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

Conditional counterexamples at rational targets

Finite supports whose scaled terminal carries tend to zero give an infinite $A$ with $X_{A} (2) = 1 / 2$ (exists_rational_half_counterexample_of_terminalScaledVanishing), and hence refute the universal statement (not_universal_of_terminalScaledVanishing; the universal statement itself has its own UniversalMersenneSubseriesIrrationality). The required finite supports have only been checked to a large finite cutoff. This restates the open half-membership question; it is not a new result, since any representation of $1 / 2$ already refutes the universal statement.

A Lean theorem also proves an unconditional finite-support obstruction at the second rational target $1 / 21$ :

F \subseteq {n : n \geq 2} ⟹ \sum_{n \in F} \frac{1}{2^{n} - 1} \neq \frac{1}{21}

for every finite $F$ (finiteErdosSum_ne_one_div_twenty_one). The denominator-order identity forces $lcm (F) = 6$ , after which the eight subsets of ${2, 3, 6}$ are exact arithmetic. This removes finite termination for that target; it does not establish achievement-set membership.

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.

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