%I #5 Mar 30 2012 17:36:08
%S 1,1,1,1,3,3,2,5,2,8,2,1,10,5,13,8,1,20,9,1,26,12,4,33,21,2,46,25,5,1,
%T 58,37,6,75,48,11,1,101,59,16,125,84,19,3,157,115,23,2,206,135,39,5,
%U 253,187,46,4,317,238,63,8,1,403,292,90,7,494,382,108,17,1,608,490,139,18
%N Triangle read by rows: T(n,k) is the number of partitions of n having exactly k doubletons (n>=0, k>=0). By a doubleton in a partition we mean an occurrence of a part exactly twice (the partition [4,(3,3),2,2,2,(1,1)] has two doubletons, shown between parentheses).
%C Apparently, rows n with p(p+1)<=n<(p+1)(p+2) have at most p+1 terms. Row sums are the partition numbers (A000041). T(n,0)=A116645(n). Sum(k*T(n,k),k>=0)=A116646(n).
%H Alois P. Heinz, <a href="/A116644/b116644.txt">Rows n = 0..614, flattened</a>
%F G.f.: G(t,x) = product(1+x^j+tx^(2j)+x^(3j)/(1x^j), j=1..infinity).
%e T(6,2) = 1 because [2,2,1,1] is the only partition of 6 with 2 doubletons.
%e Triangle starts:
%e 1;
%e 1;
%e 1, 1;
%e 3;
%e 3, 2;
%e 5, 2;
%e 8, 2, 1;
%e 10, 5;
%e 13, 8, 1;
%p g:=product(1+x^j+t*x^(2*j)+x^(3*j)/(1x^j),j=1..35): gser:=simplify(series(g,x=0,35)): P[0]:=1: for n from 1 to 24 do P[n]:=coeff(gser,x^n) od: for n from 0 to 24 do seq(coeff(P[n],t,j),j=0..degree(P[n])) od; # sequence given in triangular form
%Y Cf. A000041, A116645, A116646.
%K nonn,tabf
%O 0,5
%A _Emeric Deutsch_, Feb 20 2006
