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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Signs of the error and of the growth ratio

We keep the preceding one-based indexing. The growth defect need not have the opposite sign to $E_{n}$ : the next identity displays the correction, including its sign under the eventual Sylvester recurrence.

Proposition 7.14 (comparison of the two signs). The factor $M_{n} = D_{n} / L_{n}$ divides $G_{n} = gcd (C_{n}, D_{n})$ . Thus $G_{n} / M_{n}$ is a positive integer and $E_{n} / M_{n} = (G_{n} / M_{n}) {\tilde{e}}_{n}$ is an integer with the same sign as $E_{n}$ . The sign of the growth ratio minus one also depends on a correction term. The exact recurrences give, whenever $a_{n + 1} C_{n} C_{n + 1} \neq 0$ ,

\begin{matrix} (22) & \frac{a_{n}^{2}}{a_{n + 1}} - 1 = - \frac{E_{n}}{C_{n}} + Λ_{n}, Λ_{n} = \frac{(1 - E_{n} / C_{n}) (a_{n} - 1 + E_{n + 1} / C_{n + 1})}{a_{n + 1}}, \end{matrix}

and, under the standing positive rational-tail hypotheses, $0 < Λ_{n} < 3 / a_{n}$ for all sufficiently large $n$ . The Erdős–Straus quantity of Theorem 3 is

Z_{n}^{ES} = \frac{[a_{1}, \dots, a_{n}]}{a_{n + 1}} (\frac{a_{n + 1}^{2}}{a_{n + 2}} - 1) .

Its least common multiple includes $a_{n}$ but not the clearing denominator $q$ . It is not $Q_{n} = (P_{n} / a_{n}) γ_{n}$ from the preceding subsection, nor $L_{n} γ_{n + 1} / a_{n + 1}$ under our convention $L_{n} = lcm (q, a_{1}, \dots, a_{n - 1})$ . Being a positive multiple of the next growth defect, $Z_{n}^{ES}$ has the sign of $Λ_{n + 1} - E_{n + 1} / C_{n + 1}$ . For all sufficiently large $n$ , it is positive when $E_{n + 1} \leq 0$ ; for $E_{n + 1} > 0$ , it is negative precisely when $E_{n + 1} / C_{n + 1} > Λ_{n + 1}$ . Under the eventual Sylvester recurrence, $E_{n} = 0$ and $Λ_{n} = (a_{n} - 1) / a_{n + 1} > 0$ .

Proof. Since both $C_{n} = D_{n} x_{n}$ and $L_{n} x_{n}$ are integers, $M_{n} = D_{n} / L_{n}$ divides $C_{n}$ as well as $D_{n}$ . It therefore divides $G_{n}$ . For (22), write $θ_{n} = E_{n} / C_{n}$ ; Proposition 4.1 gives $C_{n + 1} = C_{n} (1 - θ_{n})$ and $D_{n} / C_{n} = a_{n} - 1 + θ_{n}$ , so $a_{n + 1} - 1 + θ_{n + 1} = a_{n} D_{n} / C_{n + 1} = a_{n} (a_{n} - 1 + θ_{n}) / (1 - θ_{n})$ . Multiplying the asserted identity by $a_{n + 1}$ and substituting reduces it to $a_{n + 1} + a_{n} - 1 + θ_{n + 1} = a_{n}^{2} / (1 - θ_{n})$ , which is the displayed relation with $a_{n}$ added to both sides. The bound follows from $θ_{n} \to 0$ and $a_{n + 1} \geq a_{n}^{2} / 2$ . On Sylvester’s sequence $θ_{n} = 0$ and $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ , so $a_{n}^{2} / a_{n + 1} - 1 = (a_{n} - 1) / a_{n + 1} = Λ_{n} > 0$ . ◻

A nonpositive upper limit admits positive values tending to zero; it is not an eventual pointwise sign condition. For the product quantity $Q_{n} = (P_{n} / a_{n}) γ_{n}$ , the identity $E_{n} + q Q_{n} = o (1)$ shows that eventual $E_{n} \geq 0$ implies $lim sup Q_{n} \leq 0$ , as in Koizumi’s Corollary 4(1) [12]. For the classical LCM expression $Z_{n}^{ES}$ , the corresponding integer error is also taken one index later. Its published hypotheses are those of Erdős–Straus Theorem 3 [1]. Reindexing relates it to an LCM-cleared error, not to $E_{n}$ with the unchanged product factor $q Q_{n}$ .

A comparison with $B$ -free integers.

