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Least m>0 such that gcd(2*n-1+m, 2*n-m) > 1.
3

%I #7 Jul 26 2012 13:01:27

%S 2,4,6,2,10,12,2,16,3,2,22,24,2,3,30,2,34,36,2,40,42,2,4,3,2,52,54,2,

%T 3,4,2,64,66,2,70,6,2,76,3,2,82,84,2,3,90,2,6,96,2,100,4,2,106,3,2,

%U 112,114,2,3,120,2,7,126,2,4,132,2,136,3,2,142,4,2,3,7,2,154,156

%N Least m>0 such that gcd(2*n-1+m, 2*n-m) > 1.

%H Clark Kimberling, <a href="/A214061/b214061.txt">Table of n, a(n) for n = 1..1000</a>

%e gcd(7+1, 8-1) = 1 and gcd(7+2, 8-2) > 1, so that a(4) = 2.

%t b[n_] := 2 n - 1; c[n_] := 2 n; Table[m = 1; While[GCD[b[n] + m, c[n] - m] == 1, m++]; m, {n, 150}]

%Y Cf. A214062.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Jul 25 2012