OFFSET
1,1
COMMENTS
For n > 0, there are infinitely many numbers k such that floor(n^k/k) is an odd number. For odd number n, it's obvious because k can be n^j for any j >= 0. For n = 2, k can be 3*2^j for any j >= 1.
FORMULA
a(2^j) <= 12. (This is because floor((2*2^j)^12/12) = floor(2^(12j+10)/3) = (2^(12j+10) - 1)/3 is an odd integer for all j >= 0. - Jianing Song, Feb 24 2019)
For n > 1, a(n) <= A090368(n).
PROG
(PARI) a(n) = {k=1; while((2*n)^k\k%2==0, k++); k; }
CROSSREFS
KEYWORD
nonn
AUTHOR
Jinyuan Wang, Feb 22 2019
STATUS
approved