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

Exact identities and reductions

Observation 6.26 (A finite half-value approximation and the first correction tail). The finite prefix has value

\frac{1}{2} - X_{{2, 3, 6, 7}} (2) = \frac{1}{16002} .

The excess of the full tail after rank $1$ over its dyadic value is

R_{1} - \frac{1}{2} = \sum_{n \geq 2} (\frac{1}{2^{n} - 1} - 2^{- n}) \leq \frac{1}{12} + \frac{1}{28} = \frac{5}{42} .

Indeed, expand each summand as $4^{- n} + \sum_{j \geq 3} 2^{- j n}$ ; for $n \geq 2$ , the second sum is at most $2 8^{- n}$ . Thus $\frac{1}{2} - (R_{1} - \frac{1}{2}) \geq \frac{8}{21} > \frac{3}{8}$ . The first identity is an exact rational check, whereas the two tail inequalities use a convergent-series estimate.

half_sub_four_term_prefix_eq mersenneCorrectionTail_one_le

Observation 6.27 (Period-4 sign weight vanishes on a residue class). Let $w (n)$ have successive values $1, 0, - 1, 0$ , repeated with period $4$ . For every $n \equiv 3 (\mod 4)$ ,

c_{w} (n) = \sum_{d ∣ n} w (d) = 0.

Indeed, $n$ is odd, and paired divisors $d, n / d$ have opposite values of $w$ because their product is $3$ modulo $4$ . There are no fixed pairs: such an $n$ is not a square. Thus an argument requiring a nonzero coefficient at every chosen residue must check that requirement; periodicity alone does not provide it. This cancellation does not conflict with irrationality of the full weighted series, which follows from Luca and Tachiya’s theorem [3].

intWeightedCoeff_periodFourSignWeight_eq_zero_of_mod_four_eq_three

Theorem 6.28 (The Lambert-series identity for the Möbius function). The absolutely convergent signed series satisfies

\sum_{d \geq 1} \frac{μ (d)}{2^{d} - 1} = \frac{1}{2} .

This classical identity is also recorded in [7]. Expand each denominator geometrically and interchange the absolutely convergent sums. The coefficient of $2^{- n}$ is $\sum_{d ∣ n} μ (d)$ , equal to $1$ for $n = 1$ and $0$ otherwise. This proves the identity. Its coefficients are $- 1, 0, 1$ , not indicators of a support; the identity therefore does not represent $1 / 2$ as a subseries with all coefficients in ${0, 1}$ .

Corollary 6.29 (Negative-Möbius Boolean support overshoots $1 / 2$ ). Let $N = {d \geq 2 : μ (d) = - 1}$ . Isolating the $d = 1$ term in Theorem 6.28 gives

X_{N} (2) = \frac{1}{2} + \sum_{\begin{matrix} d \geq 2 \\ μ (d) = 1 \end{matrix}} \frac{1}{2^{d} - 1} \geq \frac{1}{2} + \frac{1}{63} > \frac{1}{2},

since $μ (6) = 1$ . This rules out the particular candidate $N$ . It does not rule out other infinite supports representing $1 / 2$ .

Observation 6.30 (Dyadic safety need not survive a take). For a residual $p / (2 D)$ , taking rank $b$ subtracts $w_{b}$ and gives

\frac{p}{2 D} - \frac{1}{2^{b} - 1} = \frac{p (2^{b} - 1) - 2 D}{2 D (2^{b} - 1)} .

The residual $7 / 34$ is at most $2^{- 2}$ . Taking rank $3$ leaves $15 / 238$ , which is strictly between $2^{- 4}$ and $w_{4} = 1 / 15$ . Thus rank $4$ is skipped but fails the dyadic safety test. The displayed numerator has increased from $7$ to $15$ ; that increase does not preserve safety.

The same failure occurs with an initial denominator containing all Mersenne denominators through rank $7$ . With $D = lcm (3, 7, 15, 31, 63, 127) = 1240155$ , one has

\frac{12149}{2 D} - \frac{1}{255} = \frac{41179}{2 \cdot 21082635} .

