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%I #2 Mar 30 2012 17:36:08
%S 1,1,0,0,1,1,0,1,1,0,0,0,1,1,0,1,0,1,1,0,1,0,0,0,2,1,0,1,0,1,0,2,2,0,
%T 1,0,1,0,0,0,2,2,0,1,0,1,0,2,0,2,2,0,2,0,1,0,2,0,0,0,3,2,0,2,0,1,0,2,
%U 0,2,0,3,3,0,2,0,2,0,2,0,2,0,0,0,4,3,0,2,0,2,0,2,0,2,0,3,0,4,3,0,3,0,2,0,4
%N Triangle read by rows: T(n,k) is the number of partitions of n into distinct parts, in which the sum of the even parts is k (n>=0, 0<=k<=n).
%C Row 2n-1 has 2n-1 terms; row 2n has 2n+1 terms. Row sums yield A000009. T(n,0)=A000700(n). Columns 2n-1 contain only 0's. Sum(k*T(n,k), k=0..n)=A116684(n).
%F G.f.=product((1+x^(2j-1))(1+(tx)^(2j)), j=1..infinity).
%e T(9,6)=2 because we have [6,3] and [4,3,2].
%e Triangle starts:
%e 1;
%e 1;
%e 0,0,1;
%e 1,0,1;
%e 1,0,0,0,1;
%e 1,0,1,0,1;
%e 1,0,1,0,0,0,2
%p g:=product((1+x^(2*j-1))*(1+(t*x)^(2*j)),j=1..30): gser:=simplify(series(g,x=0,20)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(coeff(gser,x^n)) od: for n from 0 to 14 do seq(coeff(P[n],t,j),j=0..2*floor(n/2)) od; # yields sequence in triangular form
%Y Cf. A000009, A000700, A116684.
%K nonn,tabf
%O 0,25
%A _Emeric Deutsch_, Feb 22 2006
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