Reciprocal Mersenne Subseries

Will Cook

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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

Equivalent formulations of half-membership

The six conditions below are equivalent to half-membership. Their common reason is the carry identity: a fixed nonzero difference from $1 / 2$ produces a term of exponentially growing magnitude, whereas the remaining coefficient tail is only of order $\sqrt{N}$ . An absolute carry bound of this order along an unbounded sequence therefore forces the difference to vanish. The proofs below distinguish a fixed support from finite approximants whose supports may vary; their quantifier patterns alone would not justify these equivalences. No claim is made that every conceivable sufficient condition is equivalent to the goal.

Lemma 2.4 (Carry and residual value). Let $A \subseteq N$ with $1 \notin A$ , put $δ := 1 / 2 - x_{A}$ , and let $ihc (A, N)$ denote the integer half-carry at level $N$ . Then

ihc (A, N) = 2^{N + 1} δ + T (N + 1), 0 \leq T (N + 1) \leq 2 \sqrt{N + 1} + 4,

where $T (m) = \sum_{r \geq 1} c_{A} (m + r) 2^{- r}$ .

Proposition 2.5 (A cofinal carry bound). If $| ihc (A, N) | \leq C \sqrt{N + 1} + C^{'}$ for arbitrarily large $N$ , with fixed constants $C, C^{'}$ , then $δ = 0$ . For the particular greedy support $A = G$ of $1 / 2$ , a one-sided upper bound $ihc (G, N) \leq C \sqrt{N + 1} + C^{'}$ already suffices.

Proof. Lemma 2.4 gives

2^{N + 1} | δ | \leq | ihc (A, N) | + T (N + 1) .

The right side is $O (\sqrt{N + 1})$ along the stated sequence, so division by $2^{N + 1}$ yields $δ = 0$ . For $A = G$ , the greedy partial sums never exceed $1 / 2$ , hence $δ \geq 0$ (erdosSupportSeries_greedyMersenneSupport_le). Since $T (N + 1) \geq 0$ , the one-sided upper bound gives the same conclusion. Conversely, $δ = 0$ makes the carry equal to $T (N + 1)$ , so the bound holds with $C = 2$ and $C^{'} = 4$ . This converse does not assert the bound for arbitrary prescribed constants. For a prescribed $A$ , the conclusion is $x_{A} = 1 / 2$ ; it is not merely membership witnessed by some other support. For $A = G$ , this equality is equivalent to $1 / 2 \in A$ by mem_mersenneAchievementSet_iff_greedy_survival. ◻

Lemma 2.6 (The terminal bound at square depths). Suppose $x_{A} = 1 / 2$ and $1 \notin A$ . Then for every $k \geq 1$ ,

ihc (A, k^{2} - 1) = T (k^{2}) \leq 2 k + 4 = B (k^{2}) .

Proof. By Lemma 2.4 with $δ = 0$ , $ihc (A, k^{2} - 1) = T (k^{2}) \leq 2 \sqrt{k^{2}} + 4$ . At a perfect square the real and truncated square roots agree exactly, $\sqrt{k^{2}} = k = ⌊ \sqrt{k^{2}} ⌋$ , so the real envelope coincides with the discrete strip $B (n) = 2 ⌊ \sqrt{n} ⌋ + 4$ (halfStripBound) with no slack required. ◻

At the depths $M = k^{2}$ , the real square root and the integer square root agree. Thus the reverse implication for the terminal bound uses the original constant $4$ , not a relaxed constant $6$ . Truncating an achieving support at $M = k^{2}$ leaves the carry at that depth unchanged, by integerHalfCarry_inter_Iic_eq_of_succ_le. An achieving half-support excludes rank one because $w_{1} = 1 > 1 / 2$ .

Proposition 2.7 (Six equivalent membership conditions). Each of the following is logically equivalent to $1 / 2 \in A$ , not strictly weaker:

$C_{G} (N) \leq 2 \sqrt{N} + 4$ for every $N \geq 0$ ;
there are arbitrarily large positive exponents outside $G$ ;
at arbitrarily large depths $n$ , some $D \subseteq {2, \dots, n}$ satisfies $Q (D, n) = 2^{n - 1} - 1$ ; no agreement between the sets $D$ at different depths is required;
at arbitrarily large $M$ , $ihc (G, M) \leq B (M + 1)$ ;
at arbitrarily large positive depths $M$ , some $D \subseteq {2, \dots, M}$ satisfies $| ihc (D, M - 1) | \leq B (M)$ ;
the half-target greedy remainder never exceeds the full remaining tail, so no finite fatal gap exists; the seven equivalent forms in the later classification express this same condition.

For (d) and (e), the reverse directions use the square depths in Lemma 2.6. For (a), membership gives $C_{G} (N) \leq 2 \sqrt{N + 1} + 3 \leq 2 \sqrt{N} + 4$ when $N \geq 1$ , and $C_{G} (0) = 0$ . The forward implication still needs the greedy inequality $x_{G} \leq 1 / 2$ .

Scope of the equivalences. These conditions are equivalent to membership, not strictly weaker substitutes. Equivalence does not make a formulation useless as an invariant, but it does mean that proving any of them is the same problem as (H). The equivalent conditions remain unproved. In particular, mutually incompatible witnesses at different endpoints do not buy a weaker sufficient condition, and the $Π_{2}^{0}$ shape of these hypotheses is not evidence of extra room.

Scope limit. Item (a) does not extend to every $A$ with $1 \notin A$ . Dropping the greedy-specific fact $x_{G} \leq 1 / 2$ yields only the one-sided statement that the bound in (a), with $A$ in place of $G$ , is equivalent to $1 / 2 - x_{A} \leq 0$ . Thus the conclusion for a general support is a one-sided inequality, not equality with $1 / 2$ .

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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