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A255858
Least k > 0 such that gcd(k^n + 8, (k+1)^n + 8) > 1, or 0 if there is no such k.
5
1, 0, 1, 5, 1, 5238149, 1, 24747, 1, 5, 1, 1042068640211853141849723, 1, 28913777, 1, 5, 1, 9380, 1, 87940, 1, 5, 1, 677083547, 1, 1584, 1, 5, 1, 5355, 1, 257, 1, 5, 1, 118, 1, 3250, 1, 5, 1, 78609080, 1, 1807714890, 1, 5, 1
OFFSET
0,4
COMMENTS
See A118119, which is the main entry for this class of sequences.
FORMULA
a(2k)=1 for k >= 0, because gcd(1^(2k) + 8, 2^(2k) + 8) = gcd(9, 4^k-1) >= 3.
EXAMPLE
For n=1, gcd(k^n + 8, (k+1)^n + 8) = gcd(k+8, k+9) = 1, therefore a(1)=0.
For n=2*k, see formula.
For n=3, we have gcd(5^3 + 8, 6^3 + 8) = gcd(133, 224) = 7, and the pair (k,k+1)=(5,6) is the smallest which yields a GCD > 1, therefore a(3)=5.
MATHEMATICA
A255858[n_] := Module[{m = 1}, While[GCD[m^n + 8, (m + 1)^n + 8] <= 1, m++]; m]; Join[{1, 0}, Table[A255858[n], {n, 2, 10}]] (* Robert Price, Oct 15 2018 *)
PROG
(PARI) a(n, c=8, 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
AUTHOR
M. F. Hasler, Mar 08 2015
EXTENSIONS
a(11)-a(40) from Hiroaki Yamanouchi, Mar 13 2015
a(41)-a(46) from Max Alekseyev, Aug 06 2015
STATUS
approved