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.

Start here, by the editorial route

In this paper

Bounded increases

We return to the rational tails of Section 2. The Sylvester example had a constant integer numerator. To measure departure from this behaviour, put

E_{n} = D_{n} - (a_{n} - 1) C_{n} = C_{n} - C_{n + 1} .

Negative errors are rises of $C_{n}$ ; zero error gives the tail $1 / (a_{n} - 1)$ . Instead of prescribing a polynomial, we now bound the rises. Quadratic growth makes all steps small relative to $C_{n}$ ; the additional absolute bound applies only to rises, not decreases.

Theorem 4.1 (bounded negative part). Let $a, C, D : N \to N$ satisfy $a_{n} > 1$ , $C_{n} > 0$ , and

C_{n + 1} + D_{n} = a_{n} C_{n}, D_{n + 1} = a_{n} D_{n} (n \geq 0) .

Put $E_{n} = D_{n} - (a_{n} - 1) C_{n}$ . If

\frac{E_{n}}{C_{n}} ⟶ 0 and E_{n} \geq - B for all sufficiently large n

for some integer $B \geq 0$ , then $E_{n} = 0$ for all sufficiently large $n$ .

The linked theorem uses the equivalent division-free condition: for every $K \in N$ , eventually $K | E_{n} | < C_{n}$ . Under this condition, absorption either settles the theorem or leaves $C_{n} \to \infty$ . A stable gcd and a forbidden block will exclude the latter alternative.

Zero errors and descent

Write $z_{+} = max (z, 0)$ . The identities below use (1); positivity and $| E_{n} | < C_{n}$ are imposed only where stated.

Proposition 4.2 (error identities). For integer sequences satisfying (1),

C_{n + 1} = C_{n} - E_{n}, (a_{n + 1} - a_{n}^{2} + a_{n} - 1) C_{n + 1} = a_{n}^{2} E_{n} - E_{n + 1} .

Proof. The first identity follows by substituting $D_{n} = E_{n} + (a_{n} - 1) C_{n}$ into the numerator recurrence. Substituting this expression at the next index and using $D_{n + 1} = a_{n} D_{n}$ gives the second. ◻

Equation (2) now gives $E_{n} / C_{n} = 1 - C_{n + 1} / C_{n} \to 0$ , so $| E_{n} | < C_{n}$ for all sufficiently large $n$ . Koizumi proves this relative-gap limit for an eventual pseudo-greedy tail [10]. His rounding convention also gives $- C_{n} / 2 \leq E_{n} < C_{n} / 2$ , a stronger bound than we need.

Koizumi’s Lemma 3 gives zero absorption for pseudo-greedy tails, and Proposition 1(2) uses descent when their gaps are eventually nonnegative [10]. The following form separates what the two arguments require.

Theorem 4.3 (absorption and descent). Let $a, C, D : N \to N$ satisfy $a_{n} > 1$ , $C_{n} > 0$ and (1). If $| E_{n} | < C_{n}$ for every $n$ , then $E_{n} = 0$ implies $E_{n + 1} = 0$ . For any positive integer sequence $C_{n}$ with $C_{n + 1} = C_{n} - E_{n}$ , eventual nonnegativity of $E_{n}$ implies its eventual vanishing.

Proof. When $E_{n} = 0$ , the second identity in Proposition 4.2 shows that $C_{n + 1}$ divides $E_{n + 1}$ . The inequality $| E_{n + 1} | < C_{n + 1}$ then forces $E_{n + 1} = 0$ . For the second assertion, $C_{n + 1} = C_{n} - E_{n}$ makes the positive integer sequence $C_{n}$ eventually nonincreasing, hence constant. ◻

Corollary 4.4 (two zero errors). For integer sequences satisfying (1), if $E_{n} = E_{n + 1} = 0$ and $C_{n + 1} \neq 0$ , then $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ . In particular, if $a, C, D : N \to N$ also satisfy $a_{n} > 1$ and $C_{n} > 0$ , eventual zero error implies the eventual Sylvester recurrence.

Proof. Cancel the nonzero factor $C_{n + 1}$ in the second error identity. ◻

Thus a late zero is absorbing. Without late zeros, $| E_{n} | \geq 1$ and $| E_{n} | / C_{n} \to 0$ force $C_{n} \to \infty$ . The scalar update permits $C_{n} = n + 1$ , $E_{n} = - 1$ forever; excluding this behaviour requires the denominator recurrence.

