A109984 a(n) = number of steps in all Delannoy paths of length n.

0, 5, 44, 321, 2184, 14325, 91860, 580097, 3622928, 22437477, 138049020, 844881345, 5148375192, 31258302933, 189199514532, 1142148091905, 6878977097760, 41347348295877, 248082231062988, 1486116788646977

Offset: 0

Author

Emeric Deutsch, Jul 07 2005

Keywords

• nonn

Comments

A Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1).

Examples

			a(1)=5 because in the 3 (=A001850(1)) Delannoy paths of length 1, namely D, NE and EN, we have altogether five steps.
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
• Robert A. Sulanke, Objects Counted by the Central Delannoy Numbers, Journal of Integer Sequences, Volume 6, 2003, Article 03.1.5.

Crossrefs

Cf. A109983, A001850.

Programs

Haskell

a109984 = sum . zipWith (*) [0..] . a109983_row
-- Reinhard Zumkeller, Nov 18 2014

Maple

a:=n->add(k*binomial(n,2*n-k)*binomial(k,n),k=n..2*n): seq(a(n),n=0..23);

Mathematica

CoefficientList[Series[x*(5-x)/(1-6*x+x^2)^(3/2), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 18 2012 *)

Formula

a(n) = sum_{k=0..2n} k*A109983(k).

a(n) = sum_{k=n..2*n} k*binomial(n, 2*n-k)*binomial(k, n).

G.f.: z*(5-z)/(1-6*z+z^2)^(3/2).

Recurrence: (n-1)*(2*n-9)*a(n) = 4*(3*n^2-15*n+7)*a(n-1) - (n-1)*(2*n-7)*a(n-2). - Vaclav Kotesovec, Oct 18 2012

a(n) ~ sqrt(24 + 17*sqrt(2))*(3 + 2*sqrt(2))^n*sqrt(n)/(4*sqrt(Pi)). - Vaclav Kotesovec, Oct 18 2012