A382521 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where n unlabeled objects are distributed into k containers of three kinds. Containers may be left empty.

1, 3, 0, 6, 3, 0, 10, 9, 3, 0, 15, 18, 15, 3, 0, 21, 30, 36, 18, 3, 0, 28, 45, 66, 55, 24, 3, 0, 36, 63, 105, 114, 81, 27, 3, 0, 45, 84, 153, 195, 189, 108, 33, 3, 0, 55, 108, 210, 298, 348, 276, 145, 36, 3, 0, 66, 135, 276, 423, 558, 552, 405, 180, 42, 3, 0, 78, 165, 351, 570, 819, 936, 858, 549, 225, 45, 3, 0

Offset: 0

Author

Peter Dolland, Mar 30 2025

Keywords

• nonn
• tabl

Examples

			Array starts:
 0 : [1, 3,  6,  10,   15,   21,   28,    36,    45,    55,    66]
 1 : [0, 3,  9,  18,   30,   45,   63,    84,   108,   135,   165]
 2 : [0, 3, 15,  36,   66,  105,  153,   210,   276,   351,   435]
 3 : [0, 3, 18,  55,  114,  195,  298,   423,   570,   739,   930]
 4 : [0, 3, 24,  81,  189,  348,  558,   819,  1131,  1494,  1908]
 5 : [0, 3, 27, 108,  276,  552,  936,  1428,  2028,  2736,  3552]
 6 : [0, 3, 33, 145,  405,  858, 1532,  2427,  3543,  4880,  6438]
 7 : [0, 3, 36, 180,  549, 1248, 2340,  3861,  5811,  8190, 10998]
 8 : [0, 3, 42, 225,  741, 1785, 3510,  6000,  9300, 13410, 18330]
 9 : [0, 3, 45, 271,  957, 2451, 5051,  8967, 14307, 21126, 29424]
10 : [0, 3, 51, 324, 1227, 3312, 7137, 13125, 21552, 32553, 46194]
...
		

Crossrefs

Antidiagonal sums give A000716.

Alternating antidiagonal sums give A107635.

Without empty containers: A382025.

Cf. A382343, A000217, 2 kinds: A382345.

Programs

Maple

b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0,
      add(x^j*b(n-i*j, min(n-i*j, i-1))*(j+2)*(j+1)/2, j=0..n/i))))
    end:
A:= (n, k)-> coeff(b(n+k$2), x, k):
seq(seq(A(n, d-n), n=0..d), d=0..11);  # Alois P. Heinz, Mar 31 2025

Python

from sympy import binomial
from sympy.utilities.iterables import partitions
def a_row(n, length=11) :
    if n == 0 : return [ binomial( k + 2, 2) for k in range( length) ]
    t = list( [0] * length)
    for p in partitions( n):
        fact = 1
        s = 0
        for k in p :
            s += p[k]
            fact *= binomial( 2 + p[k], 2)
        if s > 0 :
            t[s] += fact
    a = list( [0] * length)
    for i in range( 1, length):
        for j in range( i, 0, -1):
            a[i] += t[j] * binomial( i - j + 2, 2)
    return a
for n in range(11): print(a_row(n))

Formula

A(0,k) = binomial(k + 2, 2) = A000217(k + 1).

A(1,k) = 3 * binomial(k + 1, 2).

A(n,1) = 3.

A(n,k) = Sum_{i=0..k} binomial(k + 2 - i, 2) * A382343(n,i) for k <= n.

A(n,k) = A382343(n+k,k).