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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The contradiction modulo seven

The three-term square condition has reduced the possibilities to two cubics. We now use one more term: the denominator at the middle zero must also come from the preceding exact update. In the plus case, choose a late block starting at $n \equiv 0 (\mod 7)$ ; in the minus case, start at $n \equiv 1 (\mod 7)$ . The four successive numerators have residues

(u_{n}, u_{n + 1}, u_{n + 2}, u_{n + 3}) \equiv {\begin{cases} (1, 6, 0, 2) & (c = 1), \\ (4, 5, 0, 1) & (c = - 1) . \end{cases}

Write $d_{0}, d_{1}$ for the residues of $v_{n}, v_{n + 1}$ and $c_{0}, c_{1}, 0, c_{3}$ for the displayed numerator residues. From $0 = a_{n + 1} c_{1} - d_{1}$ and $c_{3} = - a_{n + 1} d_{1}$ we get $d_{1}^{2} = - c_{1} c_{3} = 2$ . Thus $d_{1} \in {3, 4}$ . The first update, however, gives

d_{1} = a_{n} d_{0} = \frac{d_{0} (d_{0} + c_{1})}{c_{0}} .

Completing the square in these two cases gives, in $F_{7}$ ,

d_{1} + 2 = {\begin{cases} (d_{0} + 3)^{2} & (c = 1), \\ 2 (d_{0} - 1)^{2} & (c = - 1) . \end{cases}

Since $2 = 3^{2}$ in this field, $d_{1} + 2$ must be a square in either case. But $d_{1} \in {3, 4}$ would give $d_{1} + 2 \in {5, 6}$ , both nonsquares. The middle three terms admit $d_{1} = 3$ or $4$ ; neither extends backwards through the preceding exact update. Both cubics are excluded, proving Theorem 1.2.

Section 2 of the companion proves positive lower density of disagreement with each rational cubic of this form. That lower bound depends on the orbit and the cubic. For Theorem 1.2, exclusion of eventual equality suffices.

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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