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

The two finite costs

Let $F$ be a finite nonempty subset of $N_{> 0}$ , and write

Q = lcm (F), f_{F} (n) = # {a \in F : a ∣ n}, U_{F} (N) = \sum_{r \geq 1} 2^{- r} f_{F} (N + r) = \sum_{a \in F} \frac{2^{N mod a}}{2^{a} - 1} .

For $0 < t \leq 1$ define, with natural logarithms,

\begin{matrix} (17) & κ_{1} (F; t) = min {\sum_{d ∣ Q} \frac{c_{d}}{d} : c_{d} \geq 0, \log (1 + f_{F} (s) / t) \leq \sum_{d ∣ s} c_{d} (s ∣ Q)} . \end{matrix}

The constraints also hold at every positive integer $n$ , by replacing $n$ with $gcd (n, Q)$ . The programme is finite and feasible; a bounded cost sublevel bounds each coefficient, so its minimum is attained. For $L \geq 1$ , let $P_{L}$ be uniform sampling of $N = L m$ over a period of $U_{F} (L m)$ , for which $Q / gcd (Q, L)$ is a permissible length. This notation does not assert that the displayed period is least.

We use $K_{*} (F)$ for the infimum of the existing fractional-cover cost

\begin{matrix} (18) & K = \sum_{j} \frac{C_{j} η_{j}^{- α_{j}}}{2^{α_{j}} - 1}, C_{j} = \sum_{d \geq 1} \frac{c_{j, d}}{d}, \end{matrix}

over finite or countable families of finite sets $F_{j}$ of positive integers covering $F$ , exponents $0 < α_{j} \leq 1$ , weights $η_{j} > 0$ with $\sum_{j} η_{j} = 1$ , and coefficients $c_{j, d} \geq 0$ satisfying $f_{F_{j}} (n)^{α_{j}} \leq \sum_{d ∣ n} c_{j, d}$ for every $n \geq 1$ . The covering sets need not be disjoint or contained in $F$ . Covers of infinite cost do not lower the infimum, and a finite-cost cover exists, for example the single set $F$ with exponent one. This is the fractional cost of Theorem 1.3, not a redefinition of the logarithmic cost. The two costs optimise different majorants. The quantity $κ_{1}$ uses one logarithmic majorant of $f_{F}$ , whereas $K_{*}$ also permits splitting the support, varying the fractional exponents and allocating positive weights. The comparison below allows all of those choices.

Theorem 13.1 (arithmetic logarithmic counterexample). For every integer $H \geq 2$ and every real $A_{0} \geq 0$ , there are a squarefree positive integer $L$ and a finite nonempty set $F$ of distinct squarefree positive integers such that

min F > max {L, A_{0}}, κ_{1} (F; 1) \leq \frac{30 \log 2}{H}, P_{L} (U_{F} > 1) \geq 1 - e^{- 1} .

For these $F, L$ , some $R \geq 0$ satisfies $D_{L; R, L} 1_{{U_{F} > 1}} > 1 / 2$ . Consequently, no absolute constant $C$ satisfies

\begin{matrix} (E) & D_{L; R, M} 1_{{U_{F} > t}} \leq C (1 + L / M) κ_{1} (F; t) \end{matrix}

uniformly in finite $F$ , positive integers $L, M$ , integers $R \geq 0$ , and $0 < t \leq 1$ .

Proof. We choose the modulus so that several small divisibility conditions hold automatically. Fresh tag primes make the remaining conditions independent. Taking all least common multiples then makes the divisor count exponential in the number of tags. Its logarithm has small cost, but the shifted potential exceeds one at sufficiently many offsets to prevent the proposed uniform bound.

Choose a finite set of primes $P$ such that, with

L = \prod_{p \in P} p, D = τ (L), B = \prod_{p \in P} (1 + 1 / p),

we have $B \geq H^{2}$ . Such a set exists because the sum of reciprocal primes diverges and $\log (1 + 1 / p) \geq 1 / (2 p)$ . Here $τ$ counts positive divisors and $σ$ denotes their sum. Complementary divisors satisfy

\frac{σ (r)}{r} \frac{σ (L / r)}{L / r} = B (r ∣ L) .

Thus the set $G = {r ∣ L : σ (r) / r \geq H}$ has at least $D / 2$ elements. Also $D \geq B \geq H^{2}$ .

Choose a finite set $B$ of parent primes, all exceeding $max {L, H, D, A_{0}}$ , with

\frac{4}{D} \leq \sum_{q \in B} \frac{1}{q} \leq \frac{5}{D} .

Divergence of the prime reciprocal sum after any finite cutoff permits stopping at the first crossing of $4 / D$ ; the overshoot is less than $1 / D$ . For each $q \in B$ and $d ∣ L$ , choose a finite set $T_{q, d}$ of tag primes, all greater than $H$ , disjoint from $P$ , from $B$ , and from every other tag set, such that

\frac{4 d}{H} \leq S_{q, d} := \sum_{p \in T_{q, d}} \frac{1}{p} \leq \frac{5 d}{H} .

