A094508 Triangle read by rows: T[n,m] = number of solid partitions of n with trace m, where the trace of a solid partitions is defined as the sum of the traces of the constituent plane partitions.
1, 2, 2, 3, 4, 3, 4, 11, 6, 5, 5, 18, 19, 10, 7, 6, 33, 42, 34, 14, 11, 7, 48, 85, 80, 50, 22, 15, 8, 74, 156, 186, 128, 80, 30, 22, 9, 100, 275, 368, 318, 208, 112, 44, 30, 10, 140, 446, 725, 696, 534, 304, 165, 60, 42, 11, 180, 705, 1300, 1464, 1214, 808, 450, 228, 84, 56
Offset: 1
Examples
Table starts {1}, {2,2},{3,4,3},{4,11,6,5},..
T[4,3]=6 since these 6 solid partitions with trace 3 are:
[{{3,1}}], [{{3},{1}}], [{{2,1}},{{1}}], [{{2},{1}},{{1}}], [{{1,1}},{{1}},{{1}}], [{{1},{1}},{{1}},{{1}}]
Programs
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Mathematica
uses functions defined in A090984, A089924. solidform[q_?PartitionQ]:=Module[{}, Select[Flatten[Outer[z, Sequence@@(planepartitions/@q), 1]], And@@Apply[coversplaneQ, Partition[ #/.z->List, 2, 1], {1}]&]];tomatrix[par_]:=Block[{l=Max[Length/@ par]}, Map[PadRight[ #, l]&, par]]; Table[Length/@Split[Sort[Plus@@@Map[Tr[tomatrix[ # ]]&, Flatten[solidform/ @Partitions[n]], {2}]]], {n, 12}]
Formula
Finding a GF for the solid partitions is an open problem.
Comments