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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.
10
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
1,11
COMMENTS
Row n has length A000041(A056239(n)).
A vertical section is a partial Young diagram with at most one square in each row.
EXAMPLE
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
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}]
KEYWORD
nonn,tabf
AUTHOR
Gus Wiseman, Nov 19 2018
STATUS
approved