A191714 a(n,k) equals the number of semistandard Young tableaux with shape of a partition of n and maximal element <= k.
1, 1, 4, 1, 6, 19, 1, 9, 39, 116, 1, 12, 69, 260, 751, 1, 16, 119, 560, 1955, 5552, 1, 20, 189, 1100, 4615, 15372, 43219, 1, 25, 294, 2090, 10460, 40677, 131131, 366088, 1, 30, 434, 3740, 22220, 100562, 370909, 1168008, 3245311, 1, 36, 630, 6512, 45628, 239316, 1007083, 3570240, 11042199, 30569012, 1, 42, 882, 10868, 89420, 541926, 2596573, 10347864, 35587071, 108535130, 299662672, 1, 49, 1218, 17732, 170340, 1188341, 6466159, 28915056, 110426979, 370661885, 1117689232, 3079276708
Offset: 1
Examples
For n=3 and k=2 the SSYT are par= {3} SSYT= {{1, 1, 1}}, {{2, 1, 1}}, {{2, 2, 1}}, {{2, 2, 2}} par= {2,1} SSYT= {{2, 1}, {1}}, {{2, 2}, {1}} par= {1,1,1} SSYT= none counts 4+2+0 = 6 = a(3,2). Table begins: 1; 1, 4; 1, 6, 19; 1, 9, 39, 116; 1, 12, 69, 260, 751; 1, 16, 119, 560, 1955, 5552; 1, 20, 189, 1100, 4615, 15372, 43219; ...
Links
- Alois P. Heinz, Rows n = 1..44, flattened
- R. Stanley, Hook Lengths and Contents
Programs
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Mathematica
Needs["Combinatorica`"]; hooklength[(p_)?PartitionQ] := Block[{ferr = (PadLeft[1 + 0*Range[#1], Max[p]] &) /@ p}, DeleteCases[(Rest[FoldList[Plus, 0, #1]] &) /@ ferr + Reverse /@ Reverse[Transpose[(Rest[FoldList[Plus, 0, #1]] &) /@ Reverse[Reverse /@ Transpose[ferr]]]], 0, -1] - 1]; content[(p_)?PartitionQ]:= Block[{le= Max[p], ferr =(PadLeft[1+ 0*Range[#1], Max[p]]&) /@ p}, DeleteCases[ MapIndexed[-le+ Range[le,1,-1]- #1- Tr[#2]&, 0*ferr]*ferr,0,-1]+ le]; stanley[(p_)?PartitionQ, t_Integer] := Times @@ ((t + Flatten[content[p]])/Flatten[hooklength[p]]); Table[Tr[ stanley[#,k] &/@ Partitions[n] ] , {n,12}, {k,n}]
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