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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Zero errors and the Sylvester recurrence

To recover the Sylvester recurrence from a zero error, we eliminate $D_{n}$ between two successive updates. Write

Δ_{n} = a_{n + 1} - (a_{n}^{2} - a_{n} + 1),

the Sylvester defect. The identity below expresses $Δ_{n} C_{n + 1}$ through two consecutive errors. If both errors vanish, positivity lets us cancel $C_{n + 1}$ and recover the recurrence. Under the bound $| E_{n} | < C_{n}$ , it will also make a single zero propagate to the next index.

Theorem 5.1 (defect identity). For all $a, a^{'}, D, C \in Z$ ,

\begin{aligned} [a^{'} - (a^{2} - a + 1)] (a C - D) & = a^{2} [D - (a - 1) C] \\ - [a D - (a^{'} - 1) (a C - D)], \end{aligned}

that is, $Δ_{n} C_{n + 1} = a_{n}^{2} E_{n} - E_{n + 1}$ .

Proof. A direct expansion: both sides equal $a^{'} a C - a^{'} D - a^{3} C + a^{2} D + a^{2} C - a D - a C + D$ . ◻

No sign or growth hypothesis is needed for the identity itself. Such hypotheses enter only in its consequences below.

Example 5.2 (one defect and the error it creates). Continue Example 4.2 but replace $a_{3} = 43$ by $a_{3} = 44$ . The states $D_{3} = 42$ and $C_{3} = 1$ are unchanged, since they depend only on $a_{0}, a_{1}, a_{2}$ , and $E_{2} = 0$ still. The defect is $Δ_{2} = a_{3} - (a_{2}^{2} - a_{2} + 1) = 44 - 43 = 1$ , and $E_{3} = D_{3} - (a_{3} - 1) C_{3} = 42 - 43 = - 1$ , so the identity reads

Δ_{2} C_{3} = 1 \cdot 1 = 1 = 49 \cdot 0 - (- 1) = a_{2}^{2} E_{2} - E_{3} .

By Proposition 4.1 the numerator then rises, $C_{4} = C_{3} - E_{3} = 2$ . A single unit of defect at one index has produced a negative error of magnitude $1$ at the next, increasing the numerator from $1$ to $2$ .

Theorem 5.3 (two vanishing errors force the step). Let $a, a^{'}, D, C \in Z$ with $a C - D \neq 0$ . If

D - (a - 1) C = 0, a D - (a^{'} - 1) (a C - D) = 0,

then $a^{'} = a^{2} - a + 1$ .

Proof. Theorem 5.1 gives $[a^{'} - (a^{2} - a + 1)] (a C - D) = 0$ . The second factor is nonzero, so the first factor must vanish. ◻

The hypothesis $C_{n + 1} \neq 0$ is not removable: at $C_{n + 1} = 0$ the identity gives no information about $a^{'}$ .

Relations among three consecutive numerators

Eliminating the denominator from two successive steps gives a relation among three reduced numerators. Write $u, u_{1}, u_{2}$ for the reduced numerator coordinates, $v, v_{1}$ for the corresponding denominator coordinates, and $h, h_{1}$ for the two common factors removed in reduction. If the two steps have multipliers $a, a_{1}$ , then the exact hypotheses are

h u_{1} + v = a u, h v_{1} = a v, h_{1} u_{2} + v_{1} = a_{1} u_{1} .

Theorem 5.4 (eliminating the denominator from two steps). Let $a, a_{1}, u, u_{1}, u_{2}, v, v_{1}, h, h_{1}$ be integers satisfying

h u_{1} + v = a u, h v_{1} = a v, h_{1} u_{2} + v_{1} = a_{1} u_{1} .

Then $a^{2} u + h h_{1} u_{2} = h (a + a_{1}) u_{1}$ .

Proof. Multiply the first equation by $a$ , replace $a v$ using the second equation, and then replace $h_{1} u_{2} + v_{1}$ using the third. The remaining terms factor as the displayed right-hand side. ◻

This is the recurrence obtained by eliminating the denominator. Both cancellation factors remain in the formula. The identity adds no assumption to the two exact steps, but it still contains both multipliers and cannot by itself force $a_{1} = a^{2} - a + 1$ ; Lemma 5.6 specifies what it forgets modulo a fixed old denominator.

When neither step cancels a common factor, the same recurrence gives a square identity. Write $p, p_{1}, p_{2}$ for three consecutive numerators, $q = a p - p_{1}$ for the denominator before the first step, and $a, a_{1}$ for the two multipliers. Then

p_{2} + a^{2} p = (a + a_{1}) p_{1} .

The resulting square identity is as follows.

Theorem 5.5 (a square identity without cancellation). Let $a, a_{1}, p, p_{1}, p_{2}, q$ be integers with $q = a p - p_{1}$ and $p_{2} + a^{2} p = (a + a_{1}) p_{1}$ . Then

q^{2} + (p p_{2} - p_{1}^{2}) = (a_{1} - a) p p_{1} .

