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

Bounds that allow for cancellation and earlier decreases

Theorem 7.5 (bounds allowing for previous decreases). Under the standing hypotheses, the following are equivalent: eventual Sylvester behaviour; $\underset{n}{lim sup} A_{n} < \infty$ ; $\underset{n}{lim sup} R_{n} δ_{n} < \infty$ . Each of those two limits superior is $0$ or $+ \infty$ .

Corollary 7.6 (the critical rate). Under the standing hypotheses and $δ_{n} = O (1 / n)$ , with no convergence of $n δ_{n}$ assumed, eventual Sylvester behaviour is equivalent to $u_{n} = O (n)$ and to $(- {\tilde{e}}_{n})_{+} = O (1)$ . A counterexample at the critical rate therefore has $\underset{n}{lim sup} u_{n} / n = \infty$ and $\underset{n}{lim sup} (- {\tilde{e}}_{n})_{+} = \infty$ .

Proof of Theorem 7.5. Suppose the orbit is not eventually Sylvester and $A_{n} \leq K$ eventually, for an integer $K \geq 1$ . Absorption and $| {\tilde{e}}_{n} | / u_{n} \to 0$ give $u_{n} \to \infty$ . Negative reduced errors must occur arbitrarily late, since otherwise $h_{n} u_{n + 1} = u_{n} - {\tilde{e}}_{n}$ would make $u_{n}$ eventually nonincreasing.

Cancellation may lower $u_{s + 1}$ without a record. The first later negative error detects the loss: its magnitude is at least one, while $R_{t}$ remembers $u_{s}$ . Fix a late $s$ and let $t > s$ be that first negative-error index. At the intervening steps the reduced numerator is nonincreasing, so $u_{t} \leq u_{s + 1}$ . Since $R_{t} \geq u_{s}$ and $m_{t} = | {\tilde{e}}_{t} | \geq 1$ ,

A_{t} = \frac{R_{t} m_{t}}{u_{t}} \geq \frac{u_{s}}{u_{s + 1}} = \frac{h_{s}}{1 - {\tilde{e}}_{s} / u_{s}} .

Thus $h_{s} \leq 3 K / 2 < 2 K$ after the relative-error threshold. Also $m_{n} \leq A_{n} \leq K$ , since $R_{n} \geq u_{n}$ , and hence $u_{n + 1} \leq u_{n} + K$ .

We can now choose primes too large to be removed by cancellation. The density-one argument in the proof of Theorem 7.3 supplies infinitely many late multipliers $a_{n}$ coprime to $D_{n}$ . These multipliers are pairwise coprime. At each such step, $a_{n}$ is coprime to $v_{n}$ and $gcd (u_{n}, v_{n}) = 1$ , so $a_{n} u_{n} - v_{n}$ is coprime to both $a_{n}$ and $v_{n}$ . Hence $h_{n} = 1$ and $a_{n} ∣ v_{n + 1}$ . Only finitely many of the chosen multipliers can have all their prime divisors at most $2 K$ , since each uses a different prime from that finite set. Choose $K$ distinct primes $p_{0}, \dots, p_{K - 1} > 2 K$ at these steps. Once a chosen prime divides a reduced denominator, it divides every later one: $h_{n} v_{n + 1} = a_{n} v_{n}$ and $h_{n} < 2 K < p_{i}$ prevent its removal. All the primes therefore divide $v_{T}$ at a common later index $T$ .

Consequently every $u_{n}$ with $n \geq T$ is coprime to all the $p_{i}$ . Choose a CRT block of $K$ consecutive integers, one divisible by each $p_{i}$ , above $R_{T}$ . Divergence forces a first crossing, while $u_{n + 1} \leq u_{n} + K$ forces it to land inside the block, a contradiction. This proves that finite $lim sup A_{n}$ forces the eventual Sylvester recurrence.

The comparison $| δ_{n} - m_{n} / u_{n} | \leq 3 / a_{n}$ gives $| R_{n} δ_{n} - A_{n} | \leq 3 R_{n} / a_{n} \to 0$ . Here $R_{n} \leq H_{n}$ , $\log H_{n} = o (n)$ and $a_{n}$ grows doubly exponentially. Thus the two boundedness conditions are equivalent. Under the eventual Sylvester recurrence, $u_{n} = 1$ and $m_{n} = 0$ eventually, so $A_{n} \to 0$ and $R_{n} δ_{n} \to 0$ . Since both quantities are nonnegative, their limits superior can only be $0$ or $+ \infty$ . ◻

The estimate at the first later negative error is the lemma in the release, negative error scaled by the previous maximum after cancellation: at the first later negative error, $u_{t} \leq u_{s + 1}$ and $u_{s} u_{t} \leq R_{t} | {\tilde{e}}_{t} | u_{s + 1}$ . The exact example $15 / 134 = 1 / 10 + 1 / 85 + 1 / 5695$ , with steps in reduced fractions $(15, 134) \to (4, 335) \to (1, 5695)$ and $A_{1} = 15 / 4$ , attains equality in the displayed bound.

For Corollary 7.6, the comparison $| δ_{n} - m_{n} / u_{n} | \leq 3 / a_{n}$ and the rate $δ_{n} = O (1 / n)$ give $m_{n} / u_{n} = O (1 / n)$ . If $u_{n} = O (n)$ , then $m_{n} = O (1)$ . Conversely, $m_{n} = O (1)$ gives $u_{n + 1} \leq u_{n} + m_{n}$ , hence $u_{n} = O (n)$ and $R_{n} = O (n)$ . In either case $A_{n} = R_{n} m_{n} / u_{n} = O (1)$ , so Theorem 7.5 gives eventual Sylvester behaviour. The eventual Sylvester recurrence gives $u_{n} = 1$ and $m_{n} = 0$ eventually, which proves the converse implications.

