1. FormalConjectures
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  3. «112970»
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A generalized Stern sequence

A112970: A generalized Stern sequence, defined by the recurrence relations: $a(2n+1) = a(n)$ and $a(2n) = a(n) + a(n-2)$ with $a(0)=1$, $a(1)=1$ and $a(n)=0$ for $n \le -1$.

References:

namespace OeisA112970

a n is the generalized Stern sequence, defined by the recurrence relations: $a(2n+1) = a(n)$ and $a(2n) = a(n)$ + $a(n-2)$ with $a(0) = 1$, $a(1) = 1$ and $a(n) = 0$ for $n \le -1$.

def a (n : ℕ) : ℕ := if n = 0 then 1 else if n = 1 then 1 else let k := n / 2 if n % 2 = 1 then -- Odd case: a(2k + 1) = a(k) a k else -- Even case: a(2k) = a(k) + a(k - 2) let aPrev : ℕ := -- a(m) is 0 if m < 0. Equivalent to checking k < 2 for the argument k-2. if k < 2 then 0 else a (k - 2) a k + aPrev termination_by n@[category test, AMS 11] theorem a_0 : a 0 = 1 := ⊢ a 0 = 1 ⊢ (if 0 = 0 then 1 else if 0 = 1 then 1 else have k := 0 / 2; if 0 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) = 1; All goals completed! 🐙@[category test, AMS 11] theorem a_1 : a 1 = 1 := ⊢ a 1 = 1 ⊢ (if 1 = 0 then 1 else if 1 = 1 then 1 else have k := 1 / 2; if 1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) = 1; All goals completed! 🐙@[category test, AMS 11] theorem a_2 : a 2 = 1 := ⊢ a 2 = 1 ⊢ (if 2 = 0 then 1 else if 2 = 1 then 1 else have k := 2 / 2; if 2 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) = 1; ⊢ (if 2 = 0 then 1 else if 2 = 1 then 1 else have k := 2 / 2; if 2 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_1 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev) + aPrev) = 1; ⊢ (if 2 = 0 then 1 else if 2 = 1 then 1 else have k := 2 / 2; if 2 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_1 := (k - 2) / 2; if (k - 2) % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) + aPrev) + aPrev) = 1; All goals completed! 🐙@[category test, AMS 11] theorem a_3 : a 3 = 1 := ⊢ a 3 = 1 ⊢ (if 3 = 0 then 1 else if 3 = 1 then 1 else have k := 3 / 2; if 3 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) = 1; ⊢ (if 3 = 0 then 1 else if 3 = 1 then 1 else have k := 3 / 2; if 3 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_1 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev) + aPrev) = 1; All goals completed! 🐙@[category test, AMS 11] theorem a_4 : a 4 = 2 := ⊢ a 4 = 2 ⊢ (if 4 = 0 then 1 else if 4 = 1 then 1 else have k := 4 / 2; if 4 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) = 2; ⊢ (if 4 = 0 then 1 else if 4 = 1 then 1 else have k := 4 / 2; if 4 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_1 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then a k_1 else have aPrev := if k_1 < 2 then 0 else a (k_1 - 2); a k_1 + aPrev) + aPrev) = 2; ⊢ (if 4 = 0 then 1 else if 4 = 1 then 1 else have k := 4 / 2; if 4 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_1 := (k - 2) / 2; if (k - 2) % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then a k else have aPrev := if k < 2 then 0 else a (k - 2); a k + aPrev) + aPrev) + aPrev) = 2; ⊢ (if 4 = 0 then 1 else if 4 = 1 then 1 else have k := 4 / 2; if 4 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev) + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_1 := (k - 2) / 2; if (k - 2) % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev) + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_1 := k / 2; if k % 2 = 1 then if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev else have aPrev := if k_1 < 2 then 0 else if k_1 - 2 = 0 then 1 else if k_1 - 2 = 1 then 1 else have k := (k_1 - 2) / 2; if (k_1 - 2) % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev; (if k_1 = 0 then 1 else if k_1 = 1 then 1 else have k := k_1 / 2; if k_1 % 2 = 1 then if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev else have aPrev := if k < 2 then 0 else if k - 2 = 0 then 1 else if k - 2 = 1 then 1 else have k_2 := (k - 2) / 2; if (k - 2) % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev; (if k = 0 then 1 else if k = 1 then 1 else have k_2 := k / 2; if k % 2 = 1 then a k_2 else have aPrev := if k_2 < 2 then 0 else a (k_2 - 2); a k_2 + aPrev) + aPrev) + aPrev) + aPrev) = 2; All goals completed! 🐙

Conjectures: a(2^n)=a(2^(n+1)+1)=A033638(n). This formalizes the equality a(2^n) = a(2^(n+1)+1).

@[category research solved, AMS 11, formal_proof using lean4 at "https://github.com/KitaKen1/oeis-a112970-formal-conjectures/blob/71ae72f443bd9bee7d958f0a19d9f9ec5ab82af5/lean/OeisA112970FC.lean#L43-L46"] theorem conjecture1 (n : ℕ) : a (2^n) = a (2^(n + 1) + 1) := n:ℕ⊢ a (2 ^ n) = a (2 ^ (n + 1) + 1) All goals completed! 🐙

Conjectures: a(2^n-1)=a(32^n-1)=1. This formalizes the equality part a(2^n-1) = a(32^n-1).

@[category research solved, AMS 11, formal_proof using lean4 at "https://github.com/KitaKen1/oeis-a112970-formal-conjectures/blob/71ae72f443bd9bee7d958f0a19d9f9ec5ab82af5/lean/OeisA112970FC.lean#L56-L64"] theorem conjecture2 (n : ℕ) : a (2^n - 1) = a (3 * 2^n - 1) := n:ℕ⊢ a (2 ^ n - 1) = a (3 * 2 ^ n - 1) All goals completed! 🐙

Conjectures: a(2^n-1)=a(3*2^n-1)=1. This formalizes the value part a(2^n-1)=1.

@[category research solved, AMS 11, formal_proof using lean4 at "https://github.com/KitaKen1/oeis-a112970-formal-conjectures/blob/71ae72f443bd9bee7d958f0a19d9f9ec5ab82af5/lean/OeisA112970FC.lean#L48-L54"] theorem conjecture3 (n : ℕ) : a (2^n - 1) = 1 := n:ℕ⊢ a (2 ^ n - 1) = 1 All goals completed! 🐙end OeisA112970