A321854 Irregular triangle where T(H(u),H(v)) is the number of ways to partition the Young diagram of u into vertical sections whose sizes are the parts of v, where H is Heinz number.

1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 2, 1, 0, 0, 0, 0, 1, 1, 3, 1, 0, 2, 0, 4, 1, 0, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 1, 0, 2, 2, 5, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 4, 1, 0, 0, 0, 6, 0, 6, 1, 1, 3, 4, 6, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1

Offset: 1

Author

Gus Wiseman, Nov 19 2018

Keywords

• nonn
• tabf

Comments

Row n has length A000041(A056239(n)).

A vertical section is a partial Young diagram with at most one square in each row.

Examples

			Triangle begins:
  1
  1
  0  1
  1  1
  0  0  1
  0  2  1
  0  0  0  0  1
  1  3  1
  0  2  0  4  1
  0  0  0  3  1
  0  0  0  0  0  0  1
  0  2  2  5  1
  0  0  0  0  0  0  0  0  0  0  1
  0  0  0  0  0  4  1
  0  0  0  6  0  6  1
  1  3  4  6  1
  0  0  0  0  0  0  0  0  0  0  0  0  0  0  1
  0  0  4 10  4  8  1
The 12th row counts the following partitions of the Young diagram of (211) into vertical sections (shown as colorings by positive integers):
  T(12,7) = 0:
.
  T(12,9) = 2:    1 2   1 2
                  1     2
                  2     1
.
  T(12,10) = 2:   1 2   1 2
                  2     1
                  2     1
.
  T(12,12) = 5:   1 2   1 2   1 2   1 2   1 2
                  3     2     3     1     3
                  3     3     2     3     1
.
  T(12,16) = 1:   1 2
                  3
                  4
		

Crossrefs

Cf. A000085, A000110, A007016, A056239, A122111, A153452, A215366, A296188, A300121, A318396, A321719-A321731, A321737, A321738, A321742-A321765.

Programs

Mathematica

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@spsu[Select[foo,Complement[#,Complement[set,s]]=={}&],Complement[set,s]]]/@Cases[foo,{i,_}];
ptnpos[y_]:=Position[Table[1,{#}]&/@y,1];
ptnverts[y_]:=Select[Rest[Subsets[ptnpos[y]]],UnsameQ@@First/@#&];
Table[With[{y=Reverse[primeMS[n]]},Table[Length[Select[spsu[ptnverts[y],ptnpos[y]],Sort[Length/@#]==primeMS[k]&]],{k,Sort[Times@@Prime/@#&/@IntegerPartitions[Total[primeMS[n]]]]}]],{n,18}]