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 1-5 of 5 results.

A361030 a(n) = 20160*(3*n)!/(n!*(n+3)!^2).

Original entry on oeis.org

560, 210, 504, 2352, 15840, 135135, 1361360, 15519504, 194699232, 2636552100, 38003792400, 577037174400, 9155656500480, 150853746558690, 2568167588473200, 44990491457326800, 808333317429976800, 14853124707775823700, 278470827854627007600, 5316261259042879236000
Offset: 0

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Author

Peter Bala, Mar 01 2023

Keywords

Comments

Row 2 of square array A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 60*A000984(n) is divisible by (n + 1)*(n + 2)*(n + 3) and the result 60*(2*n)!/(n!*(n+3)!) is the super ballot number A007272(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 20160*A006480(n) is divisible by ((n + 1)*(n + 2)*(n + 3))^2.
Equivalently, the central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that (1*3*5)*A000984(n+3) is divisible by (2*n + 1)*(2*n + 3)*(2*n + 5). The result is always an even integer. In fact, (1/2)*(1*3*5)/((2*n + 1)*(2n + 3)*(2n + 5))*A000984(n+3) = A007272(n).
Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that (1*2*4*5*7*8)*A006480(n+3) is divisible by (3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n + 5)*(3*n + 7)*(3*n + 8). The result is always an integer divisible by 3.

Links

Crossrefs

Row 2 of A361027. Cf. A006480, A007272, A361028, A361029, A361031, A361038.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 560 else 3*(3*n-1)*(3*n-2)/(n+3)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Formula

a(n) = 20160/((n+1)*(n+2)*(n+3))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5*7*8) * A006480(n+3)/((3*n + 1)*(3*n + 2)*(3*n + 4)*
(3*n + 5)*(3*n + 7)*(3*n + 8)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/5)*A007272(n)*A361038(n). Using this it can be shown that a(n) is always an integer.
a(n) = (1/3)*27^(n+3)*binomial(7/3, n+3)*binomial(8/3, n+3).
a(n) ~ sqrt(3)*10080*(27^n)/(Pi*n^7).
P-recursive: (n + 3)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 560.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(7 - 54*x)*A'(x) + (9 - 6*x)*A(x) - 5040 = 0, with A(0) = 560 and A'(0) = 210.

A361028 a(n) = 2*(3*n)!/(n!*(n+1)!^2).

Original entry on oeis.org

2, 3, 20, 210, 2772, 42042, 700128, 12471030, 233716340, 4557468630, 91752013080, 1896208270320, 40055997189600, 862021408906800, 18849534808095360, 417929529573239310, 9379553386892837940, 212776905535994934750, 4873239487455972633000, 112571832160232967822300
Offset: 0

Views

Author

Peter Bala, Feb 28 2023

Keywords

Comments

Row 0 of square array A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that A000984(n) is divisible by n + 1 and the result (2*n)!/(n!*(n+1)!) is the n-th Catalan number A000108(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 2*A006480(n) is divisible by (n+1)^2, leading to the present sequence. Do these numbers have a combinatorial interpretation?

Crossrefs

Row 0 of A361027. Cf. A000108, A006480, A007226, A361029, A361030, A361031, A361033.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 2 else 3*(3*n-1)*(3*n-2)/ (n+1)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Formula

a(n) = (2/(n+1)^2) * (3*n)!/n!^3.
a(n) = (2/3) * A006480(n+1)/((3*n + 1)*(3*n + 2)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/3)*27^(n+1)*binomial(1/3, n+1)*binomial(2/3, n+1).
a(n) = 2*C(2*n,n)*C(3*n,n) - 3*C(2*n,n+1)*C(3*n,2*n) + 2*C(2*n,n)*C(3*n,2*n+2) + 2*C(2*n,n+1)*C(3*n,2*n+2). This formula shows that a(n) is an integer for all n.
a(n) = A007226(n) * A000108(n).
a(n) ~ sqrt(3)*27^n/(Pi*n^3).
P-recursive: (n + 1)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 2.
The o.g.f. A(x) satisfies the differential equation
x^2*(1 - 27*x)*A''(x) + x*(3 - 54*x)*A'(x) + (1 - 6*x)*A(x) - 2 = 0, with
A(0) = 2 and A'(0) = 3.

