A382241 Triangle read by rows: T(n,k) is the number of partitions of a 4-colored set of n objects into at most k parts with 0 <= k <= n.

1, 0, 4, 0, 10, 20, 0, 20, 60, 80, 0, 35, 170, 270, 305, 0, 56, 396, 816, 1016, 1072, 0, 84, 868, 2238, 3188, 3538, 3622, 0, 120, 1716, 5616, 9196, 10996, 11556, 11676, 0, 165, 3235, 13140, 24975, 32400, 35445, 36285, 36450, 0, 220, 5720, 28900, 63680, 90700, 104060, 108820, 110020, 110240

Offset: 0

Author

Peter Dolland, Mar 19 2025

Keywords

• nonn
• tabl

Comments

Two unrestricted unary predicates on the n objects mean four colors: The intersection, the both differences, and the complement of the union.

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

The 2-color case is A381891.

The 3-color case is A382045.

Examples

			Triangle starts:
 0 : [1]
 1 : [0,   4]
 2 : [0,  10,   20]
 3 : [0,  20,   60,    80]
 4 : [0,  35,  170,   270,    305]
 5 : [0,  56,  396,   816,   1016,   1072]
 6 : [0,  84,  868,  2238,   3188,   3538,   3622]
 7 : [0, 120, 1716,  5616,   9196,  10996,  11556,  11676]
 8 : [0, 165, 3235, 13140,  24975,  32400,  35445,  36285,  36450]
 9 : [0, 220, 5720, 28900,  63680,  90700, 104060, 108820, 110020, 110240]
10 : [0, 286, 9752, 60232, 154262, 242254, 294140, 315980, 323000, 324650, 324936]
...
		

Crossrefs

Main diagonal gives A255050.

Cf. A026820, A381891, A382045, A000292.

Programs

Maple

b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0, add(
      b(n-i*j, min(n-i*j, i-1))*binomial(i*(i^2+6*i+11)/6+j, j)*x^j, j=0..n/i))))
    end:
T:= proc(n, k) option remember;
     `if`(k<0, 0, T(n, k-1)+coeff(b(n$2), x, k))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Mar 19 2025

Python

from sympy import binomial
from sympy.utilities.iterables import partitions
from sympy.combinatorics.partitions import IntegerPartition
colors = 4 - 1   # the number of colors - 1
def a382241_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( binomial( k + colors, colors) + p[k] - 1, p[k])
        if s > 0 :
            t[s - 1] += fact
    for i in range( n - 1):
        t[i+1] += t[i]
    return [0] + t

Formula

T(n,1) = binomial(n + 3, 3) = A000292(n + 1) for n >= 1.