A116422 Triangle read by rows: T(n,k) is the number of self-conjugate partitions of n having Durfee square of size k (n>=1; 1<=k<=floor(sqrt(n))).

1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 2, 1, 0, 1, 0, 2, 0, 1, 0, 1, 0, 3, 0, 1, 0, 2, 0, 3, 0, 1, 0, 3, 0, 4, 0, 1, 1, 0, 4, 0, 0, 4, 0, 1, 1, 0, 5, 0, 0, 5, 0, 2, 1, 0, 7, 0, 0, 5, 0, 3, 1, 0, 8, 0, 0, 6, 0, 5, 1, 0, 10, 0, 1, 0, 6, 0, 6, 0, 1, 0, 12, 0, 1, 0, 7, 0, 9, 0, 1, 0, 14, 0, 2, 0, 7, 0, 11, 0, 1, 0, 16

Offset: 1

Author

Emeric Deutsch, Feb 14 2006

Keywords

• nonn
• tabf

Comments

Row n contains floor(sqrt(n)) terms (0's are possible even at the end of the rows). Row sums yield A000700. Sum_{k=1..floor(sqrt(n))} k*T(n,k) = A079499(n).

Also, number of partitions of n into k distinct odd parts. Example: T(13,3)=2 because we have [9,3,1] and [7,5,1]. - Emeric Deutsch, Feb 24 2006

Examples

			T(13,3)=2 because we have [5,3,3,1,1] and [4,4,3,2] (there is one more self-conjugate partition of 13, namely [7,1,1,1,1,1,1], having Durfee square of size 1).
Triangle starts:
1;
0;
1;
0,1;
1,0;
0,1;
1,0;
0,2;
1,0,1;
0,2,0;
		

References

• G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976 (pp. 27-28).
• G. E. Andrews and K. Eriksson, Integer Partitions, Cambridge Univ. Press, 2004 (pp. 75-78).

Links

• Álvar Ibeas, First 400 rows, flattened

Crossrefs

Cf. A000700, A079499.

Programs

Maple

g:=sum(t^k*q^(k^2)/product(1-q^(2*i),i=1..k),k=1..15): gser:=simplify(series(g,q=0,40)): for n from 1 to 33 do P[n]:=coeff(gser,q^n) od: for n from 1 to 33 do row[n]:=seq(coeff(P[n],t^j),j=1..floor(sqrt(n))) od; # yields sequence in triangular form

Mathematica

rows = 31; jmax = Floor[Sqrt[rows]]; T[n_, k_] := SeriesCoefficient[Sum[ t^j*x^(j^2)/Product[1-x^(2i), {i, 1, j}], {j, 1, jmax}], {x, 0, n}, {t, 0, k}]; Table[T[n, k], {n, 1, rows}, {k, 1, Floor[Sqrt[n]]}] // Flatten (* Jean-François Alcover, Jul 16 2017 *)

Formula

G.f.: Sum_{k=1..infinity} (t^k*x^(k^2))/Product_{i=1..k} 1-x^(2*i).

G.f.: -1 + Product_{j=1..infinity} 1+t*x^(2*j-1). - Emeric Deutsch, Feb 24 2006

T(n, k) = T(n-2*k, k) + T(n-2*k+1, k-1). If n+k is even, T(n, k) = A008284((n-k^2)/2 + k, k) = A072233((n-k^2)/2, k); 0 otherwise. - Álvar Ibeas, Jul 27 2020