%I #11 Dec 12 2021 20:12:46
%S 1,2,6,21,3,27,9,9,88263,9,99,594,198,99,99,99,99,99,99,9009,99,99,
%T 198,99,99,297,1089,99,198,198,594,198,396,693,99,99,99,297,594,99,99,
%U 99,198,99,99,99,99,99,99,99,99,396,2772,99,99,99,396,693,693,99,99,99,99
%N a(n) = GCD(reverse(n!), reverse((n+1)!)).
%C Through the first 200 terms, the largest term has 6 digits with the exception of a(99) which has 134 digits. - _Harvey P. Dale_, Dec 24 2018
%F a(n) = GCD(A004153((n+1)!), A004153(n!)).
%e Outstandingly high values arise at n = 10^k - 1 because
%e A004153(n) = A004153(n+1), a(n) = rev(n!), n! written backwards.
%e See n = 9, 99, 999, etc.
%t rd[x_] :=FromDigits[Reverse[IntegerDigits[x]]] Table[GCD[rd[w! ], rd[(w+1)! ]], {w, 1, 100}]
%t GCD@@#&/@Partition[IntegerReverse[Range[100]!],2,1] (* _Harvey P. Dale_, Dec 24 2018 *)
%o (Python)
%o from math import factorial, gcd
%o def a(n):
%o f = factorial(n)
%o return gcd(int(str(f)[::-1]), int(str(f*(n+1))[::-1]))
%o print([a(n) for n in range(1, 64)]) # _Michael S. Branicky_, Dec 12 2021
%Y Cf. A000142, A004153.
%K base,nonn
%O 1,2
%A _Labos Elemer_, Jan 25 2005