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Consider an infinite lattice on $\mathbb{Z}^d$. Neighbours of a point on the lattice are given as von-Neumann neighbours, i.e a neighbour of a point $(x, y)$ are the set

$$ \{(x+1,\ y),(x-1,\ y),(x,\ y+1),(x,\ y-1)\}. $$

Now, define the set of $k^{\mathrm{th}}$ nearest neighbours as

$$ x^{(k)}=\{\mathrm{neighbours\ of\ }x_{k-1}|x_{k-1}\in\ x^{(k-1)}\}, $$

with $x^{(0)} = x$. In this way, points can be in $x^{(k)}$ multiple times, for example, $x$ is in $x^{(2)}$ four times.

In particular, the multiplicity of $x$ in $x^{(k)}$ for even $k$ is given by ${2k \choose k}^2$.

The problem of finding the multiplicities of point in $x$ is equivalent to finding the number of lattice paths from the central point to the point being considered as a $k^{\mathrm{th}}$ neighbour given that the lattice path has length $k$.

I would be interested to know if there was a nice form to get the multiplicities of any point in a given set of $x^{(k)}$, similar to the form given for the central point $x$.

Jordan Abbott
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When $d=2$, the multiplicity of $(x+a,x+b)$ in $x^{(k)}$ is $$\binom{k}{\frac{k+a-b}2}\times \binom{k}{\frac{k+a+b}2},$$ when $a+b$ has the same parity as $k$, and is zero otherwise. I gave a combinatorial proof in this answer: https://math.stackexchange.com/a/4058039

For $d\ge 3$, there is no nice formula, but you can find some complicated formulae in the answers to Number of paths on $\mathbb Z^d$

Mike Earnest
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