Reciprocal Mersenne Subseries

Will Cook

Long paper

Reciprocal Mersenne Subseries

Erdős #257 By Will Cook 30 September 2026 Authorship, AI use and citation

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Summary. A weighted summability criterion proves irrationality of reciprocal Mersenne subseries: if h is the part of a supported on a finite nonempty prime set P, every infinite A with ∑ $h (a) / (a (b^{h (a)} - 1))$ finite has irrational ∑ $(b^{a} - 1)^{- 1}$ . The base-two condition gives the conclusion at every integer base for every infinite subset, and permits an example with divergent reciprocal sum, while excluding full support and all odd exponents. The proof averages over a finite part of the support and over dyadic observation lengths. Erdős's weaker reciprocal-summable theorem is proved in full, and a common window extends both arguments to unions of weighted and positive-cover supports. Finite-denominator and achievement-set calculations formulate tests for the targets $1 / 2$ and $1 / 21$ ; those tests do not decide either target or the universal question. Prime support at base 2 and squarefree support at power-of-two bases are attributed as existing work.

This paper covers a claim-bounded problem-specific reasoning surface for Erdős #257: selected checked premises, finite computations, closed routes, surviving obligations, and their exact evidence bands.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #257, whose universal and half-value questions remain open.

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In this paper

Dynamics

Theorem 5.16 (Exact recurrences for the three branches). Fix $n \geq 5$ . Among the quotient sums over subsets of ${2, \dots, n - 1}$ , let $D_{n}$ give the largest sum at most $T_{n}$ and let $B_{n}$ give the smallest sum strictly greater than $T_{n}$ . Thus $D_{n}$ is the greedy support already defined. Put

r = rem (n), o = Q (B_{n}, 2 n) - T_{n},

and define the two correction terms by

p^{-} = \sum_{d \in D_{n}} (2 1_{d ∣ 2 n + 1} + 1_{d ∣ 2 n + 2}), p^{+} = \sum_{d \in B_{n}} (2 1_{d ∣ 2 n + 1} + 1_{d ∣ 2 n + 2}) .

Both lie in $[0, 2 (n - 2)]$ , since $d \geq 2$ cannot divide both consecutive integers. The next remainder is

rem (n + 1) = {\begin{cases} 2^{n + 1} - 4 o - p^{+}, & if 4 o + p^{+} \leq 2^{n + 1} (U), \\ 4 r + 2^{n + 1} - p^{-}, & if not U and 4 r + 2^{n + 1} - p^{-} < 2^{n + 2} + 4 (M), \\ 4 r - 2^{n + 1} - p^{-} - 4, & otherwise (R) . \end{cases}

Consequently, for $λ_{n} = (rem (n) - 2^{n}) / 2^{n}$ ,

\begin{matrix} (III) & λ_{n + 1} = {\begin{cases} - (4 o + p^{+}) / 2^{n + 1}, & U, \\ 2 λ_{n} + 2 - p^{-} / 2^{n + 1}, & M, \\ 2 λ_{n} - (p^{-} + 4) / 2^{n + 1}, & R . \end{cases} \end{matrix}

Proof. Applying the one-step quotient identity twice gives

q (2 n + 2, d) = 4 q (2 n, d) + 2 1_{d ∣ 2 n + 1} + 1_{d ∣ 2 n + 2} .

The new capacity is $T_{n + 1} = 4 T_{n} + 2^{n + 1}$ , and the new terminal weight, at exponent $n$ , is $q (2 n + 2, n) = 2^{n + 2} + 4$ . Write $g = 2^{n + 1}$ . The old quotient sums are separated by at least $g$ , so the corrected sums remain separated by at least $4 g - 2 (n - 2) > 2 g + 4$ . This exceeds the new terminal weight. The upper word exists: already the terms at $d = 2, 3, 4$ have sum greater than $T_{n}$ for $n \geq 5$ , since $1 / 3 + 1 / 7 + 1 / 15 = 19 / 35 > 1 / 2$ and their total floor error is less than $3$ . Consequently, adding the terminal weight to a lower old word cannot pass the next corrected old word. Moreover, the increment $T_{n + 1} - 4 T_{n} = g$ is smaller than this separation, so no old word beyond the adjacent upper one can fit. It is therefore enough to compare the two adjacent old words and the terminal weight. The upper word fits precisely under the condition $U$ ; its remainder is then too small to take the terminal weight. Otherwise the lower word is used, and comparison with the terminal weight gives $M$ or $R$ . Substitution yields the displayed formulas. ◻

Only the $M$ and $R$ formulas have the asserted small correction to a fixed affine map. To rewrite $U$ in that form, let $γ_{n} = r + o - 2^{n + 1} \geq 0$ be the excess of the actual adjacent spacing over its lower bound. Then

λ_{n + 1} = 2 λ_{n} - 2 - \frac{4 γ_{n} + p^{+}}{2^{n + 1}} (U) .

There is no $O (n)$ bound on $γ_{n}$ supplied by the separation inequality. At $n = 13$ , exact calculation gives

r = 15147, o = 3997, p^{+} = 4, γ_{13} = 2760, rem (14) = 392.

Thus $λ_{14} - 2 λ_{13} + 2 = - 2761 / 4096$ . In particular, the proposed uniform error bound $| δ_{n} | \leq (2 n + 6) 2^{- n}$ in a three-branch signed-digit formula is false already at this row. The adjacent-spacing term cannot be absorbed into the divisor correction.

Iterating the $R$ formula is still legitimate along a run of right branches. Its accumulated correction and endpoint indices must be retained in a reset estimate. These exact recurrences do not identify the proposed ceiling $L_{r} < (r - 3) / 2$ with a converse to the conditional reset square-root estimate.

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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