A143216 -id:A143216 - cp's OEIS frontend

A143217 a(n) = n! * (!(n+1)) = n! * Sum_{k=0..n} k!.

1, 2, 8, 60, 816, 18480, 629280, 29806560, 1864154880, 148459288320, 14652782323200, 1754531527795200, 250496911136102400, 42032247888401971200, 8188505926989625036800, 1832839841629043799552000, 467088574163459753336832000, 134454052266325985991942144000

Offset: 0

Gary W. Adamson, Jul 30 2008

			a(4) = 816 = 4! * 34, where 34 = A003422(4) and A000142 = (1, 1, 2, 6, 24, 120, ...).
a(4) = 816 = sum of row 4 terms of triangle A143216: (24 + 24 + 48 + 144 + 576).

G. C. Greubel, Table of n, a(n) for n = 0..250

Cf. A000142, A003422, A061640.

Cf. A143216, A217238, A217239.

[Factorial(n)*(&+[Factorial(k): k in [0..n]]): n in [0..30]]; // G. C. Greubel, Jul 12 2022

Table[n!*Sum[i!, {i, 0, n}], {n, 0, 16}]

f=factorial; [f(n)*sum(f(k) for k in (0..n)) for n in (0..40)] # G. C. Greubel, Jul 12 2022

a(n) = A000142(n) * A003422(n+1), where A000142 = the factorials and A003422 = partial sums of the factorials. [Corrected by Georg Fischer, Dec 13 2022]

Equals row sums of triangle A143216.

Edited and extended by Olivier Gérard, Sep 28 2012

A358446 a(n) = n! * Sum_{k=0..floor(n/2)} 1/binomial(n-k, k).

Original entry on oeis.org

1, 1, 4, 9, 56, 190, 1704, 7644, 93120, 516240, 8136000, 53523360, 1047548160, 7961241600, 187132377600, 1611967392000, 44311886438400, 426483893606400, 13428757601280000, 142790947407360000, 5066854992138240000, 58981696577556480000, 2328441680297779200000

Offset: 0

Vladimir Kruchinin, Nov 16 2022

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..449

Cf. A003149, A143216, A344391.

egf := (2*x+1)/((x-1)*(x+1)*(x^2-x-1))-(x*log((1-x)^2*(x+1)))/(-x^2+x+1)^2:
ser := series(egf, x, 22): seq(n!*coeff(ser, x, n), n = 0..20); # Peter Luschny, Nov 17 2022

a(n):=factorial(n)*sum(1/binomial(n-k,k),k,0,floor(n/2));

def A358446(n):
    return sum(A143216(n, k) // A344391(n, k) for k in range((n+2)//2))
print([A358446(n) for n in range(23)]) # Peter Luschny, Nov 17 2022

E.g.f.: (2*x+1)/((x-1)*(x+1)*(x^2-x-1))-(x*log((1-x)^2*(x+1)))/(-x^2+x+1)^2.
a(n) ~ n! * (3 + (-1)^n)/2. - Vaclav Kotesovec, Nov 17 2022
a(n) = Sum_{k=0..floor(n/2)} A143216(n, k)/A344391(n, k). - Peter Luschny, Nov 17 2022

A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 6, 7, 9, 12, 10, 11, 13, 16, 20, 15, 16, 18, 21, 25, 30, 21, 22, 24, 27, 31, 36, 42, 28, 29, 31, 34, 38, 43, 49, 56, 36, 37, 39, 42, 46, 51, 57, 64, 72, 45, 46, 48, 51, 55, 60, 66, 73, 81, 90, 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110

Offset: 0

Peter Luschny, Dec 07 2023

If the rows of the triangle are extended for k > n, the array A144216 is created, which is symmetrical to the main diagonal and therefore contains no new information compared to this triangle.

			Triangle T(n, k) starts:
  0 |  0;
  1 |  1,  2;
  2 |  3,  4,  6;
  3 |  6,  7,  9, 12;
  4 | 10, 11, 13, 16, 20;
  5 | 15, 16, 18, 21, 25, 30;
  6 | 21, 22, 24, 27, 31, 36, 42;
  7 | 28, 29, 31, 34, 38, 43, 49, 56;
  8 | 36, 37, 39, 42, 46, 51, 57, 64, 72;
  9 | 45, 46, 48, 51, 55, 60, 66, 73, 81,  90;
 10 | 55, 56, 58, 61, 65, 70, 76, 83, 91, 100, 110;
.
Start at row 0, column 0 with 0. Go down by adding the column index in step n. At row n, restart the counting and go n steps right by adding the row index in step n, then change direction and go down again by adding the column index. After 3*n steps on this path you are at T(2*n, n) which is 2*triangular(n) + (triangular(2*n) - triangular(n)) = (5*n^2 + 3*n)/2. These are the sliced heptagonal numbers A147875 (see the illustration of Leo Tavares).
.
The equation T(n, k) = (n*(n + 1) + k*(k + 1))/2 can be extended to all n, k in ZZ.
  [n\k] ... -6  -5  -4  -3  -2  -1   0   1   2   3   4   5  ...
  -------------------------------------------------------------
  [-5] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...
  [-4] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [-3] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [-2] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [-1] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 0] ..., 15, 10,  6,  3,  1,  0,  0,  1,  3,  6, 10, 15, ...
  [ 1] ..., 16, 11,  7,  4,  2,  1,  1,  2,  4,  7, 11, 16, ...
  [ 2] ..., 18, 13,  9,  6,  4,  3,  3,  4,  6,  9, 13, 18, ...
  [ 3] ..., 21, 16, 12,  9,  7,  6,  6,  7,  9, 12, 16, 21, ...
  [ 4] ..., 25, 20, 16, 13, 11, 10, 10, 11, 13, 16, 20, 25, ...