The initial residual is at most $2^{- 7}$ . After this rank- $8$ take, ranks $9$ and $10$ are skipped, but the new residual exceeds $2^{- 10}$ . The exact numerator of that dyadic excess is $2^{9} \cdot 41179 - 21082635 = 1013 > 0$ . These are counterexamples to the stated local safety implication, not assertions that either state occurs on the real half-greedy orbit.

safeBracket_and_numeratorMonotone_not_inductive_fixture lcmSaturated_safeBracket_not_inductive_fixture

Theorem 6.31 (The half-skip dichotomy via Erdős–Borwein irrationality). The half target satisfies

\frac{1}{2} \in A ⟺ N_{> 0} ∖ G is infinite,

where $G$ is the greedy support for $1 / 2$ . In the forward direction, finitely many skipped exponents would express the full Mersenne sum as $1 / 2$ plus a finite rational sum, contradicting Erdős’s full-support irrationality theorem. The reverse implication uses the greedy tail criterion: once a remainder exceeds the whole available tail, every later exponent is selected. Infinitely many skips exclude that failure. The equivalence does not establish that infinitely many skips actually occur.

Theorem 6.32 (Equivalent descriptions of half-membership). Each of the following is equivalent to $1 / 2 \in A$ :

The integer-greedy sequence is not eventually always on branch $R$ .
There are arbitrarily large $s \geq 5$ with $s \notin D_{s + 1}$ .
Upper or middle transitions occur at arbitrarily large rows.
Some sequence $s_{j} \to \infty$ satisfies $s_{j} \notin D_{s_{j} + 1}$ for every $j$ .
There are rows $s_{j} \to \infty$ and omitted ranks $d_{j} \in {2, \dots, s_{j} - 1} ∖ D_{s_{j}}$ with $d_{j} \to \infty$ .
The positive ranks omitted by the real half-greedy rule form an infinite set.
The real half-greedy rule has no last omitted positive rank.

Items (1)–(4) differ only by the branch classification and the choice of a sequence of rows. Item (5) requires the omitted ranks themselves to tend to infinity, not merely their containing rows. The cited identification with the real greedy support supplies (6)–(7). These equivalences do not prove that any of the conditions holds.

Lemma 6.33 (The values on either side of an eventual right continuation). If the seam eventually always extends “true” (right branch) from some row $S$ on with a fixed lower prefix $u$ , the resulting cofinite-support value stays strictly below $1 / 2$ ; the matching alternative “upper competitor” word gives a strict excess above $1 / 2$ . Neither of these two cofinite continuations represents $1 / 2$ . This does not rule out an eventually-right integer orbit; it describes the two values in that case. The inequalities are used in Theorem 6.44.

Lemma 6.34 (Each Mersenne weight exceeds its remaining tail). For $n \geq 1$ , let $w_{n} = (2^{n} - 1)^{- 1}$ and $R_{n} = \sum_{j > n} w_{j}$ . Then

R_{n} = w_{n + 1} + R_{n + 1}, 2^{- n} < R_{n} \leq 2 w_{n + 1} < w_{n} .

For the upper bound, compare each tail term with $2^{1 - j} w_{n + 1}$ at index $n + j$ , $j \geq 1$ , and sum the geometric series. The last inequality follows by comparing $2 / (2^{n + 1} - 1)$ with $1 / (2^{n} - 1)$ . This strict term-versus-tail inequality is the separation hypothesis used in the greedy arguments.

It is not available for the totient weights $φ (n) / 2^{n}$ : the terms at indices $4$ and $5$ already sum to $2 / 16 + 4 / 32 = 1 / 4 = φ (3) / 2^{3}$ , and the remaining tail is positive. A general strict-tail argument therefore cannot be transferred to that sequence without a different hypothesis or proof.

Theorem 6.35 (Greedy survival and membership). With the greedy remainder $r_{n} (x)$ and complete tail $R_{n}$ ,

x \in A ⟺ x \geq 0 and r_{n} (x) \leq R_{n} for every n \geq 0.

If all inequalities hold, the nonnegative remainders tend to zero because $R_{n} \to 0$ , so the greedy partial sums converge to $x$ . Conversely, strict tail domination $w_{n} > R_{n}$ forces the greedy choices in any representation and therefore gives every inequality. The argument applies to positive summable weights satisfying that separation hypothesis; it is not a consequence of summability alone.

Lemma 6.36 (The two next-prefix intervals and their gap). Let $d \geq 0$ be an integer, let $u \subseteq {1, \dots, d}$ , and suppose the target $t$ lies in the interval $[X_{u} (2), X_{u} (2) + R_{d}]$ . At depth $d + 1$ , either $t$ lies in the lower interval

[X_{u} (2), X_{u} (2) + R_{d + 1}],

or in the upper interval

[X_{u} (2) + w_{d + 1}, X_{u} (2) + w_{d + 1} + R_{d + 1}],

or in the open gap between them. These alternatives are disjoint because $R_{d + 1} < w_{d + 1}$ , and exhaustive because $R_{d} = w_{d + 1} + R_{d + 1}$ . For $t = 1 / 2$ , Lemma 6.38 also excludes the interval endpoints, so all relevant comparisons are strict. Endpoint exclusion is not what makes the two child intervals disjoint; the strict-tail inequality does that.

