%I
%S 51,40333,434,16,1234,78607,8310,817172,473,116,22650,736546059,22,
%T 1080982,252,7809,644,1786225573
%N Least m > 0 such that gcd(m^(2n+1)+2, (m+1)^(2n+1)+2) > 1.
%C For n=0 one has gcd(m+2, m+3) = 1 for any m.
%C See A255852 for the sequence including also even exponents, for which the GCD is > 1 already for m=1 (because gcd(1^2k+2, 2^2k+2) = gcd(3, 2^2k1) = gcd(3, 4^k1) = 3), and also for m=4 (because gcd(4^2k+2, 5^2k+2) = gcd(4^2k+2, (5^k4^k)(5^k+4^k)) >= 3), etc.
%F a(n) = A255852(2n+1).
%t A255832[n_] := Module[{m = 1}, While[GCD[m^(2 n + 1) + 2, (m + 1)^(2 n + 1) + 2] <= 1, m++]; m]; Table[A255832[n], {n, 1, 10}] (* _Robert Price_, Oct 15 2018 *)
%o (PARI) a(n,c=2,L=10^6)={n=n*2+1;for(a=1,L,gcd(a^n+c,(a+1)^n+c)>1&&return(n))}
%Y Cf. A118119, A255853, A255853, ... for other variants, corresponding to different constant offsets (+1, +3, ...) in the arguments of gcd.
%K nonn,more,hard
%O 1,1
%A _M. F. Hasler_, Mar 08 2015
%E a(12)a(18) from _Max Alekseyev_, Aug 06 2015
