lemma 3.5 at $m = j + ℓ$ , where $E (j + ℓ, j) = j (j + 1) / 2 + j ℓ$ , gives

\begin{matrix} (14) & D_{j} v_{j + ℓ}^{*} = (- 1)^{j} \frac{(j + 1)^{2} (j + 2)}{2} q^{j (j + 1) / 2 + j ℓ} + O (q^{j (j + 1) / 2 + j ℓ + 1}) (j, ℓ \geq 0) . \end{matrix}

The coefficient is obtained by summing

h_{j}^{(0)} + \sum_{t = 1}^{j} h_{j - t}^{(t)} = \frac{(j + 1) (j + 2) (2 j + 3)}{6} + (\binom{j + 2}{3}) = \frac{(j + 1)^{2} (j + 2)}{2} .

Proof of Theorem 3.4. The operators $D_{j}$ act by lower unitriangular row operations, so they leave $det (v_{i + j}^{*})_{0 \leq i, j < N}$ unchanged. By (14) the entry in row $j$ and column $ℓ$ has order $j (j + 1) / 2 + j ℓ$ . For $N = 2$ , the matrix of entry orders is

(\begin{matrix} 0 & 0 \\ 1 & 2 \end{matrix}) .

The off-diagonal product is the unique term of order one. Its permutation sign cancels the negative leading coefficient of the second row, giving $6 q + O (q^{2})$ . At arbitrary rank the same argument selects the reversed permutation. For a permutation $ς$ , the corresponding Leibniz term has weight $\sum_{j} (j (j + 1) / 2 + j ς (j))$ . By the rearrangement inequality, $\sum_{j} j ς (j)$ is uniquely minimised by the reversal $ς (j) = N - 1 - j$ , the values $j$ being distinct. The minimum weight is $\sum_{j < N} j^{2} = N (N - 1) (2 N - 1) / 6$ , so exactly one Leibniz term attains it and no cancellation is possible there. The sign of the reversal is $(- 1)^{N (N - 1) / 2}$ , which cancels $\prod_{j < N} (- 1)^{j}$ , and the surviving coefficient is

\prod_{j = 0}^{N - 1} \frac{(j + 1)^{2} (j + 2)}{2} = \frac{(N!)^{2} (N + 1)!}{2^{N}} .

◻

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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