I am looking for an estimation of the number of paths of length $n$ going from 0 to 0 on the hexagonal (or honeycomb) lattice.


I can find plenty on references on self avoiding paths, but I am looking into every paths. Is it considerably more difficult? Has anyone a reference?



what is the number of paths returning to 0 on the hexagonal lattice
This is answered by Ian Agol [here](https://mathoverflow.net/a/114364/81295), with the reference "[All Roads Lead to Rome-Even in the Honeycomb World](https://www.jstor.org/stable/2684723)", Brani Vidakovic, *Amer. Statist.* 48 (1994) no. 3, 234-236.


An exact formula is
$$ p(n) = \sum\_{k=0}^m \binom{2k}{k} \binom{m}{k}^2$$
if $n= 2m$ is even, and $0$ otherwise.
This is sequence [A002893](https://oeis.org/A002893) on OEIS.


According to OEIS, the number of paths is asymptotic to
$$ p(n) \sim \frac{1}{2\pi n} 3^{n + 3/2}$$
when $n$ is even,
which agrees with the estimate given by shurtados.
In the above reference, Vidakovic proves that $p(n) \geq C \cdot {3^n}/{n}$ for some constant $C$.


