/- Copyright 2026 The Formal Conjectures Authors. Licensed under the Apache License, Version 2.0 (the "License"); you may not use this file except in compliance with the License. You may obtain a copy of the License at https://www.apache.org/licenses/LICENSE-2.0 Unless required by applicable law or agreed to in writing, software distributed under the License is distributed on an "AS IS" BASIS, WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied. See the License for the specific language governing permissions and limitations under the License. -/ import FormalConjecturesUtil

Central factorial numbers: $((2n)!!)^2$

Central factorial numbers: $a(n) = 4^n (n!)^2 = ((2n)!!)^2$.

References:

namespace OeisA2454

Central factorial numbers: $a(n) = 4^n (n!)^2$.

def a (n : ) : := 4 ^ n * n.factorial ^ 2

Value of the sequence a at 0.

@[category test, AMS 11] theorem a_0 : a 0 = 1 := a 0 = 1 All goals completed! 🐙

Value of the sequence a at 1.

@[category test, AMS 11] theorem a_1 : a 1 = 4 := a 1 = 4 All goals completed! 🐙

Value of the sequence a at 2.

@[category test, AMS 11] theorem a_2 : a 2 = 64 := a 2 = 64 All goals completed! 🐙

Value of the sequence a at 3.

@[category test, AMS 11] theorem a_3 : a 3 = 2304 := a 3 = 2304 All goals completed! 🐙

Value of the sequence a at 4.

@[category test, AMS 11] theorem a_4 : a 4 = 147456 := a 4 = 147456 All goals completed! 🐙

Let $\zeta$ be a primitive $(2n+1)$-th root of unity. Then the permanent of the $2n \times 2n$ matrix $[m(j,k)]_{j,k=1..2n}$ is $a(n)/(2n+1) = ((2n)!!)^2/(2n+1)$, where $m(j,k)$ is $1$ or $(1+\zeta^{j-k})/(1-\zeta^{j-k})$ according as $j = k$ or not.

    Zhi-Wei Sun, Dec 21 2021

@[category research open, AMS 11 15] theorem conjecture (n : ) : let N : := 2 * n let K : := N + 1 (ζ : ), IsPrimitiveRoot ζ K Matrix.permanent (fun (j k : Fin N) => if j = k then (1 : ) else let pow : := (j : ) - (k : ) (1 + ζ ^ pow) / (1 - ζ ^ pow) ) = (a n : ) / (K : ) := n:let N := 2 * n; let K := N + 1; (ζ : ), IsPrimitiveRoot ζ K (Matrix.permanent fun j k if j = k then 1 else let pow := j - k; (1 + ζ ^ pow) / (1 - ζ ^ pow)) = (OeisA2454.a n) / K All goals completed! 🐙end OeisA2454