A072811 T(n,k) = multiplicity of the k-th partition of n in Mathematica order, defined to be the count of its permutations (compositions).
1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 3, 1, 1, 2, 2, 3, 3, 4, 1, 1, 2, 2, 3, 1, 6, 4, 1, 6, 5, 1, 1, 2, 2, 3, 2, 6, 4, 3, 3, 12, 5, 4, 10, 6, 1, 1, 2, 2, 3, 2, 6, 4, 1, 6, 3, 12, 5, 3, 6, 12, 20, 6, 1, 10, 15, 7, 1, 1, 2, 2, 3, 2, 6, 4, 2, 6, 3, 12, 5, 3, 6, 12, 12, 20, 6, 1, 12, 10, 4, 30, 30, 7, 5, 20, 21, 8, 1
Offset: 0
Examples
The partitions of 4 are {4}, {3,1}, {2,2}, {2,1,1}, {1,1,1,1}, so the fourth row equals 1,2,1,3,1 since these are the counts of the permutations of these lists. Triangle begins: 1; 1; 1, 1; 1, 2, 1; 1, 2, 1, 3, 1; 1, 2, 2, 3, 3, 4, 1; 1, 2, 2, 3, 1, 6, 4, 1, 6, 5, 1;
Links
- Alois P. Heinz, Rows n = 0..26, flattened
Programs
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Mathematica
mult[li:{__Integer}] := Apply[Multinomial, Length/@Split[ Sort[li] ] ]; Table[mult/@Partitions[n], {n, 12}]
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PARI
\\ here mulp(v) computes the multiplicity of the given partition. mulp(v) = {my(p=(#v)!, k=1); for(i=2, #v, k=if(v[i]==v[i-1], k+1, p/=k!; 1)); p/k!} Row(n)={apply(mulp, vecsort([Vecrev(p) | p<-partitions(n)], , 4))} { for(n=0, 9, print(Row(n))) } \\ Peter Dolland, Nov 11 2019
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