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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In this paper

Positive Hankel determinants and a rational special value

We construct a generating function with the rational special value $1$ , although its moment determinants have the same formal orders, leading coefficients and type of fixed-base power correction as Zudilin’s. This shows why those determinant properties alone do not establish irrationality. The construction uses neither Zudilin’s recurrence nor his polynomial factorisation or cyclotomic divisors, and gives no conclusion about $F (3 / 2)$ . The arguments below have ordinary proofs only and are stated as unformalised remarks.

We choose the tail of the weight sequence to obtain the determinant asymptotic, and its first terms to fix the special value. A telescoping difference of the moments will then give the value $1$ . The cut-off below separates these two choices while leaving room to keep every weight positive.

Put

c_{k} = \frac{(k + 1)^{2} (k + 2)}{2}, d_{k} = \frac{(k + 1) (k + 2)}{2}, R (w) = \sum_{k \geq 0} c_{k} w^{k} = \frac{1 + 2 w}{(1 - w)^{4}} .

For $0 < q \leq 2 / 3$ define, with a single fixed cut-off $32$ ,

\begin{matrix} (25) & a_{k} (q) = {\begin{cases} c_{k} (1 - q)^{32}, & 1 \leq k < 32, \\ c_{k} - 12 q d_{k}, & k \geq 32, \end{cases} a_{0} (q) = 1 - \sum_{k \geq 1} a_{k} (q) q^{k} . \end{matrix}

The weight generating function $A_{q} (w) = \sum_{k \geq 0} a_{k} (q) w^{k}$ is rational in $w$ . The value $G_{q} (q)$ concerns a different generating function, formed from the moments below. To obtain $A_{q}$ , put

T_{q} (w) = R (w) - \frac{12 q}{(1 - w)^{3}} - \sum_{k = 0}^{31} (c_{k} - 12 q d_{k}) w^{k},

so that $A_{q} (w)$ equals

a_{0} (q) + (1 - q)^{32} \sum_{k = 1}^{31} c_{k} w^{k} + T_{q} (w), a_{0} (q) = 1 - (1 - q)^{32} \sum_{k = 1}^{31} c_{k} q^{k} - T_{q} (q) .

Set $μ_{m} (q) = A_{q} (q^{m + 1})$ , $ν_{m} (q) = μ_{m} (q) - q μ_{m + 1} (q)$ , and $G_{q} (z) = \sum_{m \geq 0} ν_{m} (q) z^{m}$ . The choice of $a_{0}$ makes $μ_{0} = 1$ , while the factor $q$ in $ν_{m}$ makes the sum defining $G_{q} (q)$ telescope. Independently, for $k \geq 32$ the tail satisfies $a_{k} / c_{k} = 1 - 12 q / (k + 1)$ ; at $q = 2 / 3$ this becomes $1 - 8 / (k + 1)$ , which produces the required power $N^{- 8}$ in the product of the weights. The proof below checks positivity, convergence and both determinant claims.

Remark (a rational special value). All the weights in (25) are positive for $0 < q \leq 2 / 3$ . The two moment sequences $(μ_{m})$ and $(ν_{m})$ have positive Hankel determinants of every size, $G_{q}$ is not a rational function, and $G_{q} (q) = 1$ . Their formal Hankel determinants both begin with

C_{N} q^{B_{N}}, B_{N} = \sum_{j < N} j^{2}, C_{N} = \frac{(N!)^{2} (N + 1)!}{2^{N}} (N \geq 1) .

At the fixed base $q_{0} = 2 / 3$ , with $P (q) = \prod_{j \geq 1} (1 - q^{j})$ and $M (q) = \prod_{j \geq 1} (1 - q^{j})^{- j}$ , there is an explicit $A > 0$ such that

det (μ_{i + j} (q_{0}))_{0 \leq i, j < N} \sim A M (q_{0})^{3} C_{N} q_{0}^{B_{N}} P (q_{0})^{2 N} N^{- 8 ξ}, ξ = G_{q_{0}} (q_{0}) = 1.

