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%I #7 Feb 17 2018 20:01:12
%S 1,2,3,4,2,2,1,6,5,1,3,2,1,5,4,4,2,2,1,6,5,1,3,2,1,5,2,4,2,2,1,2,5,1,
%T 3,2,1,5,5,1,2,2,1,6,5,1,2,2,1,5,5,1,2,2,1,6,2,1,3,2,1,2,5,4,2,2,1,6,
%U 5,1,3,2,1,5,5,4,2,1,1,6,5,1,3,2,1,5,2,4,2,2,1,2,2,1,3,2,1,5,5
%N a(n) is the least m > 0 such that Fibonacci(n+1+m) and n-m are not relatively prime.
%H Clark Kimberling, <a href="/A214065/b214065.txt">Table of n, a(n) for n = 1..1000</a>
%e gcd(F(5+1+1), 5-1) = 1 and gcd(F(5+1+2), 5-2) > 1, so that a(5) = 2.
%t b[n_] := Fibonacci[n]; c[n_] := n;
%t Table[m = 1; While[GCD[b[n + 1 + m], c[n] - m] == 1, m++]; m, {n, 1, 120}]
%Y Cf. A214063, A214064.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Jul 25 2012