Irrationality criteria for Lambert subseries

Will Cook

Short paper

Irrationality criteria for Lambert subseries

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

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Summary. The series $\sum_{a \in A} (b^{a} - 1)^{- 1}$ is irrational for every integer $b \geq 2$ and infinite A contained in E∪V when E satisfies a finite-prime weighted summability condition and V has a summable positive divisor cover. The two classes are incomparable, and their union reaches supports in neither class. A common average over dyadic lengths produces one index with both displacements small. Arbitrary infinite supports at base two remain unresolved.

This paper covers the weighted, divisor-cover and mixed support criteria with complete proofs and method limits; the long paper treats finite denominators, the achievement set and rational membership.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #257, which remains open.

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In this paper

Integer quotients, approximation and further questions

For $1 / 21$ we compare integer greedy remainders with the real greedy sequence; for $1 / 2$ we allow arbitrary finite approximating supports. Both tests require suitable data at unbounded depths.

Integer quotients for $1 / 21$ .

For $M, d \geq 1$ , set

q_{M} (d) = ⌊ \frac{2^{M}}{2^{d} - 1} ⌋, T_{M} = ⌊ \frac{2^{M}}{21} ⌋ .

For $d \geq 2$ , the geometric expansion gives $q_{M} (d) = \sum_{j = 1}^{⌊ M / d ⌋} 2^{M - j d}$ : the omitted fraction $2^{M mod d} / (2^{d} - 1)$ is less than $1$ . Thus $q_{M} (d)$ is the coefficient prefix contributed by exponent $d$ . Starting with remainder $T_{2 R}$ , consider $d = 2, \dots, R$ in order and subtract $q_{2 R} (d)$ whenever it does not exceed the current remainder. Let $D_{R}$ be the set of selected exponents and $s_{R}$ the final remainder. Thus

s_{R} = T_{2 R} - \sum_{d \in D_{R}} q_{2 R} (d) \geq 0.

The precision is $4^{R}$ , but $D_{R}$ records decisions only through rank $R$ . The agreement condition below also uses a second run at that same precision, continued through rank $2 R$ . For example, $R = 6$ gives $T_{12} = 195$ , $D_{6} = {5}$ and $s_{6} = 195 - ⌊ 4096 / 31 ⌋ = 63 \leq 64$ . This is one row satisfying the bound below, not evidence that such rows occur at unbounded depths.

Write $r_{n} = r_{n} (1 / 21)$ for the real greedy remainder and $A_{1 / 21}$ for its support. The condition $F_{21}$ means that there exist integers $n, R_{0}, K_{0} \geq 0$ with all of the following properties:

$r_{n} > R_{n}$ , the set of positive exponents outside $A_{1 / 21}$ is finite, and every exponent greater than $n$ belongs to $A_{1 / 21}$ ;
for every $R \geq R_{0}$ , the integer rule with target $T_{2 R}$ and weights $q_{2 R} (2), \dots, q_{2 R} (2 R)$ selects exactly the same exponents as the real greedy rule on ${2, \dots, 2 R}$ ;
for every $K \geq K_{0}$ , the interval $(K, 2 K]$ contains an exponent of $A_{1 / 21}$ .

A positive limiting remainder gives a tail violation and eventual selection of every exponent, hence the first and last clauses. The middle clause additionally requires agreement with the integer rules.

Theorem D.1 (integer-quotient tests for $1 / 21$ ). The following statements hold.

$1 / 21 \in A$ if and only if $F_{21}$ does not hold.
If there is an unbounded sequence of ranks $R$ with $s_{R} \leq 2^{R}$ , then $1 / 21 \in A$ .
On $F_{21}$ , eventually $s_{R} > 2^{R}$ , the boundary rank $R + 1$ belongs to $D_{R + 1}$ , and, for all sufficiently large $R$ ,
$\begin{aligned} D_{R + 1} & = D_{R} \cup {R + 1}, \\ s_{R + 1} & = 4 s_{R} + ⌊ \frac{4 (2^{2 R} mod 21)}{21} ⌋ - 2 c_{D_{R}} (2 R + 1) - c_{D_{R}} (2 R + 2) \\ - (2^{R + 1} + 1) . \end{aligned}$
Here $c_{D_{R}} (m) = # {d \in D_{R} : d ∣ m}$ .

Proof. Write $x = 1 / 21$ , $G = A_{1 / 21}$ and $δ = x - X_{G} (2) \geq 0$ . If $x \notin A$ , then $δ > 0$ . The real remainder tends to $δ$ , so every sufficiently small weight is selected and eventually $r_{n} > R_{n}$ . For the integer comparisons, first choose $d_{0}$ with $w_{d} < δ / 2$ for $d > d_{0}$ . Comparisons through $d_{0}$ are strict, since equality would give a finite representation of $x$ . After division by $4^{R}$ , each quotient weight differs from $w_{d}$ by less than $4^{- R}$ , and the remainders differ by at most $d / 4^{R}$ whenever earlier decisions agree. For large $R$ , the first $d_{0}$ decisions therefore agree. At later ranks $d \leq 2 R$ , the normalised remainder exceeds $δ / 2 > w_{d}$ , since $2 R / 4^{R} < δ / 2$ ; both rules select every remaining rank through $2 R$ . This proves $F_{21}$ . A represented target has $r_{n} \leq R_{n}$ at every rank, giving the converse in (1).

