A255859 Least m > 0 such that gcd(m^n+9,(m+1)^n+9) > 1, or 0 if there is no such m.

1, 0, 18, 533, 1, 32, 288, 484, 1, 364, 6, 176427, 1, 31239, 533, 8, 1, 8424432925592889329288197322308900672459420460792433, 30, 16561, 1, 4, 6, 349, 1, 32, 546, 2579, 1, 375766, 11, 5061867704425915, 1, 5620, 6, 8, 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(m^n+9, (m+1)^n+9) = gcd(m+9, m+10) = 1, therefore a(1)=0.
For n=2, we have gcd(18^2+9, 19^2+9) = gcd(333, 370) = 37, and the pair (m,m+1)=(18,19) is the smallest which yields a GCD > 1, therefore a(2)=37.
For n=4k, see formula.
		

Links

• Will Wei, Patterns that appear to hold, but don't - 8424432925592889329288197322308900672459420460792433, video (2020)

Crossrefs

Cf. A118119, A255832, A255852-A255869

Programs

Mathematica

A255859[n_] := Module[{m = 1}, While[GCD[m^n + 9, (m + 1)^n + 9] <= 1, m++]; m]; Join[{1, 0}, Table[A255859[n], {n, 2, 16}]] (* Robert Price, Oct 16 2018 *)

PARI

a(n,c=9,L=10^7,S=1)={n!=1&&for(a=S,L,gcd(a^n+c,(a+1)^n+c)>1&&return(a))}

Formula

a(4k)=1 for k>=0, because gcd(1^(4k)+9, 2^(4k)+9) = gcd(10, 16^k-1) = 5.

Extensions

a(17)-a(30) from Hiroaki Yamanouchi, Mar 12 2015

a(31)-a(36) from Max Alekseyev, Aug 06 2015