Lemma 6.37 (A greedy gap excludes every representation). Let $d \geq 0$ be an integer and $u \subseteq {1, \dots, d}$ a finite prefix. If

X_{u} (2) + R_{d + 1} < t < X_{u} (2) + w_{d + 1},

then no support agreeing with $u$ through rank $d$ represents $t$ : omitting $d + 1$ leaves value at most the lower endpoint, while including it gives value at least the upper endpoint. Other length- $d$ prefixes have disjoint containing intervals by the first-difference argument using $w_{n} > R_{n}$ . Since the displayed gap lies inside the interval for $u$ , none of those prefixes can represent $t$ either. Thus a certified strict gap excludes every representation, not just one proposed continuation.

Lemma 6.38 (No finite Mersenne sum equals one half). For every finite $u \subseteq N_{> 0}$ , $X_{u} (2) \neq 1 / 2$ : its reduced denominator is odd. Also $X_{u} (2) + R_{d} \neq 1 / 2$ for every $d \geq 0$ , since $R_{d}$ is the irrational full Mersenne sum minus a finite rational sum. These observations exclude equality at the finite-prefix and complete-tail endpoints of Lemma 6.36.

Lemma 6.39 (Straddle words are canonical: they agree with the greedy prefix). Let $u \subseteq {1, \dots, d}$ satisfy $X_{u} (2) \leq 1 / 2 \leq X_{u} (2) + R_{d}$ . Then

u = G \cap {1, \dots, d},

where $G$ is the real greedy support for $1 / 2$ . At the first disagreement, if $u$ takes rank $k$ and the greedy rule skips it, then $X_{u} (2) > 1 / 2$ . In the opposite case, the greedy remainder before $k$ is at least $w_{k}$ , whereas the remaining contribution allowed by $u$ , including $R_{d}$ , is at most $R_{k} < w_{k}$ . Both cases contradict the displayed interval. Thus there is at most one such word at each depth; the lemma does not assert its existence at every depth.

Theorem 6.40 (Last-skip iff local fatality: a pointwise criterion). Keep $G$ for the selected half-greedy support, $r_{M} (1 / 2)$ for its remainder and $R_{M}$ for the complete tail after rank $M$ . A positive rank $M$ is the last omitted rank if and only if $M \notin G$ and $r_{M} (1 / 2) > R_{M}$ . Consequently,

\frac{1}{2} \in A ⟺ r_{M} (1 / 2) \leq R_{M} for every positive M \notin G .

Thus it suffices to check the tail inequality at the ranks actually omitted by the greedy rule. This is an equivalence on that fixed orbit; it does not establish the inequality at its unboundedly many possible omitted ranks.

Lemma 6.41 (Seam upper-or-middle classification). At a row $s \geq 5$ , the terminal rank is omitted in the next row, $s \notin D_{s + 1}$ , if and only if either

4 o_{s} + p_{s}^{+} \leq 2^{s + 1},

or this inequality fails and

4 rem (s) + 2^{s + 1} - p_{s}^{-} < 2^{s + 2} + 4.

These are exactly branches $U$ and $M$ of Theorem 5.16. On either branch the new terminal weight does not fit; on $R$ it does. This is an exact classification, not a conjecture inferred from observed branch labels.

Lemma 6.42 (The exact gap at the largest omitted rank). Let $2 \leq d < s$ and $2 s < 3 d$ . Write $W_{s} (E) = \sum_{e \in E} ⌊ 4^{s} / (2^{e} - 1) ⌋$ for a finite set $E \subseteq {2, \dots, s - 1}$ . For $u \subseteq {2, \dots, d - 1}$ , put $E_{-} = u \cup {d + 1, \dots, s - 1}$ and $E_{+} = u \cup {d}$ . Then

3 W_{s} (E_{-}) + 3 \cdot 2^{s + 1} + 2 \cdot 4^{s - d} + 4 = 3 W_{s} (E_{+}) .

The correction is independent of the common prefix $u$ . For the integer-greedy rows $s \geq 5$ , a right transition preserves the largest omitted rank, whereas an upper or middle transition makes $s$ the largest omitted rank in the next row.

Proof. For $d \leq e < s$ , the inequality $2 s < 3 d$ gives

⌊ \frac{4^{s}}{2^{e} - 1} ⌋ = 2^{2 s - e} + 4^{s - e} .

