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%I #4 Mar 30 2012 18:36:40
%S 1,1,5,35,135,1755,6303,39815,132675,3322515,10561455,64566253,
%T 199681945,2391238415,7233344915
%N Numerator of a(n)/2^A005187(n-1), the n-th row sums of A096651^(3/2), with a(0)=1.
%C The denominators are 2^A005187(n-1) (for n>0), where A005187(n) is the number of 1's in binary expansion of 2n. Can the row sums of A096651^(3/2) be said to define the (3/2)-dimensional partitions of n?
%F a(n)/2^A005187(n-1) = Sum_{k=0..n} A096651(n, k)*A096742(k)/2^A005187(k-1).
%e Sequence begins: {1,1,5/2,35/8,135/16,1755/128,6303/256,...}.
%e Formed from the row sums of triangular matrix A096651^(3/2), which begins:
%e {1},
%e {0,1},
%e {0,3/2,1},
%e {0,15/8,3/2,1},
%e {0,41/16,27/8,3/2,1},
%e {0,387/128,53/16,39/8,3/2,1},
%e {0,1017/256,987/128,65/16,51/8,3/2,1},
%e {0,4715/1024,753/256,2067/128,77/16,63/8,3/2,1},
%e {0,11917/2048,29983/1024,-4503/256,3819/128,89/16,75/8,3/2,1},...
%e The denominator of each element at column n, row k, is A005187(n-k).
%Y Cf. A096651, A096742, A005187.
%K more,nonn
%O 0,3
%A _Paul D. Hanna_, Jul 06 2004
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