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%I #3 Mar 30 2012 17:36:08
%S 1,1,1,0,1,1,0,1,1,0,1,0,1,1,0,2,0,1,1,0,2,0,1,0,1,1,0,2,0,2,0,1,1,0,
%T 3,0,2,0,1,0,1,1,0,3,0,3,0,2,0,1,1,0,3,0,4,0,2,0,1,0,1,1,0,4,0,4,0,3,
%U 0,2,0,1,1,0,4,0,5,0,4,0,2,0,1,0,1,1,0,4,0,6,0,5,0,3,0,2,0,1,1,0,5,0,7,0,6
%N Triangle read by rows: T(n,k) is the number of partitions of n into odd parts such that the largest part is k (n>=1, k>=1).
%C Both rows 2n-1 and 2n have 2n-1 terms. Row sums yield A000009. T(n,2k)=0. T(n,3)=A002264(n). Sum(k*T(n,k),k>=1)=A092322(n).
%F G.f.=sum(t^(2j-1)*x^(2j-1)/product(1-x^(2i-1), i=1..j), j=1..infinity).
%e T(10,5)=3 because we have [3,3,3,1], [3,3,1,1,1,1] and [3,1,1,1,1,1,1,1].
%e Triangle starts:
%e 1;
%e 1;
%e 1,0,1;
%e 1,0,1;
%e 1,0,1,0,1;
%e 1,0,2,0,1;
%p g:=sum(t^(2*j-1)*x^(2*j-1)/product(1-x^(2*i-1),i=1..j),j=1..40): gser:=simplify(series(g,x=0,22)): for n from 1 to 16 do P[n]:=sort(coeff(gser,x^n)) od: for n from 1 to 16 do seq(coeff(P[n],t^j),j=1..2*ceil(n/2)-1) od; # yields sequence in triangular form
%Y Cf. A000009, A002264, A092322, A116856.
%K nonn,tabf
%O 1,16
%A _Emeric Deutsch_, Feb 24 2006
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