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

Unbounded upward increments

Proposition 14.6 (small increases when the prime moduli are sparse). There exist strictly increasing primes $p_{i}$ and a strictly increasing positive integer sequence $u_{n} \to \infty$ such that $gcd (u_{n}, p_{i}) = 1$ for all $i, n$ and

u_{n + 1} - u_{n} = O (\sqrt{\log \log (u_{n} + e^{e})}) = o (\log \log (u_{n} + 3)) .

Thus the bounded-increment hypothesis of Theorem 11.3 cannot be replaced by an $o (\log \log u_{n})$ bound without a quantitative restriction on the moduli.

Proof. Choose increasing primes $p_{i} > max {\exp (\exp ((i + 2)^{2})), 2^{i + 3}}$ , for $i \geq 0$ . Primes of arbitrarily large size suffice; no distribution theorem is used. Then $θ = \sum_{i} 1 / p_{i} < 1 / 4$ and $k (z) = # {i : p_{i} \leq z} \leq \sqrt{\log \log z}$ whenever $z$ is large. For integer $x$ put $L (x) = ⌈ 4 \sqrt{\log \log (x + e^{e})} + 8 ⌉$ . Since $L (x) = o (x)$ , eventually $L (x) > k (x + L (x)) / (1 - θ)$ . The elementary window bound of Proposition 7.15 leaves an integer in $[x, x + L (x))$ divisible by no $p_{i}$ . The set of such integers is therefore unbounded. Enumerate it increasingly as $(u_{n})$ and apply the same bound with $x = u_{n} + 1$ to obtain $u_{n + 1} - u_{n} \leq L (u_{n} + 1)$ . For prime moduli nondivisibility is exactly coprimality, proving every claim. Sparse moduli defeat the abstract coprimality argument, not the exact tail recurrences. The next theorem instead fixes the moduli’s double-exponential scale. ◻

We now keep the full family of moduli, rather than discarding a prefix as in Proposition 7.15. The resulting coefficient depends on $σ = \prod_{j} (1 - 1 / m_{j})$ . For a finite prefix, the corresponding product is exactly the proportion of admissible residue classes by the Chinese remainder theorem. This explains the constant in the statement before the limiting argument is made.

Theorem 14.7 (largest gaps between integers avoiding given multiples). Let $m_{0} < m_{1} < \dots$ be pairwise coprime integers at least $2$ with $ℓ (m_{j}) = j + O (1)$ , where $ℓ (x) = \log_{2} \log_{2} max (4, x)$ . Let $σ = \prod_{j} (1 - 1 / m_{j}) > 0$ and enumerate the positive integers divisible by no $m_{j}$ in increasing order as $(u_{n})$ . Then

\underset{n \to \infty}{lim sup} \frac{u_{n + 1} - u_{n}}{ℓ (u_{n})} = σ^{- 1} .

This is a statement about avoidance of multiples of whole moduli. It is not a statement about coprimality to composite $m_{j}$ , or about integer tail orbits.

Proof. Write $k (z) = # {j : m_{j} \leq z} = ℓ (z) + O (1)$ . For a fixed prefix length $T$ , set

M_{T} = \prod_{j < T} m_{j}, σ_{T} = \prod_{j < T} (1 - 1 / m_{j}), θ_{T} = \sum_{j \geq T} 1 / m_{j} .

Here $θ_{T} \to 0$ . By the Chinese remainder theorem, avoiding the first $T$ whole moduli selects precisely the fraction $σ_{T}$ of the residues modulo $M_{T}$ .

For the upper bound, fix $T$ before sending the interval to infinity, so the prefix error $M_{T}$ stays constant. An integer interval $[x, x + L)$ contains at least $σ_{T} L - M_{T}$ integers avoiding that prefix. The remaining moduli cover at most $L θ_{T} + k (x + L)$ integers. Choose $T$ large enough that $σ_{T} > θ_{T}$ , and then any $c > (σ_{T} - θ_{T})^{- 1}$ . For $L = ⌈ c ℓ (x) ⌉$ , the number left is positive for all sufficiently large $x$ , since $k (x + L) = ℓ (x) + O (1)$ . Thus the set being enumerated is unbounded, and every sufficiently late such interval meets it. Applying this at $x = u_{n} + 1$ gives

\underset{n}{lim sup} \frac{u_{n + 1} - u_{n}}{ℓ (u_{n})} \leq (σ_{T} - θ_{T})^{- 1} .

Letting $T \to \infty$ gives the upper bound $σ^{- 1}$ .

For the lower bound, fix $T$ and let the integer length $L$ tend to infinity. Among the offsets $0 \leq j < L$ , precisely $K_{L} = σ_{T} L + O (M_{T})$ avoid the first $T$ moduli. Assign to these offsets distinct moduli $m_{T}, \dots, m_{T + K_{L} - 1}$ . Solve simultaneously $x \equiv 0 (\mod M_{T})$ and $x \equiv - j$ modulo the modulus assigned to $j$ . Put $Q_{L} = \prod_{i < T + K_{L}} m_{i}$ and choose this solution in $[Q_{L}, 2 Q_{L})$ . Every integer of $[x, x + L)$ is then divisible by a modulus: either an old one or its assigned new one. Let $u$ be the greatest admissible integer below $x$ ; its successor is at least $x + L$ , so the gap is at least $L$ . The scale assumption gives

ℓ (2 Q_{L}) \leq T + K_{L} + O (1),

because $\log_{2} m_{i} = 2^{i + O (1)}$ and these upper bounds sum geometrically. Since $u < 2 Q_{L}$ , the ratio for this gap is at least $L / (T + K_{L} + O (1))$ , tending to $σ_{T}^{- 1}$ . These $u$ tend to infinity: $x \geq Q_{L} \to \infty$ and admissible integers are unbounded. Hence the limsup is at least $σ_{T}^{- 1}$ for every $T$ . Let $T \to \infty$ to finish. ◻

For the Fermat moduli $m_{j} = 2^{2^{j}} + 1$ , pairwise coprimality and the finite-product identity

\prod_{j = 0}^{N} (1 - \frac{1}{2^{2^{j}} + 1}) = \frac{2^{2^{N + 1} - 1}}{2^{2^{N + 1}} - 1}

give $σ = 1 / 2$ , so the coefficient is exactly $2$ . These results describe gaps between integers avoiding fixed moduli. They do not construct multipliers or numerator and denominator sequences satisfying the exact recurrences. In particular, the static proportion $σ$ must not be substituted for $φ (v_{T}) / v_{T}$ : the first excludes multiples of the whole moduli, while the second excludes every prime factor of the reduced denominator.

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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