A367106 Triangle read by rows where T(n,k) is the number of complete length-k integer partitions of n.

1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 2, 1, 1, 0, 0, 0, 1, 2, 1, 1, 0, 0, 0, 1, 3, 2, 1, 1, 0, 0, 0, 0, 3, 3, 2, 1, 1, 0, 0, 0, 0, 4, 5, 3, 2, 1, 1, 0, 0, 0, 0, 3, 5, 5, 3, 2, 1, 1, 0, 0, 0, 0, 4, 8, 7, 5, 3, 2, 1, 1, 0, 0, 0, 0, 2, 9, 9, 7, 5

Offset: 0

Author

Gus Wiseman, Nov 09 2023

Keywords

• nonn
• tabl

Comments

An integer partition of n is complete (ranks A325781) if every integer from 0 to n is the sum of some submultiset of the parts.

Examples

			Triangle begins:
  1
  0  1
  0  0  1
  0  0  1  1
  0  0  0  1  1
  0  0  0  2  1  1
  0  0  0  1  2  1  1
  0  0  0  1  3  2  1  1
  0  0  0  0  3  3  2  1  1
  0  0  0  0  4  5  3  2  1  1
  0  0  0  0  3  5  5  3  2  1  1
  0  0  0  0  4  8  7  5  3  2  1  1
  0  0  0  0  2  9  9  7  5  3  2  1  1
  0  0  0  0  2 11 12 11  7  5  3  2  1  1
  0  0  0  0  1 11 16 13 11  7  5  3  2  1  1
  0  0  0  0  1 14 21 19 15 11  7  5  3  2  1  1
Row n = 11 counts the following partitions (empty columns not shown):
  6311  62111  611111  5111111  41111111  311111111  2111111111  11111111111
  6221  53111  521111  4211111  32111111  221111111
  5321  52211  431111  3311111  22211111
  4421  44111  422111  3221111
        43211  332111  2222111
        42221  322211
        33311  222221
        33221
		

Crossrefs

Column k appears to have A000325(k) nonzero terms.

Column sums are A003513.

Central column T(2n,n) is A007042.

Row sums are A126796, ranks A325781.

The strict case is too sparse, row sums A188431 (complement A365831).

Grouping by maximum instead of length gives A261036.

A000041 counts integer partitions.

A108917 counts knapsack partitions, ranks A299702.

A299701 counts subset-sums of prime indices, firsts A259941.

A365924 counts incomplete partitions, ranks A365830.

Cf. A002033, A018818, A046663, A047967, A055932, A276024, A304792, A365921.

Programs

Mathematica

nmz[y_]:=Complement[Range[Total[y]],Total/@Subsets[y]];
Table[Length[Select[IntegerPartitions[n,{k}],nmz[#]=={}&]],{n,0,15},{k,0,n}]