A050168 a(0) = 1; for n > 0, a(n) = binomial(n, floor(n/2)) + binomial(n-1, floor(n/2)).

1, 2, 3, 5, 9, 16, 30, 55, 105, 196, 378, 714, 1386, 2640, 5148, 9867, 19305, 37180, 72930, 140998, 277134, 537472, 1058148, 2057510, 4056234, 7904456, 15600900, 30458900, 60174900, 117675360, 232676280, 455657715, 901620585, 1767883500

Offset: 0

Author

Clark Kimberling

Keywords

• nonn

Comments

a(n) = number of symmetric Dyck (n+1)-paths which either start UD or are prime, i.e., do not return to ground level until the terminal point. For example, a(2)=3 counts UUUDDD, UUDUDD, UDUDUD. - David Callan, Dec 09 2004

a(n) = number of symmetric Dyck (n+1)-paths that first return to ground level either right away or not until the very end, i.e., that remain Dyck paths when either the first two steps or the first and last steps are deleted. For example, a(2)=3 counts UUUDDD, UUDUDD, UDUDUD. - David Callan, Mar 02 2005

Hankel transform has g.f. (1-x(1+x)^2)/(1-x^2(1-x^2)). - Paul Barry, Sep 13 2007

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• A. V. Sills and H. Wang, On the maximal Wiener index and related questions, Discrete Applied Mathematics, Volume 160, Issues 10-11, July 2012, Pages 1615-1623. - From _N. J. A. Sloane_, Sep 21 2012

Crossrefs

Maximum element in n-th row of A029653 (generalized Pascal triangle).

Cf. A001405.

Programs

Haskell

a050168 n = a050168_list !! n
a050168_list = 1 : zipWith (+) a001405_list (tail a001405_list)
-- Reinhard Zumkeller, Mar 04 2012

Magma

m:=40; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!((1+x)/(2*x)*(Sqrt((1+2*x)/(1-2*x))-1))); // G. C. Greubel, Oct 26 2018

Mathematica

CoefficientList[(1+x)/(2x) (Sqrt[(1+2x)/(1-2x)]-1) + O[x]^34, x] (* Jean-François Alcover, Aug 04 2018, after Sergei N. Gladkovskii *)

PARI

x='x+O('x^40); Vec((1+x)/(2*x)*(sqrt((1+2*x)/(1-2*x))-1)) \\ G. C. Greubel, Oct 26 2018

Formula

Asymptotic to c*2^n/sqrt(n) where c = (3/4)*sqrt(2/Pi) = 0.598413... - Benoit Cloitre, Jan 13 2003

For n > 0: a(n) = A208976(n-1) + 1. -Reinhard Zumkeller, Mar 04 2012

Conjecture: (n+1)*a(n) + (n-3)*a(n-1) + 2*(-2*n+1)*a(n-2) + 4*(-n+3)*a(n-3) = 0. - R. J. Mathar, Nov 26 2012

G.f.: (1+x)/(2*x)*(sqrt((1+2*x)/(1-2*x))-1). - Sergei N. Gladkovskii, Jul 26 2013

G.f.: (1+x)/( W(0)*(1-2*x)*x) - (1+x)/(2*x), where W(k)= 1 + 1/(1 - 2*x/(2*x + (k+1)/(x*(2*k+1))/W(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 26 2013