For every $R$ , the rounding errors satisfy

\begin{matrix} (12) & | x - X_{D_{R}} (2) - \frac{s_{R}}{4^{R}} | \leq \frac{R + 1}{4^{R}} . \end{matrix}

Thus $s_{R} \leq 2^{R}$ at unbounded ranks gives finite subseries sums tending to $x$ . Closedness of $A$ proves (2), without compatibility between the sets $D_{R}$ ; Section B excludes a finite representation.

On $F_{21}$ , the eventual agreement gives $D_{R} = G \cap {2, \dots, R}$ and $D_{R + 1} = D_{R} \cup {R + 1}$ . The rounding bound implies $s_{R} / 4^{R} \to δ > 0$ . Finally, for $d \in D_{R}$ (so $d \geq 2$ ), use $q_{M + 2} (d) = 4 q_{M} (d) + 2 1_{d ∣ M + 1} + 1_{d ∣ M + 2}$ , together with the analogous target identity and $q_{2 R + 2} (R + 1) = 2^{R + 1} + 1$ , to obtain (3). The full comparison induction and both quotient identities are given in the companion paper’s integer-quotient proof for $1 / 21$ . ◻

In the list $q_{2 R} (2), \dots, q_{2 R} (R)$ , each weight exceeds the sum of its successors by at least $2^{R} + 1$ . Distinct subset sums are therefore separated by more than $2^{R}$ , and the greedy sum is the largest not exceeding the target. Thus, if $D \subseteq {2, \dots, R}$ and an integer $s$ satisfy $\sum_{d \in D} q_{2 R} (d) + s = T_{2 R}$ with $0 \leq s \leq 2^{R}$ , then $D = D_{R}$ and $s = s_{R}$ (uniqueness of a finite representation with the stated remainder bound). The cited finite crossing lemmas give additional consequences under their alignment hypotheses: an earlier finite prefix cannot occur, and a real greedy exponent must be skipped (the missing-prefix consequence of an aligned crossing, the real greedy skip forced by that crossing).

Approximation to $1 / 2$ by finite supports.

For $A \subseteq {2, \dots, M}$ and $1 \leq m \leq M$ , define the integer

K_{A} (m) = 2^{m - 1} - \sum_{j = 2}^{m} 2^{m - j} c_{A} (j) .

It measures the error in the truncated divisor-coefficient sum, after multiplication by $2^{m}$ . In particular, $K_{A} (1) = 1$ and $K_{A} (m + 1) = 2 K_{A} (m) - c_{A} (m + 1)$ .

Theorem D.2 (finite approximations with vanishing scaled error). Suppose there are integers $M_{j} \geq 1$ tending to infinity and sets $A_{j} \subseteq {2, \dots, M_{j}}$ such that

\frac{| K_{A_{j}} (M_{j}) |}{2^{M_{j}}} ⟶ 0.

Then $X_{A} (2) = 1 / 2$ for some infinite set $A \subseteq N_{> 0}$ .

Only the terminal coefficient error needs control: the omitted tail has the uniform bound below. Neither agreement between the supports $A_{j}$ nor bounds on earlier values of $K_{A_{j}}$ are required:

| X_{A_{j}} (2) - \frac{1}{2} | \leq \frac{| K_{A_{j}} (M_{j}) | + 2 \sqrt{M_{j}} + 4}{2^{M_{j}}} .

Indeed, $c_{A_{j}} (n) \leq τ (n) \leq 2 \sqrt{n}$ and $\sqrt{M + r} \leq \sqrt{M} + r$ bound the normalised tail by $\sum_{r \geq 1} 2^{- r} (2 \sqrt{M} + 2 r) = 2 \sqrt{M} + 4$ . Thus $X_{A_{j}} (2) \to 1 / 2$ . Closedness of $A$ gives a representing support; the odd reduced denominators of finite sums make it infinite. An $O (\sqrt{M_{j}})$ terminal error suffices, but any $o (2^{M_{j}})$ error is allowed. For a finite $D \subseteq {2, \dots, M}$ , the floor-quotient identity gives

K_{D} (M) = 2^{M - 1} - \sum_{d \in D} ⌊ \frac{2^{M}}{2^{d} - 1} ⌋ .

The exact quotient rows considered in the companion paper satisfy $\sum_{d \in D} q_{M} (d) = 2^{M - 1} - 1$ , so they have $K_{D} (M) = 1$ . The terminal criterion allows a larger error of either sign. At a fixed depth these are different conditions, although cofinal existence of either kind of approximation characterises $1 / 2 \in A$ .

