/-
Copyright 2026 The Formal Conjectures Authors.
Licensed under the Apache License, Version 2.0 (the "License");
you may not use this file except in compliance with the License.
You may obtain a copy of the License at
https://www.apache.org/licenses/LICENSE-2.0
Unless required by applicable law or agreed to in writing, software
distributed under the License is distributed on an "AS IS" BASIS,
WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
See the License for the specific language governing permissions and
limitations under the License.
-/
module
public import FormalConjecturesUtil
@[expose] public sectionnamespace OeisA51903Maximum exponent in the prime factorization of $n$.
def a (n : ℕ) : ℕ :=
(n.primeFactorsList.map (n.primeFactorsList.count ·)).foldr max 0@[category test, AMS 11]
theorem a_1 : a 1 = 0 := ⊢ a 1 = 0
All goals completed! 🐙@[category test, AMS 11]
theorem a_2 : a 2 = 1 := ⊢ a 2 = 1
All goals completed! 🐙@[category test, AMS 11]
theorem a_3 : a 3 = 1 := ⊢ a 3 = 1
All goals completed! 🐙@[category test, AMS 11]
theorem a_4 : a 4 = 2 := ⊢ a 4 = 2
All goals completed! 🐙@[category test, AMS 11]
theorem a_5 : a 5 = 1 := ⊢ a 5 = 1
All goals completed! 🐙Are there composite numbers $n > 4$ such that $n \equiv a(n) \pmod{\phi(n)}$?
Thomas Ordowski, Dec 02 2019
This question is equivalent to Lehmer's totient problem LehmerTotient.lehmer_totient; a
positive answer here falsifies the universal statement asked about in
Erdos828.erdos_828.variants.lehmer_conjecture. Any composite
$n$ with $\phi(n) \mid n - 1$ is squarefree, so $a(n) = 1$ and the condition here holds.
Conversely, let $n > 4$ be composite with $\phi(n) \mid n - a(n)$ and $e = a(n) \ge 2$, and
pick $p$ with $p^e \mid n$. Then $p^{e-1} \mid \phi(n) \mid n - e$ and $p^{e-1} \mid n$, so
$p^{e-1} \mid e$, which forces $p = 2$ and $e = 2$. Now $n = 4m$ with $m$ odd and squarefree,
and $2\phi(m) \mid 4m - 2$ makes $\phi(m)$ odd, so $m = 1$ and $n = 4$. Hence $e = 1$ and the
condition is $\phi(n) \mid n - 1$.
@[category research open, AMS 11]
theorem conjecture1 :
answer(sorry) ↔ ∃ n : ℕ, 4 < n ∧ ¬ n.Prime ∧ n.totient ∣ (n - a n) := ⊢ True ↔ ∃ n, 4 < n ∧ ¬Nat.Prime n ∧ n.totient ∣ n - a n
All goals completed! 🐙Are there odd numbers $n$ such that $a(n) > 1$ and $n \equiv a(n) \pmod{\lambda(n)}$? (Equivalently, odd numbers $n$ such that $a(n) > 1$ and $b^n \equiv b^{a(n)} \pmod n$ for all $b$.)
Thomas Ordowski, Dec 02 2019
@[category research open, AMS 11]
theorem conjecture2 :
answer(sorry) ↔ ∃ n : ℕ, Odd n ∧ 1 < a n ∧ ∀ b : ℕ, b ^ n ≡ b ^ (a n) [MOD n] := ⊢ True ↔ ∃ n, Odd n ∧ 1 < a n ∧ ∀ (b : ℕ), b ^ n ≡ b ^ a n [MOD n]
All goals completed! 🐙Are there odd numbers $n$ such that $a(n) > 1$ and $n \equiv a(n) \pmod{\operatorname{ord}_n(2)}$? (Equivalently, odd numbers $n$ such that $a(n) > 1$ and $2^n \equiv 2^{a(n)} \pmod n$.)
Thomas Ordowski, Dec 02 2019
@[category research open, AMS 11]
theorem conjecture3 :
answer(sorry) ↔ ∃ n : ℕ, Odd n ∧ 1 < a n ∧ 2 ^ n ≡ 2 ^ (a n) [MOD n] := ⊢ True ↔ ∃ n, Odd n ∧ 1 < a n ∧ 2 ^ n ≡ 2 ^ a n [MOD n]
All goals completed! 🐙end OeisA51903