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 1-3 of 3 results.

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

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.

A361038 a(n) = 1680 * (3*n)!/((2*n)!*(n+3)!).

Original entry on oeis.org

280, 210, 420, 1176, 3960, 15015, 61880, 271320, 1248072, 5965050, 29414700, 148874400, 770263200, 4061212722, 21765976680, 118336861720, 651555929640, 3627981880950, 20405547069180, 115815267149400, 662742214356600
Offset: 0

Views

Author

Peter Bala, Mar 04 2023

Keywords

Comments

Compare with the super ballot numbers A007272(n) = 60*(2*n)!/(n!*(n+3)!).

Crossrefs

Programs

  • Maple
    seq( 1680 * (3*n)!/((2*n)!*(n+3)!), n = 0..20);

Formula

a(n) = 280*binomial(3*n,n) - 228*binomial(3*n,n+1) + 54*binomial(3*n,n+2) - 5*binomial(3*n,n+3). Thus a(n) is an integer.
P-recursive: 2*(n + 3)*(2*n - 1) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 280.
a(n) ~ (27/4)^n * 840*sqrt(3/Pi)/n^(7/2).
The o.g.f. satisfies the differential equation
x^2*(27*x - 4)*A''(x) + 2*x*(27*x - 7)*A'(x) + (6*x + 6)*A(x) - 1680 = 0, with A(0) = 280 and A'(0) = 210.

A361039 a(n) = 55440 * (3*n)!/((2*n)!*(n+4)!).

Original entry on oeis.org

2310, 1386, 2310, 5544, 16335, 55055, 204204, 813960, 3432198, 15142050, 69334650, 327523680, 1588667850, 7883530578, 39904290580, 205532444040, 1075067283906, 5701114384350, 30608320603770, 166169731127400, 911270544740325
Offset: 0

Views

Author

Peter Bala, Mar 04 2023

Keywords

Comments

Compare with the super ballot numbers A348893(n) = 840*(2*n)!/(n!*(n+4)!).

Crossrefs

Programs

  • Maple
    seq(  55440 * (3*n)!/((2*n)!*(n+4)!), n = 0..20);

Formula

a(n) = 2310*binomial(3*n,n) - 2057*binomial(3*n,n+1) + 627*binomial(3*n,n+2) - 102*binomial(3*n,n+3) + 7*binomial(3*n, n+4). Thus a(n) is an integer.
P-recursive: 2*(n + 4)*(2*n - 1) = 3*(3*n - 1)*(3*n - 2)*a(n-1) with a(0) = 2310.
a(n) ~ (27/4)^n * 27720*sqrt(3/Pi)/n^(9/2).
The o.g.f. satisfies the differential equation
x^2*(27*x - 4)*A''(x) + 2*x*(27*x - 9)*A'(x) + (6*x + 8)*A(x) - 18480 = 0, with A(0) = 2310 and A'(0) = 1386.
Showing 1-3 of 3 results.