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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An eventual cubic polynomial

We use $F_{n} = n (n + 1) (n + 2)$ rather than $n^{3}$ : its successive ratio is exactly $1 + 3 / n$ , so dividing by it removes the model without introducing lower-order errors. Write (3) as $C_{n + 1} = (1 + 3 / n + ε_{n}) C_{n}$ , where $ε_{n} = o (n^{- 3})$ . Dividing out the exact model gives

z_{n} = \frac{C_{n}}{F_{n}}, \frac{z_{n + 1}}{z_{n}} = 1 + \frac{ε_{n}}{1 + 3 / n} .

The logarithmic increments are $o (n^{- 3})$ , so their absolute tails are $o (n^{- 2})$ . Hence $z_{n} \to A > 0$ with $z_{n} - A = o (n^{- 2})$ . Multiplying by $F_{n}$ gives

C_{n} = A F_{n} + δ_{n}, δ_{n} = o (n) .

An $o (n)$ error can still oscillate too much for finite differences. To control its first difference, subtract the numerator update for $C_{n}$ and the comparison update for $A F_{n}$ :

Δ δ_{n} = \frac{3}{n} δ_{n} + ε_{n} C_{n} = o (1), Δ h_{n} = h_{n + 1} - h_{n} .

Here $C_{n} = O (n^{3})$ makes the second term $o (1)$ . Three more differences still tend to zero and kill the cubic model:

Δ^{4} C_{n} = Δ^{3} (Δ δ_{n}) ⟶ 0.

Integrality turns this limit into eventual vanishing. Newton interpolation then expresses $C_{n}$ as a rational polynomial of degree at most three. Since its difference from $A F_{n}$ is $o (n)$ , the cubic, quadratic and linear coefficients agree; only a constant can remain:

\begin{matrix} (4) & C_{n} = A n (n + 1) (n + 2) + B, A \in Q_{> 0}, B \in Q . \end{matrix}

The precision matters at the subtraction step. The integer sequence $F_{n} + (- 1)^{n}$ has ratio $1 + 3 / n + O (n^{- 3})$ and never agrees eventually with a polynomial; its error has an oscillating first difference. This example concerns polynomial extraction alone. It is not asserted to satisfy the denominator recurrence.

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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