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Least m > 0 such that gcd(m^n+13, (m+1)^n+13) > 1, or 0 if there is no such m.
2

%I #17 Oct 16 2018 17:27:10

%S 1,0,26,1,5,24308100,1,329,71,1,6,59,1,135,5,1,23,7711,1,82,6,1,8,

%T 320594291825643656342,1,45,10,1,755,1107,1,4279,30269,1,5,205961,1,

%U 259,8,1,9,101975,1,6491,5,1,8

%N Least m > 0 such that gcd(m^n+13, (m+1)^n+13) > 1, or 0 if there is no such m.

%C See A118119, which is the main entry for this class of sequences.

%F a(3k) = 1 for k>=0, because 1^(3k)+13 = 14, 2^(3k)+13 = 8^k+13 = 14 (mod 7), therefore gcd(1^(3k)+13, 2^(3k)+13) >= 7.

%e For n=1, gcd(m^n+13, (m+1)^n+13) = gcd(m+13, m+14) = 1, therefore a(1)=0.

%e For n=2, gcd(26^2+13, 27^2+13) = 53, and (m, m+1) = (26, 27) is the smallest pair which yields a GCD > 1 here.

%e For n=0, n=3, n=6,... see formula.

%t A255863[n_] := Module[{m = 1}, While[GCD[m^n + 13, (m + 1)^n + 13] <= 1, m++]; m]; Join[{1, 0}, Table[A255863[n], {n, 2, 22}]] (* _Robert Price_, Oct 16 2018 *)

%o (PARI) a(n,c=13,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, A255852-A255869

%K nonn,hard

%O 0,3

%A _M. F. Hasler_, Mar 10 2015

%E a(5)-a(46) from _Hiroaki Yamanouchi_, Mar 12 2015