The Binary Totient Series

Will Cook

Long paper

The Binary Totient Series

Erdős #249 By Will Cook July 2026; revised 1 October 2026 Authorship, AI use and citation

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Summary. The dyadic totient sections through level $e \geq 1$ have an explicit rational basis of size $2^{e} + 1$ . Truncating a scaled tail difference produces an integer whose residue certifies nonintegrality when its distances from both endpoints of the residue interval exceed an explicit error bound. The record develops the resulting finite denominator exclusions, equivalent residue criteria, and sufficient exponential-sum estimates. Rational comparison sequences identify the limitations of particular size, parity, rank, and finite-prefix arguments. The short paper supplies the all-base basis theorem; here each conditional result carries the arithmetic hypothesis still needed to apply it. Those hypotheses are not established in the form required to prove irrationality of S=∑ $φ (n) 2^{- n}$ .

This paper covers a claim-bounded problem-specific reasoning surface for Erdős #249: selected checked premises, finite certificates, 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 #249, which remains open.

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In this paper

An upper bound for the unassigned terms.

Only an upper bound is needed here. We enclose the unassigned arguments in a set of smooth numbers and count that larger set. There is no claim that every smooth argument is unassigned.

Proposition 12.6 (A one-sided bound for the unassigned terms). Fix $h, s$ and choose the admissible depth $L$ minimally for each large $X$ . Put $t = L - s + 1 = O_{h, s} (\log X)$ and $y_{X} = 4 \sqrt{X} + 2 t / \sqrt{X}$ . If $n = N + t$ is unassigned, then its largest prime factor satisfies $P (n) \leq y_{X}$ . Consequently

\begin{aligned} # {N \in [X, 2 X) : N \notin A} & \leq Ψ (2 X + t - 1, y_{X}) - Ψ (X + t - 1, y_{X}) \\ = (1 - \log 2 + o (1)) X < \frac{8}{25} X \end{aligned}

for all sufficiently large $X$ . Here $Ψ (x, y)$ counts the positive integers at most $x$ whose prime factors are all at most $y$ . This is an upper bound for the unassigned count, not an asymptotic equality for that count.

Proof. Let $p = P (n)$ and $m = n / p$ . Since $n > 1$ , failure of the assignment condition means either $p \leq 2 ⌊ \sqrt{X} ⌋$ or $m > ⌊ \sqrt{X} / 2 ⌋$ . In the second case $m > \sqrt{X} / 2$ , so

p = \frac{n}{m} < \frac{2 (2 X + t)}{\sqrt{X}} = y_{X} .

The first case also gives $p \leq y_{X}$ . Counting the resulting smooth integers in $[X + t, 2 X + t)$ proves the first inequality.

For completeness, the classical fixed-parameter asymptotic $Ψ (x, x^{1 / u}) \sim x ρ (u)$ suffices; no additional short-interval estimate is needed. Indeed, for either $x = X + t - 1$ or $x = 2 X + t - 1$ , $\log y_{X} / \log x \to 1 / 2$ . For every fixed $0 < ϵ < 1$ , eventually

x^{1 / (2 + ϵ)} \leq y_{X} \leq x^{1 / (2 - ϵ)} .

Monotonicity in $y$ bounds $Ψ (x, y_{X}) / x$ between quantities tending to $ρ (2 + ϵ)$ and $ρ (2 - ϵ)$ . Letting $ϵ \to 0$ and using continuity gives the limit $ρ (2)$ . Subtracting the two endpoint asymptotics yields $(ρ (2) + o (1)) X$ . Since $ρ (2) = 1 - \log 2 < 8 / 25$ , the claimed bound follows. ◻

The Lean proof uses no smooth-number asymptotic. For large $X$ , $y_{X}^{2} > 2 X + t - 1$ , so an integer in $(X + t - 1, 2 X + t - 1]$ that is not $y_{X}$ -smooth has exactly one prime factor above $y_{X}$ , and

Ψ (2 X + t - 1, y_{X}) - Ψ (X + t - 1, y_{X}) = X - \sum_{y_{X} < p \leq 2 X + t - 1} (⌊ \frac{2 X + t - 1}{p} ⌋ - ⌊ \frac{X + t - 1}{p} ⌋) .

Mertens’ theorems, proved in Lean with explicit constants, then give $(1 - \log 2) X + O (X / \log X)$ . The written proof above cites instead the fixed- $u$ asymptotic [20] and the value $ρ (u) = 1 - \log u$ on $[1, 2]$ [20]. No effective crossover scale or exact density for the unassigned indices is asserted. The estimate uses the minimal admissible depth; it does not supply either weighted phase bound at that depth.

