}_{m,k})\subseteq \oplus_{s=r+1}^k \mathfrak p(m, s)$; the other inclusion is obvious. In particular, $\mathfrak p(m, k) = C^{k-1}({\mathfrak n }_{m,k}) \subseteq \mathfrak z({\mathfrak n }_{m,k})$. Now let $x\in \mathfrak z({\mathfrak n }_{m,k})$ and let $e$ be a generator and assume $x\notin C^{k-1}({\mathfrak n }_{m,k})$. Recall that ${\mathfrak n }_{m,k}$ is homomorphic image of the free Lie algebra $\mathfrak f_m$ so that there exist $X,E\in \mathfrak f_m$ such that $X\to x$ and $E \to e$, being $E$ a generator of $\mathfrak f_m$. Since $[x, e]=0$ then $[X,E]=0$ which says $X$and $E$ are proportional, which is impossible (see for instance Ch. 2 in [@Ba]). Thus $\mathfrak p(m, k) =  \mathfrak z({\mathfrak n }_{m,k})$.

Denote as $d_m(s)$ the dimension of $\mathfrak p(m,s)$. Inductively one gets [@Se] $$\label{dim}s\cdot d_m(s)= m^s- \sum_{r|s, r<s} r\cdot d_m(r),\qquad s\geq 1.$$

Hence for a fixed $m$, one has $d_m(1)=m$ and $d_m(2)=m(m-1)/2$.

\[ex:1\]Given an ordered set of generators $e_1,\ldots, e_m$ of a free $2$-step nilpotent Lie algebra ${\mathfrak n }_{m,2}$, a Hall basis is $$\label{ba2}
    \mathcal B=\{e_i,\,[e_j,e_k]:\,i=1,\ldots,m,\, 1\leq  k< j\leq m\}.$$ Equation (\[dim\]) asserts that $\dim{\mathfrak n }_{m,2}=d_m(1)+d_m(2)=m+m(m-1)/2$. Since $\mathfrak z({\mathfrak n }_{m,2}) = \mathfrak p(m, 2)$, we have $\dim \mathfrak z({\mathfrak n }_{m,2})= m(m-1)/2$.

\[ex:2\] For the free $3$-step nilpotent 