A103160 a(n) = GCD(reverse(n!), reverse((n+1)!)).

1, 2, 6, 21, 3, 27, 9, 9, 88263, 9, 99, 594, 198, 99, 99, 99, 99, 99, 99, 9009, 99, 99, 198, 99, 99, 297, 1089, 99, 198, 198, 594, 198, 396, 693, 99, 99, 99, 297, 594, 99, 99, 99, 198, 99, 99, 99, 99, 99, 99, 99, 99, 396, 2772, 99, 99, 99, 396, 693, 693, 99, 99, 99, 99

Offset: 1

Author

Labos Elemer, Jan 25 2005

Keywords

• base
• nonn

Comments

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

Examples

			Outstandingly high values arise at n = 10^k - 1 because
A004153(n) = A004153(n+1), a(n) = rev(n!), n! written backwards.
See n = 9, 99, 999, etc.
		

Crossrefs

Cf. A000142, A004153.

Programs

Mathematica

rd[x_] :=FromDigits[Reverse[IntegerDigits[x]]] Table[GCD[rd[w! ], rd[(w+1)! ]], {w, 1, 100}]
GCD@@#&/@Partition[IntegerReverse[Range[100]!],2,1] (* Harvey P. Dale, Dec 24 2018 *)

Python

from math import factorial, gcd
def a(n):
    f = factorial(n)
    return gcd(int(str(f)[::-1]), int(str(f*(n+1))[::-1]))
print([a(n) for n in range(1, 64)]) # Michael S. Branicky, Dec 12 2021

Formula

a(n) = GCD(A004153((n+1)!), A004153(n!)).