Plectis

Short paper

Reciprocal Sums and the Sylvester Recurrence

Summary. For strictly increasing positive integers with an2/an+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.

In this paper

Finite total relative increase

Summable relative increases first bound Cn by an infinite product, then allow only finitely many rises; integer descent finishes. No denominator recurrence is needed. Relative increases tending merely to zero still allow Cn=n+1 or Cn=(n+1)2.

Theorem 5.1 (a convergent sum of relative increases). Let Cn be positive integers and En integers satisfying Cn+1=Cn−En. If

∑n(−En)+Cn<∞,

then En=0 eventually. Neither denominator dynamics nor vanishing relative error is required.

Proof. Set δn=(−En)+/Cn. Then

CN≤C0∏n<N(1+δn)≤C0exp(∑nδn).

Choose an integer K above this bound. A strict rise of the integer sequence has size at least one, so it contributes at least 1/K to ∑δn. Hence only finitely many rises occur. Thereafter Cn is a nonincreasing positive integer sequence, and its eventual constancy gives En=0. ◻

The proof also allows positive real Cn with integer En: the values lie in C0+Z, whose bounded positive part is finite. Discrete increments are essential to this argument. With real errors, Cn=1+1/(n+1) decreases forever with zero relative-increase sum; without positivity, Cn=−n−1, En=1 does so too. The companion’s Section 13 gives the details.

On an exact reciprocal-tail orbit the conclusion gives the Sylvester recurrence. For the gap sequence of the pseudo-greedy expansion, the same criterion, with the same product bound and integer descent, appears in the Erdős Problem a Day working report on Problem #243, dated 12 August 2026 [8]; the statement above is for any positive integer sequence with Cn+1=Cn−En. This summability condition and the bound in Theorem 4.1 are different estimates. For exact reciprocal tails with |En|/Cn→0, each is equivalent to eventual vanishing of En; neither estimate has been derived here from the unrestricted problem.

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.

Cite and contact. Cite this paper by its title, author and date above, with its PDF; cite the earlier sources it uses for a mathematical result. For software, use the release citation and give the commit used. Contact Will with questions or corrections.

Prefer the manuscript?

Equations typeset from the exact TeX. Rendered from the LaTeX manuscript source with file fingerprint sha256:b393a9190d51137f. Its file fingerprint matched the published source record, so the PDF above, the LaTeX, and this page are one manuscript. Redeploying the site regenerates this page from whatever the public repository holds at that moment.