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%I #4 Mar 30 2012 17:37:15
%S 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,
%T 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,
%U 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
%N 2^a(n) divides A000009(n) but 2^(a(n)+1) does not.
%C Almost all members of A000009 are divisible by 2^k for any k, therefore almost all a(n)>k for any k.
%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.
%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>.
%F a(n) = A007814(A000009(n)) [From _Max Alekseyev_, Apr 27 2010]
%Y The 0's are in A001318. The 1's are in A114913. Least inverse A115251.
%K nonn
%O 0,7
%A _Christian G. Bower_, Jan 06 2006
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