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A214055 Least m>0 such that n!+2+m and n-m have a common divisor > 1. 1

%I #8 Jul 26 2012 12:45:30

%S 1,2,3,1,5,2,1,2,9,1,11,2,1,2,15,1,17,2,1,2,21,1,3,2,1,2,27,1,29,2,1,

%T 2,3,1,35,2,1,2,39,1,41,2,1,2,45,1,5,2,1,2,51,1,3,2,1,2,57,1,59,2,1,2,

%U 3,1,65,2,1,2,69,1,71,2,1,2,5,1,77,2,1,2

%N Least m>0 such that n!+2+m and n-m have a common divisor > 1.

%H Clark Kimberling, <a href="/A214055/b214055.txt">Table of n, a(n) for n = 1..1000</a>

%e 6!+2+1=723, 6-1=5; 6!+2+2=724, 6-2=4; so a(6)=2.

%t b[n_] := n! + 2; c[n_] := n; Table[m = 1; While[GCD[b[n] + m, c[n] - m] == 1, m++]; m, {n, 100}]

%Y Cf. A020639, A060680.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jul 22 2012

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