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

Four tested approaches to the reset estimate

Observation 10.1 (Prime steps can occur within a bounded drift window). Route as conceived. The universal recursion $K (M + 1) = 2 K (M) - (τ (M + 1) - 1)$ doubles at every step and subtracts the divisor-excess $τ (M + 1) - 1$ . At a prime $M + 1 = p$ , $τ (p) = 2$ so the subtracted term is $1$ ; the naive hope is that primality of an index forces the step to be "almost pure doubling," and that prime density (Chebyshev/PNT-scale) then bounds how long a run of near-pure-doubling steps can last, which would bound the danger run-length $L_{r}$ in the right-branch recurrence in Theorem 5.16. The upper branch has a different, adjacent-spacing correction, so a run-length argument cannot use the discarded uniform-error formula across resets.

A finite test. Use the skip set of the quotient row $n = 607$ from Section 5.4, and define the frozen-row drift

J_{607} (M) = K (M) + \sum_{d \in {Skip}_{607}} ⌊ \frac{2^{M}}{2^{d} - 1} ⌋ .

Exact integer arithmetic gives, in order at $M = 607, \dots, 617$ ,

5, 5, 4, 4, 6, 2, 3, 4, 5, 2, 3.

There are eleven values and ten transitions. The indices $607$ , $613$ , and $617$ are prime. For $M + 1 \geq 607$ the recurrence is

J_{607} (M + 1) = 2 J_{607} (M) - (τ (M + 1) - 1 - c_{{Skip}_{607}} (M + 1)) .

At a prime index the last coefficient is $1$ . The composite-index coefficients can compensate for those prime steps, as the displayed values illustrate. This table is a new finite arithmetic check; the longer earlier run search was not repeated.

Scope. Primality at some indices does not force this drift to leave a small interval immediately. The example supplies no universal run-length bound and no statistical conclusion about prime indices. It also does not exclude a proof that combines prime counts with quantitative information about the intervening divisor counts.

Observation 10.2 (Parity and perfect squares in the recurrence). Route as conceived. $K (M)$ is even exactly when $M$ is a perfect square (a true, provable fact). The hope was to use this parity law to forbid certain residues or run patterns in the digit stream of $K$ , since parity constraints are a classical first attack on digit-run questions.

What the parity calculation says. The recurrence implies $J (M + 1) \equiv δ (M + 1) (\mod 2)$ for every integer forcing sequence. For $K$ , the additional arithmetic fact that $τ (M)$ is odd exactly at squares gives the stated parity pattern. One should not call that pattern a property of every forcing sequence, or say that it excludes no residue: at each fixed index it does specify the parity.

Scope. Restating a parity consequence of a known recurrence does not add a new independent constraint. To turn it into a run-length bound would require further information about the sizes or arrangement of the forcing terms. No argument here rules out using parity together with such information.

Use of the observation. The parity law does constrain $K (M)$ at each specified index and is a useful check on any proposed run pattern. The missing implication is from this residue information to a quantitative bound on run lengths. No such implication is proved by the parity calculation alone.

Observation 10.3 (What the denominator estimate does not give). One proposed separation argument would prevent $\sum_{d \in {Skip}_{n}} x_{d}$ from approximating $C := E - \frac{3}{2} = 0.1066951524152917 \dots$ at scale $2^{- 3 n / 2}$ . An irrationality-measure bound is relevant only after relating the reduced denominator of that finite sum to $n$ .

For $n \geq 2$ and ${Skip}_{n} \subseteq {2, \dots, n - 1}$ , its reduced denominator $D_{n}$ satisfies

\log_{2} D_{n} \leq \sum_{d \in {Skip}_{n}} \log_{2} (2^{d} - 1) \leq \sum_{d \in {Skip}_{n}} d \leq \frac{n (n - 1)}{2} - 1.

The middle inequality is strict when the skip set is nonempty. This is an upper bound; it is not an asymptotic formula for $D_{n}$ . In particular, cardinality alone does not justify an asymptotic of order $n^{2} / 4$ .

Zudilin proves $μ (E) \leq 2.46497868 \dots$ [9]. Rational translation preserves the irrationality exponent, so for any fixed $ν > 2.46497868 \dots$ there is a constant $c_{ν} > 0$ such that every reduced rational $p / q$ , with $q \geq 1$ , satisfies $| C - p / q | \geq c_{ν} q^{- ν}$ . Using only the displayed denominator estimate therefore gives

| C - \sum_{d \in {Skip}_{n}} x_{d} | \geq c_{ν} 2^{- ν (n (n - 1) / 2 - 1)} .

This guaranteed lower bound is eventually smaller than $2^{- 3 n / 2}$ , so it does not prove the desired separation. A sharper estimate for the actual reduced denominators, or an approximation result using additional structure of these numerators, could change the comparison. Neither is excluded by this calculation.

Retaining only the parity of an input coefficient leaves considerable freedom. In $e^{'} = 2 e + c$ , every incoming integer $e$ admits the odd choice $c = - 1 - 2 e$ , which gives $e^{'} = - 1$ . The next proposition uses the same centred-remainder construction when more unit bits and a valuation are prescribed.

Proposition 10.4 (Centred completion of fixed-precision 2-adic data). Fix $u \geq 1$ and a finite list $(v_{i}, a_{i})_{0 \leq i < m}$ , where $v_{i} \geq 0$ are integers and the integers $a_{i}$ are odd. For every initial integer $e_{0}$ , there are integers $z_{i}, e_{i + 1}$ such that

e_{i + 1} = 2 e_{i} + 2^{v_{i}} (a_{i} + 2^{u} z_{i}), | e_{i + 1} | \leq 2^{v_{i} + u - 1} (0 \leq i < m) .

Thus one may prescribe the valuation and $u$ bits of the odd unit of each input coefficient while keeping every successor state in its stated centred interval. The initial state is arbitrary.

Proof. For a single step, put $M = 2^{v_{i} + u}$ and $R = M / 2$ , and choose

e_{i + 1} = ((2 e_{i} + 2^{v_{i}} a_{i} + R) mod M) - R .

The remainder convention $0 \leq x mod M < M$ gives $- R \leq e_{i + 1} < R$ . Also $e_{i + 1} - 2 e_{i} \equiv 2^{v_{i}} a_{i} (\mod M)$ , so this difference has the required form $2^{v_{i}} (a_{i} + 2^{u} z_{i})$ . Repeating the construction proves the finite-list statement. It uses no relation between the freely chosen quotients $z_{i}$ at different steps. ◻

The formal-source statements are fixedPrecisionTropicalNoGo, vu_step_has_centred_completion and vu_word_has_prefix_locked_completion. They concern this unrestricted congruence model, not coefficients already fixed by a divisor-count sequence. Fixed-precision data alone cannot exclude every completion in this model. The proposition does not show that the completion comes from a Mersenne support, satisfies additional arithmetic relations, or belongs to the particular greedy orbit. In particular, it is not a general obstruction to every fixed-window argument for Problem 249 or 257.

These observations have separate scopes: a finite prime-window example, a parity consequence of a known recurrence, an insufficient denominator estimate, and a completion theorem for specified fixed-precision symbols. They do not exhaust the possible approaches to the reset estimate or force every proof to use digit-carry machinery. The next section studies that machinery because it gives explicit conditional implications, not because other methods have been ruled out.

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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