A255864 Least m > 0 such that gcd(m^n+14, (m+1)^n+14) > 1, or 0 if there is no such m.

1, 0, 1, 12, 1, 15, 1, 2, 1, 1929501, 1, 13228907223310811104028677, 1, 94, 1, 11, 1, 85364353, 1, 1563, 1, 49, 1, 9258095644888888790279763522646107297983, 1, 23, 1, 66, 1

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(29) with 141 decimal digits is too large to include here (see b-file).

Examples

			For n=1, gcd(m^n+14, (m+1)^n+14) = gcd(m+14, m+15) = 1, therefore a(1)=0.
For n=0 and n=2, see formula with k=0 and k=1.
For n=3, gcd(12^3+14, 13^3+14) = 67, and (m, m+1) = (12, 13) is the smallest pair which yields a GCD > 1 here.
		

Links

• Max Alekseyev, Table of n, a(n) for n = 0..40

Crossrefs

Cf. A118119, A255832, A255852-A255869

Programs

Mathematica

A255864[n_] := Module[{m = 1}, While[GCD[m^n + 14, (m + 1)^n + 14] <= 1, m++]; m]; Join[{1, 0}, Table[A255864[n], {n, 2, 10}]] (* Robert Price, Oct 16 2018 *)

PARI

a(n,c=14,L=10^7,S=1)={n!=1 && for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1 && return(a))}

Formula

a(2k) = 1 for k>=0, because gcd(1^(2k)+14, 2^(2k)+14) = gcd(15, 4^k-1) >= 3, since 4^k-1 = 1-1 = 0 (mod 3).

Extensions

a(11)-a(40) from Max Alekseyev, Aug 06 2015