A361029 a(n) = 120*(3*n)!/(n!*(n+2)!^2).

Original entry on oeis.org

30, 20, 75, 504, 4620, 51480, 656370, 9237800, 140229804, 2259901800, 38230005450, 673210036800, 12262039956000, 229872375708480, 4417859720647350, 86767376381987400, 1736954330906081100, 35364582637561485000, 730985923118395894950, 15315895532004485418000
Offset: 0

Views

Author

Peter Bala, Mar 01 2023

Keywords

Comments

Row 1 of the square array A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result 6*(2n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 120*A006480(n) is divisible by ((n + 1)*(n + 2))^2, leading to the present sequence. Do these numbers have a combinatorial interpretation?

Crossrefs

Row 1 of A361027. Cf. A006480, A007054, A361028, A361030, A361031, A361037.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 30 else 3*(3*n-1)*(3*n-2)/(n+2)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);
  • Mathematica
    Table[120 (3n)!/(n!(n+2)!^2),{n,0,20}] (* Harvey P. Dale, Jul 02 2023 *)

Formula

a(n) = 120/((n+1)*(n+2))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5) * A006480(n+2)/((3*n + 1)*(3*n + 2)*(3*n + 4)*(3*n +
5)), where A006480(n) = (3*n)!/n!^3.
a(n) = (10*binomial(3*n,n) - 7*binomial(3*n,n+1) + binomial(3*n,n+2)) * (3*binomial(2*n,n) - 4*binomial(2*n,n+1) + binomial(2*n,n+2)), shows that a(n) is an integer for all n.
a(n) = (1/3)*27^(n+2)*binomial(4/3, n+2)*binomial(5/3, n+2).
a(n) ~ sqrt(3)*60*27^n/(Pi*n^5).
P-recursive: (n + 2)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 30.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(5 - 54*x)*A'(x) + (4 - 6*x)*A(x) - 120 = 0, with A(0) = 30 and A'(0) = 20.

A361031 a(n) = (3^3)*(1*2*4*5*7*8*10*11)*(3*n)!/(n!*(n+4)!^2).

Original entry on oeis.org

11550, 2772, 4620, 15840, 81675, 550550, 4492488, 42325920, 446185740, 5148297000, 63985977000, 846321189120, 11802213457650, 172255143129300, 2615726247519000, 41127042052404000, 666874986879730860, 11114583114662181000, 189866473537245687000, 3316382259894423720000
Offset: 0

Views

Author

Peter Bala, Mar 01 2023

Keywords

Comments

Row 3 of A361027.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 840*A000984(n) is divisible by (n + 1)*(n + 2)*(n + 3)*(n + 4) and the result 840*(2*n)!/(n!*(n+4)!) is the super ballot number A348893(n). Similarly, the de Bruijn numbers A006480(n) = (3*n)!/n!^3 have the property that 6652800 * A006480(n) is divisible by ((n + 1)*(n + 2)*(n + 3)*(n + 4))^2.

Crossrefs

Row 3 of A361027. Cf. A006480, A348893, A361028, A361029, A361030.

Programs

  • Maple
    a := proc(n) option remember; if n = 0 then 11550 else
3*(3*n-1)*(3*n-2)/(n+4)^2*a(n-1) end if; end proc:
seq(a(n), n = 0..20);

Formula

a(n) = (3^3)*(1*2*4*5*7*8*10*11)/((n+1)*(n+2)*(n+3)*(n+4))^2 * (3*n)!/n!^3.
a(n) = (1/3)*(1*2*4*5*7*8*10*11) * A006480(n+4)/((3*n + 1)*(3*n + 2)*(3*n + 4)* (3*n + 5)*(3*n + 7)*(3*n + 8)*(3*n + 10)*(3*n + 11)), where A006480(n) = (3*n)!/n!^3.
a(n) = (1/3)*27^(n+4)*binomial(10/3, n+4)*binomial(11/3, n+4).
a(n) = (1/7)*A348893(n)*A361039(n). It can be shown from this that a(n) is always an integer.
a(n) ~ sqrt(3)*3326400*(27^n)/(Pi*n^9).
P-recursive: (n + 4)^2*a(n) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 11550.
The o.g.f. A(x) satisfies the differential equation x^2*(1 - 27*x)*A''(x) + x*(9 - 54*x)*A'(x) + (16 - 6*x)*A(x) - 184800 = 0, with A(0) = 11550 and A'(0) = 2772.

