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Least m>0 such that n!!+m and n-m have a common divisor > 1.
4

%I #8 Jul 26 2012 13:50:43

%S 1,2,1,1,1,2,1,1,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,1,2,

%T 1,2,1,2,1,2,1,2,1,2,1,2,1,1,1,2,1,2,1,2,1,2,1,2,1,1,1,2,1,2,1,2,1,2,

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

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

%H Clark Kimberling, <a href="/A214054/b214054.txt">Table of n, a(n) for n = 1..1000</a>

%e For n=6: gcd(n!!+1,n-1) = 1 and gcd(n!!+2,n-2)=2, so that a(6)=2.

%t b[n_]:=n!!;c[n_]:=n;

%t t=Table[m=1;While[GCD[b[n]+m,c[n]-m]==1,m++];m,{n,1,200}] (* A214054 *)

%t Flatten[Position[t,1]] (* A214585 *)

%t Flatten[Position[t,2]] (* A214586 *)

%Y Cf. A214585, A214586.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jul 21 2012