A119309 a(n) = binomial(2*n,n) * 6^n.

1, 12, 216, 4320, 90720, 1959552, 43110144, 960740352, 21616657920, 489977579520, 11171488813056, 255928652808192, 5886359014588416, 135839054182809600, 3143703825373593600, 72933928748667371520

Offset

0,2

Comments

Number of lattice paths from (0,0) to (n,n) using three kinds of steps (1,0) and two kinds of steps (0,1). - Joerg Arndt, Jul 01 2011

Central terms of the triangles in A013620 and A038220.

Links

Indranil Ghosh, Table of n, a(n) for n = 0..400

Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.

Formula

a(n) = 6^n * A000984(n).

G.f.: 1/sqrt(1-24*x). - Zerinvary Lajos, Dec 20 2008 [Corrected by Joerg Arndt, Jul 01 2011]

D-finite with recurrence: n*a(n) +12*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Jan 20 2020

a(n) = 2^n*A098658(n) = 3^n*A059304(n). - R. J. Mathar, Jan 20 2020

From Amiram Eldar, Jul 21 2020: (Start)

Sum_{n>=0} 1/a(n) = 24/23 + 24*sqrt(23)*arcsin(1/sqrt(24))/529.

Sum_{n>=0} (-1)^n/a(n) = 24/25 - 24*arcsinh(1/sqrt(24))/125. (End)

E.g.f.: exp(12*x) * BesselI(0,12*x). - Ilya Gutkovskiy, Sep 14 2021

Example

a(3) = binomial(2*3,3) * (6^3) = 20 * 216 = 4320. - Indranil Ghosh, Mar 03 2017

Mathematica

Table[Binomial[2n, n]*(6^n), {n, 0, 15}] (* Indranil Ghosh, Mar 03 2017 *)

Prog

(PARI) /* same as in A092566 but use */
steps=[[1, 0], [1, 0], [1, 0], [0, 1], [0, 1]]; /* note repeated entries */
/* Joerg Arndt, Jun 30 2011 */
(PARI) a(n)=binomial(2*n, n)*6^n \\ Charles R Greathouse IV, Mar 03 2017
(Python)
import math
f=math.factorial
def C(n, r): return f(n)//f(r)//f(n-r)
def A119309(n): return C(2*n, n)*(6**n) # Indranil Ghosh, Mar 03 2017

Crossrefs

Cf. A000984, A013620, A038220, A059304, A098658.

Sequence in context: A116164 A268369 A091266 * A034788 A082165 A006689

Adjacent sequences: A119306 A119307 A119308 * A119310 A119311 A119312

Keyword

nonn

Author

Reinhard Zumkeller, May 14 2006

Status

approved