A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where F(n) = (1/8)*(4*n + 4)!/(n + 1)!; n, k >= 0.

Original entry on oeis.org

3, 315, 9, 46200, 280, 280, 7882875, 17325, 3675, 17325, 1466593128, 1513512, 116424, 116424, 1513512, 288592936632, 162954792, 5885880, 2134440, 5885880, 162954792, 59064793444800, 20193091776, 399072960, 67953600, 67953600, 399072960, 20193091776, 12445136556298875
Offset: 0

Views

Author

Peter Bala, Mar 04 2023

Keywords

Comments

The Catalan numbers A000108 are given by the formula Catalan(k) = (2*k)!/(k!*(k + 1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(n,k) = J(n) * (2*k)!/(k!*(k + n + 1)!), where J(n) = (2*n + 2)!/(2*(n + 1)!) = (2^n)*Product_{j = 0..n} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.
For this table we carry out an analogous construction using the numbers B(k) = (4*k)!/k!^4 = A008977(k) in place of the central binomial numbers (2*k)!/k!^2. We define sequences {B(n,k) : k >= 0}, n = 0, 1, 2, ..., by B(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where choosing F(n) = (4*n + 4)!/(8*(n + 1)!) = (1/2)*(4^n)*Product_{j = 0..n} (4*j + 1)*(4*j + 2)*(4*j + 3) appears to produce integer values for these quantities. The rows of the square array below are the sequences {B(0,k)}, {B(1,k)}, {B(2,k)}, ....
An alternative expression is B(n,k) = G(n,k) * B(n+k+1), where G(n,k) = (1/8)*Product_{j = 0..n} ( (4*j + 1)*(4*j + 2)*(4*j + 3)/((4*k + 4*j + 1)*(4*k + 4*j + 2)*(4*k + 4*j + 3)) ).

Examples

			The square array with rows n >= 0 and columns k >= 0 begins:
  n\k|          0          1          2           3           4 ...
  ----------------------------------------------------------------------
   0 |          3          9        280       17325     1513512 ...
   1 |        315        280       3675      116424     5885880 ...
   2 |      46200      17325     116424     2134440    67953600 ...
   3 |    7882875    1513512    5885880    67953600  1449322875 ...
   4 | 1466593128  162954792  399072960  3086579925 46235189000 ...
   5 |  ...
  ...
As a triangle:
 Row
  0 |             3
  1 |           315          9
  2 |         46200        280      280
  3 |       7882875      17325     3675    17325
  4 |    1466593128    1513512   116424   116424  1513512
  5 |  288592936632  162954792  5885880  2134440  5885880  162954792

Links

Crossrefs

A361033 (row 0), A361034 (row 2), A361035 (row 3).
Cf. A008977, A135573, A361027.

Programs

  • Maple
    # as a square array
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n, k), k = 0..10); end do;
# as a triangle
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n-k, k), k = 0..n); end do;
  • PARI
    T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1) \\ Andrew Howroyd, Jan 05 2024

Formula

T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1).
P-recursive: (n + k + 1)^3*T(n,k) = 4*(4*k - 1)*(4*k - 2)*(4*k - 3)*T(n,k-1) with T(n,0) = (1/8)*(4*n + 4)!/(n + 1)!^4 = (1/8)*A008977(n+1).
(n + k + 1)^3*T(n,k) = 4*(4*n + 1)*(4*n + 2)*(4*n + 3)*T(n-1,k) with T(0,k) = 3*(4*k)!/(k!*(k+1)!^3) = A361033(k).
T(n,k) = (1/2) * (1/(2*Pi))^3 * 256^(n+k+1) * Integral_{x = 0..1} (1 - x)^(n+1/4)*x^(k-1/4) dx * Integral_{x = 0..1} (1 - x)^(n+1/2)*x^(k-1/2) dx * Integral_{x = 0..1} (1 - x)^(n+3/4)*x^(k-3/4) dx.
Showing 1-5 of 5 results.

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