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

Cancellation during recovery intervals

The LCM quotient $M_{n}$ need not equal the common divisor $G_{n}$ removed by reduction to lowest terms. Since $M_{n}$ divides both $C_{n}$ and $D_{n}$ , we have $M_{n} ∣ G_{n}$ , but equality is not assumed. The following estimate concerns the successive reduction factors, over an interval where the reduced numerator regains its starting value. In the calculation below, $h_{n}$ is the factor removed at step $n$ , $c_{n}$ is a positive integer with $c_{n}^{2} ∣ h_{n}$ , and ${\tilde{e}}_{n}$ denotes the signed reduced error, as in Section 7. For the estimate itself, let $u, h, c : N \to N$ and $\tilde{e} : N \to Z$ satisfy

h_{n} u_{n + 1} = u_{n} - {\tilde{e}}_{n}, u_{n} > 0, h_{n} > 0.

Fix $K, r, L \in N$ with $K, L > 0$ , assume

K | {\tilde{e}}_{r + i} | < u_{r + i} (0 \leq i < L), u_{r} \leq u_{r + L},

and let $R$ be a finite family of moduli. Suppose every $m_{q}$ , $q \in R$ , divides $\prod_{r \leq n < r + L} c_{n}$ , and that $c_{n}^{2} ∣ h_{n}$ throughout the interval. Then

\begin{matrix} (23) & K^{L} lcm (m_{q} : q \in R)^{2} < (K + 1)^{L} . \end{matrix}

To see this, divide the recurrence by $u_{n}$ and multiply over the interval:

(\prod_{r \leq n < r + L} h_{n}) \frac{u_{r + L}}{u_{r}} = \prod_{r \leq n < r + L} (1 - \frac{{\tilde{e}}_{n}}{u_{n}}) < (1 + 1 / K)^{L} .

Recovery makes the endpoint ratio at least $1$ , giving $\prod_{r \leq n < r + L} h_{n} < (1 + 1 / K)^{L}$ . Without $u_{r} \leq u_{r + L}$ , a small endpoint ratio could conceal a large cancellation product. The least common multiple of the $m_{q}$ divides the product of the $c_{n}$ , and $c_{n}^{2} ∣ h_{n}$ at every step. Its square therefore divides, and is at most, the positive integer $\prod_{r \leq n < r + L} h_{n}$ . Multiplying by $K^{L}$ proves the claimed bound. This is the bound on cancellation over a recovery interval.

The square in (23) comes from the assumption $c_{n}^{2} ∣ h_{n}$ . If the least common multiple of the chosen moduli is at least $2^{| R |}$ , then

K^{L} 4^{| R |} < (K + 1)^{L} .

The formal consequence is the bound on the number of independent moduli. Taking logarithms gives the explicit bound

\frac{| R |}{L} < \frac{\log (1 + 1 / K)}{\log 4} \leq \frac{1}{K \log 4} .

Thus the number of these moduli is small relative to the recovery length when the relative-error bound is small. This statement requires the lower bound $2^{| R |}$ on their least common multiple; it does not apply to an arbitrary family with repeated prime factors.

For a fixed number of steps, the same estimate has a simpler consequence: a sufficiently late recovery cannot include any cancellation. If $K | {\tilde{e}}_{n} | < u_{n}$ eventually for every $K$ , then for each fixed $L > 0$ there is an $N$ such that every recovery $u_{r} \leq u_{r + L}$ with $r \geq N$ satisfies

\prod_{0 \leq i < L} h_{r + i} = 1.

Indeed, choose $K$ so large that $(1 + 1 / K)^{L} < 2$ and apply the same product estimate beyond its error threshold. The positive integer product is then smaller than $2$ , so it equals $1$ . The same $K$ works for every length $1 \leq j \leq L$ , since $(1 + 1 / K)^{j} \leq (1 + 1 / K)^{L} < 2$ . Thus, after a sufficiently late step with $h_{n} > 1$ , the numerator cannot regain its starting value at any of the next $L$ indices. This includes a recovery followed by another fall before the last endpoint. The fixed-length conclusion is formalised as absence of late cancellation during recovery intervals of fixed length. The threshold $N$ depends on $L$ : late cancellations require longer recoveries, but have not been excluded. A global contradiction needs control of those varying lengths and a lower bound on the moduli’s least common multiple; the interval estimate supplies neither.

Problem 14.3 (growth of repeated denominator factors). For every rational-tail orbit satisfying the hypotheses but not the conclusion of Problem 1.1, must

\underset{n \to \infty}{lim sup} \frac{\log M_{n}}{n} > 0 ?

Equivalently, must there be a $K \geq 1$ for which

2^{n} \leq M_{n}^{K}

at infinitely many indices?

The two displayed formulations are equivalent up to changing the positive constant; the second is not a weaker target. Section 6 already gives $1 \leq M_{n} \leq C_{n}$ , hence $\log M_{n} / n \to 0$ on every orbit under consideration. A positive answer would therefore be a contradiction, not a further compatible necessary condition. It would prove Problem #243. The missing step is to force exponential growth of $M_{n}$ from failure of the Sylvester recurrence.

There is also a more local-looking question whose content is nevertheless the entire prefix. Put $A_{n} = \prod_{j < n} a_{j}$ , so $D_{n} = D_{0} A_{n}$ . From

D_{0} A_{n} = (a_{n} - 1) C_{n} + E_{n}

one immediately obtains

gcd (A_{n}, a_{n} - 1) ∣ E_{n} .

The checked tail-height estimate and vanishing relative error make $| E_{n} |$ subexponential in $n$ .

Problem 14.4 (common factors of the prefix and $a_{n} - 1$ ). In every rational-tail orbit satisfying the hypotheses but not the conclusion of Problem 1.1, is

\underset{n \to \infty}{lim sup} \frac{\log gcd (A_{n}, a_{n} - 1)}{n} > 0 ?

Equivalently, do there exist $η > 0$ and infinitely many $n$ such that $gcd (A_{n}, a_{n} - 1) \geq e^{η n}$ ?

By Proposition 14.1, the error is nonzero eventually. The displayed divisibility therefore bounds $gcd (A_{n}, a_{n} - 1)$ by $| E_{n} |$ at every sufficiently late index. A positive answer contradicts the subexponential bound on that error. Arbitrarily long locally admissible blocks do not answer this question: the gcd uses the complete prefix.

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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