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

1, 0, 30, 5, 120, 133, 15, 14, 11, 5, 7680, 968, 18, 243, 26, 5, 9, 1844434621273219148118716000949433592399169477194046126, 8, 22173201293492286974730770140, 51, 5, 593, 5885, 41, 112, 15, 5, 23

Offset: 0

Author

M. F. Hasler, Mar 09 2015

Keywords

• nonn

Comments

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

Examples

			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.
For n=1, gcd(m^n+15, (m+1)^n+15) = gcd(m+15, m+16) = 1, therefore a(1)=0.
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.
For n=3, see formula with k=0.
		

Links

• Table of n, a(n) for n = 0..42

Crossrefs

Cf. A118119, A255832, A255852-A255869

Programs

Mathematica

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 *)

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))}

Formula

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.

Extensions

a(17)-a(36) from Hiroaki Yamanouchi, Mar 12 2015

a(37)-a(42) from Max Alekseyev, Aug 07 2015