Reciprocal Mersenne Subseries

Will Cook

Long paper

Reciprocal Mersenne Subseries

Erdős #257 By Will Cook 30 September 2026 Authorship, AI use and citation

PDFOpen the PDF, 152 pages TeXLaTeX source Source on GitHub Problem page

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 quotients and their remainder identity

The quotient rows are finite integer approximations to the real greedy rule, not identical finite stages of that rule. Their normalised weights tend to the Mersenne weights and their targets tend to $1 / 2$ . Proposition 8.1 proves agreement at each fixed prefix once the row index is sufficiently large; it does not assert simultaneous agreement at all ranks. Fix a row $n \geq 6$ . The integer quotient weights at row $n$ are

w (n, d) := ⌊ \frac{4^{n}}{2^{d} - 1} ⌋, d = 2, \dots, n - 1,

and the integer target is $T_{n} := 2^{2 n - 1} - 2^{n}$ . The integer greedy process takes ranks $d = 2, \dots, n - 1$ in ascending order against $T_{n}$ exactly as in Definition 5.11, producing a take set $D_{n} \subseteq {2, \dots, n - 1}$ and skip set ${Skip}_{n} := {2, \dots, n - 1} \ D_{n}$ . Define the integer remainder $rem (n) := T_{n} - \sum_{d \in D_{n}} w (n, d)$ and the scaled remainder deviation $Δ_{n} := rem (n) - 2^{n}$ .

Definition 5.13 (The full-support quotient recurrence). For $M \geq 2$ , define

K (M) := 2^{M - 1} - \sum_{d = 2}^{M} ⌊ \frac{2^{M}}{2^{d} - 1} ⌋ .

This depends only on $M$ , not on the selected support in a quotient row. The floor-quotient recurrence gives

K (2) = 1, K (M + 1) = 2 K (M) - (τ (M + 1) - 1) .

Here $τ (n)$ counts the positive divisors of $n$ . To see the correction term, double the quotients for $2 \leq d \leq M$ and add one for each such $d$ dividing $M + 1$ . The new endpoint $d = M + 1$ contributes one more quotient. Together these are exactly the $τ (M + 1) - 1$ divisors at least two. This is an identity for every $M \geq 2$ , not merely a checked finite recurrence.

Theorem 5.14 (Decomposition of the integer remainder). For every quotient row $n \geq 6$ ,

\begin{matrix} (I) & Δ_{n} = K (2 n) + \sum_{d \in {Skip}_{n}} ⌊ \frac{4^{n}}{2^{d} - 1} ⌋ . \end{matrix}

Here the take and skip sets, capacity and remainder are those defined at the start of this subsection. The identity separates the contribution of the full exponent range from that of the omitted ranks. It makes no claim that there are few omitted ranks; controlling their sum is a separate part of any application.
Proof. For $d = n$ the quotient is $2^{n} + 1$ , and for $n < d \leq 2 n$ it is $2^{2 n - d}$ . Hence

\sum_{d = n}^{2 n} ⌊ \frac{4^{n}}{2^{d} - 1} ⌋ = 2^{n} + 1 + \sum_{j = 0}^{n - 1} 2^{j} = 2^{n + 1} .

Substitute this in the definition of $K (2 n)$ , split ${2, \dots, n - 1} = D_{n} ⊔ {Skip}_{n}$ , and use $Δ_{n} = 2^{2 n - 1} - 2^{n + 1} - \sum_{d \in D_{n}} w (n, d)$ . This gives (I). Certification: rows $6 \leq n \leq 200$ rechecked with zero mismatches by an exact program (§5.7).
Use: (I) permits substitution of information about $K (2 n)$ . A resulting estimate for $Δ_{n}$ must also retain or bound the skip-set sum; information about $K$ alone need not control their cancellation.

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.

Prefer the manuscript?

PDFOpen the PDF, 152 pages TeXOpen the LaTeX source Problem page

Equations typeset from the exact TeX. Rendered from the LaTeX manuscript source with file fingerprint sha256:47cf6a152738f17c. Its file fingerprint matched the published source record, so the PDF above, the LaTeX, and this page are one manuscript. Redeploying the site regenerates this page from whatever the public repository holds at that moment.