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A255856
Least k > 0 such that gcd(k^n+6, (k+1)^n+6) > 1, or 0 if there is no such k.
5
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
0,3
COMMENTS
See A118119, which is the main entry for this class of sequences.
FORMULA
a(3k)=1 for k>=0, because gcd(1^(3k)+6, 2^(3k)+6) = gcd(7, 8^k-1) = 7.
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
M. F. Hasler, Mar 08 2015
EXTENSIONS
a(11)-a(36) from Hiroaki Yamanouchi, Mar 13 2015
STATUS
approved