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A255865 Least m > 0 such that gcd(m^n+15, (m+1)^n+15) > 1, or 0 if there is no such m.

%I A255865 #21 Oct 16 2018 17:24:59
%S A255865 1,0,30,5,120,133,15,14,11,5,7680,968,18,243,26,5,9,
%T A255865 1844434621273219148118716000949433592399169477194046126,8,
%U A255865 22173201293492286974730770140,51,5,593,5885,41,112,15,5,23
%N A255865 Least m > 0 such that gcd(m^n+15, (m+1)^n+15) > 1, or 0 if there is no such m.
%C A255865 See A118119, which is the main entry for this class of sequences.
%H A255865 <a href="/A255865/b255865.txt">Table of n, a(n) for n = 0..42</a>
%F A255865 a(6k+3) = 5 for k>=0, because 5^(6k+3) = 125^(2k+1), 6^(6k+3) = 216^(2k+1), and 125 = 216 = -1 (mod 7), therefore gcd(5^(6k+3)+15, 6^(6k+3)+15) >= 7.
%e A255865 For n=0, gcd(m^0+15, (m+1)^0+15) = gcd(16, 16) = 16, therefore a(0)=1, the smallest possible (positive) m-value.
%e A255865 For n=1, gcd(m^n+15, (m+1)^n+15) = gcd(m+15, m+16) = 1, therefore a(1)=0.
%e A255865 For n=2, gcd(30^2+15, 31^2+15) = 61 and (m, m+1) = (30, 31) is the smallest pair which yields a GCD > 1 here.
%e A255865 For n=3, see formula with k=0.
%t A255865 A255865[n_] := Module[{m = 1}, While[GCD[m^n + 15, (m + 1)^n + 15] <= 1, m++]; m]; Join[{1, 0}, Table[A255865[n], {n, 2, 16}]] (* _Robert Price_, Oct 16 2018 *)
%o A255865 (PARI) a(n,c=15,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}
%Y A255865 Cf. A118119, A255832, A255852-A255869
%K A255865 nonn
%O A255865 0,3
%A A255865 _M. F. Hasler_, Mar 09 2015
%E A255865 a(17)-a(36) from _Hiroaki Yamanouchi_, Mar 12 2015
%E A255865 a(37)-a(42) from _Max Alekseyev_, Aug 07 2015