A210590 Triangle of numbers generated by the Nekrasov-Okounkov formula.

1, 1, 1, 4, 5, 1, 18, 29, 12, 1, 120, 218, 119, 22, 1, 840, 1814, 1285, 345, 35, 1, 7920, 18144, 14674, 5205, 805, 51, 1, 75600, 196356, 185080, 79219, 16450, 1624, 70, 1, 887040, 2427312, 2515036, 1258628, 324569, 43568, 2954, 92, 1, 10886400, 32304240, 37012572, 21034376, 6431733, 1088409, 101178, 4974, 117, 1

Offset: 0

Author

Wouter Meeussen, Mar 24 2012

Keywords

• nonn
• tabl

Comments

Row sums are A000712, alternating sign row sums are zero (except for first row); application of the Nekrasov-Okounkov formula; see A138782.

Examples

			Table starts as:
     1;
     1,     1;
     4,     5,     1;
    18,    29,    12,    1;
   120,   218,   119,   22,   1;
   840,  1814,  1285,  345,  35,  1;
  7920, 18144, 14674, 5205, 805, 51,  1;
  ...
		

Links

• G. C. Greubel, Rows n=0..50 of triangle, flattened
• Richard P. Stanley, Some Combinatorial Properties of Hook Lengths, Contents, and Parts of Partitions arXiv:0807.0383 [math.CO], 2009.

Crossrefs

Cf. A000712, A053529, A057623, A138782, A234937.

T(2n,n) gives A338755.

Programs

Mathematica

w=9; MapIndexed[ CoefficientList[#1,t] Tr[#2-1]! &, CoefficientList[Series[Product[(1-x^i)^(-1-t), {i,w}], {x,0,w}], x]];
or alternatively:
CoefficientList[#, t] & /@ Table[1/n! Tr[(NumberOfTableaux[#1]^2 Apply[Times, t + Flatten[hooklength[#1]]^2] &) /@ Partitions[n]], {n,0,9}]
or alternatively:
Table[1/n!Tr[NumberOfTableaux[#]^2 f[ Flatten[hooklength[#]]^2,e,k,n ]&/@ Partitions[n] ],{n,0,9},{k,0,n}]
with e and f defined as:
e[n_,v_]:= Tr[Times @@@ Select[Subsets[Table[Subscript[x,j],{j,v}]],Length[#]==n&]];
f[li_List,fun_,par_,k_]:=fun[par,k]/.Thread[Array[Subscript[x,#1]&,Length[li]]->li];

Formula

E.g.f.: Product_{i=1..n} (1 - x^i)^(-1 - t).