Proposition 12.7 (The excluded-cofactor estimate). Fix $h, s$ and use the minimal admissible depth $L$ , as in Proposition 12.6; thus $t = L - s + 1 = O_{h, s} (\log X)$ . For $η \in (0, 1)$ let $B (η) = {m \geq 1 : φ (m) < η m}$ , with natural density $D (η)$ ; by Schoenberg’s theorem $D$ exists, is continuous, and $D (0 +) = 0$ [21], in the framework of [22]. Then

# {N \in A : m_{N} \in B (η)} \leq (D (η) + o (1)) X,

so a single choice of $η$ with $D (η) < 1 / 200$ meets the $\frac{1}{100} X$ bound for all large $X$ . This choice fixes $η$ before $X_{0}$ , as required. It supplies only the excluded-cofactor bound: the mean and mean-subtracted estimates must still hold for this same $η$ , and do not follow from making $η$ smaller.

Proof. For a fixed cofactor $m \leq \sqrt{X} / 2$ , each assigned index gives a prime $p$ with $X + t \leq m p < 2 X + t$ . Dropping the other assignment restrictions bounds the number of indices by the number of primes in this interval. The prime number theorem, uniformly for $X / m \geq 2 \sqrt{X}$ , and $t = O_{h, s} (\log X)$ give the upper bound

(2 + o (1)) \frac{X}{m \log X} .

The possible endpoint contribution is $O (1)$ for each $m$ , hence $o (X)$ after summing over $m \leq \sqrt{X} / 2$ . Natural density and partial summation give

\sum_{\begin{matrix} m \leq \sqrt{X} / 2 \\ m \in B (η) \end{matrix}} \frac{1}{m} = (\frac{D (η)}{2} + o (1)) \log X .

Multiplying the two estimates proves the stated bound. If $D (η) < 1 / 200$ , its right side is less than $X / 100$ for all sufficiently large $X$ . ◻

The existence of a usable cutoff can also be made explicit by an elementary estimate. For $x \geq 1$ ,

\begin{aligned} \frac{1}{x} \sum_{n \leq x} \frac{n}{φ (n)} & \leq \sum_{d \geq 1} \frac{μ (d)^{2}}{d φ (d)} = \prod_{p} (1 + \frac{1}{p (p - 1)}) \\ \leq \exp (\sum_{n \geq 2} \frac{1}{n (n - 1)}) = \exp (1) < 3. \end{aligned}

The first step uses $n / φ (n) = \sum_{d ∣ n} μ (d)^{2} / φ (d)$ and $⌊ x / d ⌋ \leq x / d$ ; the Euler product has nonnegative terms and converges by the displayed majorant. Markov’s inequality gives $# (B (η) \cap [1, x]) < 3 η x$ , hence $D (η) \leq 3 η$ . In particular, $η = 1 / 1000$ gives $D (η) \leq 3 / 1000 < 1 / 200$ , an explicit sufficient choice for the excluded-cofactor bound. This does not establish either weighted phase estimate for that choice, and it is not claimed to be an optimal cutoff. Lean checks the estimate $# (B (η) \cap [1, x]) < 3 η x$ with no further input (Lean proof) and the excluded-cofactor bound at $η = 1 / 1000$ with no input at all (Lean proof). For that bound Chebyshev’s bound replaces the prime number theorem in the proof above: the primes $p$ with $a + 1 < p \leq 2 a + 1$ divide $(\binom{2 a + 1}{a}) \leq 4^{a}$ , so $# {p prime : a < p \leq 2 a + 1} \log a \leq a \log 4 + \log a$ , the fibre bound becomes $(2 \log 4 + o (1)) X / (m \log X)$ , and the count is at most $(3 \log 4 / 1000 + o (1)) X$ with $3 \log 4 / 1000 < 0.0042$ . The proof uses $η = 1 / 1000$ only through $3 η \log 4 < 1 / 100$ , so Lean checks the bound with no input for every $η$ with $0 < η < 1 / (300 \log 4)$ (Lean proof).

The first element of $B (1 / 5)$ is $30030 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ , and $φ (30030) / 30030 = 5760 / 30030 \approx 0.191808$ . Enumeration through $3 \cdot 10^{5}$ gives $38$ elements of $B (1 / 5)$ and

\frac{1}{\log (3 \cdot 10^{5})} \sum_{\begin{matrix} m \leq 3 \cdot 10^{5} \\ m \in B (1 / 5) \end{matrix}} \frac{1}{m} \approx 2.82 \cdot 10^{- 5} .

This finite weighted proportion is not the limiting density $D (1 / 5)$ . Using $η = 1 / 5$ in the asymptotic bound would require a rigorous bound for $D (1 / 5)$ , including the integers beyond the enumeration; the finite count does not supply one.

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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