A255856 Least k > 0 such that gcd(k^n+6, (k+1)^n+6) > 1, or 0 if there is no such k.

1, 0, 2, 1, 48, 9, 1, 19, 14, 1, 2, 77364635, 1, 1943, 2, 1, 5, 299788383228819788, 1, 1578270389554680057141787800241971645032008710129107338825798, 9, 1, 2, 6676, 1, 415, 2, 1, 39, 168338080349, 1, 305, 6806, 1, 2, 9, 1

Offset: 0

Author

M. F. Hasler, Mar 08 2015

Keywords

• nonn
• hard
• more

Comments

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

Examples

			For n=1, gcd(k^n+6, (k+1)^n+6) = gcd(k+6, k+7) = 1, therefore a(1)=0.
For n=2, we have gcd(2^2+6, 3^2+6) = gcd(10, 15) = 5, and the pair (k,k+1)=(2,3) is the smallest which yields a gcd > 1, therefore a(2)=2.
For n=3k, see formula.
		

Crossrefs

Cf. A118119, A255832, A255852-A255869

Programs

Mathematica

A255856[n_] := Module[{m = 1}, While[GCD[m^n + 6, (m + 1)^n + 6] <= 1, m++]; m]; Join[{1, 0}, Table[A255856[n], {n, 2, 10}]] (* Robert Price, Oct 15 2018 *)

PARI

a(n,c=6,L=10^7,S=1)={n!=1&&for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

Python

from sympy import primefactors, resultant, nthroot_mod
from sympy.abc import x
def A255856(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+6,(x+1)**n+6)):
        for d in (a for a in sorted(nthroot_mod(-6,n,p,all_roots=True)) if pow(a+1,n,p)==-6%p):
            k = min(d,k) if k else d
            break
    return int(k) # Chai Wah Wu, May 09 2024

Formula

a(3k)=1 for k>=0, because gcd(1^(3k)+6, 2^(3k)+6) = gcd(7, 8^k-1) = 7.

Extensions

a(11)-a(36) from Hiroaki Yamanouchi, Mar 13 2015