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

PDFOpen the PDF, 61 pages TeXLaTeX source Source on GitHub Problem page

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.

Start here, by the editorial route

In this paper

Periodic negative errors

Let the negative error have periodic magnitude $e_{n} = - E_{n} > 0$ , so $e_{n + h} = e_{n}$ and $C_{n + 1} = C_{n} + e_{n}$ . Summing over a period gives $C_{n + h} = C_{n} + M$ , where the positive integer $M = \sum_{j = 0}^{h - 1} e_{j}$ is the drift. Periodicity makes this sum independent of the starting index.

At $h = 1$ this says that the magnitude is constant and that $M$ is its value, which is the situation of Section 9; Theorem 9.2 already excludes it without using the pointwise bound $e_{n} < a_{n}$ in the proof below. The new case is $h \geq 2$ .

For a constant error, the proof used the finitely many prime divisors of its magnitude. For a periodic error we use the prime divisors of the increase $M$ over one period. The required divisibility facts are as follows. Every multiplier $a_{j}$ divides each later denominator, by persistence of a multiplier. If a prime divides both $D_{n}$ and $a_{n}$ , the numerator update makes it divide $C_{n + 1}$ , the one-step divisibility implication. Once it divides both numerator and denominator, it divides every later numerator, denominator and error, by persistence of a common divisor.

Periodicity carries this divisibility back to every phase. Suppose $d ∣ M$ and $d ∣ a_{i}, a_{j}$ with $i < j$ . Then $d ∣ D_{j}$ , and $C_{j + 1} = a_{j} C_{j} - D_{j}$ gives $d ∣ C_{j + 1}$ . The common divisor persists in both sequences and in every later error. For any phase $r$ , choose $k$ with $r + k h \geq j + 1$ ; periodicity gives $d ∣ e_{r + k h} = e_{r}$ . Taking $k h \geq j + 1$ also gives $d ∣ C_{k h} = C_{0} + k M$ , hence $d ∣ C_{0}$ . This is the repeated-divisor implication; it uses the period relation and the increase over successive periods, each iterated through whole periods. A repeated prime can therefore be divided out of the entire system. This explains the induction on $M$ in the proof.

Theorem 10.1 (no periodic negative magnitude). Let $a, D, C, e : N \to N$ with $a_{n} \geq 2$ , $e_{n} > 0$ and $e_{n} < a_{n}$ for every $n$ , satisfying

D_{n + 1} = a_{n} D_{n}, C_{n + 1} = C_{n} + e_{n}, D_{n} + e_{n} = (a_{n} - 1) C_{n},

and suppose $e_{n + h} = e_{n}$ and $C_{n + h} = C_{n} + M$ for some $h > 0$ and $M > 0$ . This is impossible.

Proof. We use strong induction on the drift $M$ . A prime divisor of $M$ that recurs among the multipliers need not give an immediate contradiction. It instead divides the whole orbit, allowing us to reduce $M$ by that prime and apply the induction hypothesis.

Suppose first that no prime divisor of $M$ is a common divisor of $C_{0}$ and of every magnitude. The repeated-divisor implication applies in the contrapositive: a prime divisor of $M$ occurring in two of the multipliers would be exactly such a common divisor, so each prime divisor of $M$ occurs in at most one multiplier. On the other hand, every multiplier shares a prime with $M$ . Indeed, if $gcd (a_{j}, M) = 1$ , choose $k \geq 1$ so that $a_{j} ∣ C_{j} + k M = C_{j + k h}$ . The same multiplier divides $D_{j + k h}$ , and periodicity gives $e_{j + k h} = e_{j}$ . The identity at $j + k h$ therefore implies $a_{j} ∣ e_{j}$ , contrary to $0 < e_{j} < a_{j}$ . Thus infinitely many multipliers would require distinct prime divisors of the fixed integer $M$ , a contradiction.

