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a(0) = a(1) = 1, a(n) = n! / a(n-2).
3

%I #30 Sep 08 2022 08:46:02

%S 1,1,2,6,12,20,60,252,672,1440,5400,27720,88704,224640,982800,5821200,

%T 21288960,61102080,300736800,1990850400,8089804800,25662873600,

%U 138940401600,1007370302400,4465572249600,15397724160000,90311261040000,707173952284800,3375972620697600

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

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

%C 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

%H Joerg Arndt, <a href="/A214916/b214916.txt">Table of n, a(n) for n = 0..100</a>

%H MathOverflow, <a href="http://mathoverflow.net/questions/260279/link-of-a-power-series-by-the-bernoullis-for-a-riccati-equation-to-zonotopes">Link of a power series by the Bernoulli brothers for a Riccati equation to zonotopes?</a>, a MathOverflow question by Tom Copeland, 2017.

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

%o (Python)

%o import math

%o prpr = prev = 1

%o for n in range(2, 33):

%o print prpr,

%o cur = math.factorial(n) / prpr

%o prpr = prev

%o prev = cur

%o (Magma) [1] cat [n le 2 select n else Factorial(n) div Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Jan 25 2017

%Y Cf. A214914, A214915, A214961, A214963.

%K nonn

%O 0,3

%A _Alex Ratushnyak_, Aug 03 2012