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%I #2 Mar 30 2012 17:37:16
%S 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,
%T 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,
%U 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
%N 2^a(n) divides A001935(n) but 2^(a(n)+1) does not.
%C Almost all members of A001935 are divisible by 2^k for any k, therefore almost all a(n)>k for any k.
%H Basil Gordon and Ken Ono, <a href="http://www.math.wisc.edu/~ono/reprints/018.pdf">Divisibility of Certain Partition Functions By Powers of Primes</a>.
%H K. Alladi, <a href="http://www.ams.org/tran/1997-349-12/S0002-9947-97-01831-X/S0002-9947-97-01831-X.pdf">Partition Identities Involving Gaps and Weights</a>, Transactions of the American Mathematical Society, Vol. 349, No. 12, Dec 1997, pp. 5001-5019.
%Y The 0's are in A000217. The 1's are in A115248. Least inverse A115250.
%K nonn
%O 0,5
%A _Christian G. Bower_, Jan 17 2006
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