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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Reduced coefficients

Reduction modulo $m$ restores the boundedness missing from $φ$ . The short paper isolates a prescribed nonzero coefficient between two long zero blocks: a hypothetical rational denominator forces the right tail to vanish, after which the left block makes the nonzero centre too small. CRT creates the blocks by imposing a separate prime divisor congruent to $1 (\mod m)$ at each neighbour. We extend that proof in two directions. Changing the integer base changes only the geometric tail bound. Combining dilations is subtler: one must first ensure that their contributions cannot cancel at the chosen centre.

Proposition 1.10 (Residue series in every integer base). Let $t \geq 2$ be an integer.

For every $m \geq 3$ , $\sum_{n \geq 1} (φ (n) mod m) t^{- n}$ is irrational (radix_residue_series_irrational).
For $k \geq 1$ and $f : Z / 2^{k} Z \to Q$ , the series $\sum_{n \geq 1} f (φ (n) mod 2^{k}) t^{- n}$ is rational exactly when $f$ is constant on the even residue classes, and its value is then $(t + 1) f (1) / t^{2} + f (0) / (t^{2} (t - 1))$ (rational_zmod_radix_observable_iff, positiveRadixValue_eq_of_even_constant).

Proof. For coefficients bounded by $C$ , the right-block tail is at most $C t^{- L} / (t - 1) \leq C t^{- L}$ . Choose $L$ with $C t^{- L} < 1 / q$ ; the rational-tail lattice forces that tail to be zero. The left-block tail is then the isolated coefficient times $t^{- L - 1}$ , giving the same contradiction as in the short paper. In (b), the two exceptional terms contribute $(t + 1) f (1) / t^{2}$ and the constant tail contributes $f (0) / (t^{2} (t - 1))$ , since all totients after $n = 2$ are even. ◻

For example, in base $2$ the function modulo $4$ with $f (0) = f (2) = 5$ , $f (1) = 7$ and $f (3) = - 2$ gives the rational value $3 \cdot 7 / 4 + 5 / 4 = 13 / 2$ : the class $3$ never occurs. In contrast, the indicator of the class $2 (\mod 4)$ gives an irrational value. This classification uses the fact that the prime centres $s - 1$ , with $s$ a unit, cover every even class modulo $2^{k}$ . Modulo $6$ they cover only $0$ and $4$ . The least-residue assertion needs one nonzero accessible centre, but the same argument does not classify all rational-valued functions for a general modulus.

Proposition 1.11 (Finitely many dilations). Let $t \geq 2$ be an integer.

For every $m \geq 3$ , the numbers $1$ and $\sum_{n \geq 1} (φ (n) mod m) t^{- d n}$ , $d = 1, 2, \dots$ , are linearly independent over $Q$ (linearIndependent_one_and_least_residue_values).
Let $D$ be a finite set of positive integers, and for $d \in D$ let $k_{d} \geq 1$ and $f_{d} : Z / 2^{k_{d}} Z \to Q$ . For every $c \in Q$ , the number
$c + \sum_{d \in D} \sum_{n \geq 1} f_{d} (φ (n) mod 2^{k_{d}}) t^{- d n}$
is rational exactly when every $f_{d}$ is constant on the even residue classes (rational_mixed_moduli_with_constant_iff).

Proof. (a) A relation gives integers $c_{0}$ and $c_{d}$ , $d$ in a finite set $D$ of positive integers, with some $c_{d} \neq 0$ and $c_{0} + \sum_{N \geq 1} b_{N} t^{- N} = 0$ , where $b_{N} = \sum_{d \in D, d ∣ N} c_{d} (φ (N / d) mod m)$ is bounded. Discard the $d$ with $c_{d} = 0$ and let $δ$ be the least remaining one. Different dilations can contribute to the same exponent $N$ , so isolating a prime in one summand is not enough. Choose instead $N = δ p$ with $p > max D$ prime. A contributing $d$ divides $δ p$ ; since $p ∤ d$ , it divides $δ$ . Minimality then forces $d = δ$ . The centre now receives exactly one active contribution, while CRT will remove all contributions at its neighbours.

Given $L$ , choose distinct primes $q_{j} \equiv 1 (\mod m)$ , $0 < | j | \leq L$ , each larger than $L$ and than every element of $D$ . Impose

p \equiv - 1 (\mod m), p \equiv - j δ^{- 1} (\mod q_{j}) (0 < | j | \leq L) .

The inverse exists because $q_{j} > δ$ , and the latter residue is nonzero because $q_{j} > | j |$ . Together with the unit residue $- 1$ modulo $m$ , these congruences define a reduced progression. CRT and Dirichlet’s theorem give a prime $p$ as large as needed. At the centre, $b_{δ p} = c_{δ} (m - 2) \neq 0$ . At a neighbour $δ p + j$ , every contributing dilation $d$ is coprime to $q_{j}$ , so $q_{j} ∣ (δ p + j) / d$ and $m ∣ φ ((δ p + j) / d)$ . Hence $b_{δ p + j} = 0$ . The isolated coefficient makes $\sum_{N} b_{N} t^{- N}$ irrational, contradicting $c_{0} \in Z$ .

