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

Integer-quotient tests for one twenty-first

For $x = 1 / 21$ , nonmembership forces the real greedy sequence eventually to select every exponent. The integer quotient rows then agree with that sequence at growing cutoffs. Let $F_{21}$ denote the conjunction consisting of a fatal greedy gap, finite skipped support (hence cofinite selection), eventual alignment between the quotient rows and the rational greedy prefix, and eventual visits to every doubling block. Then

\frac{1}{21} \in A ⟺ F_{21} does not hold

(one_div_twenty_one_mem_iff_not_fatalAlignedBranch).

On the fatal branch the canonical quotient remainder $s_{R}$ is eventually strictly above the closed capacity $2^{R}$ (twentyOneFatalAlignedBranch_eventually_strict_supercapacity), and thereafter the support appends $R + 1$ while $s_{R}$ follows one explicit affine recurrence (twentyOneFatalAlignedBranch_eventually_affine_supercapacity). Conversely, closed rows $s_{R} \leq 2^{R}$ along any unbounded sequence decay after normalisation and put $1 / 21$ in the achievement set (twentyOneCofinalEvenQuotientGreedyDecay_of_closedRows; one_div_twenty_one_mem_mersenneAchievementSet_of_cofinalGreedyDecay).

We give the definitions and the complete ordinary proof, including the rounding estimate and the agreement through rank $2 R$ needed in the short paper. Put $T_{M} = ⌊ 2^{M} / 21 ⌋$ and $q_{M} (d) = ⌊ 2^{M} / (2^{d} - 1) ⌋$ . Starting with $T_{2 R}$ , apply the integer greedy rule to the weights $q_{2 R} (d)$ , in increasing order of $d$ , selecting a weight exactly when it does not exceed the current remainder. Let $D_{R}$ be the selected exponents in ${2, \dots, R}$ , and let

s_{R} = T_{2 R} - \sum_{d \in D_{R}} q_{2 R} (d) .

The alignment in $F_{21}$ refers to the full integer word through $2 R$ agreeing with the real greedy word through $2 R$ , for all sufficiently large $R$ . Its other clauses are a finite fatal gap $r_{n} (1 / 21) > R_{n}$ with all later exponents selected, and a selected exponent in every sufficiently late block $(K, 2 K]$ . On this conjunction, the eventual recurrence is

\begin{matrix} (15) & \begin{aligned} s_{R + 1} = & 4 s_{R} + ⌊ \frac{4 (4^{R} mod 21)}{21} ⌋ - 2 c_{D_{R}} (2 R + 1) - c_{D_{R}} (2 R + 2) \\ - (2^{R + 1} + 1), D_{R + 1} = D_{R} \cup {R + 1}, s_{R} > 2^{R} . \end{aligned} \end{matrix}

Proof. Put $x = 1 / 21$ , let $G = G_{1 / 21}$ , and write $δ = x - X_{G} (2) \geq 0$ . If $x \notin A$ , then $δ > 0$ and $r_{n} ↓ δ$ . Since $w_{n} \to 0$ , every sufficiently large exponent is selected. Choose $n$ beyond that point with $R_{n} < δ$ . This proves the first clause of $F_{21}$ , and cofinite support gives its third clause.

We verify the integer comparisons as well. Set $B = 2^{2 R}$ . When the integer and real rules agree through $d - 1$ , their remainders after normalisation by $B$ differ by at most $d / B$ , because each floor changes its argument by less than one. The normalised integer weight differs from $w_{d}$ by less than $1 / B$ . Each fixed real comparison is strict: equality would give a finite representation of $x$ , and hence $x \in A$ . Choose $d_{0}$ such that $w_{d} < δ / 2$ for $d > d_{0}$ . For large $R$ the finitely many comparisons through $d_{0}$ agree, and $2 R / B < δ / 2$ . At every subsequent step $d \leq 2 R$ , the normalised integer remainder is at least $δ - d / B > δ / 2 > w_{d}$ , so that both rules select $d$ . Induction proves the second clause. Conversely, a represented target is recovered by the real greedy rule, and its remainder always satisfies $r_{n} \leq R_{n}$ . Thus $F_{21}$ is impossible for a represented target, proving the membership equivalence.

For any $R$ , rounding the target and the selected weights gives

\begin{matrix} (16) & | x - X_{D_{R}} (2) - \frac{s_{R}}{4^{R}} | \leq \frac{R + 1}{4^{R}} . \end{matrix}

If $s_{R} \leq 2^{R}$ on an unbounded sequence, the finite subseries values $X_{D_{R}} (2)$ tend to $x$ . The achievement set is closed, so $x \in A$ . This proves the closed-return implication. A finite representation is excluded by the finite-denominator argument already given.

On $F_{21}$ , the integer comparison and cofinite support give $D_{R} = G \cap {2, \dots, R}$ and $D_{R + 1} = D_{R} \cup {R + 1}$ for all large $R$ . Equation (16) yields $s_{R} / 4^{R} \to δ > 0$ , and hence $s_{R} > 2^{R}$ eventually. For $d \geq 2$ , the finite geometric expansion of $q_{M} (d)$ gives

q_{M + 2} (d) = 4 q_{M} (d) + 2 1_{d ∣ M + 1} + 1_{d ∣ M + 2} .

Also

T_{2 R + 2} = 4 T_{2 R} + ⌊ \frac{4 (2^{2 R} mod 21)}{21} ⌋, q_{2 R + 2} (R + 1) = 2^{R + 1} + 1.

Subtract the contributions of $D_{R}$ and of the new exponent $R + 1$ to obtain (15). ◻

The remaining question is whether this eventual recurrence can occur for the actual greedy word, or whether closed returns occur at unbounded depths. Neither assertion is established here. Since no finite support represents $1 / 21$ , membership would give an infinite-support rational counterexample to the universal problem.

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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