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URL: https://oeis.org/A382025

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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.
3
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
0,3
COMMENTS
The 1-kind case is Euler's table A026820.
The 2-kind case is A381895.
EXAMPLE
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]
...
PROG
(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
CROSSREFS
Main diagonal gives A000716.
Sequence in context: A197270 A117940 A099093 * A137339 A230184 A132330
KEYWORD
nonn,tabl
AUTHOR
Peter Dolland, Mar 12 2025
STATUS
approved