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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Bounded increases and coprimality

For nonperiodic errors, the Chinese remainder theorem replaces the fixed period. It produces a block of consecutive forbidden values. An unbounded numerator with bounded upward steps must enter that block when it first crosses its lower end.

Lemma 11.1 (shifted blocks of consecutive multiples). Let $m_{0}, \dots, m_{B - 1}$ be pairwise coprime and at least $2$ . For every bound there is a $t$ beyond it with $m_{i} ∣ t + i$ for each $i < B$ .

Proof. Solve $t \equiv - i (\mod m_{i})$ simultaneously by the Chinese remainder theorem. Adding a sufficiently large multiple of $\prod_{i} m_{i}$ puts the solution beyond the prescribed bound. ◻

Lemma 11.1 matches each modulus $m_{i}$ to its own multiple $t + i$ inside a window of $B$ consecutive integers. Matchings of a set of integers to distinct multiples in an interval are the subject of Erdős Problem #650, solved by van Doorn, Li and Tang: for every set $S$ of $k$ positive integers, every open interval of length $2 max S$ contains distinct multiples of at least $min (k, ⌈ 2 \sqrt{k} ⌉)$ elements of $S$ , and this count is optimal [11]. Their extremal construction uses the same Chinese remaindering for moduli that need not be pairwise coprime, with their Claim 3.2 supplying the divisibility of residue differences by greatest common divisors that the generalised theorem requires [11]. The first-crossing argument below needs only the pairwise coprime case.

Example 11.2 (a block of three consecutive multiples). Take $B = 3$ and $(m_{0}, m_{1}, m_{2}) = (3, 4, 5)$ . The congruences $t \equiv 0 (\mod 3)$ , $t \equiv 3 (\mod 4)$ and $t \equiv 3 (\mod 5)$ hold exactly when $t \equiv 3 (\mod 60)$ . At $t = 3$ the block is $3, 4, 5$ with $3 ∣ 3$ , $4 ∣ 4$ and $5 ∣ 5$ ; at $t = 63$ it is $63, 64, 65$ with $3 ∣ 63$ , $4 ∣ 64$ and $5 ∣ 65$ . A sequence of natural numbers that starts at $0$ , tends to infinity and rises by at most $3$ at each step cannot step over the block ${3, 4, 5}$ : it has a first value at least $3$ , that value is at most $5$ , and every member of ${3, 4, 5}$ is divisible by one of the three moduli.

Theorem 11.3 (bounded increases and coprimality to earlier moduli). Let $u : N \to N$ tend to infinity with $u_{n + 1} \leq u_{n} + B$ for a fixed integer $B \geq 1$ . Then $u$ cannot remain coprime to infinitely many fresh pairwise coprime moduli: there is no family of pairwise coprime $m_{i} \geq 2$ , one for each index, such that $gcd (m_{i}, u_{t}) = 1$ whenever $i < t$ .

The CRT block must lie above the initial segment where some moduli are not yet available. No monotonicity of $u$ is assumed.

Proof. Choose $m_{0}, \dots, m_{B - 1}$ and use Lemma 11.1 to find $t > max (u_{0}, \dots, u_{B})$ with $m_{i} ∣ t + i$ for $0 \leq i < B$ . Since $u_{n} \to \infty$ , there is a first $n > B$ with $u_{n} \geq t$ . Minimality and the rise bound give

t \leq u_{n} \leq u_{n - 1} + B < t + B .

Hence $u_{n} = t + i$ for some $0 \leq i < B < n$ , so $m_{i} ∣ u_{n}$ . The inequality $i < n$ now permits the avoidance hypothesis, which says $gcd (m_{i}, u_{n}) = 1$ ; this contradicts $m_{i} \geq 2$ . The argument uses the first crossing of the CRT block, not monotonicity of $u$ . ◻

Only unboundedness above is used to obtain this first crossing. Divergence is stated because it holds for the reduced rational tail in the application. This distinction also explains why the weighted proof in Section 6 works with an unbounded running maximum, without assuming that its numerator tends to infinity.

Divergence forces $u$ to reach the forbidden block, and the uniform rise bound prevents it from jumping over the block. A bound only along a subsequence would leave the intervening steps unrestricted. Without a scale restriction on the moduli, Proposition 14.6 gives an unbounded coprime walk with rises $o (\log \log u_{n})$ . The same forbidden-block crossing, under a two-sided bound on the error, appears in the proof of Bado’s Theorem 5.1 [8], posted in September 2026.

In the application, $m_{i}$ is the $i$ th multiplier. Its coprimality is known only for later numerators, which explains the condition $i < t$ .

To apply the preceding theorem, we reduce $C_{n} / D_{n}$ to lowest terms and require that no further common factor be cancelled. In that case the reduced numerators and denominators still satisfy the original recurrences. The relevant coprimality is as follows.

Proposition 11.4 (persistent coprimality). Suppose that $gcd (u_{n}, v_{n}) = 1$ at every index and

u_{n + 1} + v_{n} = a_{n} u_{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 common prime divisor of $a_{n}$ and $v_{n}$ would divide both $u_{n + 1} = a_{n} u_{n} - v_{n}$ and $v_{n + 1} = a_{n} v_{n}$ , contradicting their coprimality. Hence $gcd (a_{n}, v_{n}) = 1$ . For $i < t$ , the denominator recurrence gives $a_{i} ∣ v_{t}$ . Combining this with $gcd (a_{t}, v_{t}) = 1$ proves that $a_{i}$ and $a_{t}$ are coprime; combining it with $gcd (u_{t}, v_{t}) = 1$ proves that $a_{i}$ and $u_{t}$ are coprime. ◻

These are the step coprimality, the pairwise coprimality, and the coprimality to each earlier multiplier. Applying Theorem 11.3 shows that a numerator satisfying these conditions cannot tend to infinity with uniformly bounded upward increments. This is the reduced exclusion under bounded upward increments, together with its eventual form, the version with an eventual increment bound.

