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.

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In this paper

Excluding particular final dyadic intervals

The two-sided bound $min (rem (s), overshoot (s)) \leq 2^{s}$ for all $s \geq 5$ is conditional. It propagates by induction from a local hypothesis which at every row excludes three specified integer values on the middle branch and imposes one further inequality on the right branch:

\begin{matrix} s \geq 5, no carry and a middle transition ⟹ C_{s} \notin {- 3, - 2, - 1}; \\ s \geq 5, no carry and a right transition, overshoot (s) \leq 2^{s} \\ ⟹ charge (s) \leq 2^{s + 2} . \end{matrix}

Here $C_{s} = 4 rem (s) - p_{s}^{-} - 4$ . On a middle transition, $rem (s + 1) - 2^{s + 1} = C_{s} + 4$ ; on a right transition, $rem (s + 1) = C_{s} - 2^{s + 1}$ . Thus $C_{s}$ must not be identified with the next row’s centred remainder. Erdos249257.SeamTwoSidedDyadicCellEscape

Proposition 10.5 (The upper (carries) successor needs no exceptional-cell exclusion). Let $s \geq 5$ and suppose the upper branch occurs at row $s$ , that is, the transition at row $s$ is on branch $U$ of Theorem 5.16. Then

rem (s + 1) \leq 2^{s + 1} .

Indeed, the reset identity expresses $2^{s + 1}$ as $rem (s + 1)$ plus a nonnegative reset charge. Thus the upper branch needs no exceptional-value exclusion for the next-row two-sided invariant. This conclusion concerns that induction step, not the absence of upper transitions on the orbit.

Theorem 10.6 ( $C_{D} = - 3$ is impossible at the final middle transition, $D \geq 13$ ). Consider a hypothetical final middle transition: a middle transition at row $D \geq 13$ ( $\neg carries$ , and the middle-branch inequality $4 rem (D) + gap - belowPulse < terminalWeight$ holds at $D$ ), followed by an all-right tail forever after ( $\forall s \geq D + 1$ , $seamGreedyWord (s + 1) = seamGreedyWord (s) . extend true$ ). Under these assumptions,

belowPulse (D) + 2 \leq 4 \cdot rem (D),

i.e. $C_{D} \geq - 2$ ; the cell $C_{D} = - 3$ (and every more negative value) is excluded.

Proof of the bound. The all-right-tail assumption puts the terminal-augmented finite prefix strictly above $1 / 2$ . Consequently the producer carry is below its complete incidence tail, or equivalently $Z_{D} < 1 + 1 / (2^{D} - 1)$ , where $Z_{D} = rem (D) - seamWordFloorError (D)$ . The lower bound uses the other endpoint: under the same tail assumption, the finite value at $D$ plus the complete Mersenne tail remains below $1 / 2$ . The exact scaled remainder identity therefore gives

Z_{D} > 4^{D} \sum_{e > D} \frac{1}{2^{e} - 1} - 2^{D} > \frac{1}{3} .

The last inequality follows by retaining the first two geometric channels of the complete Mersenne tail. Put $E_{D} = seamWordFloorError (D)$ . Pulse absorption gives $belowPulse (D) \leq 4 E_{D}$ , so

4 rem (D) - belowPulse (D) \geq 4 (rem (D) - E_{D}) = 4 Z_{D} > \frac{4}{3} .

The left side is an integer, hence is at least $2$ , as required. The upper threshold inequality alone would not give this lower bound.

Scope. This excludes $C_{D} = - 3$ at a middle row $D \geq 13$ under the additional all-right-tail assumption. The general induction requires exclusion of all three cells at every middle row, not just under this extra hypothesis, together with its right-branch bound. The two statements share the same coordinate $C_{D}$ but have different hypotheses. Theorem 10.6 is universal over rows satisfying the additional all-right-tail assumption; it does not assert that such a row exists. The induction hypothesis concerns every middle row, without that tail assumption. Excluding $C_{D} = - 3$ in the former statement therefore does not supply the exclusion required in the latter. Object or representation: about the object under the stated hypothesis ; a genuine arithmetic consequence of the all-right-tail assumption via the fatal-gap orbit, not a coordinate artifact.

Corollary 10.7 (Only $C_{D} \in {- 2, - 1}$ remain among the exceptional dyadic cells). At a final middle row $D \geq 13$ , Theorem 10.6 removes $- 3$ from the exceptional set ${- 3, - 2, - 1}$ , leaving $- 2, - 1$ within that set. It does not assert $C_{D} \in {- 2, - 1}$ : nonnegative values have not been excluded. Proposition 10.5 handles upper transitions in the two-sided induction; it supplies no additional restriction on a middle coordinate. For the all-middle-row induction, the exclusion of all three values still needs proof without an all-right-tail assumption, together with the separate right-branch inequality.

Remark 10.8 (A sufficient tail inequality and its source). Let $U_{D}$ be the finite set of exponents selected by the below word at row $D$ , and put $F_{D} = U_{D} \cup {D}$ . With $c_{F_{D}} (n) = # {d \in F_{D} : d ∣ n}$ , define

Θ_{D} := \sum_{j \geq 1} c_{F_{D}} (2 D + 2 + j) 2^{- j} .

This is a tail sum, not a ratio. Since $0 \leq c_{F_{D}} (n) \leq | F_{D} |$ , it satisfies $0 \leq Θ_{D} \leq | F_{D} |$ . The integer $C_{D} = 4 rem (D) - belowPulse (D) - 4$ is the producer carry in the cited source. Under the final-middle and all-right-tail assumptions, that source proves $C_{D} < Θ_{D}$ : Erdos249257.middleProducer_allRight_forces_carry_lt_tail.

An inequality in the opposite direction would exclude such a final middle transition. The following sufficient condition retains the exception already handled by Theorem 10.6:

(TD) \forall D \geq 13, (D is a middle row and C_{D} \neq - 3) ⟹ Θ_{D} < C_{D} .

Because $Θ_{D} \geq 0$ , this condition in particular excludes $C_{D} = - 2$ and $C_{D} = - 1$ . It requires an inequality at the other indicated middle rows as well, so it is not merely a reformulation of excluding two integer values. Nor does it supply the separate right-branch hypothesis used in the two-sided induction.

This is the reduced middle tail condition of Part III, and half-membership follows from it (half_mem_mersenneAchievementSet_of_middleProducerTailEscapeExceptNegThree). The inequality itself is not established here. Its role is a conditional implication, not a newly identified necessary condition.

The two more concrete sufficient bounds in Theorem 6.7 imply the tail inequality even without the exception. Their Lean forms are Erdos249257.SeamMiddleProducerCardEscape and Erdos249257.SeamMiddleProducerRowEscape. The cardinality condition is

| U_{D} | + belowPulse (D) + 5 < 4 rem (D) .

It gives $| F_{D} | = | U_{D} | + 1 < C_{D}$ , hence $Θ_{D} < C_{D}$ . The simpler row condition $D \leq rem (D)$ is stronger: $belowPulse (D) \leq 2 | U_{D} |$ and $| U_{D} | \leq D - 2$ give $| U_{D} | + belowPulse (D) + 5 \leq 3 D - 1 < 4 D$ . The corresponding half-membership implication is Erdos249257.half_mem_mersenneAchievementSet_of_middleProducerCardEscape. No converse between these sufficient conditions is claimed.

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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