A380893 Triangle read by rows: T(n,m) = number of solid partitions of n with shape of a plane partition of m.

1, 1, 3, 1, 3, 6, 1, 6, 6, 13, 1, 6, 15, 13, 24, 1, 9, 21, 37, 24, 48, 1, 9, 30, 58, 75, 48, 86, 1, 12, 39, 95, 132, 159, 86, 160, 1, 12, 54, 128, 231, 297, 299, 160, 282, 1, 15, 63, 197, 345, 552, 593, 574, 282, 500, 1, 15, 81, 251, 546, 873, 1156, 1180, 1038, 500, 859, 1, 18, 96, 345, 771, 1452, 1933, 2390, 2208, 1874, 859, 1479, 1, 18, 114, 432, 1110, 2151, 3340, 4154, 4614, 4082, 3268, 1479, 2485, 1, 21, 132, 558, 1491, 3276, 5214, 7430, 8310, 8758, 7276, 5685, 2485, 4167

Offset: 1

Author

Wouter Meeussen, Feb 07 2025

Keywords

• nonn
• tabl

Comments

A solid (or 3D) partition of n describes a piling of boxes in a corner with heights nonincreasing away from the corner, and containing integers, similarly nonincreasing, that sum to n.

The shape of a solid partitions is defined as the plane partition containing the heights of the piling, irrespective of the numerical content of the boxes.

Row sums equal A000293, T(n,n) = T(n+1,n) equals A000219;

Equals number of solid partitions with total by layer equal to partitions of n with largest part m.

Examples

			Table starts as:
  1,
  1,3
  1,3,6
  1,6,6,13
  1,6,15,13,24
  1,9,21,37,24,48
T(4,2) = 6 since the solid partitions of 4 with shapes a plane partition of 2 are:
 z[{{2,2}}], z[{{3,1}}], z[{{2},{2}}], z[{{3},{1}}], z[{{3}},{{1}}], z[{{2}},{{2}}]
with shapes equal to these plane partitions:
  {{2}}, {{2}}, {{1,1}}, {{1,1}}, {{1},{1}}, {{1},{1}}
		

Links

• Wouter Meeussen, Mma functions for plane and solid partitions

Crossrefs

Cf. A000219, A000293, A094504.

Programs

Mathematica

Table[Tr@(Count[First[lapse[#]]&/@Flatten[sols=Table[solidformBTK[par],{par,IntegerPartitions[n]}] ],#]&/@planepartitions[k]),{n,10},{k,n}] (* using functions from link above, or with the faster second program: *)
Table[ Sum[Length[solidformBTK[TransposePartition@par]],{par,IntegerPartitions[n,{k}]} ],{n,16},{k,n}] (* with transposePartition[par:{Integer..}]:=Count[par,i/;i>=#]&/@Range[Max[par]] *)