%I #3 Mar 30 2012 17:36:08
%S 1,1,2,2,2,1,2,2,2,3,2,4,2,6,2,8,2,10,2,12,1,2,14,2,2,16,4,2,18,7,2,
%T 20,10,2,22,14,2,24,20,2,26,26,2,28,34,2,30,44,2,32,54,1,2,34,66,2,2,
%U 36,80,4,2,38,94,8,2,40,110,13,2,42,128,20,2,44,146,30,2,46,166,42,2,48,188
%N Triangle read by rows: T(n,k) is the number of partitions into distinct parts having Durfee square of size k (n>=1, k>=1).
%C Row n has floor([1+sqrt(1+24n)]/6) terms. Row sums yield A000009. Sum(k*T(n,k),k>=0)=A116859(n).
%F G.f.=sum(t^k*x^(k(3k-1)/2)*(1+x^(2k))*product((1+x^j)/(1-x^j), j=1..k-1)/(1-x^k), k=1..infinity).
%e T(8,2)=4 because we have [6,2], [5,3], [5,2,1] and [4,3,1].
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 2;
%e 2,1;
%e 2,2;
%e 2,3;
%e 2,4;
%p g:=sum(t^k*x^(k*(3*k-1)/2)*(1+x^(2*k))*product((1+x^j)/(1-x^j),j=1..k-1)/(1-x^k),k=1..10): gser:=simplify(series(g,x=0,36)): for n from 1 to 32 do P[n]:=sort(coeff(gser,x^n)) od: for n from 1 to 32 do seq(coeff(P[n],t^j),j=1..floor((1+sqrt(1+24*n))/6)) od; # yields sequence in triangular form
%Y Cf. A000009, A116859.
%K nonn,tabf
%O 1,3
%A _Emeric Deutsch_, Feb 26 2006