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Least m>0 such that n!-m and 2^n-m are relatively prime.
1

%I #7 Apr 23 2013 00:13:17

%S 1,1,1,1,1,1,1,1,1,1,3,1,1,1,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%T 3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,

%U 1,1,1,1,1,1,3,1,1,1,1,1,1,1,3,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

%N Least m>0 such that n!-m and 2^n-m are relatively prime.

%H Clark Kimberling, <a href="/A214709/b214709.txt">Table of n, a(n) for n = 1..1000</a>

%e gcd(11!-1,2^11-1) = 23, gcd(11!-2, 2^11-2) = 2, gcd(11!-3,2^11-3) = 1, so a(11) = 3.

%t Table[m = 1; While[GCD[n! - m, 2^n - m] != 1, m++]; m, {n, 1, 140}]

%Y Cf. A214074, A214707.

%K nonn,easy

%O 1,11

%A _Clark Kimberling_, Jul 27 2012