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

Reduction and lower density

For $S \subseteq N$ write $\underset{―}{d} (S) = \underset{X}{lim inf} # (S \cap [1, X]) / X$ . Thus $\underset{―}{d} (S) = 0$ gives cutoffs along which the exceptional proportion tends to zero, not eventual agreement. We cannot replace $C_{n}$ by its proposed polynomial at arbitrary indices; the following lemma supplies the finite windows where such a replacement is valid. For a sequence $F$ , let $Δ F_{n} = F_{n + 1} - F_{n}$ ; higher powers of $Δ$ mean repeated forward differences. This operator is distinct from the later indexed quantity $Δ_{n}$ , the Sylvester defect.

Lemma 2.4 (periodic obstruction principle). Let $h, L$ be positive integers and let $j_{1}, \dots, j_{L}$ be fixed integer offsets. Suppose that for every sufficiently large $n$ in one residue class modulo $h$ at least one of $n + j_{1}, \dots, n + j_{L}$ lies in $S$ . Then $\underset{―}{d} (S) \geq 1 / (L h)$ .

Proof. There are $X / h + O (1)$ relevant starting indices below $X$ . Their witnesses lie below $X + O (1)$ because the offsets are fixed, and each element of $S$ serves as a witness for at most $L$ starting indices. Hence $L # (S \cap [1, X]) \geq X / h - O (1)$ , which gives the claimed lower density. ◻

Fix $A \in Q_{> 0}$ , $B \in Q$ , and suppose for contradiction that

P (n) = A n (n + 1) (n + 2) + B, S = {n : C_{n} \neq P (n)}, \underset{―}{d} (S) = 0 .

We retain this assumption through Section 2.4.

Lemma 2.5 (gcd stabilisation and the reduced polynomial). Write $G_{n} = gcd (C_{n}, D_{n})$ . Then $6 A$ is a positive integer, $G_{n}$ divides $6 A$ for every $n$ , and $G_{n}$ is eventually equal to a positive integer $g$ . On the tail where $G_{n} = g$ , put $u_{n} = C_{n} / g$ , $v_{n} = D_{n} / g$ and $Q (n) = P (n) / g$ . Then

\begin{matrix} (2) & u_{n + 1} = a_{n} u_{n} - v_{n}, v_{n + 1} = a_{n} v_{n}, gcd (u_{n}, v_{n}) = 1, gcd (u_{n}, u_{n + 1}) = 1, \end{matrix}

and there are $m \in Z_{> 0}$ and $c \in {- 1, 1}$ with

\begin{matrix} (3) & Q (n) = \frac{m}{6} n (n + 1) (n + 2) + c . \end{matrix}

Proof. Equation (1) gives $G_{n} ∣ G_{n + 1}$ and $G_{n} ∣ C_{t}$ for every $t \geq n$ . Since $\underset{―}{d} (S) = 0$ , beyond every threshold there is a block of four consecutive indices disjoint from $S$ : otherwise every block of four would meet $S$ and $\underset{―}{d} (S) \geq 1 / 4$ . On such a block $Δ^{3} C_{t} = Δ^{3} P (t) = 6 A$ , and $G_{n}$ divides each of the four values, hence divides $6 A$ . So $6 A$ is a positive integer and the divisibility chain $(G_{n})$ is bounded and stabilises at some $g$ .

On the primitive tail, $gcd (u_{n}, v_{n}) = 1$ by construction, and a prime dividing $u_{n}$ and $u_{n + 1}$ would divide $v_{n} = a_{n} u_{n} - u_{n + 1}$ , which gives $gcd (u_{n}, u_{n + 1}) = 1$ .

Since $g ∣ 6 A$ , the number $m := 6 A / g$ is a positive integer. Put $c = B / g$ , so

Q (n) = m (\binom{n + 2}{3}) + c .

Choose any sufficiently late $n \notin S$ . Then $Q (n) = u_{n} \in Z$ , and the binomial term is an integer, so $c \in Z$ . This proves (3) with $c \in Z$ . It also shows that $Q$ is integer-valued: its values at integer arguments are integers, even though its coefficients need not all be integers.

It remains to exclude $c = 0$ and every $c$ with a prime factor. Let $ℓ$ be a prime dividing $c$ , or any prime if $c = 0$ . For $n \equiv - 1 (\mod 6 ℓ)$ , write $n + 1 = 6 ℓ k$ ; then

\frac{n (n + 1) (n + 2)}{6} = ℓ k (6 ℓ k - 1) (6 ℓ k + 1), \frac{(n + 1) (n + 2) (n + 3)}{6} = ℓ k (n + 2) (n + 3),

so $ℓ$ divides both $Q (n)$ and $Q (n + 1)$ . Agreement at both indices would give $ℓ ∣ gcd (u_{n}, u_{n + 1}) = 1$ . Hence one of $n, n + 1$ lies in $S$ for every late $n \equiv - 1 (\mod 6 ℓ)$ , and Lemma 2.4 with $L = 2$ , $h = 6 ℓ$ gives $\underset{―}{d} (S) \geq 1 / (12 ℓ)$ . This contradiction leaves $c = \pm 1$ . ◻

Lemma 2.6 (coprimality and irreducibility). On the primitive tail, $gcd (a_{n}, v_{n}) = 1$ , the multipliers at distinct indices are pairwise coprime, infinitely many of them exceed $1$ , and the cubic $Q_{m, c}$ of (3) is irreducible over $Q$ .

Proof. A prime dividing $a_{n}$ and $v_{n}$ would divide $u_{n + 1}$ and $v_{n + 1}$ , against $gcd (u_{n + 1}, v_{n + 1}) = 1$ ; and every earlier multiplier divides every later $v_{n}$ , so distinct multipliers are coprime. If $a_{n} = 1$ for all large $n$ then $v_{n}$ is eventually a positive constant and $u_{n + 1} = u_{n} - v_{n}$ decreases without bound, against positivity. So infinitely many distinct primes divide late multipliers.

Suppose $Q_{m, c}$ had a rational root. Choose a prime $ℓ ∤ 6 m$ dividing a late multiplier $a_{N}$ , large enough that the root reduces modulo $ℓ$ . Then $Q_{m, c}$ vanishes modulo $ℓ$ on a full residue class, while $ℓ ∣ v_{n}$ and therefore $ℓ ∤ u_{n}$ for every $n > N$ . Agreement at any late index of that class is impossible, and Lemma 2.4 with $L = 1$ , $h = ℓ$ gives $\underset{―}{d} (S) \geq 1 / ℓ$ . ◻

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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