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

Three combined conditional arguments

Entry A has a Lean proof, and entry B is an ordinary argument about what its hypotheses would imply. The facts behind entry C, the survival equivalence of Theorem 5.10 and the one-sided certificates, are proved in Lean. None of the three establishes an unproved orbit hypothesis or decides #257.

Theorem 9.46 (A: a sufficient lower bound at every reset). Assume that $| rem (r + 1) - 2^{r + 1} | > 2^{(r + 5) / 2}$ for every upper or middle reset $r \geq 10$ . Then $1 / 2 \in A$ .

Here is the role of the constants. Put

B (r) = 2^{⌊ (r + 4) / 2 ⌋} + 2 r + 3.

For $r \geq 10$ , one has $B (r)^{2} \leq 2^{r + 5}$ . A right branch at the first crossing of the largest omitted rank $r$ would force $| Δ_{r + 1} | \leq B (r)$ , by tracing the exact affine recurrence back to that reset. The assumed strict lower bound excludes this crossing. The largest-skip induction then gives half-membership. The hypothesis concerns every actual reset beyond the threshold; neither a long finite check nor an asserted run-length asymptotic supplies it.

Remark 9.47 (B: rigidity of dangerous resets). For a dangerous reset preceded by a sufficiently long pure right run, Observation 12.9 confines the deviation at the preceding reset to at most two adjacent integers determined by the divisor-pulse stream. Applying that observation along a chain of $n$ dangerous resets gives $n - 1$ nested integer conditions, provided each intervening run meets its length threshold. The argument leaves a dangerous reset after a short run untouched. In particular, failure of square-root escape alone has not been shown to produce such a chain. No probability model is supplied and the pinned values are not excluded. This is an ordinary argument under the stated long-run hypotheses.

Theorem 9.48 (C: the one-sided finite decision boundary). The equivalence of Theorem 5.10 separates two logically different kinds of evidence. A fatal greedy gap found at a finite rank is a finite certificate that $1 / 2 \notin A$ . By contrast, membership requires survival at every rank; this paper supplies no finite certificate that the orbit survives forever and no completion theorem turning a long surviving prefix into membership.

The reset-crossing hypotheses, truncation-rung ladder, and sharp-capacity inequalities feed sufficient conditions for membership. They are not proved equivalent to one another, nor is failure to find one at a given depth evidence of survival. Consequently these searches give at most a semi-decision procedure for nonmembership of the specific value $1 / 2$ . Halting does not prove the universal statement in #257, and non-halting does not prove that $1 / 2$ is represented. In particular, neither truth nor falsity of the universal problem is shown semi-decidable here. Deeper negative search alone establishes nothing beyond the tested finite ranks; this is the quantifier boundary of Theorem 2.11.

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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