This countermodel and its denominator calculation have ordinary proofs below; they have not been Lean-checked or externally peer reviewed.

Proof. For $k \geq 32$ we have $a_{k} (q) = d_{k} (k + 1 - 12 q) > 0$ , and the smaller-index weights are plainly positive. Write $M (q) = (1 - q)^{32} \sum_{k = 1}^{31} c_{k} q^{k}$ . The generating function for the $c_{k}$ gives $R (q) - 1 \leq 6 q (1 - q)^{- 4}$ , whence $M (q) \leq 6 q (1 - q)^{28} \leq 6 / 29 < 1 / 4$ . For $k \geq 32$ ,

\frac{d}{d q} (d_{k} (k + 1 - 12 q) q^{k}) = d_{k} (k + 1) q^{k - 1} (k - 12 q) > 0.

Direct substitution into the rational expression for $T_{q}$ gives

T_{2 / 3} (2 / 3) = \frac{24794846199808}{205891132094649} < \frac{1}{4} .

Therefore $1 / 2 < a_{0} (q) = 1 - M (q) - T_{q} (q) < 1$ .

The moments have positive atomic representations

μ_{m} = \sum_{k \geq 0} a_{k} q^{k} (q^{k})^{m}, ν_{m} = \sum_{k \geq 0} a_{k} q^{k} (1 - q^{k + 1}) (q^{k})^{m} .

Infinitely many distinct atoms have positive mass. The Gram quadratic form of a nonzero polynomial of degree below $N$ is strictly positive for either measure, proving positive definiteness of every $N$ -by- $N$ Hankel matrix. Summing the geometric series in $m$ gives

G_{q} (z) = \sum_{k \geq 0} \frac{a_{k} q^{k} (1 - q^{k + 1})}{1 - z q^{k}} .

Each $z = q^{- k}$ is a genuine pole: its residue is nonzero and the other terms are analytic there. Hence $G_{q}$ is not rational. Telescoping shows $G_{q} (q) = μ_{0} (q) = A_{q} (q) = 1$ and, for $S_{n - 1} = \sum_{m < n} q^{m} ν_{m}$ , gives

1 - S_{n - 1} = q^{n} μ_{n} \in (0, q^{n}] (n \geq 0) .

Every $a_{k} (q)$ belongs to $Z [[q]]$ and has constant term $c_{k}$ , including $a_{0}$ , for which the defining sum is coefficientwise finite. The moments $μ_{m}$ and $ν_{m}$ also lie in $Z [[q]]$ , but their constant terms are $1$ : only $a_{0}$ contributes at $q = 0$ . Their rational expressions have no poles for $| q | < 1$ , so these formal series converge at $q_{0}$ . For $H_{N} (q) = det (μ_{i + j} (q))_{i, j < N}$ , the Gram determinant expansion is

H_{N} (q) = \sum_{0 \leq k_{0} < \dots < k_{N - 1}} (\prod_{i < N} a_{k_{i}} (q) q^{k_{i}}) \prod_{i < j} (q^{k_{i}} - q^{k_{j}})^{2} .

It holds first for finitely many atoms and then coefficientwise. The lowest possible degree of one summand is $\sum_{i < N} (2 N - 1 - 2 i) k_{i}$ . Its unique minimum is attained at $k_{i} = i$ and equals $B_{N}$ ; the coefficient is $\prod_{i < N} c_{i} = C_{N}$ . Replacing $a_{k}$ by $a_{k} (1 - q^{k + 1})$ proves the same leading term for $(ν_{m})$ .

