Reciprocal Sums and the Sylvester Recurrence

Will Cook

Short paper

Reciprocal Sums and the Sylvester Recurrence

Erdős #243 By Will Cook September 2026 Authorship, AI use and citation

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Summary. For strictly increasing positive integers with $a_{n}^{2} / a_{n + 1} = 1 + 3 / n + o (n^{- 3})$ , the reciprocal sum is irrational. The argument extracts a cubic polynomial from an integer tail numerator, restricts its coefficients using a square in a cubic field, and excludes the remaining cases modulo seven. Under rationality and the quadratic growth limit, separate criteria using bounded numerator increases and weighted first crossings force the eventual Sylvester recurrence. The unrestricted Erdős–Graham question remains open.

This paper covers cubic-rate irrationality, nonintegral regular-rate exclusion and sufficient tail criteria for the Sylvester recurrence; the unrestricted Erdős–Graham question remains open.

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 the constant term

We now use the denominator recurrence. Both updates preserve common divisors, so $G_{n} = gcd (C_{n}, D_{n})$ divides $G_{n + 1}$ and every later numerator. On the polynomial tail it therefore divides the third difference $6 A$ , a fixed positive integer. The fourth difference gave polynomiality; the nonzero third difference now bounds cancellation. Thus $G_{n}$ stabilises at a positive integer $g$ .

Dividing by this fixed $g$ gives $u_{n} = C_{n} / g$ , $v_{n} = D_{n} / g$ in lowest terms without changing the recurrences. Their coprimality persists as follows.

Proposition 3.1 (persistent coprimality). Let $a_{n}, u_{n}, v_{n}$ be integer sequences with $u_{n} > 0$ , $v_{n} \geq 0$ and $gcd (u_{n}, v_{n}) = 1$ satisfying

u_{n + 1} = a_{n} u_{n} - v_{n}, v_{n + 1} = a_{n} v_{n} .

Then $gcd (a_{n}, v_{n}) = 1$ . The $a_{n}$ are pairwise coprime, and $gcd (a_{i}, u_{t}) = 1$ whenever $i < t$ .

Proof. A prime dividing both $a_{n}$ and $v_{n}$ would divide $u_{n + 1}$ and $v_{n + 1}$ , contrary to reducedness. For $i < t$ , the factor $a_{i}$ divides $v_{t}$ . Reducedness gives $gcd (a_{i}, u_{t}) = 1$ , and $gcd (a_{t}, v_{t}) = 1$ gives $gcd (a_{i}, a_{t}) = 1$ . ◻

In the cubic case, adjacent numerators are also coprime, since $gcd (u_{n}, u_{n + 1}) = gcd (u_{n}, v_{n}) = 1$ . Write the polynomial for $u_{n}$ as

\begin{matrix} (5) & Q (n) = \frac{m}{6} n (n + 1) (n + 2) + c . \end{matrix}

The third difference $m = 6 A / g$ is a positive integer. Since $n (n + 1) (n + 2) / 6 = (\binom{n + 2}{3})$ is an integer, any sufficiently late value $u_{n} = Q (n)$ gives

c = u_{n} - m (\binom{n + 2}{3}) \in Z .

Adjacent coprimality now restricts this integer $c$ . If a prime $ℓ$ divides $c$ , choose an arbitrarily late $n \equiv - 1 (\mod 6 ℓ)$ . Both three-factor products contain $n + 1$ , so division by $6$ leaves a factor $ℓ$ , even for $ℓ = 2, 3$ . Thus $ℓ$ divides both $Q (n)$ and $Q (n + 1)$ , contradicting adjacent coprimality. This also excludes $c = 0$ . Thus

m \in N_{> 0}, c \in {1, - 1} .

Irreducibility follows from persistence of denominator primes. The pairwise coprime multipliers $a_{n} > 1$ supply infinitely many distinct primes; each divides every later $v_{t}$ and no later $u_{t}$ . A rational root of $Q$ would reduce to a root modulo all but finitely many primes. Choose a persistent prime outside those exceptions. Every sufficiently late index in the root’s residue class would give $ℓ ∣ Q (t) = u_{t}$ , contradicting reducedness. Thus $Q$ has no rational root and, being cubic, is irreducible.

Set $κ = m / 6$ , $η = 6 c / m$ , and

f (T) = T^{3} - T + η, Q (n) = κ f (n + 1) .

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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