A383352 -id:A383352 - cp's OEIS frontend

A383351 Triangle read by rows: T(n, k) is the number of partitions of a 2-colored set of n objects into k parts where 0 <= k <= n, and each part is one of 2 kinds.

1, 0, 4, 0, 6, 10, 0, 8, 24, 20, 0, 10, 53, 60, 35, 0, 12, 88, 164, 120, 56, 0, 14, 144, 348, 370, 210, 84, 0, 16, 208, 672, 904, 700, 336, 120, 0, 18, 299, 1174, 1998, 1870, 1183, 504, 165, 0, 20, 400, 1952, 3952, 4524, 3360, 1848, 720, 220

Offset: 0

Peter Dolland, Apr 24 2025

			Triangle starts:
 0 : [1]
 1 : [0,  4]
 2 : [0,  6,  10]
 3 : [0,  8,  24,   20]
 4 : [0, 10,  53,   60,   35]
 5 : [0, 12,  88,  164,  120,   56]
 6 : [0, 14, 144,  348,  370,  210,   84]
 7 : [0, 16, 208,  672,  904,  700,  336,  120]
 8 : [0, 18, 299, 1174, 1998, 1870, 1183,  504,  165]
 9 : [0, 20, 400, 1952, 3952, 4524, 3360, 1848,  720, 220]
10 : [0, 22, 534, 3052, 7394, 9834, 8652, 5488, 2724, 990, 286]
...

Main diagonal gives A000292(n+1).
Partial row sums are A383352.
Cf. A382342 (1-colored), A382339 (1-kind), A008284 (1-colored, 1-kind).

from sympy import binomial
from sympy.utilities.iterables import partitions
def calc_w(k , m):
    s = 0
    for p in partitions(m, m=k+1):
        fact = 1
        j = k + 1
        for x in p :
            fact *= binomial(j, p[x]) * (x + 1) ** p[x]
            j -= p[x]
        s += fact
    return s
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 *= calc_w(k, p[k])
        if s > 0 :
            t[s - 1] += fact
    return [0] + t

T(n,n) = binomial(n + 3, 3) = A000292(n + 1).
T(n,1) = 2*n + 2 for n >= 1.
T(n,k+1) = A383352(n,k+1) - A383352(n,k) for 0 <= k < n.

A383353 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where n 2-colored objects are distributed into k containers of two kinds. Containers may be left empty.

Original entry on oeis.org

1, 2, 0, 3, 4, 0, 4, 8, 6, 0, 5, 12, 22, 8, 0, 6, 16, 38, 40, 10, 0, 7, 20, 54, 92, 73, 12, 0, 8, 24, 70, 144, 196, 112, 14, 0, 9, 28, 86, 196, 354, 376, 172, 16, 0, 10, 32, 102, 248, 512, 760, 678, 240, 18, 0, 11, 36, 118, 300, 670, 1200, 1554, 1136, 335, 20, 0

Offset: 0

Peter Dolland, Apr 24 2025

			Array starts:
 0 : [1,  2,   3,    4,     5,     6,     7,      8,      9,     10,     11, ...]
 1 : [0,  4,   8,   12,    16,    20,    24,     28,     32,     36,     40, ...]
 2 : [0,  6,  22,   38,    54,    70,    86,    102,    118,    134,    150, ...]
 3 : [0,  8,  40,   92,   144,   196,   248,    300,    352,    404,    456, ...]
 4 : [0, 10,  73,  196,   354,   512,   670,    828,    986,   1144,   1302, ...]
 5 : [0, 12, 112,  376,   760,  1200,  1640,   2080,   2520,   2960,   3400, ...]
 6 : [0, 14, 172,  678,  1554,  2640,  3810,   4980,   6150,   7320,   8490, ...]
 7 : [0, 16, 240, 1136,  2936,  5436,  8272,  11228,  14184,  17140,  20096, ...]
 8 : [0, 18, 335, 1826,  5315, 10674, 17216,  24262,  31473,  38684,  45895, ...]
 9 : [0, 20, 440, 2812,  9136, 19984, 34192,  50248,  67024,  84020, 101016, ...]
10 : [0, 22, 578, 4186, 15188, 36024, 65512, 100488, 138188, 176878, 215854, ...]
...

Alois P. Heinz, Rows n = 0..200, flattened

Antidiagonal sums give A161870.
Cf. A383351, A383352.
Cf. A382345 (1-color), A381891 (1-kind), A026820 (1-color, 1-kind).
Cf. A278710.

b:= proc(n, i) option remember; expand(`if`(n=0 or i=1, (n+1)*x^n,
      add(b(n-i*j, min(n-i*j, i-1))*binomial(i+j, j)*x^j, j=0..n/i)))
    end:
g:= proc(n, k) option remember;
      `if`(k<0, 0, g(n, k-1)+coeff(b(n$2), x, k))
    end:
A:= (n, k)-> add(add(g(j, h)*g(n-j, k-h), h=0..k), j=0..n):
seq(seq(A(n, d-n), n=0..d), d=0..10);  # Alois P. Heinz, May 05 2025

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

A(0,k) = k + 1.
A(1,k) = 4*k.
A(2,k+1) = 6 + 16 * k.
A(n,1) = 2 + 2 * n.
A(n,n+k) = A(n,n) + k * A383352(n,n).
A(n,k) = Sum_{i=0..k} (k + 1 - i) * A383351(n,i) for 0 <= k <= n.
Sum_{k=0..n} (-1)^k*T(n-k,k) = A278710(n). - Alois P. Heinz, May 05 2025

cp's OEIS Frontend

A383351 Triangle read by rows: T(n, k) is the number of partitions of a 2-colored set of n objects into k parts where 0 <= k <= n, and each part is one of 2 kinds.

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