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.

Start here, by the editorial route

In this paper

Propagation of nonintegral tail differences

Write

S = \sum_{n \geq 1} φ (n) 2^{- n}, R_{N} = \sum_{j \geq 1} φ (N + j) 2^{- j}, Δ_{h} (N) = R_{N + h} - R_{N} .

The identity

2^{N} S = \sum_{n = 1}^{N} φ (n) 2^{N - n} + R_{N}

has an integer finite sum. Subtracting the identities at $N + h$ and $N$ gives

\begin{matrix} (6) & Δ_{h} (N) \equiv 2^{N} (2^{h} - 1) S (\mod) Z . \end{matrix}

In particular, a nonintegral tail difference excludes every rational value whose denominator divides $2^{N} (2^{h} - 1)$ .

The finite test.

For $h, N, L \in N$ , let

D_{h, N, L} = \sum_{j = 0}^{L - 1} (φ (N + h + 1 + j) - φ (N + 1 + j)) 2^{L - 1 - j}, B = N + h + L + 2.

All residues below are least nonnegative residues. Write $K (h, N, L)$ for

B < D_{h, N, L} mod 2^{L} < 2^{L} - B .

The omitted terms form another difference of scaled tails:

2^{L} Δ_{h} (N) - D_{h, N, L} = R_{N + h + L} - R_{N + L} .

Since $0 \leq R_{M} \leq \sum_{j \geq 1} (M + j) 2^{- j} = M + 2$ , their absolute difference is bounded by $N + h + L + 2$ . Thus

\begin{matrix} (7) & | 2^{L} Δ_{h} (N) - D_{h, N, L} | \leq B . \end{matrix}

If $K (h, N, L)$ holds, the interval $[D_{h, N, L} - B, D_{h, N, L} + B]$ lies strictly between two consecutive multiples of $2^{L}$ . It contains $2^{L} Δ_{h} (N)$ , so $Δ_{h} (N) \notin Z$ . Conversely, write $‖ x ‖_{R / Z} = min_{z \in Z} | x - z |$ for the distance to the nearest integer. If $Δ_{h} (N) \notin Z$ , choose $L$ so that $2^{L} ‖ Δ_{h} (N) ‖_{R / Z} > 2 B$ . Such a choice exists: $h, N$ are fixed, the distance is positive, and $B = N + h + L + 2$ grows only linearly in $L$ . The error estimate then puts the residue of $D_{h, N, L}$ more than $B$ from either endpoint. Thus nonintegrality is equivalent to a certificate at some unrestricted depth. This is a one-sided finite test: success proves nonintegrality, but failure at a particular depth does not prove integrality. The converse guarantees eventual success only after a fixed nonintegral difference has been supplied.

Theorem A.1 (Propagation of one nonintegral tail difference). Fix $d \geq 1$ and $N \geq 0$ . If $Δ_{d} (N) \notin Z$ , then every sufficiently late pair ${t, t + 1}$ contains an $m$ such that $K (m d, N, m d)$ holds. Consequently

\exists t \geq 1 : K (t d, N, t d) ⟺ Δ_{d} (N) \notin Z .

Requiring this for every $d \geq 1, N \geq 0$ is equivalent to $S \notin Q$ .

Unrestricted completeness cannot simply set $L$ equal to the shift: it supplies a depth only after the shift is fixed. The theorem couples them by varying both along multiples of $d$ . The recurrence below prevents two adjacent failures. Its starting hypothesis is still just one nonintegral difference; only the requirement for every $d, N$ is equivalent to irrationality.

Proof. Set $F_{t} = Δ_{t d} (N)$ . Equation (6) yields $F_{t + 1} \equiv 2^{d} F_{t} + F_{1} (\mod) Z$ . This recurrence expresses the fixed $F_{1}$ in terms of two adjacent differences. A failed certificate makes one of those differences close to an integer: if $K (t d, N, t d)$ fails, (7) gives

‖ F_{t} ‖_{R / Z} \leq ϵ_{t}, ϵ_{t} = 2 (N + 2 t d + 2) 2^{- t d} .

If the test also fails at $t + 1$ , the triangle inequality on $R / Z$ and the recurrence give

‖ F_{1} ‖_{R / Z} \leq ‖ F_{t + 1} ‖_{R / Z} + 2^{d} ‖ F_{t} ‖_{R / Z} \leq ϵ_{t + 1} + 2^{d} ϵ_{t} .

The recurrence has transferred two shrinking errors to the same fixed number $F_{1}$ . Its distance from $Z$ is positive, whereas $ϵ_{t + 1} + 2^{d} ϵ_{t} \to 0$ ; hence adjacent failures are impossible for large $t$ . Conversely, integral $F_{1}$ makes every $F_{t}$ integral by the recurrence, so soundness excludes every certificate.

Finally, (6) is nonintegral for every $d, N$ when $S$ is irrational. If $S = a / (2^{c} v)$ with $v$ odd, take $N = c$ and $d = φ (v)$ to make it integral. ◻

For instance $D_{1, 12, 16} = - 143140$ , with residue $53468$ modulo $2^{16}$ and $B = 31$ , verifies the hypothesis at $d = 1, N = 12$ . More explicitly,

Δ_{1} (12) \in [\frac{- 143171}{65536}, \frac{- 143109}{65536}] \subset (- 3, - 2) .

This enclosure proves nonintegrality without evaluating the infinite tails. The theorem gives infinitely many certificates with depth equal to the shift for this pair, not for every $d, N$ .

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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