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.

Start here, by the editorial route

In this paper

Whether one half is represented

Definition 12.2 (Half membership).

HALF := \frac{1}{2} \in A .

$HALF$ is a single instance of the negation of $U 257$ : any witness $A$ with $0 \notin A$ , $X_{A} (2) = 1 / 2$ is infinite (finite_boolSupport_ne_half; finite supports have odd-denominator value, so can never equal a dyadic rational), hence a genuine counterexample to $U 257$ . Thus $HALF \Rightarrow \neg U 257$ . The reverse implication is not established: failure of half-membership would not decide $U 257$ if false, since $U 257$ could still fail at a different target $t \in Q$ . $HALF$ is open. It is logically equivalent, by an unconditional Lean theorem, to a purely combinatorial statement about one fixed orbit:

HALF ⟺ the set of ranks skipped by the greedy rule for \frac{1}{2} is infinite

(half_mem_mersenneAchievementSet_iff_greedySkippedSupport_infinite; the reverse direction holds for every target, and the forward direction uses only that the target is rational, so the same theorem holds verbatim for every rational $t \geq 0$ in place of $1 / 2$ ).

Conditions equivalent to half-membership

The following three conditions are equivalent to half-membership. They express the same unresolved assertion in different terms. The proofs of equivalence identify the content still to be established; they do not compare the difficulty of possible proofs.

Proposition 12.3 (Infinitely many positive skips are equivalent to half-membership). Define

CPGS := \forall N, \exists c \geq N, c skipped by the rational half-greedy orbit \land 0 < r (1 / 2) (c - 1)

The positivity conjunct is unconditionally true for every skipped $c$ (an odd-denominator parity fact), hence deletable without changing the Prop’s meaning. What survives, $\forall N \exists c \geq N$ skipped, is exactly $(greedyMersenneSkippedSupport (1 / 2)) . Infinite$ , so

CPGS ⟺ HALF .

The implication cofinalExactLocalMersenneHalfRows_of_positiveHalfGreedySkips therefore gives an equivalent formulation, not a proof of membership.

Proposition 12.4 (Terminal carry bounds are equivalent to half-membership). Consider finite sets $D \subseteq {2, \dots, M}$ at arbitrarily large depths $M$ with

| ihc (D, M - 1) | \leq 2 ⌊ \sqrt{M} ⌋ + 4.

This is the exact terminal-strip condition. It is equivalent to $1 / 2 \in A$ : the forward direction follows from the displayed finite-approximation estimate; for the converse, truncate an achieving support at $M = k^{2}$ and apply Lemma 2.6. Thus constant $4$ already suffices at cofinally many depths. The relaxed constant $6$ gives a bound at every depth, using $\sqrt{M} \leq ⌊ \sqrt{M} ⌋ + 1$ , but that relaxation is not needed for the cofinal statement.

The existence condition is therefore not logically weaker than half-membership. An equivalent formulation may still be useful for finding a proof.

Observation 12.5. Propositions 12.3 and 12.4 do not decide half-membership. They show that the proposed existence conditions are equivalent to it, not that those formulations have no possible use. The upper bound on the centred carry in greedy_half_infinite_of_mobiusCenteredHalfCarry_sqrtBound has the same equivalence for the greedy support $A = G$ . For arbitrary $A$ with $1 \notin A$ , an upper bound alone yields only $x_{A} \geq 1 / 2$ , as the identity $ihc (A, N) = 2^{N + 1} (1 / 2 - x_{A}) + T (N + 1)$ shows. The greedy inequality $x_{G} \leq 1 / 2$ is what turns that conclusion into equality.

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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