Reciprocal Sums and the Sylvester Recurrence

Will Cook

Short paper

Reciprocal Sums and the Sylvester Recurrence

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

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Summary. For strictly increasing positive integers with $a_{n}^{2} / a_{n + 1} = 1 + 3 / n + o (n^{- 3})$ , the reciprocal sum is irrational. The argument extracts a cubic polynomial from an integer tail numerator, restricts its coefficients using a square in a cubic field, and excludes the remaining cases modulo seven. Under rationality and the quadratic growth limit, separate criteria using bounded numerator increases and weighted first crossings force the eventual Sylvester recurrence. The unrestricted Erdős–Graham question remains open.

This paper covers cubic-rate irrationality, nonintegral regular-rate exclusion and sufficient tail criteria for the Sylvester recurrence; the unrestricted Erdős–Graham question remains open.

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

A constant-negative error $E_{n} = - m$ with $C_{0} = c$ gives $C_{n} = c + n m$ and the identity

\begin{matrix} (14) & D_{n} + m = (a_{n} - 1) C_{n} . \end{matrix}

In the case $m = c = 1$ , where (14) reads $D_{n} + 1 = (a_{n} - 1) (n + 1)$ , each multiplier is determined by its predecessor. The recursion is $a_{n + 1} = num (n, a_{n}) / (n + 2)$ , where

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

the forced numerator. We stop at the first nonintegral quotient. This procedure checks divisibility only; it does not impose the positivity conditions required of an infinite reciprocal-tail sequence. The formal exact-division condition requires exact division at each step.

Example A.1 (a nonintegral second step). Starting from $a_{0} = 3$ , the first numerator $num (0, 3) = 6$ is divisible by $2$ , giving $a_{1} = 3$ . The next numerator $num (1, 3) = 13$ is not divisible by $3$ , 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 [10]; for the forced numerator that product is a factorial.

Theorem A.2 (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. Division by $i + 2$ consumes that factor of the modulus; no modular inverse is required. Define $M (0, i) = 1$ and $M (h + 1, i) = (i + 2) M (h, i + 1)$ . Induct on $h$ , allowing any starting index $i$ : inputs congruent modulo $M (h, i)$ either both complete $h$ integral steps or neither does. The case $h = 0$ is immediate.

If $a \equiv b (\mod (i + 2) M (h, i + 1))$ , the integer polynomial $num (i, \cdot)$ preserves this congruence. Its values are therefore either both divisible by $i + 2$ or neither is. In the latter case both recursions stop. In the former, exact division leaves the quotients congruent modulo $M (h, i + 1)$ , and induction supplies the remaining $h$ steps. Finally $M (h, 0) = (h + 1)!$ . ◻

For $h = 1$ , the condition is $2 ∣ a^{2} - 2 a + 3$ , or equivalently that $a$ is odd. This illustrates Theorem A.2 with modulus $2! = 2$ .

The enumeration in Example 9.1 of the companion, over $2 \leq a_{0} < 5000$ , gives at most $17$ successful updates ( $18$ values including $a_{0}$ ). The seed $a_{0} = 1$ is excluded because it is fixed and violates $a_{n} \geq 2$ . This finite calculation establishes only the stated range. The infinite constant-negative exclusion requires the separate argument under the hypotheses given there.

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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