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import FormalConjecturesUtilSquare odd-indexed terms in the recurrence $a(n) = 8a(n-1) - 8a(n-2) + 8a(n-3) - a(n-4)$
A113254: Corresponds to $m = 8$ in a family of 4th-order linear recurrence sequences.
The sequence $a(n)$ is defined by the initial conditions $a(0)=-1, a(1)=4, a(2)=176, a(3)=3136$, and the linear recurrence relation $a(n) = -4 * a (n-1) + 256 * a (n-3) + 4096 * a (n-4)$ for $n \ge 4$.
References:
arxiv/2605.22763 Advancing Mathematics Research with AI-Driven Formal Proof Search by George Tsoukalas et al.
namespace OeisA113254open Nat IntThe sequence $a(n)$ is defined by the initial conditions $a(0)=-1, a(1)=4, a(2)=176, a(3)=3136$, and the linear recurrence relation $a(n) = -4 * a (n-1) + 256 * a (n-3) + 4096 * a (n-4)$ for $n \ge 4$.
def a (n : ℕ) : ℤ :=
match n with
| 0 => -1
| 1 => 4
| 2 => 176
| 3 => 3136
| n' + 4 => -4 * a (n' + 3) + 256 * a (n' + 1) + 4096 * a n'@[category test, AMS 11]
lemma a_0 : a 0 = -1 := ⊢ a 0 = -1 All goals completed! 🐙@[category test, AMS 11]
lemma a_1 : a 1 = 4 := ⊢ a 1 = 4 All goals completed! 🐙@[category test, AMS 11]
lemma a_2 : a 2 = 176 := ⊢ a 2 = 176 All goals completed! 🐙@[category test, AMS 11]
lemma a_3 : a 3 = 3136 := ⊢ a 3 = 3136 All goals completed! 🐙@[category test, AMS 11]
lemma a_4 : a 4 = -15616 := ⊢ a 4 = -15616 All goals completed! 🐙Conjecture: $a(8, 2n+1)$ is a perfect square for all $n$ (see A113249).
A formal proof has been found with the methods described in arxiv/2605.22763.
@[category research solved, AMS 11, formal_proof using formal_conjectures at
"https://github.com/mo271/formal-conjectures/blob/a32396489dcb8f86c3549b93aa358ac6a10a3a1f/FormalConjectures/OEIS/113254.wip.lean#L130"]
theorem a_odd_is_square : ∀ n : ℕ, IsSquare (a (2 * n + 1)) := ⊢ ∀ (n : ℕ), IsSquare (a (2 * n + 1))
All goals completed! 🐙end OeisA113254