Subtract the sum over $d < e < s$ from the term at $d$ . The powers of $2$ leave $2^{s + 1}$ ; the powers of $4$ leave $(2 \cdot 4^{s - d} + 4) / 3$ . The common prefix $u$ cancels. The branch statements follow from the support updates in Theorem 5.16: the right branch appends $s$ to the lower word, and the other two branches omit $s$ . ◻

For example, $s = 5$ , $d = 4$ and $u = {2}$ give $W_{s} (E_{-}) = 341$ and $W_{s} (E_{+}) = 409$ ; the displayed identity is $3 \cdot 341 + 204 = 3 \cdot 409$ . Both the factor $3$ on the right and the restriction $2 s < 3 d$ are essential to this formula.

Theorem 6.43 (Reduction to the nearest dyadic boundary). Let $d, E$ be nonnegative integers with $E \leq 2^{d + 1}$ , and let $j_{*}$ be the largest $j \in {0, \dots, d}$ for which $E \leq 2^{d - j + 1}$ . The set is nonempty, so $j_{*}$ is well defined, including when $E = 0$ . Then

\begin{aligned} \forall j \in {0, \dots, d}, 2^{d - j + 1} < E or E + 2 (d + j) \leq 2^{d - j + 1} \\ ⟺ E + 2 (d + j_{*}) \leq 2^{d - j_{*} + 1} . \end{aligned}

For $j > j_{*}$ the first alternative holds by maximality. For $j \leq j_{*}$ , the left side of the required inequality increases with $j$ and its right side decreases, so the condition at $j_{*}$ implies all the others. This is an elementary reduction from $d + 1$ inequalities to one. Specialized to the seam reset charge (where the actual upper-reset condition gives $E \leq 2^{d + 1}$ , as explained in Proposition 6.80), the reduced hypothesis at the single index $j_{*}$ is proved logically equivalent to Theorem 6.10’s band-avoidance hypothesis.

Theorem 6.44 (Excluding minus three at a final middle transition). Suppose that row $D \geq 13$ is a middle transition in Theorem 5.16 and every transition at a row $s \geq D + 1$ is right. Then

C_{D} := 4 rem (D) - p_{D}^{-} - 4 \neq - 3.

This excludes one particular value under the stated tail assumption; it neither excludes every final middle transition nor proves the all-middle-row hypothesis of Theorem 6.9.

To see the contradiction, complete $D_{D}$ by all ranks greater than $D$ . The resulting support $B = D_{D} \cup {D + 1, D + 2, \dots}$ has $X_{B} (2) < 1 / 2$ under the all-right assumption. The exact carry identity of the Lean proof (margin) gives $C_{B} (2 D + 1) = C_{D} + 3$ . If $C_{D} = - 3$ , this is zero. The recurrence $C_{B} (N + 1) = 2 C_{B} (N) + 1 - c_{B} (N + 2)$ and nonnegativity from Theorem 6.11 first force $C_{B} (2 D + 2) = 0$ : indeed $2 D + 3 \in B$ , so $c_{B} (2 D + 3) \geq 1$ . Next $D + 2, 2 D + 4 \in B$ imply $c_{B} (2 D + 4) \geq 2$ , giving $C_{B} (2 D + 3) < 0$ , a contradiction. The stronger conclusion $C_{D} \geq - 2$ is proved in Theorem 10.6.

Lemma 6.45 (The signed position of a skipped greedy prefix). Let $s \geq 5$ be omitted by the real greedy support $G$ , and put

H_{s} = G \cap {2, \dots, s - 1}, f_{s} = \sum_{d \in H_{s}} ⌊ \frac{4^{s}}{2^{d} - 1} ⌋ - T_{s} .

Exactly one of the following holds:

\begin{array}{lll} f_{s} < 0 : & H_{s} = D_{s}, & f_{s} = - rem (s), rem (s) \geq 1; \\ f_{s} = 0 : & H_{s} = D_{s}, & rem (s) = 0; \\ f_{s} > 0 : & H_{s} = B_{s}, & f_{s} = o_{s} . \end{array}

Thus the real greedy prefix is one of the two adjacent integer words. This classifies its signed distance $f_{s}$ from $T_{s}$ ; it does not exclude the negative case. Nor does it identify $f_{s}$ with the different coordinate $C_{s} = 4 rem (s) - p_{s}^{-} - 4$ used in the final-middle argument.

