A382522 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where n unlabeled objects are distributed into k containers of four kinds. Containers may be left empty.

1, 4, 0, 10, 4, 0, 20, 16, 4, 0, 35, 40, 26, 4, 0, 56, 80, 80, 32, 4, 0, 84, 140, 180, 124, 42, 4, 0, 120, 224, 340, 320, 184, 48, 4, 0, 165, 336, 574, 660, 535, 248, 58, 4, 0, 220, 480, 896, 1184, 1200, 800, 332, 64, 4, 0, 286, 660, 1320, 1932, 2284, 1956, 1176, 416, 74, 4, 0

Offset: 0

Author

Peter Dolland, Mar 31 2025

Keywords

• nonn
• tabl

Examples

			Array starts:
 0 : [1, 4, 10,  20,   35,    56,    84,   120,    165,    220,    286]
 1 : [0, 4, 16,  40,   80,   140,   224,   336,    480,    660,    880]
 2 : [0, 4, 26,  80,  180,   340,   574,   896,   1320,   1860,   2530]
 3 : [0, 4, 32, 124,  320,   660,  1184,  1932,   2944,   4260,   5920]
 4 : [0, 4, 42, 184,  535,  1200,  2284,  3892,   6129,   9100,  12910]
 5 : [0, 4, 48, 248,  800,  1956,  3968,  7088,  11568,  17660,  25616]
 6 : [0, 4, 58, 332, 1176,  3080,  6618, 12364,  20892,  32776,  48590]
 7 : [0, 4, 64, 416, 1616,  4560, 10368, 20280,  35536,  57376,  87040]
 8 : [0, 4, 74, 520, 2187,  6580, 15778, 32196,  58414,  97012, 150570]
 9 : [0, 4, 80, 628, 2848,  9140, 23088, 49172,  92352, 157808, 250720]
10 : [0, 4, 90, 752, 3660, 12440, 33002, 73188, 142160, 249740, 406036]
...
		

Crossrefs

Antidiagonal sums give A023003.

Without empty containers: A382041.

Cf. A382344, A000292, 2 kinds: A382345, 3 kinds: A382521.

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))*binomial(j+3, 3), 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..10);  # Alois P. Heinz, Mar 31 2025

Mathematica

b[n_, i_] := b[n, i] = Expand[If[n == 0, 1, If[i < 1, 0, Sum[x^j*b[n-i*j, Min[n-i*j, i-1]]*Binomial[j+3, 3], {j, 0, n/i}]]]];
A[n_, k_] := Coefficient[b[n+k, n+k], x, k];
Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Aug 07 2025, after Alois P. Heinz *)

Python

from sympy import binomial
from sympy.utilities.iterables import partitions
def a_row(n, length=11) :
    if n == 0 : return [ binomial( k + 3, 3) 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( 3 + p[k], 3)
        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 + 3, 3)
    return a
for n in range(11): print(a_row(n))

Formula

A(0,k) = binomial(k + 3, 3) = A000292(k + 1).

A(1,k) = 4 * binomial(k + 2, 3).

A(n,1) = 4.

A(n,k) = Sum_{i=0..k} binomial(k + 3 - i, 3) * A382344(n,i) for k <= n.

A(n,k) = A382344(n+k,k).