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Numbers k such that gcd(k!+1, k^k+1) > 1.
5

%I #22 Feb 19 2024 22:53:04

%S 1,3,23,39,51,63,95,99,131,183,191,215,239,251,299,303,315,363,371,

%T 411,419,431,443,495,543,575,659,683,711,743,755,791,831,891,911,935,

%U 975,1019,1031,1055,1071,1143,1155,1191,1211,1223,1251,1275,1295,1355

%N Numbers k such that gcd(k!+1, k^k+1) > 1.

%C See A116892 for the corresponding values of the GCD. See also comments in A116891.

%H Nick Hobson, <a href="/A116893/b116893.txt">Table of n, a(n) for n = 1..10000</a> (first 1832 terms from Antti Karttunen)

%H Nick Hobson, <a href="/A116893/a116893.c.txt">C program</a>.

%e gcd(1!+1, 1^1+1) = 2, gcd(2!+1, 2^2+1) = 1 and gcd(3!+1, 3^3+1) = 7, so 1 and 3 are the first two terms of the sequence.

%t Select[Range[1500], (GCD[ #!+1, #^#+1] > 1)&]

%o (PARI) isok(n) = gcd(n! + 1, n^n + 1) != 1; \\ _Michel Marcus_, Jul 22 2018

%o (C) See Links section.

%Y Cf. A014566, A038507, A067658, A116891, A116892, A116894.

%K easy,nonn

%O 1,2

%A _Giovanni Resta_, Mar 01 2006