%I
%S 1,0,2,22,128,12,2,81,1,5982,2,11417025,32768,70471611388086,2,26,1,
%T 1019,2,12168420936538713481747,48,128,2,788,1,131711329,2,91,13,
%U 2920553219286322570768516629247,2,237,1,22,2,108,27,9404578,2,2859801,1,41772125,2
%N Least m > 0 such that gcd(m^n+16, (m+1)^n+16) > 1, or 0 if there is no such m.
%C See A118119, which is the main entry for this class of sequences.
%F a(4k+2) = 2 for k>=0, because 2^(4k+2) = 4^(2k+1), 3^(4k+2) = 9^(2k+1), and 4 = 9 = 1 (mod 5), therefore gcd(2^(4k+2)+16, 3^(4k+2)+16) >= 5.
%e For n=0, gcd(m^0+16, (m+1)^0+16) = gcd(16, 16) = 16, therefore a(0)=1, the smallest possible (positive) mvalue.
%e For n=1, gcd(m^n+16, (m+1)^n+16) = gcd(m+15, m+16) = 1, therefore a(1)=0.
%e For n=2, see formula with k=0.
%e For n=3, gcd(22^3+16, 23^3+16) = 31 and (m, m+1) = (22, 23) is the smallest pair which yields a GCD > 1 here.
%t A255866[n_] := Module[{m = 1}, While[GCD[m^n + 16, (m + 1)^n + 16] <= 1, m++]; m]; Join[{1, 0}, Table[A255866[n], {n, 2, 10}]] (* _Robert Price_, Oct 16 2018 *)
%o (PARI) a(n,c=16,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}
%Y Cf. A118119, A255832, A255852A255869
%K nonn,hard
%O 0,3
%A _M. F. Hasler_, Mar 09 2015
%E a(11)a(42) from _Max Alekseyev_, Aug 06 2015