For a fixed family $B = {m_{i}}$ , the integers divisible by none of the $m_{i}$ are called $B$ -free integers. With pairwise coprime moduli and $\sum_{i} 1 / m_{i} < \infty$ , this is the standard setting of [21]. The next proposition is an elementary interval count in that setting. It tests what can follow from avoiding whole multiples alone. Unlike the reciprocal-tail argument, it imposes all the moduli from the outset and has no denominator recurrence or cancellation factors.

Proposition 7.15 (an elementary interval bound). Let $m_{0} < m_{1} < \dots$ be pairwise coprime integers at least $2$ with $θ = \sum_{i} 1 / m_{i} < 1$ . For all integers $x \geq 1$ and $L \geq 1$ satisfying $L > k / (1 - θ)$ , where $k = # {i : m_{i} \leq x + L}$ , the interval $[x, x + L)$ contains an integer divisible by no $m_{i}$ . If also $ℓ (m_{i}) = i + O (1)$ , then for every $ϵ > 0$ there are an index $T$ and a strictly increasing sequence of positive integers $(u_{n})$ such that

m_{i} ∤ u_{n} for every i \geq T and every n, u_{n + 1} - u_{n} \leq (1 + ϵ) ℓ (u_{n}) eventually .

Proof. The interval $[x, x + L)$ contains exactly $L$ integers. Each lies in $[1, \infty)$ and is smaller than $x + L$ , so a modulus exceeding $x + L$ divides none of them. Each of the $k$ remaining moduli divides at most $L / m_{i} + 1$ of them, so the covered count is at most $L θ + k < L$ . This interval count itself does not use pairwise coprimality; that condition is needed for the exact CRT proportions in the later gap theorem.

For the second assertion, choose a tail of the family and reindex it so that $θ < ϵ / (1 + ϵ)$ . If $k (z) = # {i : m_{i} \leq z}$ , the hypothesis $ℓ (m_{i}) = i + O (1)$ gives $k (z) \leq ℓ (z) + C$ for all large $z$ and some constant $C$ . Choose

(1 - θ)^{- 1} < ρ < 1 + ϵ, L (y) = ⌈ ρ ℓ (y) ⌉ .

Then $L (y) = O (ℓ (y)) = o (y)$ , and the definition of $ℓ$ gives $ℓ (y + 1 + L (y)) = ℓ (y) + o (1)$ . Consequently, for all large integers $y$ ,

\frac{k (y + 1 + L (y))}{1 - θ} \leq \frac{ℓ (y) + C + o (1)}{1 - θ} < ρ ℓ (y) \leq L (y),

while $L (y) \leq (1 + ϵ) ℓ (y)$ . The first part also supplies an admissible $u_{0}$ beyond this threshold. Apply it successively with $x = u_{n} + 1$ and length $L (u_{n})$ , and choose $u_{n + 1}$ in the resulting window. The sequence is strictly increasing, avoids every retained modulus, and has the required rise bound. ◻

The small-increment sequence avoids only the retained moduli $m_{i}$ , $i \geq T$ . Discarding the prefix makes their reciprocal sum small. For the full family, Theorem 14.7 instead gives the coefficient $\prod_{i} (1 - 1 / m_{i})^{- 1} > 1$ for the increasing enumeration of integers avoiding all the moduli.

Proposition 7.15 concerns avoidance of whole multiples, not coprimality to a composite modulus. The distinction matters: no integer in $[2, 5)$ is coprime to $30$ , although none is divisible by $30$ . Thus this proposition alone does not disprove a coprime-walk extension of Theorem 11.3. Proposition 14.6 below gives a separate counterexample using sparse prime moduli. Under the additional scale $ℓ (m_{j}) = j + O (1)$ , Theorem 14.7 determines the exact maximal-gap coefficient for these $B$ -free integers. Its proof uses a finite sieve estimate and the Chinese remainder theorem, not an orbit construction. Neither static construction supplies an exact reciprocal-tail orbit.

Conditions on a possible counterexample.

Together, these criteria give necessary conditions on a counterexample. For every fixed $δ \in (0, 1)$ , it must have

(1 - δ) ℓ (C_{n}) \leq - E_{n} = o (C_{n}) at infinitely many n .

It must also have a divergent sum of relative increases, unbounded negative error scaled by the previous maximum, and infinitely many record jumps of the reduced numerator of size at least three, with $h_{n} = 1$ . These are necessary conditions, not a construction of a counterexample. The residue-saturation lemma shows precisely one limitation of the fixed-old-modulus method; it is not an impossibility theorem about every future approach. A useful next target is interaction between new prime support, three consecutive numerators and the global bound on numerator growth.

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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Signs of the error and of the growth ratio

A comparison with BB-free integers.

Conditions on a possible counterexample.

About this paper

A comparison with $B$ -free integers.