login
A255859
Least m > 0 such that gcd(m^n+9,(m+1)^n+9) > 1, or 0 if there is no such m.
4
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
0,3
COMMENTS
See A118119, which is the main entry for this class of sequences.
FORMULA
a(4k)=1 for k>=0, because gcd(1^(4k)+9, 2^(4k)+9) = gcd(10, 16^k-1) = 5.
EXAMPLE
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.
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 *)
PROG
(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))}
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
M. F. Hasler, Mar 08 2015
EXTENSIONS
a(17)-a(30) from Hiroaki Yamanouchi, Mar 12 2015
a(31)-a(36) from Max Alekseyev, Aug 06 2015
STATUS
approved