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module
public import FormalConjecturesUtilNumber of refactorable numbers (A033950) $\le 10^n$
A number $k$ is refactorable if its number of divisors, $\tau(k)$, divides $k$.
References:
[Co99] Colton, S., Refactorable numbers - a machine invention. J. Integer Seq. 2 (1999), Article 99.1.2.
[Ze02] Zelinsky, J., Tau numbers: a partial proof of a conjecture and other results. J. Integer Seq. 5 (2002), Article 02.2.8.
[Sp85] Spiro, C., How often is the number of divisors of n a divisor of n? J. Number Theory 21 (1985), 81--100.
@[expose] public sectionnamespace OeisA111291open Nat Finset RealHelper function: number of refactorable numbers $\le m$.
def countRefactorableNat (m : ℕ) : ℕ :=
(Icc 1 m).filter (fun k => k.divisors.card ∣ k) |>.card
a n is the number of refactorable numbers $\le 10^n$.
A number $k$ is refactorable if its number of divisors, $\tau(k)$, divides $k$.
def a (n : ℕ) : ℕ :=
countRefactorableNat (10 ^ n)@[category test, AMS 11]
theorem a_0 : a 0 = 1 := ⊢ a 0 = 1 All goals completed! 🐙@[category test, AMS 11]
theorem a_1 : a 1 = 4 := ⊢ a 1 = 4 All goals completed! 🐙@[category test, AMS 11]
theorem a_2 : a 2 = 16 := ⊢ a 2 = 16 All goals completed! 🐙@[category test, AMS 11]
theorem a_3 : a 3 = 92 := ⊢ a 3 = 92
All goals completed! 🐙 --this was previously proven using `native_decide`
countRefactorable x is the number of refactorable numbers $\le x$.
noncomputable def countRefactorable (x : ℝ) : ℕ :=
if _hx : x ≥ 1 then
countRefactorableNat (Int.toNat (floor x))
else
0Simon Colton conjectures that the number of refactorable numbers less than $x$ is at least $\frac{x}{2\log x}$. This is an asymptotic claim, so we state it for sufficiently large $x$.
In this form it is a theorem: Zelinsky [Ze02, Theorem 7] proved that for every $k$ the number of
refactorable numbers $\le n$ exceeds $k \pi(n)$ for all sufficiently large $n$, and
$\pi(n) \sim n / \log n$. It also follows from Spiro's asymptotic [Sp85], by which the count is
$\frac{x}{\sqrt{\log x}} (\log \log x)^{-1 + o(1)}$. See colton_conjecture for the
pointwise conjecture that remains open.
@[category research solved, AMS 11]
theorem conjecture : ∀ᶠ x in Filter.atTop,
(countRefactorable x : ℝ) ≥ x / (2 * Real.log x) := ⊢ ∀ᶠ (x : ℝ) in Filter.atTop, ↑(countRefactorable x) ≥ x / (2 * Real.log x)
All goals completed! 🐙Colton's conjecture [Co99] as stated by Zelinsky [Ze02]: for every $n$, the number of refactorable numbers $\le n$ is at least half the number of primes $\le n$, i.e. $\pi(n) \le 2,T(n)$.
Zelinsky [Ze02] proved this for all sufficiently large $n$ (see conjecture), with an explicit
bound of $7.42 \cdot 10^{13}$ beyond which it holds; the remaining range is open.
@[category research open, AMS 11]
theorem colton_conjecture : ∀ n : ℕ, Nat.primeCounting n ≤ 2 * countRefactorableNat n := ⊢ ∀ (n : ℕ), n.primeCounting ≤ 2 * countRefactorableNat n
All goals completed! 🐙Colton's conjecture holds for $n \le 500$.
@[category test, AMS 11]
theorem colton_conjecture_le_500 : ∀ n ≤ 500, Nat.primeCounting n ≤ 2 * countRefactorableNat n := ⊢ ∀ n ≤ 500, n.primeCounting ≤ 2 * countRefactorableNat n
All goals completed! 🐙 --this was previously proven using `native_decide`end OeisA111291