Consider a "Christmas tree lattice" composed of equilateral triangles, as shown. The lattice extends down infinitely.
Start at the top vertex and take a random walk. At each step, move to a randomly chosen neighboring vertex, each with equal probability. Re-visiting vertices is allowed.
What is the probability of returning to the top?
Context
I was reading about Polya's random walk constants, and with all the Christmas trees appearing these days, this question naturally arose.
My thoughts
Let $p(k)$ be the probability of returning to the top for the first time on the $k$th step, and let $P(n)=\sum\limits_{k=2}^n p(k)$. We are looking for $P(\infty)$.
According to my calculations, based on counting paths:
$p(2)=\frac14$
$p(3)=\frac{1}{4^2}=\frac{1}{16}$
$p(4)=\frac{2}{4^3}+\frac{1}{4^2\cdot3}=\frac{5}{96}$
$p(5)=\frac{3}{4^4}+\frac{4}{4^3\cdot 3}=\frac{25}{768}$
$p(6)=\frac{6}{4^5}+\frac{21}{4^4 6}+\frac{12}{4^3 6^2}+\frac{4}{4^2 6^3}=\frac{179}{6912}$
Calculating gets progressively harder.
$P(6)=\frac14+\frac{1}{16}+\frac{5}{96}+\frac{25}{768}+\frac{179}{6912}=\approx 0.423032$.
I guess $P(\infty)<1$, because whenever the path touches the boundary of the tree, the probability of going down is double the probability of going up.
If we have a triangular lattice (instead of a Christmas tree), the probability of returning to the origin is presumably $1$.
Edit
In the comments, @user has provided values of $P(n)$ for $n=5,6,50,100, 150, 200, 250,300$. (I got the same results for $n=5,6$.)
Here is a graph of $P(n)$ against $n$, for $n=2,6,50,100,150,200,250,300$.
The probability to come back to the top is one. Just take 5 more copies of the tree and tile the plane with them, getting a tiling of the plane by triangles. With a good definition of what happens on the boundaries of the tree, we have a random walk in the plane with expectation zero which will be null recurrent. Rigorous details can be painful, but I am pretty sure of the result.
Edit. What follows is not a complete proof, since some argument (martingale? coupling? harmonic functions?) is missing. Denote $w=e^{i\pi/3}$ and consider the lattice $E=\mathbb{Z}+w\mathbb{Z}+w^2\mathbb{Z}.$
We consider $(U_n)$ iid such that $\Pr(U_n=u)=1/6$ for $u=\pm 1,\pm w,\pm w^2.$ We consider $(X_n)$ iid such that $\Pr(X_n=x)=1/4$ if $x=\pm 1$ and $\Pr(X_n=x)=1/8$ if $x=\pm w,\ \pm w^2.$ We consider $(Y_n)$ iid such that $\Pr(Y_n=y)=1/4$ if $y=\pm w$ and $\Pr(Y_n=y)=1/8$ if $y=\pm 1,\ \pm w^2.$ We consider $(Z_n)$ iid such that $\Pr(Z_n=z)=1/4$ if $z=\pm w^2$ and $\Pr(Z_n=z)=1/8$ if $z=\pm 1,\ \pm w.$
Now we consider the Markov chain $(S_n)$ on $E$ such that $$S_{n+1}=s+U_{n+1} $$ if $s=S_n=a+wb+cw^2$ is such that either $a=b=c=0$ or $a,b,c$ are such that at least either $a,b\neq 0,$ or $b,c\neq 0$, or $c,a\neq 0.$
In other cases $S_{n+1}=s+X_{n+1}$ if $s\in \mathbb{Z}\setminus \{0\}$, $S_{n+1}=s+Y_{n+1}$ if $s\in w\mathbb{Z}\setminus \{0\}$,$S_{n+1}=s+Z_{n+1}$ if $s\in w^2\mathbb{Z}\setminus \{0\}.$
This chain is so close to the null recurrent random walk $$S'_n=u+U_1+\cdots+U_n$$ that one guess that it should also be null recurrent.
Finally coming back to the initial chain $C_n$ on the Chrismas tree it is clear that $C_n$ is obtained by folding $E$ on the tree in an obvious way.
Final edit. I found the solution in the theorem page 133 of the book'Random walks and electric networks' by Peter Doyle and Laurie Snell, Carus Mathematical Monographs 1984, which shows that $S_n$ is recurrent. Indeed $S'_n$ is recurrent and the transition probabilities $p_{ij}$ and $p'(ij)$ of these two Markov chains on $E$ are such that there exist $u>0$ and $v$ satisfiying $up_{ij}<p'_{ij}<vp_{ij}.$ The proof of the fact that the random walk on the Christmas tree will hit the top an infinite number of times is complete.