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%I #19 Jun 05 2021 06:10:07
%S 1,0,1,1,0,1,0,1,0,2,2,0,1,0,2,0,2,0,2,0,3,3,0,2,0,2,0,4,0,3,0,4,0,3,
%T 0,5,5,0,3,0,4,0,4,0,6,0,5,0,6,0,6,0,5,0,8,7,0,5,0,6,0,8,0,6,0,10,0,7,
%U 0,10,0,9,0,10,0,8,0,12,11,0,7,0,10,0,12,0,12,0,10,0,15,0,11,0,14,0,15,0
%N Triangular array read by rows: T(n,k) is the number of partitions of n in which sum of odd parts is k, for k=0,1,...,n; n>=0.
%C (Sum over row n) = A000041(n) = number of partitions of n.
%C Reversal of this array is array in A113686.
%C From _Gary W. Adamson_, Apr 11 2010: (Start)
%C Let M = an infinite lower triangular matrix with A000041 interleaved with zeros: (1, 0, 1, 0, 2, 0, 3, 0, 5, ...) and Q = A000009 diagonalized with the rest zeros.
%C Then A113685 = M*Q. That row sums of the triangle (deleting prefaced zeros) = A000041 is equivalent to the identity: p(x) = p(x^2) * A000009(x). (End)
%F G.f.: G(t,x) = 1/Product_{j>=1} (1 - t^(2j-1)*x^(2j-1))*(1-x^(2j)). - _Emeric Deutsch_, Feb 17 2006
%e First 5 rows:
%e 1;
%e 0, 1;
%e 1, 0, 1;
%e 0, 1, 0, 2;
%e 2, 0, 1, 0, 2;
%e 0, 2, 0, 2, 0, 3.
%e The partitions of 5 are 5, 1+4, 2+3, 1+1+3, 1+2+2, 1+1+1+2, 1+1+1+1+1.
%e The sums of odd parts are 5,1,3,5,1,3,5, respectively, so that the numbers of 0's, 1's, 2s, 3s, 4s, 5s are 0,2,0,2,0,3, which is row 5 of the array.
%p g := 1/product((1-t^(2*j-1)*x^(2*j-1))*(1-x^(2*j)),j=1..20):
%p gser := simplify(series(g,x=0,22)):
%p P[0] := 1: for n from 1 to 14 do P[n] := coeff(gser,x^n) od:
%p for n from 0 to 14 do seq(coeff(P[n],t,j),j=0..n) od;
%p # yields sequence in triangular form - _Emeric Deutsch_, Feb 17 2006
%Y Cf. A000041, A113686, A066967.
%K nonn,tabl
%O 0,10
%A _Clark Kimberling_, Nov 05 2005
%E More terms from _Emeric Deutsch_, Feb 17 2006
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