A114912 2^a(n) divides A000009(n) but 2^(a(n)+1) does not.

0, 0, 0, 1, 1, 0, 2, 0, 1, 3, 1, 2, 0, 1, 1, 0, 5, 1, 1, 1, 6, 2, 0, 3, 1, 1, 0, 6, 1, 8, 3, 2, 1, 6, 9, 0, 2, 3, 5, 1, 0, 2, 1, 1, 3, 11, 8, 1, 1, 6, 1, 0, 1, 10, 1, 1, 2, 0, 3, 6, 7, 2, 1, 9, 2, 3, 2, 1, 13, 1, 0, 5, 9, 1, 1, 1, 1, 0, 1, 3, 9, 2, 6, 1, 1, 6, 6, 1, 1, 1, 1, 11, 0, 5, 6, 1, 2, 8, 6, 1, 0, 1

Offset

0,7

Comments

Almost all members of A000009 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..101.

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

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

Formula

a(n) = A007814(A000009(n)) [From Max Alekseyev, Apr 27 2010]

Crossrefs

The 0's are in A001318. The 1's are in A114913. Least inverse A115251.

Sequence in context: A263401 A369814 A239928 * A231723 A342720 A029274

Adjacent sequences: A114909 A114910 A114911 * A114913 A114914 A114915

Keyword

nonn

Author

Christian G. Bower, Jan 06 2006

Status

approved