TITLE: what is the number of paths returning to 0 on the hexagonal lattice
QUESTION [7 upvotes]: 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?

REPLY [11 votes]: This is answered by Ian Agol here, with the reference "All Roads Lead to Rome-Even in the Honeycomb World", 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 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 $C$.