A stable gcd and a forbidden block

Proposition 4.5 (gcd stabilisation). Let $a, C, D : N \to N$ satisfy $a_{n} > 1$ , $C_{n} > 0$ and (1). If some fixed integer $B \geq 1$ satisfies $- B \leq E_{n} < 0$ at infinitely many indices, then $G_{n} = gcd (C_{n}, D_{n})$ is eventually constant. Dividing $C_{n}, D_{n}$ by its eventual value gives coprime sequences satisfying the same recurrences.

Proof. We have $G_{n} ∣ G_{n + 1}$ and $G_{n} ∣ E_{n}$ . Given $n$ , choose $t \geq n$ with $- B \leq E_{t} < 0$ . Then $G_{n} \leq G_{t} \leq - E_{t} \leq B$ . Hence the positive divisibility chain stabilises. Division by its stable value preserves the recurrences and gives coprime numerators and denominators. ◻

There are two distinct uses of the bound. Arbitrarily late negative errors in a fixed interval suffice to stabilise the gcd. The crossing argument below needs every sufficiently late upward step bounded; otherwise a single large rise could jump over the forbidden block.

A sequence avoiding multiples of $2$ and $3$ cannot cross the pair $6 k + 2, 6 k + 3$ from below with upward steps at most two: its first value at least $6 k + 2$ would be one of those two integers. Decreases before that crossing do not help. For a general step bound $B$ , the Chinese remainder theorem supplies a forbidden block of length $B$ .

Lemma 4.6 (consecutive multiples). For pairwise coprime integers $m_{0}, \dots, m_{B - 1} \geq 2$ and every lower bound, there is a larger $t$ such that $m_{i} ∣ t + i$ for each $i < B$ .

Proof. Solve $t \equiv - i (\mod m_{i})$ by the Chinese remainder theorem and add multiples of $\prod_{i} m_{i}$ to exceed the prescribed bound. ◻

Theorem 4.7 (Chinese remainder theorem and first crossing). Let $u : N \to N$ tend to infinity and let $B \geq 1$ be an integer with $u_{n + 1} \leq u_{n} + B$ for every $n$ . There is no sequence of pairwise coprime integers $m_{i} \geq 2$ for which $gcd (m_{i}, u_{t}) = 1$ whenever $i < t$ .

Proof. Apply Lemma 4.6 to $m_{0}, \dots, m_{B - 1}$ , choosing $t > max (u_{0}, \dots, u_{B})$ . There is a first $n > B$ with $u_{n} \geq t$ . Then

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

Thus $u_{n} = t + i$ for some $i < B < n$ , and $m_{i} ∣ u_{n}$ , contrary to $gcd (m_{i}, u_{n}) = 1$ . ◻

Bado proves a two-sided bounded-error criterion by this forbidden-block argument [7]. The gcd stabilisation above allows us to use only the lower error bound.

Proof of Theorem 4.1. Suppose the conclusion fails. Absorption excludes late zeros; $| E_{n} | \geq 1$ and the relative limit then give $C_{n} \to \infty$ . Integer descent forces infinitely many late negative errors, where $1 \leq - E_{n} \leq B$ . Thus $B \geq 1$ and Proposition 4.5 gives a stable gcd $g$ . Restrict to a tail on which the gcd is $g$ and $E_{n} \geq - B$ , then reindex that tail. Dividing by $g$ gives $u_{n} = C_{n} / g \to \infty$ and

u_{n + 1} - u_{n} = - E_{n} / g \leq B .

Proposition 3.1 makes the multipliers $a_{i} \geq 2$ pairwise coprime and coprime to every later numerator, contradicting Theorem 4.7. ◻

Corollary 4.8. Under Theorem 4.1, the multipliers satisfy $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ eventually.

Proof. Theorem 4.1 gives eventual zero error; positivity of $C_{n}$ and Corollary 4.4 give the recurrence. ◻

The condition in the original sequence

Let $γ_{n} = a_{n}^{2} / a_{n + 1} - 1$ . For $t_{n} = P_{n} / a_{n}$ ,

t_{n + 1} - t_{n} = t_{n} γ_{n} .

We bound upward increments, not values or decreases. The relation $C_{n} \sim q t_{n}$ does not control increments; instead an exact identity will give $E_{n} > - q t_{n} γ_{n}$ , turning an upper increment bound into a lower error bound.

