Reciprocal Sums and the Sylvester Recurrence: Further Results and Proofs

Will Cook

Long paper

Reciprocal Sums and the Sylvester Recurrence: Further Results and Proofs

Erdős #243 By Will Cook September 2026 Authorship, AI use and citation

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Summary. Integer finite differences exclude nonintegral regular rates and reduce the cubic rate to a polynomial. A square condition in the cubic field and congruences modulo seven then prove irrationality. For rational reciprocal sums with $a_{n + 1} / a_{n}^{2}$ tending to one, the record derives sufficient Sylvester-recurrence criteria from persistent divisors, bounded numerator increases and weighted crossings of new maxima. The arithmetic estimates needed to apply these criteria to every rational reciprocal sum remain open.

This paper covers cubic and nonintegral regular-rate irrationality, integer-tail arithmetic and sufficient Sylvester-recurrence criteria, with their separate hypotheses and remaining estimates.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #243, which remains open.

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In this paper

The two remaining cubics fail modulo seven

Lemma 2.10 (two incompatible four-term residue patterns). For $Q_{12, 1}$ , every sufficiently late four-index window beginning at $n \equiv 0 (\mod 7)$ contains an exceptional index. For $Q_{12, - 1}$ , the same holds for windows beginning at $n \equiv 1 (\mod 7)$ . In either case, $\underset{―}{d} (S) \geq 1 / 7$ . The specified starting residue classes are part of the assertion; no claim is made for every block of four consecutive indices.

Proof. Reduce modulo $7$ . For $c = 1$ the values of $Q_{12, 1}$ at $n = 0, 1, 2, 3$ are $1, 13, 49, 121$ , that is

\begin{matrix} (14) & (1, 6, 0, 2), \end{matrix}

and for $c = - 1$ the values at $n = 1, 2, 3, 4$ are $11, 47, 119, 239$ , that is

\begin{matrix} (15) & (4, 5, 0, 1) . \end{matrix}

Each four-term pattern recurs with period $7$ .

Index a proposed occurrence locally by $0, 1, 2, 3$ , and use $a_{i}$ for the multiplier at local index $i$ . Let $(c_{0}, c_{1}, 0, c_{3})$ be the residues of $u$ and $(d_{0}, d_{1}, d_{2}, d_{3})$ those of $v$ . Thus $c_{i + 1} = a_{i} c_{i} - d_{i}$ and $d_{i + 1} = a_{i} d_{i}$ modulo $7$ . The middle zero gives $a_{1} c_{1} = d_{1}$ , then $d_{2} = a_{1} d_{1}$ and $c_{3} = - d_{2}$ , so

\begin{matrix} (16) & d_{1}^{2} = c_{1} a_{1} d_{1} = c_{1} d_{2} = - c_{1} c_{3} . \end{matrix}

For (14) this is $- 6 \cdot 2 = 2$ , and for (15) it is $- 5 \cdot 1 = 2$ . The square roots of $2$ modulo $7$ are $3$ and $4$ , so $d_{1} \in {3, 4}$ in both cases.

The first step gives $a_{0} = (c_{1} + d_{0}) / c_{0}$ and hence $d_{1} = d_{0} (d_{0} + c_{1}) / c_{0}$ , which is legitimate because $c_{0} \in {1, 4}$ is invertible. For (14) this is $d_{1} = d_{0} (d_{0} + 6)$ , with values

(0, 0, 2, 6, 5, 6, 2) at d_{0} = 0, 1, \dots, 6,

and for (15) it is $d_{1} = 2 d_{0} (d_{0} + 5)$ , with values

(0, 5, 0, 6, 2, 2, 6) .

The incompatibility also follows without enumerating $d_{0}$ . Completing the square gives

d_{1} + 2 = (d_{0} + 3)^{2} (c = 1), d_{1} + 2 = 2 (d_{0} - 1)^{2} (c = - 1) .

Since $2$ is a square modulo $7$ , either update makes $d_{1} + 2$ a square. The required values $d_{1} = 3, 4$ instead give the nonsquares $5, 6$ . The two displayed lists provide the same exhaustive check: both images are ${0, 2, 5, 6}$ , disjoint from ${3, 4}$ .

Hence every sufficiently late window $[7 k, 7 k + 3]$ for the plus profile, and every sufficiently late window $[1 + 7 k, 4 + 7 k]$ for the minus profile, contains an element of $S$ . Windows within each family are pairwise disjoint. There are $X / 7 + O (1)$ such windows contained in $[0, X]$ , so choosing one exceptional index in each gives $# (S \cap [0, X]) \geq X / 7 - O (1)$ and therefore $\underset{―}{d} (S) \geq 1 / 7$ . This stronger local bound is not promoted to a uniform lower bound for arbitrary cubic profiles. ◻

Proof of Theorem 2.3. Assume $\underset{―}{d} (S) = 0$ . Lemmas 2.5 and 2.6 put the primitive tail in the shape (3) with $c = \pm 1$ and $Q_{m, c}$ irreducible. Each excluded branch already contradicts the assumption by producing a positive, branch-dependent lower density, namely $1 / (12 ℓ)$ or $1 / ℓ$ for the relevant prime $ℓ$ . Proposition 2.8 and Lemma 2.9 force $m = 12$ , and Lemma 2.10 then gives $\underset{―}{d} (S) \geq 1 / 7$ , the final contradiction. ◻

The two densities play different roles. If the field element is not a square, Chebotarev supplies one witnessing prime; its periodic obstruction gives a positive density of disagreement indices. No lower bound uniform over cubic profiles is needed.

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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