Example 11.5 (coprimality in Sylvester’s sequence). Put $u_{n} = 1$ and $v_{n} = D_{n} = 1, 2, 6, 42, 1806, \dots$ as in Example 4.2, with multipliers $a_{n} = v_{n} + 1 = 2, 3, 7, 43, 1807, \dots$ . Then $gcd (u_{n}, v_{n}) = 1$ , $u_{n + 1} + v_{n} = 1 + v_{n} = a_{n} u_{n}$ and $v_{n + 1} = a_{n} v_{n}$ , so Proposition 11.4 applies and gives the classical fact that Sylvester’s numbers are pairwise coprime. Its numerator is constant, so it satisfies every hypothesis of the exclusion just stated except divergence; since the tail exists, that hypothesis cannot be dropped.

The fraction $C_{n} / D_{n}$ in Section 4 need not be in lowest terms. To use the preceding argument, we must show that the common factor stops changing. One uniform bound on the negative magnitudes at arbitrarily late indices suffices: each gcd divides every later gcd, and at a negative index it also divides the nonzero magnitude. Thus a bound at arbitrarily late negative indices bounds the entire gcd sequence. The next proposition justifies dividing by its eventual value.

Proposition 11.6 (the tail gcd stabilises). Let $a, D, C : N \to N$ satisfy $C_{n + 1} + D_{n} = a_{n} C_{n}$ and $D_{n + 1} = a_{n} D_{n}$ , and put $E_{n} = D_{n} - (a_{n} - 1) C_{n}$ . Suppose some fixed integer $B \geq 1$ satisfies $- B \leq E_{n} < 0$ at infinitely many indices. Then $gcd (C_{n}, D_{n})$ is eventually constant. Dividing $C_{n}, D_{n}$ by its eventual value gives coprime sequences satisfying the same recurrences.

Proof. First, $C_{n} > 0$ at every index. If $C_{n} = 0$ , the nonnegative recurrence forces $C_{n + 1} = D_{n} = 0$ , and both sequences then vanish at all later indices. This contradicts the existence of arbitrarily late negative errors.

The two recurrences give $G_{n} ∣ G_{n + 1}$ for $G_{n} = gcd (C_{n}, D_{n})$ , as in the gcd divisibility relation. Also $G_{n} = gcd (C_{n}, | E_{n} |)$ , since $D_{n} = E_{n} + (a_{n} - 1) C_{n}$ ; at a negative index this is the gcd identity with the negative magnitude. For any $n$ , choose $t \geq n$ with $- B \leq E_{t} < 0$ . Then

G_{n} \leq G_{t} \leq | E_{t} | \leq B .

The entire positive divisibility chain is therefore bounded and eventually constant, by stabilisation of a bounded divisibility chain. In Lean this is the gcd stabilisation result. Dividing both sequences by the eventual value preserves the recurrences and gives coprime numerator and denominator at every later index. ◻

Other negative errors may exceed $B$ in magnitude. Gcd stabilisation uses the bounded witnesses, whereas the CRT application needs a bound on every upward increment.

Vanishing relative error also gives sparsity of the strict gcd increases. The next proposition asserts long finite intervals of constancy, not constancy on an infinite tail. For $a, C, D : N \to N$ satisfying (19) with $C_{n} > 0$ , put

G_{n} = gcd (C_{n}, D_{n}), Γ (N) = # {0 \leq j < N : G_{j} < G_{j + 1}} .

Proposition 11.7 (vanishing relative error makes strict gcd changes sparse). Let $a, C, D : N \to N$ satisfy $C_{n + 1} + D_{n} = a_{n} C_{n}$ and $D_{n + 1} = a_{n} D_{n}$ with $C_{n} > 0$ , and put $E_{n} = D_{n} - (a_{n} - 1) C_{n}$ , $G_{n} = gcd (C_{n}, D_{n})$ and $Γ (N) = # {0 \leq j < N : G_{j} < G_{j + 1}}$ . If $| E_{n} | / C_{n} \to 0$ , then $Γ (N) = o (N)$ . Moreover, for every starting bound $B$ and block length $L$ , some $n \geq B$ satisfies

G_{n} = G_{n + 1} = \dots = G_{n + L} .

The first assertion is the sublinear strict-growth theorem; the second is the arbitrarily late constant-block theorem. The proof first converts vanishing relative error into subexponential growth of $C_{n}$ . Each strict divisibility increase of $G_{n}$ contributes a factor of at least $2$ , so $2^{Γ (N)} G_{0} \leq G_{N} \leq C_{N}$ . Taking logarithms gives $Γ (N) = o (N)$ . For $L = 0$ the second assertion is immediate. For $L \geq 1$ , if every block of $L$ successive transitions beyond $B$ contained a strict increase, disjoint such blocks would give $Γ (N) \geq (N - B) / L - O (1)$ , a contradiction.

The starting index may depend on the block length; arbitrarily long constant blocks do not give an infinite constant tail. Proposition 11.6 needs additional bounded negative witnesses, which Proposition 11.7 does not supply.

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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