cp's OEIS Frontend

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))).

%I A116422 #12 Aug 01 2020 11:39:06
%S A116422 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,
%T A116422 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,
%U A116422 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
%N 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))).
%C A116422 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).
%C A116422 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
%D A116422 G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976 (pp. 27-28).
%D A116422 G. E. Andrews and K. Eriksson, Integer Partitions, Cambridge Univ. Press, 2004 (pp. 75-78).
%H A116422 Álvar Ibeas, <a href="/A116422/b116422.txt">First 400 rows, flattened</a>
%F A116422 G.f.: Sum_{k=1..infinity} (t^k*x^(k^2))/Product_{i=1..k} 1-x^(2*i).
%F A116422 G.f.: -1 + Product_{j=1..infinity} 1+t*x^(2*j-1). - _Emeric Deutsch_, Feb 24 2006
%F A116422 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
%e A116422 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).
%e A116422 Triangle starts:
%e A116422 1;
%e A116422 0;
%e A116422 1;
%e A116422 0,1;
%e A116422 1,0;
%e A116422 0,1;
%e A116422 1,0;
%e A116422 0,2;
%e A116422 1,0,1;
%e A116422 0,2,0;
%p A116422 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
%t A116422 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 *)
%Y A116422 Cf. A000700, A079499.
%K A116422 nonn,tabf
%O A116422 1,13
%A A116422 _Emeric Deutsch_, Feb 14 2006