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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Comparison with Duverney’s reduced parameters.

For the positive reciprocal series, the eventually reduced parameters in Duverney’s Theorem 3.1 and Section 5.2 [2] can be taken as follows. The symbols $p_{n}, q_{n}$ in this comparison are Duverney’s auxiliary integers, not the fixed numerator and denominator of the full reciprocal sum:

p_{n} = u_{n}, q_{n} = w_{n} = a_{n} u_{n} - v_{n} = h_{n} u_{n + 1}, E_{n} = G_{n} (p_{n} - q_{n}) .

Indeed, $gcd (u_{n}, w_{n}) = gcd (u_{n}, v_{n}) = 1$ . Since $a_{n} v_{n} = a_{n}^{2} u_{n} - a_{n} w_{n}$ , reduction of the next tail gives

h_{n} = gcd (w_{n}, a_{n} v_{n}) = gcd (w_{n}, a_{n}^{2}), p_{n + 1} ∣ q_{n} .

For the gcd equality, subtract the multiple $a_{n} w_{n}$ and use the coprimality of $u_{n}, w_{n}$ ; the divisibility follows from $q_{n} = h_{n} p_{n + 1}$ . Eliminating $v_{n + 1}$ then gives Duverney’s recurrence:

a_{n + 1} = \frac{p_{n}}{q_{n}} a_{n}^{2} - a_{n} + \frac{q_{n + 1}}{p_{n + 1}} .

Thus $q_{n} / p_{n + 1} = h_{n}$ is precisely the cancellation factor, and $q_{n} / p_{n} \to 1$ . This identifies the reduced parameters, not every possible unreduced choice in the original theorem. An upper bound on $q_{n} - p_{n}$ controls the increase before cancellation in the reduced coordinates. To transfer that bound directly to $(- E_{n})_{+} = G_{n} (q_{n} - p_{n})_{+}$ would also require control of $G_{n}$ .

An old prime power can be lost only when the multiplier supplies exactly its exponent and subtraction cancels more. The formula below separates this case from unequal exponents.

Lemma 7.1 (the denominator valuation transition). Let $u, v, a$ be positive integers, $gcd (u, v) = 1$ , and $w = a u - v > 0$ . Put $h = gcd (w, a v)$ and $v^{'} = a v / h$ . For a prime $p$ , write $r = ν_{p} (a)$ , $s = ν_{p} (v)$ and $t = ν_{p} (w)$ . Then

ν_{p} (v^{'}) = {\begin{cases} max (r, s), & r \neq s, \\ max (0, 2 s - t), & r = s . \end{cases}

In particular $ν_{p} (v^{'}) \leq max (r, s)$ . A strict loss relative to $s$ requires $r = s \geq 1$ and $t > s$ .

Proof. Always $ν_{p} (v^{'}) = r + s - min (t, r + s)$ . If $r \neq s$ , primitivity implies $t = min (r, s)$ : when $s > 0$ , $u$ is a $p$ -adic unit, and when $s = 0 < r$ , $v$ is a unit. If $r = s > 0$ , both terms in $a u - v$ are divisible by $p^{s}$ , so $t \geq s$ ; substituting gives the second case. If $r = s = 0$ , the same formula gives zero without needing $u$ to be a unit. ◻

Cancellation is measured against the enlarged denominator $a v$ , not against $v$ . Thus $h > 1$ need not mean that an old prime power is lost. For example, $(u, v, a) = (2, 15, 9)$ gives $w = 3$ , $h = 3$ and $(u^{'}, v^{'}) = (1, 45)$ . The numerator before cancellation exceeds $u$ by only $w - u = 1$ , yet the $3$ -adic valuation of the reduced denominator rises from $1$ to $2$ . This is a finite exact step, not an infinite counterexample.

Corollary 7.2 (persistence of a prime power). Suppose $p^{k} ∣ v_{s}$ . If $w_{n} < p^{k + 1}$ at every step from $s$ through $t - 1$ , then $p^{k} ∣ v_{t}$ .

Proof. At a first loss of divisibility by $p^{k}$ , the current exponent is some $j \geq k$ and the valuation lemma forces $ν_{p} (w_{n}) > j$ . This would give $w_{n} \geq p^{j + 1} \geq p^{k + 1}$ , contrary to the hypothesis. ◻

The bound on $w_{n}$ protects the old prime power despite varying cancellation factors. It is needed only until the crossing under study.

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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