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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.
2

%I #17 Jan 28 2019 02:26:00

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

%N 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 a(2) = 1 because 2^2+1 = 5 and 3^2+1 = 2*5 => gcd(..) = 5 and q = 1;

%e a(53) = 3 because

%e 5^53 + 1 = 2 * 3 * 107 * 28838378869 * 599659003321309822423087;

%e 6^53 + 1 = 7 * 107 * 97351567 * 33685364386033 * 71080464397105403;

%e 7^53 + 1 = 2^3 * 107 * 345449549 * 35416476134069*58902316970027001503;

%e 8^53 + 1 = 3^2 * 107 * 6043 * 28059810762433 * 4475130366518102084427698737 => gcd(..) = 107 and q=3.

%p 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 Cf. A186710, A118119.

%K nonn

%O 2,4

%A _Michel Lagneau_, Feb 26 2011

%E a(55), a(56) corrected by _R. J. Mathar_, Mar 07 2011