Lemma 6.46 (Spacing of reverse-carry words). For $i = 1, 2$ , let integer sequences $a_{i}, b_{i}, u_{i}$ satisfy

b_{i} (m) + 2 u_{i} (m) = a_{i} (m) + u_{i} (m + 1) .

Let $k, L \geq 0$ be integers. Suppose that at index $k$ the coefficients agree and $b_{1} (k) - b_{2} (k) = 1$ . Suppose also that both coefficients and bits agree at indices $k + 1, \dots, k + L$ . Then

u_{1} (k + L + 1) - u_{2} (k + L + 1) = 2^{L} (2 (u_{1} (k) - u_{2} (k)) + 1) .

The factor in parentheses is an odd integer. If the two terminal carries have absolute values at most $B_{1}, B_{2}$ , respectively, then $2^{L} \leq B_{1} + B_{2}$ . For a common bound $B$ the conclusion is $2^{L} \leq 2 B$ , not $2^{L} \leq B$ . The formula follows by subtracting the two recurrences: the initial bit difference gives the odd factor, and each subsequent agreement doubles it. An application must supply the stated agreements and terminal bounds.

Lemma 6.47 (Linear functionals factoring through one value). Let $V$ be a vector space over $Q$ , let $ev : V \to Q$ be linear, and choose $e \in V$ with $ev (e) = 1$ . Suppose that each linear functional $ℓ_{j} : V \to Q$ vanishes on $\ker (ev)$ . Then $v - ev (v) e \in \ker (ev)$ gives $ℓ_{j} (v) = ℓ_{j} (e) ev (v)$ . Consequently any finite evaluation matrix $(ℓ_{j} (v_{i}))_{i, j}$ is an outer product and has rank at most one. Every square minor of order at least two therefore vanishes. This excludes determinant arguments formed from these particular functionals, not determinant methods with additional independent information.

Theorem 6.48 (Two-thirds band: exact localisation of post-take skip-unsafety). Write a positive residual as $1 / R$ . A skipped rank $k$ passes the sufficient dyadic test precisely when $R \geq 2^{k}$ . Suppose a weight at rank $b$ is taken without exhausting the residual: with $q = 2^{b} - 1$ , assume $0 < R < q$ . The new reciprocal residual is $R q / (q - R)$ . If the next take is at rank $c \geq b + 2$ , put $m = 2^{c - 1}$ . The last skipped rank is dyadically unsafe exactly when

m - 1 < \frac{R q}{q - R} < m, equivalently \frac{q (m - 1)}{q + m - 1} < R < \frac{q m}{q + m} .

Clearing the positive denominators proves the equivalence. The interval has width $q^{2} / ((q + m) (q + m - 1))$ .

For a single skipped rank, $c = b + 2$ and $m = 2 q + 2$ . The width is then $q^{2} / ((3 q + 1) (3 q + 2)) < 1 / 9$ , and the interval lies inside $2 q < 3 R < 2 q + 2 / 3$ . In this single-skip case, the cited corollaries exclude integral $R$ and show that a reduced pre-take residual $p / (2 D)$ with $p, D, q$ odd can be unsafe only if $p \geq 7$ . The latter conclusion uses the divisibility by $4$ of $6 D - 2 p q$ under the unsafe-band inequalities.

These are local transition statements, not bounds on the actual half-greedy orbit. Dyadic safety is only sufficient for the current tail test. The odd-coprime data $(p, D, b) = (17, 41, 3)$ give an unsafe single-skip example; no conclusion that the half-greedy orbit avoids such data follows from the band calculation.

Theorem 6.49 (A weaker sufficient test against the remaining tail). Let $k, u, L$ be positive integers, and put $a = 2 L - (2^{k} - 1) u$ . Suppose $a > 0$ , equivalently that $ρ = u / (2 L) < w_{k}$ and the greedy rule skips weight $w_{k}$ . The dyadic sufficient test $ρ \leq 2^{- k}$ is equivalent to $u \leq a$ . The weaker sufficient condition

2 u \leq 3 a

ensures $ρ < R_{k}$ , by comparison with $2^{- k} + (3 \cdot 4^{k})^{- 1} + (7 \cdot 8^{k})^{- 1} < R_{k}$ . This conclusion excludes a fatal tail-mass deficit at the current step; it does not assert that $ρ$ is representable by the remaining weights, or that all future greedy steps survive.

The containment of these sufficient conditions is strict even for valid rational data: $(k, u, L, a) = (2, 7, 13, 5)$ gives $1 / 4 < 7 / 26 < R_{2}$ . In contrast, the scalar pair $(u, a) = (3, 2)$ in the linked inequality lemma does not arise from integral $L$ under the present parity relation. Unit numerators are nonfatal at such a skipped step, since then $a \geq 1$ . Conversely, a fatal step requires $3 a < 2 u$ , hence $u \geq 2$ , or $u \geq 3$ when $u$ is odd.

