A115247 2^a(n) divides A001935(n) but 2^(a(n)+1) does not.

0, 0, 1, 0, 2, 1, 0, 2, 4, 1, 0, 1, 1, 6, 1, 0, 2, 1, 4, 1, 6, 0, 2, 4, 1, 2, 2, 8, 0, 1, 1, 2, 1, 4, 6, 1, 0, 3, 4, 1, 2, 7, 1, 6, 1, 0, 1, 4, 6, 2, 1, 1, 2, 1, 1, 0, 1, 1, 8, 2, 4, 2, 6, 4, 3, 1, 0, 2, 4, 7, 3, 1, 1, 4, 1, 1, 6, 1, 0, 2, 1, 2, 1, 6, 1, 2, 4, 2, 7, 8, 6, 0, 2, 4, 2, 1, 1, 6, 4, 4

Offset

0,5

Comments

Almost all members of A001935 are divisible by 2^k for any k, therefore almost all a(n)>k for any k.

Links

Table of n, a(n) for n=0..99.

Basil Gordon and Ken Ono, Divisibility of Certain Partition Functions By Powers of Primes.

K. Alladi, Partition Identities Involving Gaps and Weights, Transactions of the American Mathematical Society, Vol. 349, No. 12, Dec 1997, pp. 5001-5019.

Crossrefs

The 0's are in A000217. The 1's are in A115248. Least inverse A115250.

Sequence in context: A144106 A104558 A206022 * A204163 A122542 A227341

Adjacent sequences: A115244 A115245 A115246 * A115248 A115249 A115250

Keyword

nonn

Author

Christian G. Bower, Jan 17 2006

Status

approved