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

PDFOpen the PDF, 26 pages TeXLaTeX source Source on GitHub Problem page

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.

Start here, by the editorial route

In this paper

Rational membership

The preceding results start from a support and prove irrationality. We now fix a target and ask whether any support represents it. The first recurrence is a necessary condition on a rational value; the greedy recurrence later gives an exact membership criterion. Neither supplies the missing arithmetic information for $1 / 2$ or $1 / 21$ .

Rational values and scaled tails

Suppose $X_{A} (2) = p / v$ , where $p \in Z$ and $v \geq 1$ is an integer. Multiplying the coefficient-series identity by $v 2^{N}$ gives

z_{N} := v \sum_{r \geq 1} c_{A} (N + r) 2^{- r} = 2^{N} p - v \sum_{n = 1}^{N} c_{A} (n) 2^{N - n} \in Z .

Separating the first term of the tail then gives $z_{N + 1} = 2 z_{N} - v c_{A} (N + 1)$ . All the coefficients $c_{A} (n)$ are divisor counts of the same set $A$ .

For nonempty $A$ , every $a_{0} = min A$ consecutive integers contain a multiple of $a_{0}$ , so zero runs of $c_{A}$ have length at most $a_{0} - 1$ . For infinite $A$ , choose $k$ exponents and a common multiple $L$ . Then $c_{A} (L) \geq k$ , and the scaled tail satisfies $\sum_{r \geq 1} c_{A} (L - 1 + r) 2^{- r} \geq k / 2$ . Thus scaled tails are unbounded.

The compatibility condition on the coefficients is expressed by Dirichlet convolution: $μ * c_{A} = 1_{A}$ , where $μ$ is the Möbius function. A putative recurrence with integral forcing must therefore satisfy $(μ * c_{A}) (n) \in {0, 1}$ for all $n$ . The telescoping proof of the integer-recurrence criterion and the Möbius inversion argument are given under Bounds for a general coefficient sequence in the companion paper. The example $A = {2}$ there has value $1 / 3$ . The correspondence permits finite supports; infinitude remains a separate requirement for a counterexample to Problem 1.1.

The achievement set

Let $w_{n} = (2^{n} - 1)^{- 1}$ and

A = {\sum_{n \geq 1} ε_{n} w_{n} : ε_{n} \in {0, 1}}, R_{N} = \sum_{n > N} w_{n} .

The estimate

2^{- N} < R_{N} \leq 2^{- N} + \frac{2}{3} 4^{- N} < w_{N} (N \geq 1)

follows by writing $w_{n} = 2^{- n} + 4^{- n} / (1 - 2^{- n})$ and using $n \geq N + 1 \geq 2$ for the middle bound. If two selectors first differ at $n$ , their sums differ by at least $w_{n} - R_{n} > 0$ , proving uniqueness. Since every weight exceeds its tail, Hornich’s theorem [15], as proved by Nitecki [10], shows that the coding image is a Cantor set of measure $lim_{N} 2^{N} R_{N}$ ; the estimate above gives $2^{N} R_{N} \to 1$ , so $λ (A) = 1$ . Concretely, fixing the first $N$ digits gives $2^{N}$ disjoint closed intervals of length $R_{N}$ . These interval unions decrease to $A$ , and continuity of Lebesgue measure from above gives the same limit. Kovač–Tao record this strict-tail inequality and the Cantor conclusion in the fixed-base setting [4]. The geometry section of the companion paper also treats restricted supports. For any allowed set $J$ , removing exponents can only decrease the tail of each retained weight, so the same inequality gives injectivity. If $J$ omits exactly the finite set $F$ , its achievement set has measure $2^{- | F |}$ ; if $J$ omits infinitely many exponents, that measure is zero. The tail comparison, injectivity and measure dichotomy have separate source statements.

The greedy recurrence

For $x \geq 0$ , the greedy rule starts with $r_{0} = x$ and, at rank $n \geq 1$ , selects $n$ when $r_{n - 1} \geq (2^{n} - 1)^{- 1}$ , subtracting that weight if selected. Let $A_{x}$ be the resulting support and $c_{x} (n) = # {a \in A_{x} : a ∣ n}$ . Set

P_{0} = 0, P_{N + 1} = 2 P_{N} + c_{x} (N + 1), Q_{N} = ⌊ 2^{N} x ⌋ - P_{N} .

