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import FormalConjecturesUtilRecurrence with bitwise XOR
The sequence is defined by $a(0) = 0$, $a(1) = 1$, and for $n \ge 0$, $$a(n+2) = (a(n+1) \mathbin{\mathrm{XOR}} (n+2)) - a(n),$$ where $\mathrm{XOR}$ is the bitwise exclusive-or operator on integers.
References:
namespace OeisA182510Defining recurrence for $a(n)$.
def a : ℕ → ℤ
| 0 => 0
| 1 => 1
| n + 2 => Int.xor (a (n + 1)) (n + 2 : ℤ) - a n@[category test, AMS 11]
theorem a_0 : a 0 = 0 := ⊢ a 0 = 0 All goals completed! 🐙@[category test, AMS 11]
theorem a_1 : a 1 = 1 := ⊢ a 1 = 1 All goals completed! 🐙@[category test, AMS 11]
theorem a_2 : a 2 = 3 := ⊢ a 2 = 3 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 = -8 := ⊢ a 4 = -8 All goals completed! 🐙@[category test, AMS 11]
theorem a_5 : a 5 = -2 := ⊢ a 5 = -2 All goals completed! 🐙@[category test, AMS 11]
theorem a_6 : a 6 = 0 := ⊢ a 6 = 0 All goals completed! 🐙Conjecture: the sequence contains 8 zeros.
@[category research open, AMS 11]
theorem conjecture1 : Set.ncard {n : ℕ | a n = 0} = 8 := ⊢ {n | a n = 0}.ncard = 8
All goals completed! 🐙Conjecture: more positive terms than negative.
@[category research open, AMS 11]
theorem conjecture2 :
∃ d_pos d_neg : ℝ,
({n : ℕ | 0 < a n}).HasDensity d_pos ∧
({n : ℕ | a n < 0}).HasDensity d_neg ∧
d_neg < d_pos := ⊢ ∃ d_pos d_neg, {n | 0 < a n}.HasDensity d_pos ∧ {n | a n < 0}.HasDensity d_neg ∧ d_neg < d_pos
All goals completed! 🐙end OeisA182510