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A186713 For the starting base k = A118119(n), a(n) is the largest value q such that gcd(k^n+1, (k+1)^n+1, ..., (k+q)^n+1) > 1.

%I A186713 #17 Jan 28 2019 02:26:00
%S A186713 1,1,1,2,1,2,2,1,1,1,1,1,1,2,1,1,1,1,1,1,1,1,1,2,1,1,1,1,1,2,1,1,1,2,
%T A186713 1,1,1,1,2,1,1,1,1,1,1,1,1,2,1,1,1,3,1,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,
%U A186713 1,1,1,1,1,2,1,2
%N A186713 For the starting base k = A118119(n), a(n) is the largest value q such that gcd(k^n+1, (k+1)^n+1, ..., (k+q)^n+1) > 1.
%e A186713 a(2) = 1 because 2^2+1 = 5 and 3^2+1 = 2*5 => gcd(..) = 5 and q = 1;
%e A186713 a(53) = 3 because
%e A186713 5^53 + 1 = 2 * 3 * 107 * 28838378869 * 599659003321309822423087;
%e A186713 6^53 + 1 = 7 * 107 * 97351567 * 33685364386033 * 71080464397105403;
%e A186713 7^53 + 1 = 2^3 * 107 * 345449549 * 35416476134069*58902316970027001503;
%e A186713 8^53 + 1 = 3^2 * 107 * 6043 * 28059810762433 * 4475130366518102084427698737 => gcd(..) = 107 and q=3.
%p A186713 A186713 := proc(n) local k ,g,q; k := A118119(n) ; for q from 1 do g := igcd(seq((k+i)^n+1,i=0..q)) ; if g=1 then return q-1 ; end if; end do: end proc: # _R. J. Mathar_, Mar 07 2011
%Y A186713 Cf. A186710, A118119.
%K A186713 nonn
%O A186713 2,4
%A A186713 _Michel Lagneau_, Feb 26 2011
%E A186713 a(55), a(56) corrected by _R. J. Mathar_, Mar 07 2011