Peter Dolland - cp's OEIS frontend

A384105 Triangle read by rows: T(n,k) is the number of binary relations on a set of n objects, exactly k of which are self referencing, 0 <= k <= n.

1, 1, 1, 3, 4, 3, 16, 36, 36, 16, 218, 752, 1104, 752, 218, 9608, 45960, 90416, 90416, 45960, 9608, 1540944, 9133760, 22692704, 30194176, 22692704, 9133760, 1540944, 882033440, 6154473664, 18425858880, 30679088480, 30679088480, 18425858880, 6154473664, 882033440

Offset: 0

Peter Dolland, May 19 2025

Also the number of essentially different simple digraphs on a node set A of size n with a distinguished subset B of size k, where elements are indistinguishable within B and within A \ B.

			Triangle starts:
            1
            1,              1
            3,              4,              3
           16,             36,             36,              16
          218,            752,           1104,             752,             218
         9608,          45960,          90416,           90416,           45960, ...
      1540944,        9133760,       22692704,        30194176,        22692704, ...
    882033440,     6154473664,    18425858880,     30679088480,     30679088480, ...
1793359192848, 14334221970688, 50138592081152, 100240050239744, 125284653092864, ...
...

Peter Dolland, Python calculation of T(n,k)

Cf. A000273 (edge cases), A000595 (row sums), A353996, A328874, A383617.

T(n,k) = T(n,n-k).
T(n,0) = T(n,n) = A000273(n).
T(n,1) = T(n,n-1) = A353996(n+1) = A329874(n,4).
Sum_{k=0..n} T(n,k) = A000595(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

A383617 Triangle read by rows: T(n,k) is the number of binary relations on a set of n objects, k of which are picked out, 0 <= k <= n.

Original entry on oeis.org

1, 2, 2, 10, 16, 10, 104, 272, 272, 104, 3044, 11456, 16960, 11456, 3044, 291968, 1432608, 2842304, 2842304, 1432608, 291968, 96928992, 578431232, 1441700480, 1920352256, 1441700480, 578431232, 96928992, 112282908928, 784780122880, 2351993457920, 3918054495616, 3918054495616, 2351993457920, 784780122880, 112282908928

Offset: 0

Peter Dolland, May 02 2025

The row sums are the number of simple digraphs with n 4-colored nodes. The colors result from the four cases combining the property self-referencing (yes/no) with "picked out" (yes/no).

			Triangle starts:
            1;
            2,            2;
           10,           16,            10;
          104,          272,           272,           104;
         3044,        11456,         16960,         11456,          3044;
       291968,      1432608,       2842304,       2842304,       1432608,  291968;
     96928992,    578431232,    1441700480,    1920352256,    1441700480, ...
 112282908928, 784780122880, 2351993457920, 3918054495616, 3918054495616, ...
...
Example n=2, k=1: The both objects are differentiated. As a consequence all binary relations on two different objects have to be counted: These are the subsets of the cross product of the objects set with itself. This contains four pairs, so the number of subsets is 2^4 = 16.

Peter Dolland, Python calculation of T(n,k)

Cf. A000595 (edge cases), A353996 (row sums), A329874 (4th column = row sums).

T(n,k) = T(n,n-k).
T(n,0) = T(n,n) = A000595(n).
Sum_{k=0..n} T(n,k) = A353996(n+1) = A329874(n,4).

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.

Original entry on oeis.org

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.

A383352 Triangle read by rows: T(n, k) is the number of partitions of a 2-colored set of n objects into at most k parts where 0 <= k <= n, and each part is one of 2 kinds.

Original entry on oeis.org

1, 0, 4, 0, 6, 16, 0, 8, 32, 52, 0, 10, 63, 123, 158, 0, 12, 100, 264, 384, 440, 0, 14, 158, 506, 876, 1086, 1170, 0, 16, 224, 896, 1800, 2500, 2836, 2956, 0, 18, 317, 1491, 3489, 5359, 6542, 7046, 7211, 0, 20, 420, 2372, 6324, 10848, 14208, 16056, 16776, 16996

Offset: 0

Peter Dolland, Apr 24 2025

			Triangle starts:
 0 : [1]
 1 : [0,  4]
 2 : [0,  6,  16]
 3 : [0,  8,  32,   52]
 4 : [0, 10,  63,  123,   158]
 5 : [0, 12, 100,  264,   384,   440]
 6 : [0, 14, 158,  506,   876,  1086,  1170]
 7 : [0, 16, 224,  896,  1800,  2500,  2836,  2956]
 8 : [0, 18, 317, 1491,  3489,  5359,  6542,  7046,  7211]
 9 : [0, 20, 420, 2372,  6324, 10848, 14208, 16056, 16776, 16996]
10 : [0, 22, 556, 3608, 11002, 20836, 29488, 34976, 37700, 38690, 38976]
...

Row sums of A383351.

Cf. A381895 (1-color), A381891 (1-kind), A026820 (1-color, 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
    for i in range(n - 1):
        t[i + 1] += t[i]
    return [0] + t

T(n,k) = Sum_{i=0..k} A383351(n,i).

T(n,1) = 2*n + 2 for n >= 1.

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.

Original entry on oeis.org

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

Peter Dolland, Mar 31 2025

			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]
...

Antidiagonal sums give A023003.
Without empty containers: A382041.
Cf. A382344, A000292, 2 kinds: A382345, 3 kinds: A382521.

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

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 *)

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))

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).

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

Peter Dolland, Mar 30 2025

			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]
...

