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.

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

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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

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.