A255855 Least k > 0 such that gcd(k^n+5, (k+1)^n+5) > 1, or 0 if there is no such k.

1, 0, 1, 5, 1, 533360, 1, 55, 1, 7, 1, 796479131355665831357, 1, 41, 1, 5, 1, 3775, 1, 42296, 1, 7, 1, 653246700175064613889, 1, 21, 1, 5, 1, 1619, 1, 42842, 1, 7, 1, 2945, 1, 323371, 1, 5, 1, 1102221, 1, 633524110177, 1, 7, 1

Offset

0,4

Comments

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

Links

Table of n, a(n) for n=0..46.

Formula

a(2k)=1 for k>=0, because gcd(1^(2k)+5,2^(2k)+5) = gcd(6,4^k-1) = 3.

Example

For n=1, gcd(k^n+5, (k+1)^n+5) = gcd(k+5, k+6) = 1, therefore a(1)=0.
For n=2k, see formula.
For n=3, we have gcd(5^3+5, 6^3+5) = 13, and the pair (k,k+1)=(5,6) is the smallest which yields a GCD > 1, therefore a(3)=5.

Mathematica

A255855[n_] := Module[{m = 1}, While[GCD[m^n + 5, (m + 1)^n + 5] <= 1, m++]; m]; Join[{1, 0}, Table[A255855[n], {n, 2, 10}]]

Prog

(PARI) a(n, c=5, L=10^7, S=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 A255855(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+5, (x+1)**n+5)):
        for d in (a for a in sorted(nthroot_mod(-5, n, p, all_roots=True)) if pow(a+1, n, p)==-5%p):
            k = min(d, k) if k else d
            break
    return int(k) # Chai Wah Wu, May 08 2024

Crossrefs

Cf. A118119, A255832.

Cf. A255852, A255853, A255854, A255856, A255857, A255858, A255869.

Sequence in context: A075266 A094096 A009826 * A255858 A238798 A112871

Adjacent sequences: A255852 A255853 A255854 * A255856 A255857 A255858

Keyword

nonn,hard

Author

M. F. Hasler, Mar 08 2015

Extensions

a(11)-a(46) from Hiroaki Yamanouchi, Mar 12 2015

Status

approved