Remove powers of 2 from A000069. Let b(n) be n-th term of the remaining sequence. Then a(n) is the least number m such that (b(n))^m is in A001969
3, 3, 2, 3, 3, 2, 3, 4, 2, 3, 3, 4, 2, 3, 3, 2, 3, 2, 4, 4, 2, 3, 3, 5, 2, 3, 4, 2, 4, 2, 2, 3, 4, 2
3 seqfan posts
Tue May 25 12:51:09 CEST 2010 [seqfan] Re: A178307