A382025 Triangle read by rows: T(n, k) is the number of partitions of n with at most k parts where 0 <= k <= n, and each part is one of three kinds.

1, 0, 3, 0, 3, 9, 0, 3, 12, 22, 0, 3, 18, 36, 51, 0, 3, 21, 57, 87, 108, 0, 3, 27, 82, 148, 193, 221, 0, 3, 30, 111, 225, 330, 393, 429, 0, 3, 36, 144, 333, 528, 681, 765, 810, 0, 3, 39, 184, 460, 808, 1106, 1316, 1424, 1479, 0, 3, 45, 225, 630, 1182, 1740, 2163, 2439, 2574, 2640

Offset: 0

Author

Peter Dolland, Mar 12 2025

Keywords

• nonn
• tabl

Comments

The 1-kind case is Euler's table A026820.

The 2-kind case is A381895.

Examples

			Triangle starts:
 0 : [1]
 1 : [0, 3]
 2 : [0, 3,  9]
 3 : [0, 3, 12,  22]
 4 : [0, 3, 18,  36,  51]
 5 : [0, 3, 21,  57,  87,  108]
 6 : [0, 3, 27,  82, 148,  193,  221]
 7 : [0, 3, 30, 111, 225,  330,  393,  429]
 8 : [0, 3, 36, 144, 333,  528,  681,  765,  810]
 9 : [0, 3, 39, 184, 460,  808, 1106, 1316, 1424, 1479]
10 : [0, 3, 45, 225, 630, 1182, 1740, 2163, 2439, 2574, 2640]
...
		

Crossrefs

Main diagonal gives A000716.

Cf. A026820, A381895.

Programs

Python

from sympy import binomial
from sympy.utilities.iterables import partitions
from sympy.combinatorics.partitions import IntegerPartition
kinds = 3 - 1   # the number of part kinds - 1
def a382025_row( n):
    if n == 0 : return [1]
    t = list( [0] * n)
    for p in partitions( n):
        p = IntegerPartition( p).as_dict()
        fact = 1
        s = 0
        for k in p :
            s += p[k]
            fact *= binomial( kinds + p[k], kinds)
        if s > 0 :
            t[s - 1] += fact
    for i in range( n - 1):
        t[i+1] += t[i]
    return [0] + t