Lemma 6.50 (Summability of one gap length per level). For $n \geq 1$ , put $g_{n} = w_{n} - R_{n} > 0$ . The sum of one gap length per level satisfies, for $N \geq 0$ ,

\sum_{n > N} g_{n} \leq \frac{2}{9} 4^{- N} + \frac{3}{7} 8^{- N} .

This follows by summing the per-level upper bounds geometrically. It is not the measure of the union of all gaps: level $n$ has $2^{n - 1}$ translated gaps of length $g_{n}$ . The corresponding weighted sum and its geometric interpretation are given after Theorem 6.118. Neither estimate decides membership of a specified point such as $1 / 2$ .

Lemma 6.51 (The effect of adding one divisor). Let $N \geq 0$ and let finite supports $D_{0}, D_{1}$ differ only by $D_{1} = D_{0} \cup {N + 1}$ , with $N + 1 \notin D_{0}$ . For every positive integer $m$ ,

c_{D_{1}} (m) - c_{D_{0}} (m) = 1_{N + 1 ∣ m} .

In particular, the difference is $1$ at $m = 2 (N + 1)$ . This follows directly from the definition of a divisor count. Its use in another coefficient sequence requires proving that sequence has the same support-incidence description.

Theorem 6.52 (Uniqueness of an integer recurrence with a vanishing scaled limit). For any nonnegative-integer coefficient sequence $c : N \to N$ with $c (n) \leq n$ : the binary coefficient series $X_{c} = \sum_{n \geq 1}^{'} c (n) / 2^{n}$ is rational iff there exists a positive integer multiplier $v$ and an integer orbit $u : N \to Z$ satisfying the exact carry recurrence $u (N + 1) = 2 u (N) - v \cdot c (N + 1)$ together with the condition $u (N) / 2^{N} \to 0$ . Every integer sequence satisfying both conditions obeys $u (N) = v \cdot T_{c} (N)$ exactly, where $T_{c} (N) = \sum_{j \geq 1} c (N + j) / 2^{j}$ is the scaled tail. For each fixed $v$ there is at most one such sequence. Positivity of the orbit alone is deliberately not used as an equivalent criterion: a homogeneous $2^{N}$ -scaled perturbation can be added to any orbit without breaking the recurrence, so the limit condition cannot be dropped. The choices $c = c_{A}$ for Problem 257 and $c = φ$ for Problem 249 both satisfy $0 \leq c (n) \leq n$ . The telescoping argument is given again, with an example, in Theorem 6.105.

Lemma 6.53 (Changing finitely many support elements). Let $b \geq 2$ be an integer, and let $A, B \subseteq N_{\geq 1}$ . If $X_{A} (b)$ is irrational and $A, B$ have finite symmetric difference, then $X_{B} (b)$ is irrational too: the two sums differ by a finite sum of rational numbers. This observation supplies the finite modifications in Theorem 6.21.

Lemma 6.54 (An integer test for the interval between the dyadic and Mersenne weights). For positive integers $p, L$ and $n \geq 0$ , write the residual as $p / (2 L)$ and put $E = 2^{n} p - L$ . Clearing positive denominators gives

\frac{p}{2 L} \leq \frac{1}{2^{n + 1}} ⟺ E \leq 0, \frac{1}{2^{n + 1}} < \frac{p}{2 L} < \frac{1}{2^{n + 1} - 1} ⟺ 0 < E < \frac{p}{2} .

The first test concerns the dyadic bound, not the greedy selection threshold itself. The second identifies precisely the interval in which a Mersenne weight is skipped but the dyadic safety test fails.

Lemma 6.55 (The odd denominator survives dyadic subtraction). Let $r / D$ be a reduced fraction with $D > 0$ odd, let $p \in Z$ , and let $c \geq 0$ be an integer. The reduced denominator of $p / 2^{c} - r / D$ is divisible by $D$ and divides $2^{c} D$ . Indeed, its unreduced numerator $p D - 2^{c} r$ is coprime to $D$ , since $gcd (r, D) = gcd (2^{c}, D) = 1$ . Reduction can therefore remove only powers of $2$ from the displayed denominator. No property of Mersenne weights is used.

