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

Author

Christian G. Bower, Jan 06 2006

Keywords

• nonn

Comments

Almost all members of A000009 are divisible by 2^k for any k, therefore almost all a(n)>k for any k.

Links

• Amiram Eldar, Table of n, a(n) for n = 0..10000
• Krishnaswami 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, The Ramanujan Journal, Vol. 1, No. 1 (1997), pp. 25-34; alternative link.

Crossrefs

Cf. A000009, A007814.

Cf. A001318 (positions of 0's), A114913 (positions of 1's), A115251 (least inverse).

Programs

Mathematica

a[n_] := IntegerExponent[PartitionsQ[n], 2]; Array[a, 100, 0] (* Amiram Eldar, Aug 24 2024 *)

Formula

a(n) = A007814(A000009(n)). - Max Alekseyev, Apr 27 2010