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%I #2 Mar 30 2012 17:36:08
%S 1,0,1,1,0,0,0,1,0,1,1,0,0,0,1,0,1,0,1,0,1,2,0,0,0,1,0,1,0,2,0,1,0,1,
%T 0,1,2,0,0,0,1,0,1,0,2,0,2,0,2,0,1,0,1,0,2,3,0,0,0,2,0,1,0,2,0,2,0,3,
%U 0,2,0,2,0,1,0,2,0,2,4,0,0,0,2,0,2,0,2,0,2,0,3,0,4,0,3,0,2,0,2,0,2,0,2,0,3
%N Triangle read by rows: T(n,k) is the number of partitions of n into distinct parts, in which the sum of the odd parts is k (n>=0, 0<=k<=n).
%C Row sums yield A000009. T(2n,0)=A000009(n), T(2n-1,0)=0. T(2n,1)=0, T(2n+1,1)=A000009(n), T(n,2)=0. T(n,n)=A000700(n). Sum(k*T(n,k), k=0..n)=A116682(n).
%F G.f.=product((1+tx^(2j-1))(1+x^(2j)), j=1..infinity).
%e T(10,4)=2 because we have [6,3,1] and [4,3,2,1].
%e Triangle starts:
%e 1;
%e 0,1;
%e 1,0,0;
%e 0,1,0,1;
%e 1,0,0,0,1;
%e 0,1,0,1,0,1;
%p g:=product((1+(t*x)^(2*j-1))*(1+x^(2*j)),j=1..30): gser:=simplify(series(g,x=0,20)): P[0]:=1: for n from 1 to 15 do P[n]:=sort(coeff(gser,x^n)) od: for n from 0 to 15 do seq(coeff(P[n],t,j),j=0..n) od; # yields sequence in triangular form
%Y Cf. A000009, A035457, A036469, A116676.
%K nonn,tabl
%O 0,22
%A _Emeric Deutsch_, Feb 22 2006
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