Lemma 6.56 (A guaranteed divisor of a reduced denominator). Let $a \in Z$ , $D \in N_{> 0}$ and $m, C, h \in N$ . Write $den (x)$ for the positive reduced denominator of a rational number $x$ . If $m ∣ D$ and $gcd (m, | a |) = 1$ , then $m ∣ den (a / D)$ . If $C ∣ D$ and $gcd (C, | a |) = 1$ , then

\frac{C}{gcd (C, h)} ∣ den (h a / D) .

Thus scaling may remove part of the guaranteed divisor. Indeed, reduction divides $D$ by $gcd (D, | a |)$ , so no prime power in $m$ can be lost. After scaling, a prime $p ∣ C$ can lose at most $v_{p} (h)$ powers from the guaranteed divisor. This is a lower bound on the reduced denominator, not an equality: for $D = 12$ , $C = 6$ , $a = 1$ , $h = 2$ , it guarantees the divisor $3$ , while the reduced denominator is $6$ . The lemma feeds Theorem 6.24 directly.

Lemma 6.57 (Commuting prime-power differences). For a function $g : N_{> 0} \to Z$ , a positive integer $p$ and an integer $e \geq 1$ , define

(Δ_{p, e} g) (n) = g (p^{e} n) - g (p^{e - 1} n) .

These operators commute. If $p \neq q$ are primes, $e, f \geq 1$ , $A \subseteq N_{> 0}$ , $n \geq 1$ and $gcd (n, p q) = 1$ , then

(Δ_{q, f} Δ_{p, e} c_{A}) (n) = \sum_{d ∣ n} 1_{A} (p^{e} q^{f} d) .

Thus the mixed difference counts precisely the support elements whose $p$ - and $q$ -adic exponents are $e$ and $f$ in this divisor sum.

Proof. Expanding either order gives the same four terms:

\begin{aligned} g (p^{e} q^{f} n) - g (p^{e - 1} q^{f} n) \\ - g (p^{e} q^{f - 1} n) + g (p^{e - 1} q^{f - 1} n) . \end{aligned}

This commutation needs no primality or coprimality. For $g = c_{A}$ , write each divisor of $p^{e} q^{f} n$ uniquely as $p^{i} q^{j} d$ , with $d ∣ n$ , $0 \leq i \leq e$ and $0 \leq j \leq f$ . The two differences cancel all terms except $i = e$ , $j = f$ . ◻

For $A = {12}$ , $p = 2$ , $e = 2$ , $q = 3$ , $f = 1$ , $n = 1$ , the four counts are $1, 0, 0, 0$ , so the mixed difference is $1$ . This coefficient identity alone does not transfer rationality through the subsequence operation $g (n) \mapsto g (p n)$ .

Definition 6.58 (The Mersenne achievement set and the support-series object). For a support $A \subseteq N_{\geq 1}$ , its Mersenne subsum is

x_{A} = \sum_{n \geq 1} \frac{1_{A} (n)}{2^{n} - 1}, A = {x_{A} : A \subseteq N_{\geq 1}} .

Finite, empty, and infinite supports are all allowed in $A$ . More generally, for an integer base $b \geq 2$ , $X_{A} (b) = \sum_{n \geq 1} 1_{A} (n) / (b^{n} - 1)$ and $c_{A} (n) = \sum_{d ∣ n} 1_{A} (d) = (1_{A} * 1) (n)$ for $n \geq 1$ , where $*$ denotes Dirichlet convolution. Full support gives the series in Theorem 6.14. The linked definitions implement these positive-index sums.

mersenneAchievementSet erdosSupportSeries supportCoeff

Proposition 6.59 (Achievement-set topology: compact, closed, perfect, measure exactly one). $A$ is compact (continuous image of the binary-sequence Cantor space $N \to Fin 2$ under the product topology, via the positive-index digit map), hence closed; it is also perfect, totally disconnected, and nowhere dense, with Lebesgue measure exactly $1$ . The compactness/closedness argument (binary coding $\to$ Cantor space $\to$ continuous image) is a fully generic technique for characterizing the achievement set of any absolutely convergent digit-weighted series, not specific to Mersenne denominators ; reusable for a $φ (n) / 2^{n}$ subsum set after checking summability. No separation of successive weights is needed for compactness or closedness; the stronger topological conclusions require their own hypotheses. Closedness alone is what powers every “limit of a sequence of achieved points is achieved” argument in this catalogue (e.g. Theorem 6.2, Theorem 6.5).

Theorem 6.60 (Nonmembership of one half and a finite fatal gap). The value $1 / 2$ is not in $A$ if and only if there are an integer $d \geq 0$ and a finite set $u \subseteq {1, \dots, d}$ such that

X_{u} (2) + R_{d + 1} < \frac{1}{2} < X_{u} (2) + w_{d + 1} .

