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

1,3

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.

Conjecture: If A000219(k) == 0 (mod 3) then all terms of the k-th column of this triangle are also congruent to 0 (mod 3). - John Tyler Rascoe, Feb 08 2026

The above conjecture is false, it fails at column 14 see the Mathematics Stack Exchange link. - John Tyler Rascoe, Mar 16 2026

Links

Table of n, a(n) for n=1..105.

Mathematics Stack Exchange, Proving a conjecture about solid partitions with a fixed number of parts, 2026.

Wouter Meeussen, Mma functions for plane and solid partitions

Example

Table starts as:
  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
  ...
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}}

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]] *)

Crossrefs

Columns k = 1..7: A000012, A168237, A387996, A387997, A387998, A389773, A389774.

Cf. A000219, A000293, A094504, A393239.

Sequence in context: A131111 A336574 A128549 * A055885 A181425 A174505

Adjacent sequences: A380890 A380891 A380892 * A380894 A380895 A380896

Keyword

nonn,tabl

Author

Wouter Meeussen, Feb 07 2025

Status

approved