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

%I #7 Aug 06 2012 15:51:35

%S 1,3,1,15,1,3,1,3,1,3,1,3,1,3,2,3,1,3,1,3,1,3,1,3,1,3,2,5,1,5,1,3,1,3,

%T 1,3,1,5,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,1,1,3,3,3,1,3,1,3,1,3,1,3,1,3,

%U 2,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,1,3,2

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

%H Clark Kimberling, <a href="/A214073/b214073.txt">Table of n, a(n) for n = 1..1000</a>

%e gcd(63,35) = 7, gcd(62,34) = 2, gcd(61,33) = 1, so a(6) = 3.

%t b[n_] := 2^n; c[n_] := n^2;

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

%Y Cf. A071222.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jul 26 2012