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

A factorial modulus for integral recursion

In the case $m = c = 1$ of Section 9, the identity (5.1) becomes $D_{n} + 1 = (a_{n} - 1) (n + 1)$ . It determines the recursion $a_{n + 1} = num (n, a_{n}) / (n + 2)$ , where

num (n, a) = (n + 1) a^{2} - (n + 2) a + (n + 3),

the forced numerator. We stop at the first nonintegral quotient. The formal survival predicate requires exact division at each step. For $a_{0} = 3$ , as in Example 9.1, the first quotient is $num (0, 3) / 2 = 3$ , but the next quotient is $num (1, 3) / 3 = 13 / 3$ , so the recursion stops. Direct iteration can produce very large intermediate integers. To decide whether the first $h$ divisions are exact, however, it suffices to know the initial value modulo $(h + 1)!$ . Computing a pseudo-greedy orbit through residues modulo a shrinking product modulus is Koizumi’s method [12]; for the forced numerator that product is a factorial.

Theorem B.1 (factorial residue reduction). Let $h$ be a nonnegative integer and let $a, b$ be integers with $a \equiv b (\mod (h + 1)!)$ . The first $h$ steps of the recursion $a_{n + 1} = num (n, a_{n}) / (n + 2)$ , starting at index zero, are integral for $a_{0} = a$ if and only if they are integral for $a_{0} = b$ .

Proof. Each exact division consumes a congruence factor. Retain the remaining factors by setting $M (0, i) = 1$ and $M (h + 1, i) = (i + 2) M (h, i + 1)$ . Thus $M (h, 0) = (h + 1)!$ . We prove the stronger statement that, at index $i$ , congruent inputs modulo $M (h, i)$ give the same answer to whether the next $h$ quotients are all integral. There is nothing to prove for $h = 0$ .

For the inductive step, suppose $a \equiv b (\mod (i + 2) M (h, i + 1))$ . Since $num (i, \cdot)$ is an integer polynomial, its values at $a$ and $b$ are congruent modulo that product. In particular, one is divisible by $i + 2$ if and only if the other is. If neither is divisible, both recursions stop. Otherwise their quotients are congruent modulo $M (h, i + 1)$ , so the induction hypothesis applies to the remaining $h$ updates at index $i + 1$ . Taking $i = 0$ proves the claim. ◻

The formal factorial residue reduction uses the shrinking-modulus induction, polynomial congruence, and cancellation after exact division. Its modulus is identified by the ascending-factorial formula and its factorial value at the initial index.

For $h = 1$ , the modulus is $2! = 2$ . The first quotient is integral if and only if $2 ∣ a^{2} - 2 a + 3$ , or equivalently if $a$ is odd. For instance, $num (0, 3) = 6$ and $num (0, 4) = 11$ . Thus the parity of $a$ decides the first step, as Theorem B.1 asserts in this case.

Exact enumeration for $2 \leq a_{0} < 5000$ gives a maximum of $17$ successful updates, or $18$ multiplier values including $a_{0}$ . Nine seeds attain it, the least being $1501$ . The seed $a_{0} = 1$ is excluded: it gives $a_{n} = 1$ at every step and violates the hypothesis $a_{n} \geq 2$ . The test certifies exact division, not a positive orbit. Even a surviving seed for every finite horizon would not supply one integer seed surviving forever. Theorem 9.2 proves the infinite exclusion; the residue reduction also applies to other polynomial-division recurrences.

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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