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

Author

Christian G. Bower, Jan 17 2006

Keywords

• nonn

Comments

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

Links

• 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.