Reciprocal Mersenne Subseries

Will Cook

Long paper

Reciprocal Mersenne Subseries

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

PDFOpen the PDF, 152 pages TeXLaTeX source Source on GitHub Problem page

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.

Start here, by the editorial route

In this paper

Doubling the endpoint

Theorem 6.95 (The binary bound after doubling). Let $n \geq 6$ and $D \subseteq {2, \dots, n}$ satisfy $2 \in D$ , $Q (D, n) = 2^{n - 1} - 1$ , and $X_{D} (2) < 1 / 2$ . Then

S (D, 1, 2 n - 1) < 2^{n - 1} .

The missing quotient can therefore be filled using only the ranks $n + 1, \dots, 2 n - 1$ , without changing $D$ .

Proof. Write $F_{M} = \sum_{d \in D} {2^{M} / (2^{d} - 1)}$ and $P = 2^{n - 1}$ . The exact row at $n$ gives $2^{n} (1 / 2 - X_{D} (2)) = 1 - F_{n}$ . At $M = 2 n - 1$ this becomes

S (D, 1, 2 n - 1) = P (1 - F_{n}) + F_{2 n - 1} - 1.

The below-half assumption makes this an ordinary integer difference, so no truncation is needed. Since $2 \in D$ , one has $F_{n} \geq 1 / 3$ ; also $F_{2 n - 1} \leq | D | \leq n - 1$ . Therefore

S (D, 1, 2 n - 1) \leq \frac{2}{3} P + n - 2 < P,

using $3 (n - 2) < 2^{n - 1}$ for $n \geq 6$ . The ranks $n + 1, \dots, 2 n - 1$ supply exactly $n - 1$ binary digits, with total capacity $2^{n - 1} - 1$ . The binary expansion of the nonnegative missing quotient gives the required extension. ◻

For example, $n = 6$ and $D = {2, 3, 6}$ give $X_{D} (2) = 31 / 63$ , $Q (D, 11) = 1006$ and missing quotient $1023 - 1006 = 17$ . The ranks $7$ and $11$ supply $16 + 1$ . The resulting exact row $E = {2, 3, 6, 7, 11}$ has $X_{E} (2) - 1 / 2 = 13955 / 32756094 > 0$ . Thus the extension is exact at its integer target, but need not remain below half in real value.

Proposition 6.96 (Every exact sum contains exponent two). Let $n \geq 3$ and let $D \subseteq {2, \dots, n}$ satisfy $Q (D, n) = 2^{n - 1} - 1$ . Then $2 \in D$ . Thus the rank-two hypothesis in Theorem 6.95 is automatic for its exact rows. The interval restriction on $D$ is part of the assertion, not an assumption about arbitrary finite quotient sums.

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.

Cite and contact. Cite this paper by its title, author and date above, with its PDF; cite the earlier sources it uses for a mathematical result. For software, use the release citation and give the commit used. Contact Will with questions or corrections.

Prefer the manuscript?

PDFOpen the PDF, 152 pages TeXOpen the LaTeX source Problem page

Equations typeset from the exact TeX. Rendered from the LaTeX manuscript source with file fingerprint sha256:47cf6a152738f17c. Its file fingerprint matched the published source record, so the PDF above, the LaTeX, and this page are one manuscript. Redeploying the site regenerates this page from whatever the public repository holds at that moment.