A300185 Irregular triangle read by rows: T(n, {j,k}) is the number of partitions of n that have exactly j parts equal to k; 1 <= j <= n, 1 <= k <= n.

1, 0, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 2, 1, 1, 0, 1, 2, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 2, 1, 1, 0, 1, 3, 1, 1, 0, 0, 0, 2, 1, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 4, 2, 2, 1, 1, 0, 1, 4, 2, 1, 0, 0, 0, 0, 4, 1, 0, 0, 0, 0, 0, 3, 0

Offset: 1

Author

J. Stauduhar, Feb 27 2018

Keywords

• nonn
• tabf

Comments

Row sums = A027293.

If superfluous zeros are removed from the right side of each row, the row lengths = 1,2,1,3,1,1,4,2,... = A010766.

Sum of each N X N block of rows = 1,2,4,7,12,19,... = A000070.

The sum of the partitions of n that are over-counted in each block of N x N rows = A000070(n) - A000041(n) = A058884(n), n >= 1.

Concatenation of first row from each N X N block = A116598.

As noted by Joerg Arndt in A116598, the first row from each N X N block in reverse converges to A002865. Two sequences emerge from alternating second rows in reverse: for 2n, converges to even-indexed terms in A027336, and for 2n+1, converges to odd-indexed terms in A027336.

Counting the rows in each N X N block where columns j=2 > 0 and j=3 through j=n are all zeros produces A008615(n), n > 0.

Examples

			      \ j  1 2 3 4 5
     k
n
1:   1     1
2:   1     0 1
     2     1 0
3:   1     1 0 1
     2     1 0 0
     3     1 0 0
4:   1     1 1 0 1
     2     1 1 0 0
     3     1 0 0 0
     4     1 0 0 0
5:   1     2 1 1 0 1
     2     2 1 0 0 0
     3     2 0 0 0 0
     4     1 0 0 0 0
     5     1 0 0 0 0
.
.
.
		

Links

• J. Stauduhar, Table of n, a(n) for n = 1..10000
• Jerome Kelleher and Barry O'Sullivan, Generating All Partitions: A Comparison Of Two Encodings, arXiv:0909.2331 [cs.DS], 2009-2014.
• J. Stauduhar, Original Python program.

Crossrefs

Cf. A000041, A000070, A008615, A010766, A027293, A027336, A058884, A116598.

Programs

Mathematica

Array[With[{s = IntegerPartitions[#]}, Table[Count[Map[Count[#, k] &, s], j], {k, #}, {j, #}]] &, 7] // Flatten (* Michael De Vlieger, Feb 28 2018 *)

Python

# See Stauduhar link.