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A115247
2^a(n) divides A001935(n) but 2^(a(n)+1) does not.
2
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
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
KEYWORD
nonn
AUTHOR
Christian G. Bower, Jan 17 2006
STATUS
approved