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%I #4 Jun 13 2013 14:00:14
%S 1,1,2,2,3,3,1,4,1,4,2,5,2,1,5,4,1,6,4,2,6,6,3,7,6,5,7,9,6,8,9,10,8,
%T 12,12,9,12,17,9,16,21,10,16,28,10,20,33,1,11,20,44,1,11,25,51,2,12,
%U 25,64,3,12,30,75,5,13,30,92,6,1,13,36,105,10,1,14,36,128,12,2,14,42,145,18,3
%N Triangle read by rows: T(n,k) is the number of partitions of n into odd parts and having Durfee square of size k (n>=1, k>=1).
%C Rows from (2n-1)^2 to 2n(2n+1)-1 have 2n-1 terms; rows from 2n(2n+1) to (2n+1)^2-1 have 2n terms. Row sums yield A000009. sum(k*T(n,k),k>=1)=A116465.
%D G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976 (pp. 27-28).
%D G. E. Andrews and K. Eriksson, Integer Partitions, Cambridge Univ. Press, 2004 (pp. 75-78).
%F G.f.=G(t,x)=sum(t^(2i)*x^(4i^2+2i)/product(1-x^(2j),j=1..2i)/product(1-x^(2j-1),j=1..i),i=1..infinity)+ sum(t^(2i-1)*x^((2i-1)^2)/product(1-x^(2j),j=1..2i-1)/product(1-x^(2j-1),j=1..i),i=1..infinity).
%e T(10,2)=4 because the only partitions of 10 into odd parts and having Durfee square of size 2 are [7,3], [5,5], [5,3,1,1] and [3,3,1,1,1,1].
%e Triangle starts:
%e 1;
%e 1;
%e 2;
%e 2;
%e 3;
%e 3,1;
%e 4,1;
%e 4,2;
%e 5,2,1;
%e 5,4,1;
%p g:=sum(t^(2*i)*x^(4*i^2+2*i)/product(1-x^(2*j),j=1..2*i)/product(1-x^(2*j-1),j=1..i),i=1..20)+ sum(t^(2*i-1)*x^((2*i-1)^2)/product(1-x^(2*j),j=1..2*i-1)/product(1-x^(2*j-1),j=1..i),i=1..20): gser:=simplify(series(g,x=0,32)): for n from 1 to 30 do P[n]:=coeff(gser,x^n) od: for n from 1 to 30 do seq(coeff(P[n],t^j),j=1..6) od; # yields sequence in triangular form (with several 0's at the end of each row)
%Y Cf. A116465.
%K nonn,tabf
%O 1,3
%A _Emeric Deutsch_ and _Vladeta Jovovic_, Feb 18 2006
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