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a(n) = gcd(k^n + 1, (k+1)^n + 1) for the smallest k at which the GCD exceeds 1.
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%I #17 Sep 16 2017 00:33:47

%S 5,7,17,11,5,29,17,19,25,23,17,53,145,61,353,137,5,191,41,43,5,47,97,

%T 11,265,19,337,59,25,5953,257,67,5,29,17,223,5,157,17,83,145,173,89,

%U 19,5,283,353,29,12625,307,17,107,5,121,1921,229,5,709,241,367,5,817,769,521,5,269,137,139,725,853,55969,293,745,61,17,29,265

%N a(n) = gcd(k^n + 1, (k+1)^n + 1) for the smallest k at which the GCD exceeds 1.

%C For k=0, the GCD equals 1. Increasing k, the GCD first exceeds 1 at k = A118119(n), and that GCD is a(n).

%e a(2) = 5 because 2^2 + 1 = 5 and 3^2+1 = 2*5;

%e a(3) = 7 because 5^3 + 1 = 2*3^2*7 and 6^3 + 1 = 7*31.

%p A186710 := proc(n) local k ,g; for k from 1 do g := igcd(k^n+1,(k+1)^n+1) ; if g>1 then return g ; end if; end do: end proc: # _R. J. Mathar_, Mar 07 2011

%Y Cf. A118119.

%K nonn

%O 2,1

%A _Michel Lagneau_, Feb 26 2011