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.

A026016 a(n) = binomial(2*n-1, n) - binomial(2*n-1, n+3).

Original entry on oeis.org

1, 3, 10, 34, 117, 407, 1430, 5070, 18122, 65246, 236436, 861764, 3157325, 11622015, 42961470, 159419670, 593636670, 2217608250, 8308432140, 31212003420, 117544456770, 443690433654, 1678353186780, 6361322162444, 24155384502452, 91882005146652
Offset: 1

Views

Author

Keywords

Comments

Number of (s(0), s(1), ..., s(2n-1)) such that s(i) is a nonnegative integer and |s(i) - s(i-1)| = 1 for i = 1,2,...,n, s(0) = 2, s(2n-1) = 3. Also a(n) = T(2n-1,n-1), where T is the array defined in A026009.
Number of integer lattice paths from (0,2) to (n-1,n+2) that do not cross the main diagonal.

Crossrefs

Programs

  • Magma
    [Binomial(2*n-1, n) - Binomial(2*n-1, n+3): n in [1..30]]; // Vincenzo Librandi, Jun 21 2013
    
  • Maple
    a:= n-> binomial(2*n-1, n) -binomial(2*n-1, n+3): seq(a(n), n=1..27); #  Zerinvary Lajos, Dec 10 2007
  • Mathematica
    Table[Binomial[2 n - 1, n] - Binomial[2 n - 1, n + 3], {n, 1, 40}] (* Vincenzo Librandi, Jun 21 2013 *)
  • PARI
    a(n) = binomial(2*n-1, n) - binomial(2*n-1, n+3) \\ Charles R Greathouse IV, Oct 23 2023
  • Sage
    [binomial(2*n-1, n) - binomial(2*n-1, n+3) for n in (2..30)] # G. C. Greubel, Mar 19 2021
    

Formula

Expansion of (1+x^1*C^3)*C^2, where C = (1-(1-4*x)^(1/2))/(2*x) is g.f. for Catalan numbers, A000108.
(n+3)*a(n) +(-7*n-9)*a(n-1) +2*(7*n-4)*a(n-2) +4*(-2*n+5)*a(n-3)=0. - R. J. Mathar, Jun 20 2013
From G. C. Greubel, Mar 19 2021: (Start)
G.f.: (1-x)*( 1 -4*x +2*x^2 - (1-2*x)*sqrt(1-4*x) )/(2*x^3).
E.g.f.: -1 + (exp(2*x)/x^2)*( x*(1+x)*BesselI(0, 2*x) - (2 +x +2*x^2)*Bessel(1, 2*x) ).
a(n) = C(n) + Sum_{j=0..n-2} C(n-j-2)*(C(j+3) - 2*C(j+2)), where C(n) are the Catalan numbers. (End)
a(n) = C(n+2) -3*C(n+1) +2*C(n) = 6*((n^2+1)/((n+2)*(n+3)))*C(n). - G. C. Greubel, Mar 22 2021

Extensions

Better description from Darko Marinov (marinov(AT)lcs.mit.edu), May 17 2001