Here $P_{N} = \sum_{j = 1}^{N} 2^{N - j} c_{x} (j)$ truncates the coefficient series (6), not the selected Lambert weights. Thus $Q_{N}$ measures coefficient truncation, whereas $r_{N}$ is the real remainder after $N$ greedy decisions. The bound $P_{N} \leq 2^{N} X_{A_{x}} (2) \leq 2^{N} x$ gives $Q_{N} \geq 0$ , and

\begin{matrix} (10) & Q_{N + 1} = 2 Q_{N} + β_{N} - c_{x} (N + 1), β_{N} = ⌊ 2^{N + 1} x ⌋ - 2 ⌊ 2^{N} x ⌋ \in {0, 1} . \end{matrix}

The recurrence is integral for every real target. Write $δ = x - X_{A_{x}} (2) \geq 0$ . The coefficient expansion gives

Q_{N} = ⌊ 2^{N} δ + \sum_{r \geq 1} c_{x} (N + r) 2^{- r} ⌋ .

For a represented target, $δ = 0$ and $Q_{N} = O (\sqrt{N})$ , leaving too little room for unit increases throughout a long window. For an unrepresented target, $2^{N} δ$ eventually dominates the divisor count in the recurrence and forces strict increase.

Theorem C.1 (membership and nonincreasing integer remainders). For every real $x \geq 0$ , the following are equivalent:

\begin{matrix} x \in A; \\ \forall K \geq 0 \exists N \geq K : Q_{N + 1} \leq Q_{N}; \\ \forall K \geq 0 \exists N \in [K, K + 2 ⌊ \sqrt{K} ⌋ + 12) : Q_{N + 1} \leq Q_{N} . \end{matrix}

Proof. For a represented target, the remaining sum is at least $w_{n}$ exactly when $n$ is selected, since $R_{n} < w_{n}$ . Hence the greedy rule recovers the support, and

0 \leq Q_{N} \leq \sum_{r \geq 1} c_{x} (N + r) 2^{- r} \leq 2 \sqrt{N} + 4,

since $c_{x} (n) \leq τ (n) \leq 2 \sqrt{n}$ and $\sqrt{N + r} \leq \sqrt{N} + \sqrt{r} \leq \sqrt{N} + r$ . Put $s = ⌊ \sqrt{K} ⌋$ and $T = 2 s + 12$ . Strict increase at all $T$ steps would imply $Q_{K + T} \geq T$ . But $K + T < (s + 4)^{2}$ gives $Q_{K + T} < 2 s + 12 = T$ . This proves the window condition and hence cofinal nonincreases.

Conversely, if $x \notin A$ , then $δ > 0$ and $Q_{N} \geq 2^{N} δ - 1$ , which eventually exceeds $c_{x} (N + 1) \leq N + 1$ . Equation (10) then makes $Q_{N}$ strictly increasing, contradicting cofinal nonincreases. ◻

The nonincreases must occur along the target’s own greedy sequence. For $x = 0$ , no exponent is selected and $Q_{N} = 0$ at every rank. A finite subseries sum is likewise represented and satisfies the criterion. The target $x = 3 / 4$ lies strictly between $R_{1} < 2 / 3$ and $w_{1} = 1$ , so it is not represented and its integer remainders eventually increase strictly. For $1 / 2$ and $1 / 21$ , the theorem does not establish which alternative occurs.

An ordinary extension gives a uniform subpower window: for each $0 < ε < 1$ , its length is $⌈ C_{ε} (K + 1)^{ε} ⌉$ for a suitable constant $C_{ε}$ . The linked formal proof establishes the square-root window. Indeed, $τ (n) \leq A_{ε} n^{ε}$ gives $Q_{N} \leq D_{ε} (N + 1)^{ε}$ for represented targets, where $D_{ε} = A_{ε} \sum_{r \geq 1} r^{ε} 2^{- r}$ . Choose $C_{ε} > D_{ε} (C_{ε} + 3)^{ε}$ , which is possible because $ε < 1$ . For $T = ⌈ C_{ε} (K + 1)^{ε} ⌉$ , strict increase at all $T$ steps would give

T \leq Q_{K + T} \leq D_{ε} (K + T + 1)^{ε} \leq D_{ε} (C_{ε} + 3)^{ε} (K + 1)^{ε} < T,

a contradiction. The converse still follows from exponential growth when $x$ is not represented. The shorter window does not decide membership of a specified target.

At $x = 1 / 2$ , the digits $β_{N}$ vanish for $N \geq 1$ ; at $x = 1 / 21$ they are six-periodic. Thus the unresolved arithmetic input is

\begin{matrix} (11) & \forall K \exists N \geq K : c_{x} (N + 1) \geq Q_{N} + β_{N}, x \in {1 / 2, 1 / 21} . \end{matrix}

Both $Q_{N}$ and $c_{x}$ must arise from the same greedy support $A_{x}$ . Counterexamples to fixed-multiplier schedules, with their finite phase masks, are retained in the companion paper. They concern those schedules; they do not decide either target.

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, 26 pages TeXOpen the LaTeX source Problem page

Equations typeset from the exact TeX. Rendered from the LaTeX manuscript source with file fingerprint sha256:2676b3e9f6a5ba6f. 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.