A382045 Triangle read by rows: T(n,k) is the number of partitions of a 3-colored set of n objects into at most k parts with 0 <= k <= n.

1, 0, 3, 0, 6, 12, 0, 10, 28, 38, 0, 15, 66, 102, 117, 0, 21, 126, 249, 309, 330, 0, 28, 236, 562, 788, 878, 906, 0, 36, 396, 1167, 1845, 2205, 2331, 2367, 0, 45, 651, 2292, 4128, 5289, 5814, 5982, 6027, 0, 55, 1001, 4272, 8703, 12106, 13881, 14602, 14818, 14873, 0, 66, 1512, 7608, 17634, 26616, 32088, 34608, 35556, 35826, 35892

Offset: 0

Author

Peter Dolland, Mar 13 2025

Keywords

• nonn
• tabl

Comments

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

The 2-color case is A381891.

Examples

			Triangle starts:
 0 : [1]
 1 : [0,  3]
 2 : [0,  6,   12]
 3 : [0, 10,   28,   38]
 4 : [0, 15,   66,  102,   117]
 5 : [0, 21,  126,  249,   309,   330]
 6 : [0, 28,  236,  562,   788,   878,   906]
 7 : [0, 36,  396, 1167,  1845,  2205,  2331,  2367]
 8 : [0, 45,  651, 2292,  4128,  5289,  5814,  5982,  6027]
 9 : [0, 55, 1001, 4272,  8703, 12106, 13881, 14602, 14818, 14873]
10 : [0, 66, 1512, 7608, 17634, 26616, 32088, 34608, 35556, 35826, 35892]
...
		

Links

• Alois P. Heinz, Rows n = 0..150, flattened

Crossrefs

Main diagonal gives A217093.

Cf. A026820, A381891, A000217.

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+3)/2+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 13 2025

Mathematica

b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[b[n - i*j, Min[n - i*j, i - 1]]*Binomial[i*(i + 3)/2 + j, j]*x^j, {j, 0, n/i}]]]];
T[n_, k_] := T[n, k] = If[k < 0, 0, T[n, k-1] + Coefficient[b[n, n], x, k]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Mar 21 2025, after Alois P. Heinz *)

Python

from sympy import binomial
from sympy.utilities.iterables import partitions
from sympy.combinatorics.partitions import IntegerPartition
colors = 3 - 1   # the number of colors - 1
def t_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 + 2, 2) = A000217(n + 1) for n >= 1.