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import FormalConjecturesUtilCharacterization of Carmichael numbers via squarefree denominators $a(n)$
$a(n)$ is the denominator of $F(n) = \operatorname{num}(B_{n-1})/n + \operatorname{den}(B_{n-1})/n^2$.
A composite number $n$ has squarefree $a(n)$ if and only if $n$ is a Carmichael number.
References:
arxiv/2605.22763 Advancing Mathematics Research with AI-Driven Formal Proof Search by George Tsoukalas et al.
namespace OeisA309132open Rat Nat$a(n)$ is the denominator of $F(n)$ = A027641(n-1)/n + A027642(n-1)/n^2.
def a (n : ℕ) : ℕ :=
if n = 0 then 0
else
let n_q : ℚ := n
let B_nm1 : ℚ := bernoulli (n - 1)
let F_n : ℚ := (B_nm1.num : ℚ) / n_q + (B_nm1.den : ℚ) / (n_q * n_q)
F_n.denDefinition of a Carmichael number $n$: a composite number s.t. $b^{n-1} \equiv 1 \pmod n$ for all $b$ coprime to $n$.
def IsCarmichaelNumber (n : ℕ) : Prop :=
(¬ Nat.Prime n ∧ n > 1) ∧ (∀ b : ℕ, Nat.gcd b n = 1 → b ^ (n - 1) ≡ 1 [MOD n])Helper definition for "composite number"
def IsComposite (n : ℕ) : Prop := ¬ Nat.Prime n ∧ n > 1@[category test, AMS 11]
lemma a_1 : a 1 = 1 := ⊢ a 1 = 1 ⊢ (if 1 = 0 then 0
else
have n_q := ↑1;
have B_nm1 := bernoulli (1 - 1);
have F_n := ↑B_nm1.num / n_q + ↑B_nm1.den / (n_q * n_q);
F_n.den) =
1; All goals completed! 🐙@[category test, AMS 11]
lemma a_2 : a 2 = 1 := ⊢ a 2 = 1 ⊢ (if 2 = 0 then 0
else
have n_q := ↑2;
have B_nm1 := bernoulli (2 - 1);
have F_n := ↑B_nm1.num / n_q + ↑B_nm1.den / (n_q * n_q);
F_n.den) =
1; All goals completed! 🐙@[category test, AMS 11]
lemma a_3 : a 3 = 1 := ⊢ a 3 = 1 ⊢ (if 3 = 0 then 0
else
have n_q := ↑3;
have B_nm1 := bernoulli (3 - 1);
have F_n := ↑B_nm1.num / n_q + ↑B_nm1.den / (n_q * n_q);
F_n.den) =
1; All goals completed! 🐙@[category test, AMS 11]
lemma a_4 : a 4 = 16 := ⊢ a 4 = 16 ⊢ (if 4 = 0 then 0
else
have n_q := ↑4;
have B_nm1 := bernoulli (4 - 1);
have F_n := ↑B_nm1.num / n_q + ↑B_nm1.den / (n_q * n_q);
F_n.den) =
16; All goals completed! 🐙@[category test, AMS 11]
lemma a_5 : a 5 = 1 := ⊢ a 5 = 1 ⊢ (if 5 = 0 then 0
else
have n_q := ↑5;
have B_nm1 := bernoulli (5 - 1);
have F_n := ↑B_nm1.num / n_q + ↑B_nm1.den / (n_q * n_q);
F_n.den) =
1; All goals completed! 🐙Conjecture: composite numbers $n$ such that $a(n)$ is squarefree are only the Carmichael numbers (A002997). - Thomas Ordowski, Jul 15 2019
A formal proof has been found with the methods described in arxiv/2605.22763.
@[category research solved, AMS 11, formal_proof using formal_conjectures at
"https://github.com/mo271/formal-conjectures/blob/a32396489dcb8f86c3549b93aa358ac6a10a3a1f/FormalConjectures/OEIS/309132.wip.lean#L353"]
theorem carmichael_iff_squarefree_a :
∀ (n : ℕ), (IsComposite n ∧ Squarefree (a n)) ↔ IsCarmichaelNumber n := ⊢ ∀ (n : ℕ), IsComposite n ∧ Squarefree (a n) ↔ IsCarmichaelNumber n
All goals completed! 🐙end OeisA309132