Antidiagonal sums give A000716.
Alternating antidiagonal sums give A107635.
Without empty containers: A382025.
Cf. A382343, A000217, 2 kinds: A382345.

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

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))

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).

A382345 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where n unlabeled objects are distributed into k containers of two kinds. Containers may be left empty.

Original entry on oeis.org

1, 2, 0, 3, 2, 0, 4, 4, 2, 0, 5, 6, 7, 2, 0, 6, 8, 12, 8, 2, 0, 7, 10, 17, 18, 11, 2, 0, 8, 12, 22, 28, 26, 12, 2, 0, 9, 14, 27, 38, 46, 34, 15, 2, 0, 10, 16, 32, 48, 66, 64, 46, 16, 2, 0, 11, 18, 37, 58, 86, 100, 94, 56, 19, 2, 0, 12, 20, 42, 68, 106, 136, 152, 124, 70, 20, 2, 0

Offset: 0

Peter Dolland, Mar 29 2025

			Array starts:
 0 : [1, 2,  3,   4,   5,   6,   7,    8,    9,   10,   11]
 1 : [0, 2,  4,   6,   8,  10,  12,   14,   16,   18,   20]
 2 : [0, 2,  7,  12,  17,  22,  27,   32,   37,   42,   47]
 3 : [0, 2,  8,  18,  28,  38,  48,   58,   68,   78,   88]
 4 : [0, 2, 11,  26,  46,  66,  86,  106,  126,  146,  166]
 5 : [0, 2, 12,  34,  64, 100, 136,  172,  208,  244,  280]
 6 : [0, 2, 15,  46,  94, 152, 217,  282,  347,  412,  477]
 7 : [0, 2, 16,  56, 124, 214, 316,  426,  536,  646,  756]
 8 : [0, 2, 19,  70, 167, 302, 464,  640,  825, 1010, 1195]
 9 : [0, 2, 20,  84, 212, 406, 648,  922, 1212, 1512, 1812]
10 : [0, 2, 23, 100, 271, 542, 899, 1314, 1766, 2236, 2717]
...

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

Antidiagonal sums give A000712.
Alternating antidiagonal sums give A073252.
Without empty containers: A381895.
Cf. A382342.

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+1), 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 29 2025

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]]*(j + 1), {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, 11}] // Flatten (* Jean-François Alcover, Apr 07 2025, after Alois P. Heinz *)

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

A(0,k) = k + 1.
A(1,k) = 2*k.
A(2,k+1) = 2 + 5 * k.
A(n,1) = 2.
A(n,k) = Sum_{i=0..k} (k + 1 - i) * A382342(n,i) for k <= n.
A(n,n+k) = A(n,n) + k * A000712(n).
A(n,k) = A382342(n,k) + 2 * A(n,k-1) - A(n,k-2) for 2 <= k <= n.
A(n,k) = A382342(n+k,k). - Alois P. Heinz, Mar 31 2025

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

Peter Dolland, Mar 28 2025

			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]
...

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

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

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 *)

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

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

A382343 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 3 kinds.

Original entry on oeis.org

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

Offset: 0

Peter Dolland, Mar 27 2025

			Triangle starts:
 0 : [1]
 1 : [0, 3]
 2 : [0, 3,  6]
 3 : [0, 3,  9,  10]
 4 : [0, 3, 15,  18,  15]
 5 : [0, 3, 18,  36,  30,  21]
 6 : [0, 3, 24,  55,  66,  45,  28]
 7 : [0, 3, 27,  81, 114, 105,  63,  36]
 8 : [0, 3, 33, 108, 189, 195, 153,  84,  45]
 9 : [0, 3, 36, 145, 276, 348, 298, 210, 108,  55]
10 : [0, 3, 42, 180, 405, 552, 558, 423, 276, 135, 66]
...

Main diagonal gives A000217(n+1).
Row sums give A000716.
Cf. A008284 (1-kind), A382342 (2-kind).
Cf. A022598, A382025.

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:
T:= (n, k)-> coeff(b(n$2), x, k):
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Mar 27 2025

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]]*(j+2)*(j+1)/2, {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, Jul 30 2025, after Alois P. Heinz *)

from sympy import binomial
from sympy.utilities.iterables import partitions
kinds = 3 - 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

T(n,n) = binomial(n + 2, 2) = A000217(n + 1).
T(n,1) = 3 for n >= 1.
T(n,k) = A382025(n,k) - A382025(n,k-1) for 1 <= k <= n.
Sum_{k=0..n} (-1)^k * T(n,k) = A022598(n). - Alois P. Heinz, Mar 27 2025

cp's OEIS Frontend

User: Peter Dolland

A384105 Triangle read by rows: T(n,k) is the number of binary relations on a set of n objects, exactly k of which are self referencing, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

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.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Python

Formula

A383617 Triangle read by rows: T(n,k) is the number of binary relations on a set of n objects, k of which are picked out, 0 <= k <= n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Formula

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.

Views

Author

Keywords

Examples

Crossrefs

Programs

Python

Formula

A383352 Triangle read by rows: T(n, k) is the number of partitions of a 2-colored set of n objects into at most k parts where 0 <= k <= n, and each part is one of 2 kinds.

Views

Author

Keywords

Examples

Crossrefs

Programs

Python

Formula

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.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

Python

Formula

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.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Python

Formula

A382345 Square array A(n,k), n>=0, k>=0, read by antidiagonals downwards, where n unlabeled objects are distributed into k containers of two kinds. Containers may be left empty.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Mathematica