At $q_{0}$ set $ε = 3^{- 32}$ and $b_{0} = a_{0} (q_{0}) > 0$ . For $k \geq 32$ , $a_{k} (q_{0}) / c_{k} = (k - 7) / (k + 1)$ ; all these ratios lie in $[ε, 1]$ . Hence for fixed $h$ , $a_{k + h} / a_{k} \to 1$ as $k \to \infty$ , and uniformly $a_{k + h} / a_{k} \leq 3^{32} (1 + h)^{3}$ . These are precisely the weight hypotheses of Theorem 3.6, whose conclusion is

H_{N} (q_{0}) \sim M (q_{0})^{3} q_{0}^{B_{N}} P (q_{0})^{2 N} \prod_{k < N} a_{k} (q_{0}) .

The finite product has the exact form, for $N \geq 32$ ,

\frac{\prod_{k < N} a_{k} (q_{0})}{C_{N}} = b_{0} ε^{31} \frac{32!}{24!} \frac{(N - 8)!}{N!} .

Thus $A = b_{0} ε^{31} 32! / 24!$ , and $N^{8} (N - 8)! / N! \to 1$ . The displayed exponent is exactly $- 8 ξ$ because $ξ = 1$ ; it is asserted at this one fixed base, not as a uniform special-value formula for this family. ◻

The following calculation is retained with its ordinary proof. It has no Lean proof in the supplied record.

Remark (exact denominators of the calibrated moments). Let $q_{0} = 2 / 3$ and $δ_{n} = 3^{n + 1} - 2^{n + 1}$ . For every $n \geq 1$ , the reduced denominators are

den μ_{n} (q_{0}) = 3^{31 n + 62} δ_{n}^{4}, den (q_{0}^{n} μ_{n} (q_{0})) = 3^{32 n + 62} δ_{n}^{4} .

The least integer clearing both rational coefficients in $μ_{n} = q_{0}^{- n} ξ - q_{0}^{- n} S_{n - 1}$ is $2^{n} 3^{31 n + 62} δ_{n}^{4}$ . At $n = 0$ the moment and tail equal one.

Proof. At $q_{0}$ , split the rational seed into a pole part and a polynomial:

A_{q_{0}} (w) = \frac{- 7 + 10 w}{(1 - w)^{4}} + Q (w), Q (w) = p_{0} + \sum_{k = 1}^{31} p_{k} w^{k},

where $p_{0} = b_{0} + 7$ and $p_{k} = (3^{- 32} - 1) c_{k} + 8 d_{k}$ . These coefficients have only powers of three in their denominators. The finite formulas give $v_{3} (p_{0}) \geq - 63$ , $v_{3} (p_{k}) \geq - 32$ for $1 \leq k \leq 30$ , and $v_{3} (p_{31}) = - 31$ : in the last calculation $c_{31} = 16896$ has $v_{3} (c_{31}) = 1$ and the other numerator terms are multiples of $3^{32}$ . At $w = (2 / 3)^{n + 1}$ , $n \geq 1$ , the term $p_{31} w^{31}$ has valuation $- 31 n - 62$ , uniquely lower than all other terms of $Q (w)$ . The pole part equals

\frac{3^{3 n + 3} (- 7 \cdot 3^{n + 1} + 10 \cdot 2^{n + 1})}{δ_{n}^{4}}

and has positive $3$ -adic valuation. Hence $v_{3} (μ_{n}) = - 31 n - 62$ .

Every prime $ℓ ∣ δ_{n}$ differs from $2, 3$ . Its numerator in the displayed pole part is congruent to $3 \cdot 2^{n + 1} (\mod δ_{n})$ and is coprime to $δ_{n}$ . The polynomial part is $ℓ$ -integral, so it cannot cancel the pole part’s $ℓ$ -denominator of order $4 v_{ℓ} (δ_{n})$ . There are no other denominator primes. Multiplying by $q_{0}^{n} = 2^{n} / 3^{n}$ adds exactly $n$ factors of three, proving the second formula. In the affine error, $q_{0}^{- n} = 3^{n} / 2^{n}$ has denominator $2^{n}$ , coprime to the odd denominator of $μ_{n}$ . Subtraction does not reduce either negative valuation, proving the stated least common denominator. ◻

Each individual $μ_{n}$ has logarithmic height linear in $n$ : $\log den μ_{n} = (35 n + 66) \log 3 + o (1)$ . The determinant has a different scale.

