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.

A005317 a(n) = (2^n + C(2*n,n))/2.

Original entry on oeis.org

1, 2, 5, 14, 43, 142, 494, 1780, 6563, 24566, 92890, 353740, 1354126, 5204396, 20066492, 77575144, 300572963, 1166868646, 4537698722, 17672894044, 68923788698, 269129985796, 1052051579012, 4116719558104, 16123810230158, 63205319996092, 247959300028484
Offset: 0

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Author

N. J. A. Sloane and Peter Fishburn

Keywords

Comments

Hankel transform is A008619. - Paul Barry, Nov 13 2007
a(n) is the number of lattice paths from (0,0) to (n,n) using E(1,0) and N(0,1) as steps that horizontally cross the diagonal y = x with even many times. For example, a(2) = 5 because there are 6 paths in total and only one of them horizontally crosses the diagonal with odd many times, namely, NEEN. - Ran Pan, Feb 01 2016

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    [(2^n+Binomial(2*n,n))/2: n in [0..26]];  // Bruno Berselli, Jun 20 2011
    
  • Maple
    f := n->(2^n+binomial(2*n,n))/2;
  • Mathematica
    Table[(2^n + Binomial[2 n, n])/2, {n, 0, 26}] (* Michael De Vlieger, Feb 01 2016 *)
  • Maxima
    makelist(sum((-1)^k*binomial(2*n,n-2*k),k,0,floor(n/2)),n,0,26); /* Bruno Berselli, Jun 20 2011 */
    
  • PARI
    a(n)=(2^n+binomial(2*n,n))/2 \\ Charles R Greathouse IV, Dec 20 2011

Formula

From Simon Plouffe, Feb 18 2011: (Start)
G.f.: (1/2)*(-4*x+1+(-(4*x-1)*(2*x-1)^2)^(1/2))/(4*x-1)/(2*x-1).
Recurrence: 0 = (-24-28*n-8*n^2)*a(n+1) + (18+22*n+6*n^2)*a(n+2) + (-3-4*n-n^2)*a(n+3), a(0)=1, a(1)=2, a(2)=5. (End)
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*C(2*n, n-2*k), n > 0. - Mircea Merca, Jun 20 2011
E.g.f.: exp(2*x)*(1+BesselI(0,2*x))/2 = G(0)/2; G(k) = 1 + (k)!/(P-2*x*(2*k+1)*(P^2)/(2*x*(2*k+1)*P+(k+1)^2*k!/G(k+1))), where P:=((2*k)!)/(2^k)/((k)!) (continued fraction). - Sergei N. Gladkovskii, Dec 20 2011
a(n) = Sum_{r=0..n} k*(k+1)/2 where k=C(n,r). - J. M. Bergot, Sep 04 2013
a(n) = binomial(2*n,n) - A108958(n). - Ran Pan, Feb 01 2016
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024