FormalConjectures.OEIS.«114216»

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import FormalConjecturesUtil

Largest odd divisor of $a(n-1) + \textrm{prime}(n)$

$a(0)=0$; thereafter $a(n)$ = largest odd divisor of $a(n-1) + \textrm{prime}(n)$.

References:

A114216

namespace OeisA114216

The primary defining sequence a. $a(n)$ is the largest odd divisor of $a(n-1) + \textrm{prime}(n)$.

noncomputable def a (n : ℕ) : ℕ :=
  match n with
  | 0 => 0
  | n' + 1 =>
    let pN : ℕ := Nat.nth Nat.Prime n'
    let prevA : ℕ := a n'
    let sumVal : ℕ := prevA + pN
    let nu2 : ℕ := padicValNat 2 sumVal
    sumVal / (2 ^ nu2)

@[category test, AMS 11]
theorem a_0 : a 0 = 0 := by⊢ a 0 = 0
  rflAll goals completed! 🐙



@[category test, AMS 11]
theorem a_1 : a 1 = 1 := by⊢ a 1 = 1
  change (0 + Nat.nth Nat.Prime 0) / 2 ^ padicValNat 2 (0 + Nat.nth Nat.Prime 0) = 1⊢ (0 + Nat.nth Nat.Prime 0) / 2 ^ padicValNat 2 (0 + Nat.nth Nat.Prime 0) = 1
  rw [Nat.nth_prime_zero_eq_two⊢ (0 + 2) / 2 ^ padicValNat 2 (0 + 2) = 1⊢ (0 + 2) / 2 ^ padicValNat 2 (0 + 2) = 1]⊢ (0 + 2) / 2 ^ padicValNat 2 (0 + 2) = 1
  native_decideAll goals completed! 🐙



@[category test, AMS 11]
theorem a_2 : a 2 = 1 := by⊢ a 2 = 1
  change (a 1 + Nat.nth Nat.Prime 1) / 2 ^ padicValNat 2 (a 1 + Nat.nth Nat.Prime 1) = 1⊢ (a 1 + Nat.nth Nat.Prime 1) / 2 ^ padicValNat 2 (a 1 + Nat.nth Nat.Prime 1) = 1
  rw [a_1,⊢ (1 + Nat.nth Nat.Prime 1) / 2 ^ padicValNat 2 (1 + Nat.nth Nat.Prime 1) = 1⊢ (1 + 3) / 2 ^ padicValNat 2 (1 + 3) = 1 Nat.nth_prime_one_eq_three⊢ (1 + 3) / 2 ^ padicValNat 2 (1 + 3) = 1⊢ (1 + 3) / 2 ^ padicValNat 2 (1 + 3) = 1]⊢ (1 + 3) / 2 ^ padicValNat 2 (1 + 3) = 1
  native_decideAll goals completed! 🐙



@[category test, AMS 11]
theorem a_3 : a 3 = 3 := by⊢ a 3 = 3
  change (a 2 + Nat.nth Nat.Prime 2) / 2 ^ padicValNat 2 (a 2 + Nat.nth Nat.Prime 2) = 3⊢ (a 2 + Nat.nth Nat.Prime 2) / 2 ^ padicValNat 2 (a 2 + Nat.nth Nat.Prime 2) = 3
  rw [a_2,⊢ (1 + Nat.nth Nat.Prime 2) / 2 ^ padicValNat 2 (1 + Nat.nth Nat.Prime 2) = 3⊢ (1 + 5) / 2 ^ padicValNat 2 (1 + 5) = 3 Nat.nth_prime_two_eq_five⊢ (1 + 5) / 2 ^ padicValNat 2 (1 + 5) = 3⊢ (1 + 5) / 2 ^ padicValNat 2 (1 + 5) = 3]⊢ (1 + 5) / 2 ^ padicValNat 2 (1 + 5) = 3
  native_decideAll goals completed! 🐙



@[category test, AMS 11]
theorem a_4 : a 4 = 5 := by⊢ a 4 = 5
  change (a 3 + Nat.nth Nat.Prime 3) / 2 ^ padicValNat 2 (a 3 + Nat.nth Nat.Prime 3) = 5⊢ (a 3 + Nat.nth Nat.Prime 3) / 2 ^ padicValNat 2 (a 3 + Nat.nth Nat.Prime 3) = 5
  rw [a_3,⊢ (3 + Nat.nth Nat.Prime 3) / 2 ^ padicValNat 2 (3 + Nat.nth Nat.Prime 3) = 5⊢ (3 + 7) / 2 ^ padicValNat 2 (3 + 7) = 5 Nat.nth_prime_three_eq_seven⊢ (3 + 7) / 2 ^ padicValNat 2 (3 + 7) = 5⊢ (3 + 7) / 2 ^ padicValNat 2 (3 + 7) = 5]⊢ (3 + 7) / 2 ^ padicValNat 2 (3 + 7) = 5
  native_decideAll goals completed! 🐙



@[category test, AMS 11]
theorem a_5 : a 5 = 1 := by⊢ a 5 = 1
  change (a 4 + Nat.nth Nat.Prime 4) / 2 ^ padicValNat 2 (a 4 + Nat.nth Nat.Prime 4) = 1⊢ (a 4 + Nat.nth Nat.Prime 4) / 2 ^ padicValNat 2 (a 4 + Nat.nth Nat.Prime 4) = 1
  rw [a_4,⊢ (5 + Nat.nth Nat.Prime 4) / 2 ^ padicValNat 2 (5 + Nat.nth Nat.Prime 4) = 1⊢ (5 + 11) / 2 ^ padicValNat 2 (5 + 11) = 1 Nat.nth_prime_four_eq_eleven⊢ (5 + 11) / 2 ^ padicValNat 2 (5 + 11) = 1⊢ (5 + 11) / 2 ^ padicValNat 2 (5 + 11) = 1]⊢ (5 + 11) / 2 ^ padicValNat 2 (5 + 11) = 1
  native_decideAll goals completed! 🐙

Is $a(33900)$ the last term equal to $1$?

@[category research open, AMS 11]
theorem conjecture :
  answer(sorry) ↔ ∀ n > 33900, a n ≠ 1 := by⊢ True ↔ ∀ n > 33900, OeisA114216.a n ≠ 1
  sorryAll goals completed! 🐙

end OeisA114216