A255866 Least m > 0 such that gcd(m^n+16, (m+1)^n+16) > 1, or 0 if there is no such m.

1, 0, 2, 22, 128, 12, 2, 81, 1, 5982, 2, 11417025, 32768, 70471611388086, 2, 26, 1, 1019, 2, 12168420936538713481747, 48, 128, 2, 788, 1, 131711329, 2, 91, 13, 2920553219286322570768516629247, 2, 237, 1, 22, 2, 108, 27, 9404578, 2, 2859801, 1, 41772125, 2

Offset: 0

Author

M. F. Hasler, Mar 09 2015

Keywords

• nonn
• hard

Comments

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

Examples

			For n=0, gcd(m^0+16, (m+1)^0+16) = gcd(16, 16) = 16, therefore a(0)=1, the smallest possible (positive) m-value.
For n=1, gcd(m^n+16, (m+1)^n+16) = gcd(m+15, m+16) = 1, therefore a(1)=0.
For n=2, see formula with k=0.
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.
		

Crossrefs

Cf. A118119, A255832, A255852-A255869

Programs

Mathematica

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 *)

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))}

Formula

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.

Extensions

a(11)-a(42) from Max Alekseyev, Aug 06 2015