The following calculation is retained with its ordinary proof. It has no Lean proof in the supplied record.

Remark (determinant height and row denominators). For $D_{N} = den H_{N} (q_{0})$ in lowest terms,

\underset{N \to \infty}{lim inf} \frac{\log D_{N}}{N^{3}} \geq \frac{\log (3 / 2)}{3}, \log D_{N} = O (N^{3}) .

Already for $N = 2$ , multiplying $H_{2} = μ_{2} - μ_{1}^{2}$ by $den μ_{1} den μ_{2}$ leaves reduced denominator $625$ .

Proof. The positive Heine expansion above gives a bound independent of the asymptotic theorem. Since $0 < a_{k} (q_{0}) \leq c_{k}$ , write $k_{i} = i + s_{i}$ . Dropping the ordering restriction and bounding $c_{i + s_{i}} / c_{i} \leq (1 + s_{i})^{3}$ and each residual Vandermonde factor by one gives

0 < H_{N} (q_{0}) \leq C_{N} q_{0}^{B_{N}} \prod_{j \geq 0} \sum_{s \geq 0} (s + 1)^{3} q_{0}^{(2 j + 1) s} .

For $A (t) = \sum_{s \geq 0} (s + 1)^{3} t^{s} = (1 + 4 t + t^{2}) / (1 - t)^{4}$ , $\log A (t) \leq A (t) - 1 \leq 8 t / (1 - t)^{4}$ . The product is therefore at most $\exp (3888 / 5)$ . Since $D_{N} H_{N}$ is a positive integer,

\log D_{N} \geq B_{N} \log (3 / 2) - \log C_{N} - 3888 / 5.

Here $B_{N} \sim N^{3} / 3$ and $\log C_{N} = O (N \log N)$ . Conversely,

E_{N} = 3^{31 N (N - 1) + 62 N} \prod_{0 \leq i, j < N} δ_{i + j}^{4}

clears every term in the determinant expansion, by the preceding denominator formula. Thus $D_{N} ∣ E_{N}$ ; using $\log δ_{m} \leq (m + 1) \log 3$ and $\sum_{i, j < N} (i + j + 1) = N^{3}$ gives the asserted cubic upper bound.

For the assertion about $H_{2}$ , $den μ_{1} = 3^{93} 5^{4}$ and $den μ_{2} = 3^{124} 19^{4}$ . At five, $v_{5} (μ_{2}) = 0$ and $v_{5} (μ_{1}^{2}) = - 8$ , while the proposed product contributes only $5^{4}$ . At three and nineteen that product clears the denominator, and no other prime occurs. The remaining denominator is therefore $5^{4} = 625$ . ◻

Zudilin’s determinantal criterion retains a divisibility condition on its successive clearing integers [11]; the calculation for $H_{2}$ rules out replacing it with their numerical ordering, not the criterion itself. The usual integer-polynomial height criterion likewise needs a common coefficient denominator for the same polynomial being evaluated. The model’s exact partial errors alone cannot supply one cheaply. In fact, with $α_{m} = q_{0}^{- m}$ and $β_{m} = α_{m} - μ_{m} (q_{0})$ , the pencil $det (α_{i + j} X - β_{i + j})_{i, j < N}$ is affine in $X$ because $(α_{i + j})$ has rank one. At $X = 1$ it is the positive rational $H_{N} (q_{0})$ ; after primitive integral clearing its value is a positive integer. Matching Hankel order, positivity, fixed-base asymptotic and a small rational value thus leaves the rational-base denominator problem exactly where it was. The Christoffel-type reweighting $a_{k} q^{k} \mapsto a_{k} q^{k} (1 - q^{k + 1})$ is standard moment machinery, not a new transform [31]; this example’s claim is the explicit combination and its arithmetic obstruction.

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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