Minimum of { k > 0 : [2^n / 3^k] mod 6 = 3 } if such k exists, 0 otherwise.
0, 0, 0, 0, 0, 2, 1, 0, 3, 0, 0, 3, 1, 3, 0, 0, 2, 0, 1, 5, 4, 12, 7, 2, 1, 11, 0, 15, 10, 4, 1, 4
1 seqfan posts
Fri Mar 6 21:27:30 CET 2009 [seqfan] Re: An arithmetic conjecture