Columns: A000217, A000124, A152950, A166136, A121263.
Diagonals: A002378, A000290, A002061, A059100, A027689.
Cf. A147875 (T(2*n, n)), A016061 (row sums), A367965 (alternating row sums), A143216 (the multiplicative equivalent), A144216 (extended array).

T := (n, k) -> (n*(n + 1) + k*(k + 1)) / 2:
for n from 0 to 10 do seq(T(n, k), k = 0..n) od;

Module[{n=1},NestList[Append[#+n,n*++n]&,{0},10]] (* or *)
Table[(n(n+1)+k(k+1))/2,{n,0,10},{k,0,n}] (* Paolo Xausa, Dec 07 2023 *)

# A purely additive construction:
from functools import cache
@cache
def a_row(n: int) -> list[int]:
    if n == 0: return [0]
    row = a_row(n - 1) + [0]
    for k in range(n): row[k] += n
    row[n] = row[n - 1] + n
    return row

Recurrence: T(n, n) = n + T(n, n-1) starting with T(0, 0) = 0.
For k <> n: T(n, k) = n + T(n-1, k).
T(n, k) = t(n) + t(k), where t(n) are the triangular numbers A000217.
G.f.: (x + x*(2 - 5*x + x^2)*y + x^4*y^2)/((1 - x)^3*(1 - x*y)^3). - Stefano Spezia, Dec 07 2023

A368045 Triangle read by rows. T(n, k) = (k(k + 1)(2k + 1) + n(n + 1)(2n + 1)) / 6.

Original entry on oeis.org

0, 1, 2, 5, 6, 10, 14, 15, 19, 28, 30, 31, 35, 44, 60, 55, 56, 60, 69, 85, 110, 91, 92, 96, 105, 121, 146, 182, 140, 141, 145, 154, 170, 195, 231, 280, 204, 205, 209, 218, 234, 259, 295, 344, 408, 285, 286, 290, 299, 315, 340, 376, 425, 489, 570

Offset: 0

Peter Luschny, Dec 09 2023

Consider a sequence-to-triangle transformation a -> T, where a is a 0-based sequence and T a regular (0, 0)-based triangular array. The transformation is recursively defined, starting with T(0, 0) = 0, and T(n, n) = a(n) + T(n, n - 1) for n > 0. For k <> n let T(n, k) = a(n) + T(n-1, k).
If a(n) = 1, then T = A051162; if a(n) = n, then T = A367964 (generalizing the triangular numbers); if a(n) = n^2, then T is this triangle.
In the multiplicative form of the transformation, T(0, 0) is set to 1, and the operation '+' is replaced by '*'. For instance, a(n) = 2 is then mapped to T = A368043 and a(n) = n to A143216.

			Triangle T(n, k) starts:
  [0] [  0]
  [1] [  1,   2]
  [2] [  5,   6,  10]
  [3] [ 14,  15,  19,  28]
  [4] [ 30,  31,  35,  44,  60]
  [5] [ 55,  56,  60,  69,  85, 110]
  [6] [ 91,  92,  96, 105, 121, 146, 182]
  [7] [140, 141, 145, 154, 170, 195, 231, 280]
  [8] [204, 205, 209, 218, 234, 259, 295, 344, 408]
  [9] [285, 286, 290, 299, 315, 340, 376, 425, 489, 570]

Cf. A000330 (T(n,0)), A056520 (T(n,1)), A005900 (T(n-1,n)), A006331 (T(n,n)), A094952 (T(2*n,n)), A368046 (row sums), A368047 (alternating row sums).
Cf. A051162 (transform of n^0), A367964 (transform of n^1), this sequence (transform of n^2).
Cf. A368043, A143216.

Module[{n=1},NestList[Append[#+n^2,Last[#]+2(n++^2)]&,{0},10]] (* or *)
Table[(k(k+1)(2k+1)+n(n+1)(2n+1))/6,{n,0,10},{k,0,n}] (* Paolo Xausa, Dec 10 2023 *)

from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0: return [0]
    row = Trow(n - 1) + [0]
    for k in range(n): row[k] += n * n
    row[n] = row[n - 1] + n * n
    return row
print([k for n in range(10) for k in Trow(n)])

T(n, k) = A000330(k) + A000330(n).

cp's OEIS Frontend

A143217 a(n) = n! * (!(n+1)) = n! * Sum_{k=0..n} k!.

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A358446 a(n) = n! * Sum_{k=0..floor(n/2)} 1/binomial(n-k, k).

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A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n*(n + 1) + k*(k + 1)) / 2.

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A368045 Triangle read by rows. T(n, k) = (k*(k + 1)*(2*k + 1) + n*(n + 1)*(2*n + 1)) / 6.

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A367964 Triangle of 2-parameter triangular numbers, read by rows. T(n, k) = (n(n + 1) + k(k + 1)) / 2.

A368045 Triangle read by rows. T(n, k) = (k(k + 1)(2k + 1) + n(n + 1)(2n + 1)) / 6.