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

%I #8 Jan 14 2020 19:58:42

%S 1,2,1,4,1,6,1,8,1,10,1,2,1,14,1,16,1,18,1,20,1,2,1,24,1,4,1,28,1,30,

%T 1,2,1,34,1,36,1,38,1,5,1,2,1,44,1,46,1,4,1,50,1,2,1,5,1,56,1,58,1,60,

%U 1,2,1,64,1,66,1,5,1,4,1,2,1,6,1,76,1,78,1,80,1,2,1,84,1,86,1

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

%H Clark Kimberling, <a href="/A214060/b214060.txt">Table of n, a(n) for n = 1..1000</a>

%e gcd(23+1,12-1) = 1 and gcd(23+2,12-2) > 1, so that a(12) = 2.

%t b[n_] := 2*n - 1; c[n_] := n;

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

%t lmg[n_]:=Module[{m=1},While[GCD[2n-1+m,n-m]<2,m++];m]; Array[lmg,90] (* _Harvey P. Dale_, Jan 14 2020 *)

%Y Cf. A214057, A214061, A214062.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jul 24 2012