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

The object

In this part, write $x_{n} = w_{n} = (2^{n} - 1)^{- 1}$ for every positive integer $n$ , and $x_{A} = X_{A} (2) = \sum_{n \in A} x_{n}$ for $A \subseteq N_{> 0}$ . These series converge by comparison with $\sum_{n} 2^{- n}$ . Exponent $1$ contributes the rational number $x_{1} = 1$ , so deleting or adjoining it does not affect irrationality. It cannot occur in a representation of $1 / 2$ ; those calculations therefore start at exponent $2$ .

Definition 5.1 (Universal Problem #257). Is $x_{A} = \sum_{n \in A} 1 / (2^{n} - 1)$ irrational for every infinite $A \subseteq N_{> 0}$ ? The quantifier ranges over all infinite sets of positive exponents. It is OPEN.

The support theorems prove irrationality for specified infinite families. The following question instead asks whether one particular rational number is represented. A positive answer would disprove the universal assertion; a negative answer would settle only that target. The later analysis also treats $1 / 21$ .

Definition 5.2 (The distinguished half-value question). Define the Mersenne achievement set $A := {y \in R : \exists A \subseteq N, 0 \notin A, y = \sum_{n \in A} x_{n}}$ , the set of reals realizable as a subset sum of the positive-index Mersenne weights ${x_{n}}_{n \geq 1}$ . Is $1 / 2 \in A$ ? This is OPEN.

Observation 5.3 (Why the half-value question controls universal #257). Any finite subset sum of ${x_{n}}$ has odd reduced denominator, since every $2^{n} - 1$ is odd; hence no finite $A$ can sum to $1 / 2$ (positiveMersenneSupportValue_coe_finset_ne_half). Consequently, if $1 / 2 \in A$ , the witnessing support $A$ is necessarily infinite, and $x_{A} = 1 / 2$ is rational. Such an $A$ is then a single infinite set for which the universal-#257 sum is rational – an outright refutation of universal #257. Conversely, universal #257 being true forces $1 / 2 \notin A$ as one instance among uncountably many. The half-value question is therefore one concrete test of the negation, not a restatement of the universal problem. Excluding $1 / 2$ would not exclude every other rational value, even though the support theorems already settle many infinite families. Representing $1 / 2$ would give an infinite-support counterexample to the universal assertion.

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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