FormalConjectures.OEIS.«72200»

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import FormalConjecturesUtil

Smallest factorial containing exactly $n$ 6's

The sequence $a(n)$ gives the smallest $k$ such that the decimal expansion of $k!$ contains exactly $n$ occurrences of the digit '6', or $0$ if no such $k$ exists.

References:

A072200

namespace OeisA72200open Classical in

Smallest $k$ such that $k!$ contains exactly $n$ 6's in base 10, or 0 if no such $k$ exists.

noncomputable def a (n : ℕ) : ℕ :=
  if h : ∃ k, (Nat.digits 10 k.factorial).count 6 = n then
    Nat.find h
  else
    0

Value of the sequence a at 1.


@[category test, AMS 11]
theorem a_1 : a 1 = 3 := by⊢ a 1 = 3
  classical
  dsimp [a]⊢ (if h : ∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1 then Nat.find h else 0) = 3
  split_ifs with hposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ Nat.find h = 3negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ False
  ·posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ Nat.find h = 3 rw [Nat.find_eq_iffposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ List.count 6 (Nat.digits 10 (Nat.factorial 3)) = 1 ∧ ∀ n < 3, ¬List.count 6 (Nat.digits 10 n.factorial) = 1posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ List.count 6 (Nat.digits 10 (Nat.factorial 3)) = 1 ∧ ∀ n < 3, ¬List.count 6 (Nat.digits 10 n.factorial) = 1]posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ List.count 6 (Nat.digits 10 (Nat.factorial 3)) = 1 ∧ ∀ n < 3, ¬List.count 6 (Nat.digits 10 n.factorial) = 1
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ False exact (h ⟨3, byh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 1⊢ List.count 6 (Nat.digits 10 (Nat.factorial 3)) = 1 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 2.


@[category test, AMS 11]
theorem a_2 : a 2 = 15 := by⊢ a 2 = 15
  classical
  dsimp [a]⊢ (if h : ∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2 then Nat.find h else 0) = 15
  split_ifs with hposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ Nat.find h = 15negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ False
  ·posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ Nat.find h = 15 rw [Nat.find_eq_iffposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ List.count 6 (Nat.digits 10 (Nat.factorial 15)) = 2 ∧ ∀ n < 15, ¬List.count 6 (Nat.digits 10 n.factorial) = 2posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ List.count 6 (Nat.digits 10 (Nat.factorial 15)) = 2 ∧ ∀ n < 15, ¬List.count 6 (Nat.digits 10 n.factorial) = 2]posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ List.count 6 (Nat.digits 10 (Nat.factorial 15)) = 2 ∧ ∀ n < 15, ¬List.count 6 (Nat.digits 10 n.factorial) = 2
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ False exact (h ⟨15, byh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 2⊢ List.count 6 (Nat.digits 10 (Nat.factorial 15)) = 2 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 3.


@[category test, AMS 11]
theorem a_3 : a 3 = 23 := by⊢ a 3 = 23
  classical
  dsimp [a]⊢ (if h : ∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3 then Nat.find h else 0) = 23
  split_ifs with hposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ Nat.find h = 23negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ False
  ·posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ Nat.find h = 23 rw [Nat.find_eq_iffposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ List.count 6 (Nat.digits 10 (Nat.factorial 23)) = 3 ∧ ∀ n < 23, ¬List.count 6 (Nat.digits 10 n.factorial) = 3posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ List.count 6 (Nat.digits 10 (Nat.factorial 23)) = 3 ∧ ∀ n < 23, ¬List.count 6 (Nat.digits 10 n.factorial) = 3]posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ List.count 6 (Nat.digits 10 (Nat.factorial 23)) = 3 ∧ ∀ n < 23, ¬List.count 6 (Nat.digits 10 n.factorial) = 3
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ False exact (h ⟨23, byh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 3⊢ List.count 6 (Nat.digits 10 (Nat.factorial 23)) = 3 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 4.


@[category test, AMS 11]
theorem a_4 : a 4 = 26 := by⊢ a 4 = 26
  classical
  dsimp [a]⊢ (if h : ∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4 then Nat.find h else 0) = 26
  split_ifs with hposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ Nat.find h = 26negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ False
  ·posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ Nat.find h = 26 rw [Nat.find_eq_iffposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ List.count 6 (Nat.digits 10 (Nat.factorial 26)) = 4 ∧ ∀ n < 26, ¬List.count 6 (Nat.digits 10 n.factorial) = 4posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ List.count 6 (Nat.digits 10 (Nat.factorial 26)) = 4 ∧ ∀ n < 26, ¬List.count 6 (Nat.digits 10 n.factorial) = 4]posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ List.count 6 (Nat.digits 10 (Nat.factorial 26)) = 4 ∧ ∀ n < 26, ¬List.count 6 (Nat.digits 10 n.factorial) = 4
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ False exact (h ⟨26, byh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 4⊢ List.count 6 (Nat.digits 10 (Nat.factorial 26)) = 4 decide +nativeAll goals completed! 🐙⟩).elim

Value of the sequence a at 5.


@[category test, AMS 11]
theorem a_5 : a 5 = 32 := by⊢ a 5 = 32
  classical
  dsimp [a]⊢ (if h : ∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5 then Nat.find h else 0) = 32
  split_ifs with hposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ Nat.find h = 32negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ False
  ·posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ Nat.find h = 32 rw [Nat.find_eq_iffposh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ List.count 6 (Nat.digits 10 (Nat.factorial 32)) = 5 ∧ ∀ n < 32, ¬List.count 6 (Nat.digits 10 n.factorial) = 5posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ List.count 6 (Nat.digits 10 (Nat.factorial 32)) = 5 ∧ ∀ n < 32, ¬List.count 6 (Nat.digits 10 n.factorial) = 5]posh:∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ List.count 6 (Nat.digits 10 (Nat.factorial 32)) = 5 ∧ ∀ n < 32, ¬List.count 6 (Nat.digits 10 n.factorial) = 5
    decide +nativeAll goals completed! 🐙
  ·negh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ False exact (h ⟨32, byh:¬∃ k, List.count 6 (Nat.digits 10 k.factorial) = 5⊢ List.count 6 (Nat.digits 10 (Nat.factorial 32)) = 5 decide +nativeAll goals completed! 🐙⟩).elim

It is conjectured that $a(24) = 0$ since no factorial less than $10000$ contained just 24 sixes.

@[category research open, AMS 11]
theorem conjecture : a 24 = 0 := by⊢ a 24 = 0
  sorryAll goals completed! 🐙

end OeisA72200