A255863 Least m > 0 such that gcd(m^n+13, (m+1)^n+13) > 1, or 0 if there is no such m.

1, 0, 26, 1, 5, 24308100, 1, 329, 71, 1, 6, 59, 1, 135, 5, 1, 23, 7711, 1, 82, 6, 1, 8, 320594291825643656342, 1, 45, 10, 1, 755, 1107, 1, 4279, 30269, 1, 5, 205961, 1, 259, 8, 1, 9, 101975, 1, 6491, 5, 1, 8

Offset: 0

Author

M. F. Hasler, Mar 10 2015

Keywords

• nonn
• hard

Comments

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

Examples

			For n=1, gcd(m^n+13, (m+1)^n+13) = gcd(m+13, m+14) = 1, therefore a(1)=0.
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.
For n=0, n=3, n=6,... see formula.
		

Crossrefs

Cf. A118119, A255832, A255852-A255869

Programs

Mathematica

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 *)

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))}

Formula

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.

Extensions

a(5)-a(46) from Hiroaki Yamanouchi, Mar 12 2015