(b) Each $f_{d}$ that is constant on the even classes contributes a rational number, by Proposition 1.10(b) in base $t^{d}$ . Remove these contributions, lift the remaining functions to one modulus $2^{K}$ , subtract their values at zero, and clear a common denominator. None of these operations changes rationality. For the least remaining dilation $δ$ , choose an even residue $r$ with $f_{δ} (r) \neq f_{δ} (0)$ . Use the construction in (a), now with $q_{j} \equiv 1 (\mod 2^{K})$ and $p \equiv r + 1 (\mod 2^{K})$ . The central residue is allowed because $r + 1$ is odd. Minimality again leaves just the nonzero contribution from $δ$ at the centre, while every neighbouring totient is zero modulo $2^{K}$ and hence contributes zero after the subtraction. The same bounded integer-coefficient argument applies. ◻

Remark 1.12 (Slowly growing dyadic moduli). Let $κ : N \to N$ satisfy $2^{κ (n)} = o (\log n)$ , and put $a_{n} = φ (n) mod 2^{κ (n)}$ . Then $\sum_{n \geq 1} a_{n} 2^{- n}$ is rational exactly when $a_{n} = 0$ for all large $n$ . In particular the sum is irrational when $κ (3^{k}) \geq 2$ for infinitely many $k$ .

Proof. If $a_{n}$ vanishes eventually the sum is finite. Suppose instead that $a_{n} \neq 0$ infinitely often and that the sum equals $u / v$ . Then $T_{N} = \sum_{j \geq 1} a_{N + j} 2^{- j}$ lies in $v^{- 1} Z$ for every $N$ . For $1 \leq j \leq L$ let $q_{j}$ be a prime factor of the Fermat number $2^{2^{j}} + 1$ ; then $q_{j} \equiv 1 (\mod 2^{j + 1})$ , and the $q_{j}$ are distinct because distinct Fermat numbers are coprime. Put $Q = q_{1} \dots q_{L}$ , so $Q < 2^{2^{L + 1}}$ . These factors give a size bound as well as the needed divisibility: $\log Q = O (2^{L})$ . The factor $2^{- L}$ in the remaining tail will offset this growth, leaving the arbitrarily small constant from $2^{κ (n)} = o (\log n)$ . Given $M$ , the Chinese remainder theorem gives $N \in [M, M + Q)$ with $q_{j} ∣ N + j$ for $1 \leq j \leq L$ . Then $2^{j + 1} ∣ φ (N + j)$ , so $2^{j} ∣ a_{N + j}$ : either $κ (N + j) \leq j$ and $a_{N + j} = 0$ , or $2^{j}$ divides both $φ (N + j)$ and the modulus. Hence $T_{N}$ differs from an integer by $\sum_{j > L} a_{N + j} 2^{- j}$ , which is positive because $a_{n} \neq 0$ infinitely often, and at most $\sum_{j > L} 2^{κ (N + j) - j}$ . Choose $ϵ > 0$ with $3 ϵ < 1 / v$ , then $M$ with $2^{κ (n)} \leq ϵ \log n$ for $n \geq M$ , then $L \geq 2$ so large that $Q \geq M$ . Then $N < M + Q \leq 2 Q$ and $\log (N + j) \leq \log (2 Q) + j \leq 2^{L + 1} + j$ , so the tail is at most $ϵ 2^{- L} (2^{L + 1} + L + 2) \leq 3 ϵ < 1 / v$ . It is a positive element of $v^{- 1} Z$ below $1 / v$ , a contradiction. Finally $φ (3^{k}) = 2 \cdot 3^{k - 1}$ is not divisible by $4$ , so $a_{3^{k}} \neq 0$ whenever $κ (3^{k}) \geq 2$ . ◻

Remark 1.13 (Earlier results and the reduced coefficients). Proposition 1.10(a), and with it the first assertion of the short paper’s residue theorem, already follows from Erdős’s Lemma 1 of [6]: if integers $a_{k} \geq 0$ have $\sum_{k \leq n} a_{k} = O (n)$ , infinitely many $a_{k} \neq 0$ , and $\underset{n}{lim inf} # {k \leq n : a_{k} > 0} / n = 0$ , then $\sum_{k} a_{k} t^{- k}$ is irrational for every integer $t \geq 2$ . Erdős notes that this is the case $b_{k} = 0$ of his Lemma 4 (stated pp. 215–216, proved pp. 216–218). For $a_{n} = φ (n) mod m$ the hypotheses hold: $a_{n} \neq 0$ forces $n$ to have no prime factor $p \equiv 1 (\mod m)$ , and such $n$ have density zero because $\sum_{p \equiv 1 (m)} 1 / p$ diverges, while $a_{p} = m - 2$ for every prime $p \equiv - 1 (\mod m)$ . Erdős’s Lemma 4 $^{'}$ (p. 218, proof omitted there) and Theorem 3 of Kaneko, Suzuki and Tachiya [7] give the same conclusion; the latter paper proves Lemma 4 $^{'}$ as its Theorem A (pp. 2 and 18). The same argument covers Proposition 1.10(b) and Proposition 1.11 when the values $f (r) - f (0)$ , respectively the coefficients $c_{d}$ , have one sign. With both signs present, Lemmas 4 and 4 $^{'}$ and Theorem 3 of [7] require an additional spacing condition on the negative terms, which is not checked here; the isolated coefficient needs none. Theorem 3 of [7], with $b = 0$ and $z_{n} = \log_{2} \log x_{n}$ , also gives the irrational direction of Remark 1.12 once the $n < x$ with $v_{2} (φ (n)) < \log_{2} \log x$ are shown to number $o (x / \log \log x)$ ; this record does not write out that density estimate. Wong’s base-equals-modulus theorem [5] is the case $t = m$ of Proposition 1.10(a).

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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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