A214916 a(0) = a(1) = 1, a(n) = n! / a(n-2).

1, 1, 2, 6, 12, 20, 60, 252, 672, 1440, 5400, 27720, 88704, 224640, 982800, 5821200, 21288960, 61102080, 300736800, 1990850400, 8089804800, 25662873600, 138940401600, 1007370302400, 4465572249600, 15397724160000, 90311261040000, 707173952284800, 3375972620697600

Offset: 0

Author

Alex Ratushnyak, Aug 03 2012

Keywords

• nonn

Comments

a(n) is least k > a(n-1) such that k*a(n-2) is a factorial.

Two periodic subsets of these numbers appear in the coefficients of a series involved in a solution of a Riccati-type differential equation addressed by the Bernoulli brothers: z = 1 - x^4/12 + x^8/672 - x^12/88704 + ... = 1 - 2 * x^4/4! + 60 * x^8/8! - 5400 * x^12/12! + ... . See the MathOverflow question. - Tom Copeland, Jan 24 2017

Links

• Joerg Arndt, Table of n, a(n) for n = 0..100
• MathOverflow, Link of a power series by the Bernoulli brothers for a Riccati equation to zonotopes?, a MathOverflow question by Tom Copeland, 2017.

Crossrefs

Cf. A214914, A214915, A214961, A214963, A007662.

Programs

Magma

[1] cat  [n le 2 select n  else  Factorial(n)  div  Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jan 25 2017

Python

import math
prpr = prev = 1
for n in range(2, 33):
    print(prpr, end=', ')
    cur = math.factorial(n) // prpr
    prpr = prev
    prev = cur

Formula

a(0) = a(1) = 1, for n>=2, a(n) = n! / a(n-2).

a(n) = n!!!! * (n-1)!!!! = A007662(n) * A007662(n-1), for n >= 1. - David Radcliffe, Jun 05 2025