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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A382344 Triangle read by rows: T(n, k) is the number of partitions of n into k parts where 0 <= k <= n, and each part is one of 4 kinds.

Original entry on oeis.org

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

Views

Author

Peter Dolland, Mar 28 2025

Keywords

Examples

			Triangle starts:
 0 : [1]
 1 : [0, 4]
 2 : [0, 4, 10]
 3 : [0, 4, 16,  20]
 4 : [0, 4, 26,  40,   35]
 5 : [0, 4, 32,  80,   80,   56]
 6 : [0, 4, 42, 124,  180,  140,   84]
 7 : [0, 4, 48, 184,  320,  340,  224,  120]
 8 : [0, 4, 58, 248,  535,  660,  574,  336,  165]
 9 : [0, 4, 64, 332,  800, 1200, 1184,  896,  480, 220]
10 : [0, 4, 74, 416, 1176, 1956, 2284, 1932, 1320, 660, 286]
...

Crossrefs

Main diagonal gives A000292(n+1).
Row sums give A023003.
Cf. A008284 (1-kind), A382342 (2-kind), A382343 (3-kind).
Cf. A022599, A382041.

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:
T:= (n, k)-> coeff(b(n$2), x, k):
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Mar 28 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}]]]];
T[n_, k_] := Coefficient[b[n, n], x, k];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Aug 07 2025, after Alois P. Heinz *)
  • Python
    from sympy import binomial
from sympy.utilities.iterables import partitions
kinds = 4 - 1   # the number of part kinds - 1
def t_row( n):
    if n == 0 : return [1]
    t = list( [0] * n)
    for p in partitions( n):
        fact = 1
        s = 0
        for k in p :
            s += p[k]
            fact *= binomial( kinds + p[k], kinds)
        if s > 0 :
            t[s - 1] += fact
    return [0] + t

Formula

T(n,n) = binomial(n + 3, 3) = A000292(n + 1).
T(n,1) = 4 for n >= 1.
T(n,k) = A382041(n,k) - A382041(n,k-1) for 1 <= k <= n.
Sum_{k=0..n} (-1)^k * T(n,k) = A022599(n). - Alois P. Heinz, Mar 28 2025

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.

Original entry on oeis.org

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

Views

Author

Peter Dolland, Mar 30 2025

Keywords

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).
Showing 1-2 of 2 results.

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