The critical-rate assumption is doing real work: a bound on $(- {\tilde{e}}_{n})_{+}$ alone does not directly control $R_{n} / u_{n}$ after a decrease. The preceding corollary supplies the missing control from $δ_{n} = O (1 / n)$ . No such bound is derived here from the unrestricted quadratic limit.

Lemma 7.7 (large odd prime powers in the reduced denominator). Under the standing hypotheses, for every fixed $A > 0$ and every sufficiently large $n$ , the reduced denominator $v_{n}$ has an odd prime-power divisor $Q = p^{k}$ with

Q > (H_{n} + 2)^{A}, H_{n} = max_{j \leq n} C_{j} .

The prime $p$ may depend on $n$ ; no stable-gcd or prime-arrival assumption is imposed.

Proof. The denominator grows too fast for all its odd prime-power factors to remain small compared with $H_{n}$ . Indeed, the estimate for the rational tail gives $C_{n + 1} / C_{n} \to 1$ , hence $\log H_{n} = o (n)$ . Since $x_{n} = u_{n} / v_{n} \leq 2 / a_{n}$ and $u_{n} \geq 1$ , $v_{n} \geq a_{n} / 2$ . Also $v_{n} ∣ L_{n} = lcm (q, a_{0}, \dots, a_{n - 1})$ , so the $2$ -primary part of $v_{n}$ is at most $max (q, a_{n - 1}) = a_{n - 1}$ for all large $n$ . Its odd part $W_{n}$ therefore satisfies

W_{n} \geq \frac{a_{n}}{2 a_{n - 1}} \geq \frac{a_{n - 1}}{4}, \log W_{n} \geq c 2^{n - 1} - O (1)

for some $c > 0$ . Put $B_{n} = (H_{n} + 2)^{A} = \exp (o (n))$ . If every exact odd prime-power factor of $W_{n}$ were at most $B_{n}$ , then $W_{n} ∣ lcm (1, \dots, ⌊ B_{n} ⌋)$ , giving

\log W_{n} \leq B_{n} \log B_{n} = \exp (o (n)) .

This contradicts the preceding exponential lower bound for all sufficiently large $n$ . An offending exact prime-power factor is the required $Q$ . ◻

This uses only an elementary factorial bound, not prime distribution. The prime may change with $n$ : choose one large prime power at a late start and protect it until the first forbidden crossing.

Theorem 7.8 (unit record increments). Under the standing hypotheses, if $R_{n + 1} - R_{n} \leq 1$ for all large $n$ , then $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ for all large $n$ . Hence the sequence is eventually Sylvester if and only if $# {n : R_{n + 1} - R_{n} \geq 2}$ is finite. No hypothesis is placed on decreases in $u_{n}$ , on record-setting jumps, or on the cancellation factors $h_{n}$ .

Proof. Suppose the orbit is not eventually Sylvester. Absorption gives ${\tilde{e}}_{n} \neq 0$ on a late tail, so vanishing relative error and integrality give $u_{n} \to \infty$ . Choose $s$ so large that the unit-increment bound and $| {\tilde{e}}_{n} | < u_{n} / 2$ hold for every $n \geq s$ , and that Lemma 7.7 applies with $A = 3$ . Since $H_{s} \geq R_{s} \geq 1$ , it supplies an odd $Q = p^{k} ∣ v_{s}$ with $Q > (H_{s} + 2)^{3} > 4 (R_{s} + 2)$ and $Q \geq 16$ . Let $Y$ be the least multiple of $p$ above $R_{s}$ . Then $Y \leq R_{s} + p < Q / 4 + p \leq 5 Q / 4$ . At the first $t > s$ with $u_{t} \geq Y$ , the integer maximum rises from at most $Y - 1$ by at most one, forcing $u_{t} = Y$ regardless of $u_{t - 1}$ . Before $t$ , one has $w_{n} < 3 u_{n} / 2 < 15 Q / 8 < p Q = p^{k + 1}$ . Corollary 7.2 therefore gives $p^{k} ∣ v_{t}$ . But $p ∣ u_{t} = Y$ , contradicting $gcd (u_{t}, v_{t}) = 1$ . The converse follows because the eventual Sylvester recurrence gives $u_{n} = 1$ eventually. ◻

The plectis-erdos-lean repository contains the unit-increment landing lemma and the one-unit record-increment implication; they assume the existence of the prime power that Lemma 7.7 supplies.

The two-unit criterion discussed below bounds the actual jump at each record step, rather than the increase in the running maximum. As numerical bounds on individual steps, neither condition implies the other: the first allows $s_{n} = 2$ , while $s_{n} \leq 1$ permits large jumps after an earlier decrease. This is a comparison of the bounds, not a claim that they remain logically independent under all the standing tail hypotheses; there both criteria force eventual Sylvester behaviour. The finite example $(8, 177) \to (7, 4071) \to (10, 2373393)$ with multipliers $23$ and $583$ has records $8$ and $10$ , increment $2$ , jump $3$ and $gcd (8, 10) = 2$ , which is why the parity argument for crossing an odd level does not transfer from actual jumps to increments of the running maximum.

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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