Otherwise some prime $p$ divides $M$ , divides $C_{0}$ , and divides every magnitude. Then $p ∣ C_{n}$ for every $n$ , since $C_{n + 1} = C_{n} + e_{n}$ and both summands on the right are divisible by $p$ , and the same identity $D_{n} + e_{n} = (a_{n} - 1) C_{n}$ then gives $p ∣ D_{n}$ for every $n$ . Divide $D$ , $C$ and $e$ by $p$ . The multipliers are untouched, so $a_{n} \geq 2$ and $e_{n} < a_{n}$ persist and the magnitudes stay positive; the three recurrences are homogeneous in $(D, C, e)$ and so survive; the period is still $h$ ; and the drift becomes $M / p < M$ . The inductive hypothesis applies. ◻

The first of the two cases above is the exclusion when no prime of $M$ divides all phases, the induction is the exclusion at every scale, and the eventual form is the eventual exclusion.

The bound $e_{n} < a_{n}$ rules out $a_{j} ∣ e_{j}$ . It follows eventually, without rescaling: on an infinite orbit with periodic positive magnitudes, bounded errors and exponential denominator growth suffice. The identity gives $C_{n} > 0$ , and periodicity gives

C_{n} = \frac{M}{h} n + O (1), sup_{n} e_{n} < \infty,

and $D_{0}$ cannot be zero: otherwise all $D_{n}$ vanish and the defining identity gives $e_{n} = (a_{n} - 1) C_{n} \geq C_{n}$ , contradicting these bounds. Thus $D_{n} \geq 2^{n} D_{0}$ , and

a_{n} - 1 = \frac{D_{n} + e_{n}}{C_{n}} ⟶ \infty .

Consequently $e_{n} < a_{n}$ eventually. Applying the eventual exclusion therefore rules out periodic positive magnitudes without an independent smallness assumption. The numbered theorem retains the pointwise bound used by its checked phase-by-phase proof. The later bounded-negative theorem also excludes this case.

Remark. For the tail numerators of [12], the absolute error satisfies $| E_{n} | < a_{n}$ eventually. Indeed, under the dictionary in Section 3, Lemma 4(2) gives $- C_{n} / 2 \leq E_{n} < C_{n} / 2$ , hence $| E_{n} | \leq C_{n} / 2$ . The proof of Lemma 4(3) gives $C_{n + 1} \leq \frac{3}{2} C_{n}$ [12], hence $C_{n} \leq C_{N} (3 / 2)^{n - N}$ beyond a pseudo-greedy starting index $N$ . On the other hand, convergence of $\sum 1 / a_{n}$ gives $a_{n} \to \infty$ , and $a_{n + 1} / a_{n}^{2} \to 1$ then gives $a_{n + 1} \geq 2 a_{n}$ eventually. Thus $| E_{n} | / a_{n} \to 0$ . On a negative-error tail this gives $e_{n} < a_{n}$ .

For integer solutions of (19), the same conclusion follows when $a_{n} > 1$ , $C_{n} > 0$ and $E_{n} / C_{n} \to 0$ hold: the calculation in Section 4 already derives the reciprocal series and its growth from these hypotheses. Summability is not an independent assumption in that setting. The bound $e_{n} < a_{n}$ is still only eventual, whereas Theorem 10.1 states it at every index; its eventual form, cited above, is therefore the applicable one. For a periodic magnitude, boundedness of $e_{n}$ and divergence of $a_{n}$ already suffice.

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.

Cite and contact. Cite this paper by its title, author and date above, with its PDF; cite the earlier sources it uses for a mathematical result. For software, use the release citation and give the commit used. Contact Will with questions or corrections.

Prefer the manuscript?

PDFOpen the PDF, 61 pages TeXOpen the LaTeX source Problem page

Equations typeset from the exact TeX. Rendered from the LaTeX manuscript source with file fingerprint sha256:92512a9aa35a79c5. Its file fingerprint matched the published source record, so the PDF above, the LaTeX, and this page are one manuscript. Redeploying the site regenerates this page from whatever the public repository holds at that moment.