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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Successive scaled remainders

Consecutive partial sums differ by one term. After clearing the powers of a rational base, this identity gives a recurrence for the scaled differences from a rational number. We derive it for arbitrary coefficients; the differences are tails only when the series converges to the chosen rational number.

Let $r, s, B, ξ$ be rationals with $r \neq 0$ and let $c : N \to Q$ be arbitrary. Define the prefix and the scaled remainder by

\begin{matrix} (*) & P_{N} = \sum_{m = 0}^{N - 1} \frac{c (m + 1) s^{m + 1}}{r^{m + 1}}, U_{N} = B r^{N} (ξ - P_{N}) . \end{matrix}

Thus $P_{N}$ is the partial sum through index $N$ , and $U_{N}$ is its scaled difference from $ξ$ . These are the rational-base partial sum and the scaled remainder.

Theorem 7.1 (recurrence for the scaled remainder). Let $r, s, B, ξ \in Q$ with $r \neq 0$ , let $c : N \to Q$ , and let $P_{N}$ and $U_{N}$ be as in (∗). Then for every $N$ ,

U_{N + 1} = r U_{N} - B c (N + 1) s^{N + 1} .

Proof. Expanding $P_{N + 1} = P_{N} + c (N + 1) s^{N + 1} / r^{N + 1}$ and $r^{N + 1} = r^{N} \cdot r$ in the definition of $U_{N + 1}$ and clearing the denominator $r^{N + 1}$ , which is nonzero, gives the identity. ◻

For a reduced positive base $r / s > 1$ , the forcing term $B c (N + 1) s^{N + 1}$ contains the denominator power absent at integer bases. Its size gives a useful bound on two consecutive remainders, although it need not bound each remainder separately.

Theorem 7.2 (the forcing term). Let $s, B$ be natural numbers and $c : N \to N$ .

If $s \geq 2$ , $B \geq 1$ and $c (N + 1) \geq 1$ , then $2^{N + 1} \leq B c (N + 1) s^{N + 1}$ .
If $s = 1$ , then $B c (N + 1) s^{N + 1} = B c (N + 1)$ .

Proof. For the first part, $2^{N + 1} \leq s^{N + 1} = 1 \cdot s^{N + 1} \leq B c (N + 1) s^{N + 1}$ , using $B c (N + 1) \geq 1$ . The second part is the definition with $s = 1$ . ◻

At $s = 1$ the forcing term is $B c (N + 1)$ , so it grows only as fast as the coefficient; for the divisor function this is $O (N^{ε})$ for every $ε > 0$ . This comparison does not itself construct a bounded sequence of remainders. At $s \geq 2$ the same term is at least $2^{N + 1}$ whenever the coefficient is nonzero.

Example 7.3. Take $B = 1$ , $c = τ$ and $N = 9$ , so that the coefficient is $τ (10) = 4$ . At $s = 1$ the forcing term is $4$ . At $s = 2$ it is $4 \cdot 2^{10} = 4096$ , and part (1) of Theorem 7.2 already guarantees at least $2^{10} = 1024$ without knowing the coefficient at all.

A large forcing term cannot be balanced by two small consecutive remainders. For a natural index $N$ , with $B, s$ positive integers, $s \geq 2$ and $c (N + 1) \geq 1$ , the recurrence and triangle inequality give

2^{N + 1} \leq B c (N + 1) s^{N + 1} = | r U_{N} - U_{N + 1} | \leq (1 + | r |) max {| U_{N} |, | U_{N + 1} |} .

If these assumptions hold at every index, the full sequence is unbounded. They do not imply $| U_{N} | \to \infty$ under the stated algebraic hypotheses. For example, take $r = - 2$ , $s = 2$ , $B = 1$ , $c (n) = 1$ and $ξ = 0$ . Directly from the defining partial sums,

U_{2 j} = 0, U_{2 j + 1} = - 2^{2 j + 1} (j \geq 0) .

The forcing term grows exponentially while every even remainder is zero. This is an algebraic example, not a convergent Lambert series: it shows that the adjacent-pair bound does not control every subsequence. It concerns the displayed normalisation only and supplies no small nonzero integer forms.

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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