Normed Lie products need not form a basis for graded component of free Lie algebra

Statement for right-normed Lie products
Suppose $$n \ge 2$$ and $$r \ge 4$$ are integers. Suppose $$L$$ is a free Lie ring on $$n$$ variables $$x_1,x_2,\dots,x_n$$. Consider all the right-normed Lie products in $$L$$ of weight $$r$$ with inputs coming from the generating set, i.e., for any function $$f:\{1,2,\dots,r\} \to \{ 1,2,\dots,n\}$$ satisfying $$f(r - 1) < f(r)$$ consider the product:

$$[x_{f(1)},[x_{f(2)},[x_{f(3)},\dots,[x_{f(r-1)},x_{f(r)}]\dots]]]$$

The condition $$f(r - 1) < f(r)$$ takes into account alternation in the last two variables.

Then, these normed Lie products do not form a freely generating set for the weight $$r$$ graded component of the free Lie algebra. Further, even if we look at the free $$\mathbb{Q}$$-Lie algebra instead, the normed Lie products do not form a basis for that algebra.

Extent of divergence
For any $$r$$, the actual rank of the graded component for a free Lie ring on $$n$$ variables is given by formula for dimension of graded component of free Lie algebra, which gives a number-theoretically defined polynomial of degree $$r$$ in $$n$$. The number of right-normed Lie product, taking into account alternation in the last variable, is $$(n^r - n^{r-1})/2$$. We compare these polynomials for small values of $$r$$.

Proof
The Jacobi identity in $$x,y,[x,y]$$ to obtain:

$$[x,[y,[x,y]] = [y,[x,[x,y]]$$

Thus, although there are four right-normed Lie products up to skew symmetry in the last two variables, the rank of the free abelian group is only three.