A116891 a(n) = gcd(n! + 1, n^n + 1).
2, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 47, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 79, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 103, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 127, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 191, 1, 1, 1, 199, 1, 1
Offset: 1
Examples
a(3) = gcd(3! + 1, 3^3 + 1) = gcd(7,28) = 7.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..80001
Programs
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Mathematica
Table[GCD[n! + 1, n^n + 1], {n, 101}] (* Robert G. Wilson v, Mar 09 2006 *)
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PARI
A116891(n) = gcd(n!+1,(n^n)+1); \\ Antti Karttunen, Jul 22 2018
Comments