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A085643
Irregular triangle T(n,k) = A048996(n,k) * A118851(n,k), read by rows, 1 <= k <= A000041(n).
5
1, 1, 2, 1, 3, 4, 1, 4, 6, 4, 6, 1, 5, 8, 12, 9, 12, 8, 1, 6, 10, 16, 9, 12, 36, 8, 12, 24, 10, 1, 7, 12, 20, 24, 15, 48, 27, 36, 16, 72, 32, 15, 40, 12, 1, 8, 14, 24, 30, 16, 18, 60, 72, 48, 54, 20, 96, 54, 144, 16, 20, 120, 80, 18, 60, 14, 1, 9, 16, 28, 36, 40, 21, 72, 90, 48, 60, 144, 27, 24, 120, 144, 192, 216, 96, 25, 160, 90, 360, 80, 24, 180, 160, 21, 84, 16, 1
OFFSET
0,3
COMMENTS
Rows have A000041(n) entries, with partitions in Abramowitz and Stegun order (A036036).
This triangle is a refinement of the triangle A078812.
Terms are the product of the number of permutations of the parts and the products of the parts of the associated permutation.
From Wolfdieter Lang, May 31 2018 (Start)
T(n, k) is the sum of the number of products of the block lengths of all the A048996(n, k) set partition of [n] := {1,2, ..., n} with m = m(n, k) blocks consisting of consecutive numbers corresponding to the k-th partition of n in A-St order. Because the block structure depends only on the exponents (signature) of the underlying partition this leads to the product of the two array entries. Equivalently, one can consider compositions. Then T(n, k) gives the sum of the products of the parts of all A048996(n, k) compositions originating from the k-th partition of n.
See also a comment of Kevin Long, May 11 2018, on A001906. (End)
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..2713 (rows 0..20)
Wolfdieter Lang, Rows n = 1..10, and more.
EXAMPLE
Triangle begins:
0 | 1;
1 | 1;
2 | 2, 1;
3 | 3, 4, 1;
4 | 4, 6, 4, 6, 1;
5 | 5, 8, 12, 9, 12, 8, 1;
6 | 6, 10, 16, 9, 12, 36, 8, 12, 24, 10, 1;
7 | 7, 12, 20, 24, 15, 48, 27, 36, 16, 72, 32, 15, 40, 12, 1;
...
When the signatures (A025487) are listed in A036035 order the number of permutations of the exponents begin
1, 1 1, 1 2 1, 1 2 1 3 1, 1 2 2 3 3 4 1, ... b(n) and the products begin
1, 2 1, 3 2 1, 4 3 4 2 1, 5 4 6 3 4 2 1, ... c(n) thus we can write
1, 2 1, 3 4 1, 4 6 4 6 1, 5 8 12 9 12 8 1, ... a(n) = b(n)*c(n).
For n = 5, k = 4: the partition is (1^2, 3^1) = [1,1,3] with m = m(n,k) = 3. The A048996(5, 4) = 3 compositions are 1 + 1 + 3, 1 + 3 + 1 and 3 + 1 + 1. The corresponding three consecutive 3-block partitions of [5] := {1, 2, ..., 5} are {1}, {2}, {3,4,5} and {1}, {2,3,4}, {5} and {1,2,3}, {4}, {5}, Therefore, a(5, 4) = 1*1*3 + 1*3*1 + 3*1*1 = 3*3 = 9. For the compositions one has the same sum from the products of the parts. - Wolfdieter Lang, May 31 2018
PROG
(PARI)
C(sig)={my(S=Set(sig)); vecprod(sig)*(#sig)!/prod(k=1, #S, (#select(t->t==S[k], sig))!)}
Row(n)={apply(C, [Vecrev(p) | p<-partitions(n)])}
{ for(n=0, 7, print(Row(n))) } \\ Andrew Howroyd, Oct 07 2025
CROSSREFS
Cf. A000041 (row lengths), A001906 (row sums), A025487, A036035, A036036, A048996, A078812, A118851.
Sequence in context: A233529 A104471 A174828 * A305309 A174829 A132110
KEYWORD
nonn,look,tabf
AUTHOR
Alford Arnold, Aug 15 2003
EXTENSIONS
a(0)=1 prepended and order of terms corrected by Andrew Howroyd, Oct 07 2025
STATUS
approved