The same stopping argument works successively: each new reciprocal is less than $1 / H \leq d / H$ , and only finitely many primes have been excluded. No bound on the size of these prime sets is needed.

Let $I_{q} = {(d, p) : d ∣ L, p \in T_{q, d}}$ , and define

F_{q} = {q lcm {d p : (d, p) \in I} : I \subseteq I_{q}}, F = ⋃_{q \in B} F_{q},

where the least common multiple of the empty set is one. Every tag prime occurs in just one labelled pair. Its presence in the least common multiple therefore recovers whether that pair was selected. Hence different subsets give distinct exponents. All exponents are squarefree; different parent primes give disjoint sets, since no parent prime occurs in any other set. Their minimum is greater than $max {L, A_{0}}$ . The nonempty tag sets with $d = L$ also show $L ∣ Q$ .

For an integer $n \geq 1$ put $Z_{q} (n) = \sum_{(d, p) \in I_{q}} 1_{{d p ∣ n}}$ . The least-common-multiple encoding gives the exact identity

f_{F_{q}} (n) = 1_{{q ∣ n}} 2^{Z_{q} (n)} .

Indeed, when $q ∣ n$ , a selected least common multiple divides $n$ exactly when every selected pair does. Consequently

\log (1 + f_{F_{q}} (n)) \leq (\log 2) (1_{{q ∣ n}} + \sum_{(d, p) \in I_{q}} 1_{{q d p ∣ n}}) .

All moduli in this majorant divide $lcm (F_{q})$ . Its cost is at most

\begin{matrix} (19) & \frac{\log 2}{q} (1 + \sum_{d ∣ L} \frac{S_{q, d}}{d}) \leq \frac{\log 2}{q} (1 + 5 D / H) . \end{matrix}

Now sample $N$ uniformly modulo $Q$ , subject to $L ∣ N$ . For $r \in G$ , the event $q ∣ N + r$ has probability $1 / q$ . For a fixed parent these events are disjoint as $1 \leq r \leq L < q$ . Conditional on one such event, the tag-prime events remain independent by the Chinese remainder theorem. Since $d ∣ N + r$ is equivalent to $d ∣ r$ , the mean of $Z_{q} (N + r)$ is

μ_{r} = \sum_{d ∣ r} S_{q, d} \geq \frac{4}{H} σ (r) \geq 4 r,

and its variance is at most $μ_{r}$ . Chebyshev’s inequality gives

P_{L} (Z_{q} (N + r) < r ∣ q ∣ N + r) \leq \frac{μ_{r}}{(μ_{r} - r)^{2}} \leq \frac{4}{9 r} \leq \frac{4}{9} .

Thus success has conditional probability at least $1 / 2$ . On success the $r$ th term of $U_{F_{q}} (N)$ is at least $2^{- r} 2^{r} = 1$ . Later positive terms, for example those supplied by the exponent $q \in F_{q}$ , make $U_{F_{q}} (N) > 1$ strictly. The event $E_{q}$ of success for some $r \in G$ consequently satisfies

P_{L} (E_{q}) \geq \frac{| G |}{2 q} \geq \frac{D}{4 q} .

Different parent groups depend on disjoint prime coordinates, so the $E_{q}$ are independent under this conditional uniform measure. It follows that

P_{L} (U_{F} > 1) \geq 1 - \prod_{q \in B} (1 - P_{L} (E_{q})) \geq 1 - \exp (- \frac{D}{4} \sum_{q \in B} \frac{1}{q}) \geq 1 - e^{- 1} .

Since $f_{F} = \sum_{q} f_{F_{q}}$ and $\log (1 + \sum_{q} x_{q}) \leq \sum_{q} \log (1 + x_{q})$ for $x_{q} \geq 0$ , we may sum the positive majorants in (19):

κ_{1} (F; 1) \leq \frac{5 \log 2}{D} (1 + 5 D / H) \leq \frac{30 \log 2}{H} .

Since $L ∣ Q$ , write $P_{0} = Q / L$ and $p = P_{L} (U_{F} > 1)$ . The bounded periodic sequence $1_{{U_{F} (L m) > 1}}$ satisfies, for every integer $T \geq 1$ ,

| \frac{1}{T} \sum_{m = 1}^{T} 1_{{U_{F} (L m) > 1}} - p | \leq \frac{P_{0}}{T} .

Summing over the dyadic window gives

| D_{L; R, M} 1_{{U_{F} > 1}} - p | \leq \frac{2 P_{0}}{M 2^{R}} .

Take $M = L$ and $2^{R} \geq 16 P_{0}$ . The error is at most $1 / 8$ , whereas $p \geq 1 - e^{- 1} > 5 / 8$ , so the mean exceeds $1 / 2$ . A putative bound (E) would instead give at most $60 C \log 2 / H$ . Choosing an integer $H > 120 C \log 2$ contradicts it. ◻

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.