A255869 Least m > 0 such that gcd(m^n+19, (m+1)^n+19) > 1, or 0 if there is no such m.

1, 0, 3, 2408, 1, 3976, 608, 28, 1, 88, 23, 464658, 1, 319924724, 3, 7, 1, 1628, 138, 2219409, 1, 6, 5, 594, 1, 872, 3, 92, 1, 392, 65, 2278155, 1, 3755866, 4793, 13, 1, 7873, 3, 614294, 1, 448812437, 5

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.

a(43) <= 8153777984244162781089834. - Max Alekseyev, Aug 06 2015

Examples

			For n=0 and n=4, see formula with k=0 resp. k=1.
For n=1, gcd(m^n+19, (m+1)^n+19) = gcd(m+19, m+20) = 1, therefore a(1)=0.
For n=2, gcd(3^2+19, 4^2+19) = 7 and (m,m+1) = (3,4) is the smallest pair which yields a GCD > 1 here.
		

Crossrefs

Cf. A118119, A255832, A255852-A255868

Programs

Mathematica

A255869[n_] := Module[{m = 1}, While[GCD[m^n + 19, (m + 1)^n + 19] <= 1, m++]; m]; Join[{1, 0}, Table[A255869[n], {n, 2, 12}]] (* Robert Price, Oct 16 2018 *)

PARI

a(n,c=19,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 A255869(n):
    if n == 0: return 1
    k = 0
    for p in primefactors(resultant(x**n+19,(x+1)**n+19)):
        for d in (a for a in nthroot_mod(-19,n,p,all_roots=True) if pow(a+1,n,p)==-19%p):
            k = min(d,k) if k else d
    return k # Chai Wah Wu, May 07 2024

Formula

a(4k) = 1 for k>=0, because gcd(1^(4k)+19, 2^(4k)+19) = gcd(20, 16^k-1) >= 5 since 16 = 1 (mod 5).

Extensions

a(13)-a(40) from Hiroaki Yamanouchi, Mar 12 2015

a(41)-a(42) from Max Alekseyev, Aug 06 2015