%I
%S 1,1,1,0,1,2,2,0,0,0,1,3,0,1,4,0,0,0,0,0,1,5,0,1,6,0,1,0,0,0,0,0,1,8,
%T 0,1,0,1,10,0,1,0,0,0,0,0,0,0,1,12,0,2,0,1,15,0,2,0,0,0,0,0,0,0,0,0,1,
%U 18,0,2,0,1,0,1,22,0,3,0,1,0,0,0,0,0,0,0,0,0,1,27,0,3,0,1,0,1,32,0,4,0,1,0
%N Triangle read by rows: T(n,k) is the number of partitions of n into odd parts such that the smallest part is k (n>=1, k>=1).
%C Row 2n1 has 2n1 terms; row 4n+2 has 2n+1 terms; row 4n has 2n1 terms. Row sums yield A000009. T(n,2k)=0. Sum(k*T(n,k),k>=1)=A092314(n)
%F G.f.=sum(t^(2j1)*x^(2j1)/product(1x^(2i1), i=j..infinity), j=1..infinity).
%e T(12,3)=2 because we have [9,3] and [3,3,3,3].
%e Triangle starts:
%e 1;
%e 1;
%e 1,0,1;
%e 2;
%e 2,0,0,0,1;
%e 3,0,1;
%e 4,0,0,0,0,0,1
%p g:=sum(t^(2*j1)*x^(2*j1)/product(1x^(2*i1),i=j..20),j=1..30): gser:=simplify(series(g,x=0,20)): for n from 1 to 17 do P[n]:=sort(coeff(gser,x^n)) od: d:=proc(n) if n mod 2 = 1 then n elif n mod 4 = 2 then n/2 else n/21 fi end: for n from 1 to 17 do seq(coeff(P[n],t^j),j=1..d(n)) od; # yields sequence in triangular form; d(n) is the degree of the polynomial P[n]
%Y Cf. A000009, A092314, A116799.
%K nonn,tabf
%O 1,6
%A _Emeric Deutsch_, Feb 24 2006