Corollary 4.9 (bounded increments of $P_{n} / a_{n}$ ). Let $a_{1} < a_{2} < \dots$ be positive integers, $a_{n + 1} / a_{n}^{2} \to 1$ , and $\sum_{n \geq 1} 1 / a_{n} \in Q$ . Put $P_{n} = \prod_{j < n} a_{j}$ . If

\underset{n \to \infty}{lim sup} \frac{P_{n}}{a_{n}} (\frac{a_{n}^{2}}{a_{n + 1}} - 1) < + \infty,

then $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ for all sufficiently large $n$ .

Proof. For the canonical integer tail, direct substitution gives

\begin{matrix} (9) & E_{n} + q \frac{P_{n}}{a_{n}} γ_{n} = q P_{n} (\frac{1}{a_{n + 1}} - (a_{n} - 1) \sum_{k \geq n + 2} \frac{1}{a_{k}}) . \end{matrix}

The tail estimate bounds the subtracted term by $4 (a_{n} - 1) / a_{n + 1}^{2} < 1 / a_{n + 1}$ eventually. The positive remainder gives $E_{n} > - q t_{n} γ_{n}$ , so the assumed upper bound bounds $E_{n}$ below. Only the remainder’s sign is needed here. The canonical tails satisfy the exact recurrences and, by (2), $E_{n} / C_{n} \to 0$ . Theorem 4.1 and Corollary 4.8 finish. ◻

Koizumi’s proof of Corollary 4(1) uses this product-weighted comparison [10]; that corollary assumes a nonpositive upper limit, whereas the argument here allows any finite upper limit. A corresponding LCM comparison appears below. For $a_{n} = b^{2^{n - 1}}$ , $b \geq 2$ , the increments of $P_{n} / a_{n}$ are zero. Since the sequence does not eventually satisfy the Sylvester recurrence, the corollary gives irrationality. For a Sylvester sequence starting at $a_{1} > 1$ , $a_{n} - 1 = (a_{1} - 1) P_{n}$ and the increments tend to zero.

Dividing by $n$ removes the nonsummable $1 / n$ term. A summable upper error then suffices, without a lower bound on $γ_{n}$ .

Corollary 4.10 (an inclusive one-sided $1 / n$ bound). Let $a_{1} < a_{2} < \dots$ be positive integers with $a_{n + 1} / a_{n}^{2} \to 1$ and $\sum_{n} 1 / a_{n} \in Q$ . Suppose that for some $K \geq 0$ and $ε > 0$ ,

γ_{n} := a_{n}^{2} / a_{n + 1} - 1 \leq \frac{1}{n} + \frac{K}{n^{1 + ε}} for all large n .

Then the sequence is eventually Sylvester. In particular, the conclusion holds under the pointwise eventual bound $γ_{n} \leq 1 / n$ .

Proof. The exact model $n$ has successive ratio $1 + 1 / n$ . Thus $(t_{n + 1} / (n + 1)) / (t_{n} / n) = (1 + γ_{n}) / (1 + 1 / n) \leq 1 + K / [n^{ε} (n + 1)]$ . The convergent product of these upper bounds bounds $t_{n} / n$ above; no positive limit is needed. Hence $t_{n} = O (n)$ and $t_{n} γ_{n} \leq t_{n} (γ_{n})_{+} = O (1)$ , so Corollary 4.9 applies. ◻

The weaker condition $lim sup n (γ_{n})_{+} \leq 1$ does not suffice for the estimate $t_{n} = O (n)$ . For example, the scalar ratios $1 + γ_{n} = 1 + (1 + 1 / \log n) / n$ give a product of order $n \log n$ . This scalar product refutes the intermediate linear estimate under the weaker rate, not the Sylvester conclusion for reciprocal tails.

For positive limiting increments, take $a_{1} = 4$ and $a_{n + 1} = ⌈ n a_{n}^{2} / (n + 1) ⌉$ , beginning $4, 8, 43, 1387$ . Here $a_{n} \geq 2 \cdot 2^{2^{n - 1}}$ and

0 \leq 1 + \frac{1}{n} - \frac{a_{n}^{2}}{a_{n + 1}} < \frac{4}{a_{n}^{2}} .

Consequently $t_{n} \sim K n$ and $t_{n + 1} - t_{n} \to K > 0$ . The corollary proves irrationality because $γ_{n} \sim 1 / n$ is incompatible with the eventual Sylvester recurrence.

For the introduction’s $3 / n$ ceiling example, division by $n (n + 1) (n + 2)$ and summability of the rounding errors give $t_{n} \sim K_{3} n^{3}$ and $t_{n + 1} - t_{n} \sim 3 K_{3} n^{2}$ , with $K_{3} > 0$ . Its irrationality follows from Theorem 1.2, beyond the bounded-increase criterion.

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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