%I
%S 1,2,1,4,1,3,1,8,1,10,1,3,1,14,1,16,1,3,1,5,1,2,1,3,1,26,1,28,1,3,1,4,
%T 1,6,1,2,1,15,1,5,1,2,1,5,1,17,1,3,1,50,1,8,1,2,1,56,1,58,1,3,1,2,1,6,
%U 1,3,1,31,1,70,1,3,1,4,1,6,1,3,1,5,1,2,1
%N Least m>0 such that 2^n1+m and nm have a common divisor > 1.
%H Clark Kimberling, <a href="/A214057/b214057.txt">Table of n, a(n) for n = 1..1000</a>
%e gcd(2^611,61) = gcd(62,5) = 1
%e gcd(2^612,62) = gcd(61,4) = 1
%e gcd(2^613,63) = gcd(60,3) = 3,
%e so that a(6) = 3.
%t b[n_] := 2^n1; c[n_] := n;
%t Table[m = 1; While[GCD[b[n] + m, c[n]  m] == 1, m++]; m, {n, 1, 100}]
%Y Cf. A214056.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jul 22 2012
