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.

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A048619 a(n) = LCM(binomial(n,0), ..., binomial(n,n)) / binomial(n,floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 3, 3, 4, 2, 10, 5, 30, 15, 7, 7, 56, 28, 252, 126, 60, 30, 330, 165, 396, 198, 286, 143, 2002, 1001, 15015, 15015, 7280, 3640, 1768, 884, 15912, 7956, 3876, 1938, 38760, 19380, 406980, 203490, 99484, 49742, 1144066, 572033, 1961256, 980628
Offset: 0

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Author

Keywords

Examples

			If n=10 then A002944(10)=2520, A001405(10)=252, the quotient a(10)=10.
		

Crossrefs

Programs

  • Magma
    [Lcm([1..n+1]) div (Floor((n+3)/2)*Binomial(n+1,Floor((n+3)/2))): n in [0..50]]; // Vincenzo Librandi, Jul 10 2019
  • Mathematica
    Table[Apply[LCM, Binomial[n, Range[0, n]]]/Binomial[n, Floor[n/2]], {n, 0, 48}] (* Michael De Vlieger, Jun 29 2017 *)
  • PARI
    {A048619(n) = lcm(vector(n+1, i, i)) / binomial(n+1, (n+1)\2) / ((n+2)\2);}
    

Formula

a(n) = A002944(n)/A001405(n).
a(n) = lcm(1..n+1)/(floor((n+3)/2)*binomial(n+1,floor((n+3)/2))). - Paul Barry, Jul 03 2006
a(n) = lcm(1,2,...,n+1) / (ceiling((n+1)/2)*binomial(n+1,floor((n+1)/2))) = A003418(n+1) / A100071(n+1). - Max Alekseyev, Oct 23 2015
a(n) = A263673(n+1) / A110654(n+1) = A180000(n+1) / A152271(n). - Max Alekseyev, Oct 23 2015
a(2*n-1) = A068553(n) = A068550(n)/n.

Extensions

Definition corrected and a(0)=1 prepended by Max Alekseyev, Oct 23 2015

A068550 a(n) = lcm{1, ..., 2n} / binomial(2n, n).

Original entry on oeis.org

1, 1, 2, 3, 12, 10, 30, 105, 56, 252, 1260, 330, 1980, 2574, 2002, 15015, 240240, 61880, 15912, 151164, 38760, 406980, 4476780, 1144066, 13728792, 24515700, 6249100, 84362850, 21474180, 5462730, 81940950, 1270084725, 645122400
Offset: 0

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Author

N. J. A. Sloane, Mar 23 2002

Keywords

Comments

Known to be always an integer.

Crossrefs

Bisection of A180000 and A263673.

Programs

  • Mathematica
    a[0] = 1; a[n_] := (LCM @@ Range[2*n])/Binomial[2*n, n]; Array[a, 33, 0] (* Amiram Eldar, Mar 06 2022 *)
  • PARI
    a(n) = lcm([1..2*n])/binomial(2*n, n); \\ Michel Marcus, Mar 06 2022

Formula

a(n) = A099996(n) / A000984(n) = A003418(2*n) / A001405(2*n) = A180000(2*n) = A263673(2*n).
a(n) = n * A068553(n) = n * A048619(2*n-1).

Extensions

a(0)=1 prepended by Max Alekseyev, Oct 23 2015
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