Conversely, if $X_{A} (2) = 1 / 2$ , then $1 \notin A$ , and the prefixes $A_{M} = A \cap {2, \dots, M}$ satisfy

K_{A_{M}} (M) = \sum_{r \geq 1} c_{A} (M + r) 2^{- r} \leq 2 \sqrt{M} + 4.

The right-hand side uses the full support $A$ : truncation leaves its coefficients through $M$ unchanged. Thus the existence hypothesis of Theorem D.2 is equivalent to $1 / 2 \in A$ . Existence at unbounded depths remains unproved; establishing it would refute Problem 1.1.

Finite families with a shared prefix.

Put $B (m) = 2 ⌊ \sqrt{m} ⌋ + 4$ . For $0 \leq K \leq M$ , consider a family $A_{1}, \dots, A_{B (M)} \subseteq {2, \dots, M}$ with

1 \leq K_{A_{k}} (m) \leq B (m) (1 \leq m \leq M), K_{A_{k}} (M) = k .

Require that all the $A_{k}$ agree on ${1, \dots, K}$ and that some integer $E \geq B (M)$ satisfies

\sum_{j = K + 1}^{M} 1_{A_{k}} (j) 2^{M - j} + k = E (1 \leq k \leq B (M)) .

Thus the binary suffixes have consecutive values $E - k$ . The hypothesis asks for a whole family at each available depth, not merely one approximant.

Theorem D.3 (unbounded shared-prefix families represent one half). Suppose that for every $N$ there are $M, K$ with $max {N, 1} \leq M$ , $0 \leq K \leq M$ , and a family satisfying all the conditions in the preceding paragraph. Then $X_{A} (2) = 1 / 2$ for some infinite set $A \subseteq N_{> 0}$ .

At each available depth $M$ , choose the member with terminal carry $1$ . Its scaled terminal error is $2^{- M}$ , so Theorem D.2 applies along unbounded depths. The other shared-prefix conditions are not used in this compactness step. No construction at unbounded depths is established here.

The tests above leave three kinds of arithmetic information to be supplied: returns of the actual greedy sequence, finite approximants at unbounded depths, or an exclusion of a possible final skip. The following questions specify those missing inputs.

Problem D.4 (membership of 1/21 in the Mersenne achievement set). For the greedy remainder of $1 / 21$ , prove that arbitrarily large $N$ satisfy

2^{N + 1} r_{N} (1 / 21) < \frac{2^{N + 1}}{2^{N + 1} - 1}, equivalently r_{N} (1 / 21) < \frac{1}{2^{N + 1} - 1} .

Equivalently, exclude the condition $F_{21}$ defined above. One sufficient route is to rule out the eventual recurrence in Theorem D.1(3) together with $s_{R} > 2^{R}$ ; another is to prove $s_{R} \leq 2^{R}$ at arbitrarily large ranks. Neither sufficient route is claimed to be necessary by itself.

Problem D.5 (bounded returns of the scaled remainder). For each of the targets $x = 1 / 2$ and $x = 1 / 21$ , does the greedy remainder $r_{N} (x)$ return to one bounded interval after scaling by $2^{N}$ ?

\exists B < \infty \forall K \exists N \geq K : 2^{N} r_{N} (x) \leq B .

Problem D.6 (arithmetic tests for the greedy sequence). Can a finite-memory, $2$ -adic or discrepancy argument prove $s_{R} \leq 2^{R}$ at arbitrarily large ranks, or rule out the eventual recurrence in Theorem D.1(3)? Such an argument must use the divisor counts of the actual greedy support. Can a bounded window of $R mod 6$ , residues of $s_{R}$ , endpoint divisor counts and the finite set of eventual skips force a decrease or a contradiction? Alternatively, can one show that these bounded-memory data cannot distinguish the actual sequence from sequences that remain above $2^{R}$ but need not come from a support?

Problem D.7 (final-skip Diophantine exclusion). Let $E = \sum_{n \geq 1} (2^{n} - 1)^{- 1}$ . If $1 / 21 \notin A$ , let $M$ be the last skipped exponent, $S_{M}$ its finite skipped prefix, and

a_{M} = \frac{1}{21} + \sum_{d \in S_{M}} \frac{1}{2^{d} - 1} .

Write ${gap}_{M} = (2^{M} - 1)^{- 1} - R_{M} > 0$ . Can the arithmetic restrictions on this last skipped exponent be used to prove $| E - a_{M} | \geq {gap}_{M}$ , contradicting

0 < a_{M} - E < {gap}_{M} ?

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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Integer quotients, approximation and further questions

Integer quotients for 1/211/21.

Approximation to 1/21/2 by finite supports.

Finite families with a shared prefix.

About this paper

Integer quotients for $1 / 21$ .

Approximation to $1 / 2$ by finite supports.