Possibilities for dimensions of $\mathfrak{m}^i/\mathfrak{m}^{i+1}$ for a local ring Let $R$ be a local commutative ring with maximal ideal $\mathfrak{m}$, and denote by $k$ the residue field $R/\mathfrak{m}$. Then we can look at the sequence of $k$-vectorspaces
$$R/\mathfrak{m}, \mathfrak{m}/\mathfrak{m}^2, \mathfrak{m}^2/\mathfrak{m}^3, \mathfrak{m}^3/\mathfrak{m}^4 \ldots$$
This gives rise to a sequence of dimensions $(\dim_k \mathfrak{m}^i/\mathfrak{m}^{i+1})_{i\geq0}$ (we say $\mathfrak{m}^0=R$). For example, if $R=\mathbb{Z}_p$, the ring of $p$-adic integers, then the sequence is identically $1$.

Question1: What (kind of) sequences can we acquire?

My motivation for this question is that I am looking at groups that act on rooted trees that are constructed from local rings, in a way that all nodes at distance $i$ from the root have the same degree, namely $\left|\mathfrak{m}^i/\mathfrak{m}^{i+1}\right|+1$. If the given sequence has distinct patterns (or distinct lack of patterns), this will imply extra symmetry (or lack of symmetry) in my tree.
 A: Yves Cornulier's suggestion about the values of $\dim(M^i/M^{i+1})$ is correct. Either they are all infinite or there is a finitely generated local ring $(S,N)$ such that for large $i$, $ \dim(M^i/M^{i+1})=\dim(N^i/N^{i+1})$.
Claim 1: If $\dim(M^k/M^{k+1})<\infty$ for some $k$, then $\dim(M^l/M^{l+1})<\infty$ for all $l\geq k$.
If $\dim(M^k/M^{k+1})<\infty$ then there exist $a_1,\dots, a_n\in M^k$ s.t. $$M^k=\langle a_1,\dots, a_n\rangle +M^{k+1}.$$
We can assume that each $a_i$ is a product $a_{i,1}a_{i,2}\cdots a_{i,k}$ where $a_{i,j}\in M$. (Replace an $a_i$ by its summands).
Then
$$M^{k+1} =\langle s_1s_2\cdots s_{k+1} \mid s_i \in M\rangle \\
=\langle s a_i \mid s\in M, i=1,\dots,n\rangle + M^{k+2} \\
\subseteq \langle (sa_{i,1}\cdots a_{i,k-1})a_{j,k}\mid s\in M, i,j=1,\dots,n\rangle + M^{k+2} \\
\subseteq M^k\cdot \langle a_{j,k}\mid j=1,\dots,n\rangle + M^{k+2} \\
=\langle a_ia_{j,k}\mid i,j=1,\dots,n\rangle + M^{k+2}
$$
Hence, $M^{k+1}/M^{k+2}$ is finite dimensional.
Claim 2: For large $l$, $\dim(M^l/M^{l+1})$ equals the corresponding dimensions in a finitely generated local ring.
From the proof above one also sees that if one takes $N=\langle a_{i,j} \mid i=1,\dots, n, j=1,\dots k\rangle$, then
$M^l/M^{l+1}$ is generated by (residue classes of) elements in $N^l$ for all $l\geq k$.
Let 
$$S:=R/M \oplus N/(M^2\cap N)\oplus N^2/(M^3\cap N^2)\oplus \cdots$$
Then $S$ is a finitely generated subring of the associated graded ring of $R$, and the dimension of $N^l/(M^{l+1}\cap N^l)$ is equal to the dimension of $M^l/M^{l+1}$ for all $l\geq k$.
