A382025 - OEIS

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

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OFFSET

0,3

COMMENTS

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

The 2-kind case is A381895.

LINKS

Table of n, a(n) for n=0..65.

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.

Cf. A026820, A381895.

Sequence in context: A197270 A117940 A099093 * A137339 A230184 A132330

Adjacent sequences: A382022 A382023 A382024 * A382026 A382027 A382028

KEYWORD

nonn,tabl

AUTHOR

Peter Dolland, Mar 12 2025

STATUS

approved