A179314 Triangle T(n,k) read by rows: number of unordered twice partitions derived from the k-th partition of n listed in Abramowitz and Stegun order, 1 <= k <= A000041(n).

1, 1, 2, 1, 3, 2, 1, 5, 3, 3, 2, 1, 7, 5, 6, 3, 3, 2, 1, 11, 7, 10, 6, 5, 6, 4, 3, 3, 2, 1, 15, 11, 14, 15, 7, 10, 6, 9, 5, 6, 4, 3, 3, 2, 1, 22, 15, 22, 21, 15, 11, 14, 15, 15, 12, 7, 10, 6, 9, 5, 5, 6, 4, 3, 3, 2, 1, 30, 22, 30, 33, 35, 15, 22, 21, 15, 21, 30, 10, 11, 14, 15, 15, 12, 12, 7, 10, 6, 9, 5, 5, 6, 4, 3, 3, 2, 1

Offset

0,3

Comments

This triangle is a refinement of the triangle A061260.

Parts of size s and multiplicity m can be partitioned into sub-partitions in binomial(m + A000041(s) - 1, m) ways. - Andrew Howroyd, Oct 04 2025

Links

Andrew Howroyd, Table of n, a(n) for n = 0..2713 (rows 0..20)

Formula

T(n,1) = A000041(n).

Example

Triangle begins:
  0 | 1;
  1 | 1;
  2 | 2, 1;
  3 | 3, 2, 1;
  4 | 5, 3, 3, 2, 1;
  5 | 7, 5, 6, 3, 3, 2, 1;
  6 | 11, 7, 10, 6, 5, 6, 4, 3, 3, 2, 1;
  7 | 15, 11, 14, 15, 7, 10, 6, 9, 5, 6, 4, 3, 3, 2, 1;
  ...
T(6,4) = 6 corresponds to the partition {3,3} which has twice partitions {1+1+1, 1+1+1}, {1+1+1, 1+2}, {1+1+1, 3}, {1+2, 1+2}, {1+2, 3}, {3, 3}.
A061260(6,3) = 15 can be derived from partitions 4+1+1, 3+2+1, and 2+2+2 corresponding to T(6,5) + T(6,6) + T(6,7) = 5 + 6 + 4 = 15 cases.

Prog

(PARI)
C(sig)={my(S=Set(sig)); prod(k=1, #S, my(c=#select(t->t==S[k], sig)); binomial(c+numbpart(S[k])-1, c)) }
Row(n)={apply(C, [Vecrev(p) | p<-partitions(n)])}
{ for(n=0, 7, print(Row(n))) } \\ Andrew Howroyd, Oct 04 2025

Crossrefs

Cf. A000041 (row lengths), A001970 (row sums), A036036, A055884 (has same row sums), A061260, A119443 (ordered case).

Sequence in context: A383334 A386214 A304100 * A204927 A119441 A347227

Adjacent sequences: A179311 A179312 A179313 * A179315 A179316 A179317

Keyword

nonn,look,tabf

Author

Alford Arnold, Jul 19 2010

Extensions

Edited, a(0)=1 prepended and a(45) onwards from Andrew Howroyd, Oct 04 2025

Status

approved