One implication is Lemma 6.37. For the other, follow the greedy prefixes: nonmembership forces a first failure of the tail inequality, hence one of these gaps. The endpoint equalities are excluded by Lemma 6.38.

A finite prefix and strict gap provide a nonmembership witness. The tail inequality can be certified by rational truncation bounds as in Theorem 2.11. Membership, in contrast, asserts that no such witness exists; failure to find one in a finite search is not a proof. The reduction uses $w_{n} > R_{n}$ , not compactness alone. The totient weights fail this hypothesis, as the example after Lemma 6.34 shows.

Definition 6.61 (Separated integer values and bounded perturbations). Let $α$ be a set, let $s : α \to N$ be injective, and let $p : α \to N$ . Suppose integers $g \geq 1$ and $B \geq 0$ satisfy

s (x) < s (y) ⟹ s (x) + g \leq s (y), 0 \leq p (x) \leq B < 3 g .

The updated values are $t (x) = 4 s (x) + p (x)$ . For a capacity $C \geq 0$ , an adjacent pair consists of $x_{-}, x_{+} \in α$ with $s (x_{-})$ the largest value at most $C$ and $s (x_{+})$ the smallest value greater than $C$ . Existence of both values is part of the assumption; it is not guaranteed for an arbitrary family. Set

r = C - s (x_{-}), o = s (x_{+}) - C, p_{-} = p (x_{-}), p_{+} = p (x_{+}) .

The inequality $B < 3 g$ preserves the order of the updated values. The next result uses the stronger bound $B < g$ to ensure that the selected updated value is admissible at capacity $4 C + g$ . The linked structures encode these data.

PerturbedFamily AdjacentCut

Theorem 6.62 (Perturbed-family maximality and the three-branch recurrence). Use the family and adjacent pair of Definition 6.61, and assume in addition that $B < g$ . At capacity $C^{'} = 4 C + g$ , the largest admissible updated value is $t (x_{+})$ if $4 o + p_{+} \leq g$ , and $t (x_{-})$ otherwise. After this choice, apply the take-if-possible rule to an additional weight $W = 2 g + 4$ . The resulting remainder is

{\begin{cases} g - 4 o - p_{+}, & 4 o + p_{+} \leq g, \\ 4 r + g - p_{-}, & 4 o + p_{+} > g and 4 r + g - p_{-} < W, \\ 4 r - g - p_{-} - 4, & 4 o + p_{+} > g and 4 r + g - p_{-} \geq W . \end{cases}

Proof. The lower candidate satisfies $t (x_{-}) \leq 4 C + B < C^{'}$ , and $t (x_{+}) \leq C^{'}$ is exactly the test $4 o + p_{+} \leq g$ . Order preservation excludes earlier values from improving the selected one. Any value after $x_{+}$ has $s (x) \geq s (x_{+}) + g > C + g$ , and hence $t (x) > C^{'}$ . Thus no other candidate is missed.

On the first branch the remainder is at most $g < W$ , so the extra weight is not taken. The other two branches compare the lower candidate’s remainder $4 r + g - p_{-}$ with $W$ . All displayed differences are nonnegative on their respective branches. ◻

The theorem describes a two-stage rule, not automatically the maximum of ${t (x) + ε W : x \in α, ε \in {0, 1}}$ below $C^{'}$ . For example, take old values $4, 5, 7$ , $g = 1$ , $B = 0$ , $p = 0$ and $C = 6$ . Then $C^{'} = 25$ and $W = 6$ . The rule chooses $20$ and cannot add $6$ , but $16 + 6 = 22$ is a larger admissible extended value. A sufficient extra hypothesis for global maximality is $4 g - B \geq W$ : every earlier updated value plus $W$ is then at most the selected updated value. In the Mersenne application $g = 2^{s + 1}$ , $B = 2 (s - 2)$ and $s \geq 5$ , so this inequality holds. Thus the concrete recurrence is unaffected, but an application to another weight system must check this separation as well.

The hypothesis $B < g$ is additional; it does not follow from $B < 3 g$ in Definition 6.61. The weaker bound alone does not ensure admissibility of the chosen value. Take $g = 2$ , $B = 3$ , old values $0, 2, 4$ , perturbations $0, 3, 0$ , and $C = 2$ . The rule selects the old value $2$ , whose update is $11 > C^{'} = 10$ , although the updated value $0$ is admissible. This example explains why the stronger hypothesis must accompany the maximality statement.

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