0,2

Number of walks of length 2n on the 4-regular tree beginning and ending at some fixed vertex. The generating function for the corresponding sequence for the m-regular tree is 2*(m-1)/(m-2+m*sqrt(1-4*(m-1)*x)). When m=2 this reduces to the usual generating function for the central binomial coefficients. - Paul Boddington, Nov 11 2003

Main diagonal of the array A(0,j)=A(i,0)=1 for i,j>=0 and for i,j>=1 A(i,j)=min{A(i,j-1)+3*A(i-1,j); 3*A(i,j-1)+A(i-1,j)}. - Benoit Cloitre, Aug 05 2004

Hankel transform is A133461. - Philippe Deléham, Dec 01 2007

Also the number of length 2n words over an alphabet of size 4 that can be built by repeatedly inserting doublets into the initially empty word.

The sequence {b(n)}, where b(2n)=a(n), b(2n+1)=0, is the cogrowth function of the free group of rank 2. - Murray Elder, Jun 28 2016

Robert Israel, Table of n, a(n) for n = 0..837

Murray Elder, Table of n, b(n) for n = 0..200, where b(n) is mentioned in a comment above.

Libor Caha and Daniel Nagaj, The pair-flip model: a very entangled translationally invariant spin chain, arXiv:1805.07168 [quant-ph], 2018.

M. Elder and A. Rechnitzer, T. Wong, On the cogrowth of Thompson's group F, Groups, Complexity, Cryptology 4(2) (2012), 301-320.

Pakawut Jiradilok and Supanat Kamtue, Transportation Distance between Probability Measures on the Infinite Regular Tree, arXiv:2107.09876 [math.CO], 2021.

J. Novak, Three lectures on free probability, arXiv preprint arXiv:1205.2097, 2012. - N. J. A. Sloane, Oct 15 2012

Gregory Quenell, Combinatorics of free product graphs, Contemp. Math (1994) 257-281 (Eq. 19).

Ian M. Wanless, Counting Matchings and Tree-Like Walks in Regular Graphs, Combinatorics, Probability and Computing (2010) 19, 463-480 (Lemma 3.1).

a(n) = Sum_{k=0..n} A039599(n,k)*3^(n-k). - Philippe Deléham, Aug 25 2007

From Paul Barry, Sep 15 2009: (Start)

G.f.: 1/(1-4x*c(3x)), c(x) the g.f. of A000108;

G.f.: 1/(1-4x/(1-3x/(1-3x/(1-3x/(1-3x/(1-.... (continued fraction);

G.f.: 1/(1-4x-12x^2/(1-6x-9x^2/(1-6x-9x^2/(1-6x-9x^2/(1-... (continued fraction).

Integral representation: a(n) = (2/Pi)*Integral_{x=0..12} x^n*sqrt(x*(12-x))/(16-x). (End)

a(0) = 1; a(n) = (4/n) * Sum_{j=0..n-1} C(2*n,j) * (n-j) * 3^j for n > 0.

a(n) = upper left term in M^n, M = an infinite square production matrix as follows:

4, 4, 0, 0, 0, 0, ...

3, 3, 3, 0, 0, 0, ...

3, 3, 3, 3, 0, 0, ...

3, 3, 3, 3, 3, 0, ...

3, 3, 3, 3, 3, 3, ...

...

- Gary W. Adamson, Jul 15 2011

D-finite with recurrence: n*a(n) + 2*(9-14*n)*a(n-1) + 96*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011

P-recurrence confirmed using differential equation (-96*x+10)*g(x) + (-192*x^2+28*x-1)*g'(x) - 6 = 0 satisfied by the generating function. - Robert Israel, Jul 06 2015

a(n) ~ 3*12^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Jun 29 2013

a(n) are special values of the hypergeometric function 2F1, in Maple notation: a(n) = 3*12^n*GAMMA(n+1/2)*hypergeom([1,n+1/2],[n+2],3/4)/(4*sqrt(Pi)*(n+1)!), n=0,1,... . - Karol A. Penson, Jul 06 2015

a(2)=28 because there are 4*4=16 walks whose second step is to return to the starting vertex and 4*3=12 walks whose second step is to move away from the starting vertex.

a:= n-> `if`(n=0, 1, 4/n*add(binomial(2*n, j) *(n-j)*3^j, j=0..n-1)):

seq(a(n), n=0..20);

# Alternative:

f:= gfun:-rectoproc({(-192*n-288)*a(n+1)+(28*n+66)*a(n+2)+(-n-3)*a(n+3)=0, a(0)=1, a(1)=4, a(2)=28}, a(n), remember):

map(f, [$0..50]); # Robert Israel, Jul 06 2015

CoefficientList[ Series[3/(1 + 2Sqrt[1 - 12x]), {x, 0, 19}], x] (* Robert G. Wilson v, Nov 12 2003 *)

(PARI) x='x+O('x^66); Vec(3/(1+2*sqrt(1-12*x))) \\ Joerg Arndt, Sep 06 2014

Cf. A089022.

4th column of A183135.

Sequence in context: A152599 A199579 A089023 * A229652 A229651 A229650

Adjacent sequences: A035607 A035608 A035609 * A035611 A035612 A035613

Edited by Alois P. Heinz, Jan 20 2011

approved