cp's OEIS Frontend

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.

Showing 1-4 of 4 results.

A382342 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 two kinds.

Original entry on oeis.org

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

Views

Author

Peter Dolland, Mar 27 2025

Keywords

Examples

			Triangle starts:
 0 : [1]
 1 : [0, 2]
 2 : [0, 2,  3]
 3 : [0, 2,  4,  4]
 4 : [0, 2,  7,  6,  5]
 5 : [0, 2,  8, 12,  8,   6]
 6 : [0, 2, 11, 18, 17,  10,  7]
 7 : [0, 2, 12, 26, 28,  22, 12,  8]
 8 : [0, 2, 15, 34, 46,  38, 27, 14,  9]
 9 : [0, 2, 16, 46, 64,  66, 48, 32, 16, 10]
10 : [0, 2, 19, 56, 94, 100, 86, 58, 37, 18, 11]
  ...

Links

Crossrefs

Row sums give A000712.
Cf. A008284 (1-kind case), A022597, A381895, 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+1), j=0..n/i))))
    end:
T:= (n, k)-> coeff(b(n$2), x, k):
seq(seq(T(n, k), k=0..n), n=0..11);  # Alois P. Heinz, Mar 27 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]]*(j + 1), {j, 0, n/i}]]]];
T[n_, k_] := Coefficient[b[n, n], x, k];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 11}] // Flatten (* Jean-François Alcover, Apr 19 2025, after Alois P. Heinz *)
  • Python
    from sympy.utilities.iterables import partitions
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 *= 1 + p[k]
        if s > 0 :
            t[s - 1] += fact
    return [0] + t

Formula

T(n,n) = n + 1.
T(n,1) = 2 for n >= 1.
T(n,k) = A381895(n,k) - A381895(n,k-1) for 1 <= k <= n.
Sum_{k=0..n} (-1)^k * T(n,k) = A022597(n). - Alois P. Heinz, Mar 27 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).

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

Views

Author

Peter Dolland, Apr 24 2025

Keywords

Examples

			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, ...]
...

Links

Crossrefs

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

Programs

  • Maple
    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
  • Python
    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

Formula

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

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

Views

Author

Peter Dolland, Mar 31 2025

Keywords

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).
Showing 1-4 of 4 results.

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