Proof. Substitute $q = a p - p_{1}$ and $p_{2} = (a + a_{1}) p_{1} - a^{2} p$ :

\begin{aligned} q^{2} + p p_{2} - p_{1}^{2} & = (a p - p_{1})^{2} + p ((a + a_{1}) p_{1} - a^{2} p) - p_{1}^{2} \\ = (a_{1} - a) p p_{1} . \end{aligned}

◻

The term $q^{2}$ is nonnegative, while $p p_{2} - p_{1}^{2}$ measures the difference between the product of the outer numerators and the square of the middle one. The right side records the change in multiplier, so the identity gives a necessary algebraic constraint without asserting a sign or a global monotonicity. It is checked as the square identity when no factor is cancelled; no global exclusion of approximate solutions is claimed.

The reduced error update must also be compatible with the denominator recurrence. The following calculation gives the exact compatibility condition. Here $e, e_{1}$ are signed errors in reduced coordinates, not the positive magnitudes denoted by $e_{n}$ in the all-negative case. Let $Δ$ be a proposed value of the Sylvester defect. If

h u_{1} = u - e, h e_{1} = a^{2} e - Δ (u - e),

then the denominator equality

h ((a_{1} - 1) u_{1} + e_{1}) = a ((a - 1) u + e)

holds precisely when the following difference vanishes:

\begin{aligned} h ((a_{1} - 1) u_{1} + e_{1}) - a ((a - 1) u + e) \\ = [a_{1} - (a^{2} - a + 1) - Δ] (u - e) . \end{aligned}

Consequently, when $u - e \neq 0$ ,

h ((a_{1} - 1) u_{1} + e_{1}) = a ((a - 1) u + e) ⟺ a_{1} = a^{2} - a + 1 + Δ .

The equivalence with the denominator recurrence is formalised using the factored difference between the two recurrences. The hypothesis $u - e \neq 0$ is essential: if $u - e = 0$ , the factorised mismatch cannot identify $a_{1}$ .

The identities hold under the local hypotheses displayed in their statements. It is their proposed global applications, not the identities, that would need additional information, such as control of cancellation or a sign estimate for $p p_{2} - p_{1}^{2}$ . No such information is deduced here for an arbitrary rational near-quadratic tail.

Lemma 5.6 (saturation modulo an old denominator). Let $M \geq 2$ . Any finite word $r_{0}, \dots, r_{k}$ of units modulo $M$ is compatible with the cancellation-free recurrences modulo $M$ , with all reduced denominator residues equal to zero. Consequently the eliminated two-step identity alone imposes no further restriction on such unit words when the multiplier residues are free.

Proof. Set $v_{i} \equiv 0 (\mod M)$ and choose $a_{i} \equiv r_{i + 1} r_{i}^{- 1} (\mod M)$ . Then $r_{i + 1} + v_{i} \equiv a_{i} r_{i}$ and $v_{i + 1} \equiv a_{i} v_{i}$ . Eliminating $v_{i}$ gives the two-step identity automatically. ◻

This is finite residue compatibility, not existence of a positive integer orbit, much less a rational tail with near-quadratic growth. For $M ∣ v_{T}$ on a tail with no further common-factor cancellation, the actual $u_{n}$ are indeed units modulo $M$ . Useful further restrictions must therefore retain information discarded here: for example multiplier size, changing moduli, or primes outside the old denominator. The square restriction used in the cubic proof concerns precisely a prime at which a middle numerator vanishes.

Theorem 5.7 (sequence form). Let $a, D, C : N \to Z$ satisfy $D_{n + 1} = a_{n} D_{n}$ and $C_{n + 1} = a_{n} C_{n} - D_{n}$ . If $E_{n} = 0$ for all sufficiently large $n$ and $C_{n + 1} \neq 0$ for all sufficiently large $n$ , then $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ for all sufficiently large $n$ .

Take the larger of the two thresholds and apply Theorem 5.3 at each later index. This is the final step of the criteria below that force $E_{n}$ to vanish eventually.

The same identity shows that a zero error persists whenever $| E_{n} | < C_{n}$ at the subsequent indices.

Theorem 5.8 (zero is absorbing). Let $a, C, D : N \to N$ satisfy $C_{n + 1} + D_{n} = a_{n} C_{n}$ and $D_{n + 1} = a_{n} D_{n}$ , and put $E_{n} = D_{n} - (a_{n} - 1) C_{n}$ . Suppose that $| E_{n} | < C_{n}$ for every $n$ . If $E_{n} = 0$ then $E_{n + 1} = 0$ .

Proof. Putting $E_{n} = 0$ in Theorem 5.1 gives $Δ_{n} C_{n + 1} = - E_{n + 1}$ , so $C_{n + 1}$ divides $E_{n + 1}$ . A multiple of $C_{n + 1}$ of absolute value smaller than $C_{n + 1}$ is zero, and $| E_{n + 1} | < C_{n + 1}$ by hypothesis. ◻

Beyond an index from which $| E_{n} | < C_{n}$ always holds, either $E_{n}$ reaches zero and remains zero, or it is never zero. A zero at an earlier index need not persist: in Example 5.2, the altered multiplier gives $E_{3} = - 1$ and $C_{3} = 1$ , so the required strict inequality fails. Sections 9 and 10 address the special all-negative case. The mixed-sign analysis in Sections 12 and 13 separates arbitrarily late negative errors from an eventually nonnegative tail.

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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