Picture hanging problem

Problem statement
The picture hanging problem is the following problem.

Denote by $$F_n$$ the free group on generators $$x_1,x_2,\dots,x_n$$. For $$1 \le i \le n$$, define $$\pi_{n,i}$$ as the endomorphism of $$F_n$$ that sends $$x_i$$ to the identity and fixes $$x_j, j \ne i$$. Note that $$\pi_{n,i}$$ is a retraction.

Consider the intersection of the kernels of all these maps:

$$\bigcap_{i=1}^n \operatorname{Ker} (\pi_{n,i})$$

The picture hanging problem asks for a word (in terms of the $$x_i$$s and their inverses) that defines an element of this intersection. One variant of the problem asks for a word of minimum possible length, or at least for a word that is short in length.

Loosely speaking, it is asking for the minimum length of a word such that, if any one of the $$x_i$$s in the word is replaced by the identity element, the word becomes trivial.

Solution
We can show that for any $$n$$, there is a word of length somewhere between $$n^2$$ and $$(n + 1)^2$$ that solves the picture hanging problem.

More explicitly, if $$f(n)$$ is the minimum possible length of a word solving the picture hanging problem, we have the following:

$$\! f(n) \le 2 \min_{a + b = n} (f(a) + f(b))$$

(Probably, equality holds above).

In particular, we get (with equality holding?):

$$\! f(2m) \le 4f(m)$$

and (with equality holding?):

$$\! f(2m + 1) \le 2f(m) + 2f(m+1)$$

Proof of recurrence relations
We demonstrate the inequality:

$$\! f(n) \le 2 \min_{a + b = n} (f(a) + f(b))$$

This is equivant to proving that if $$a + b = n$$ then:

$$f(n) \le 2(f(a) + f(b))$$

Note that $$f(a)$$ is the length of a word that solves the problem for the free group with $$a$$ generators. Let $$u$$ be such a word in terms of the generators $$x_1,x_2,\dots,x_a$$. $$f(b)$$ is the length of a word that solves the problem for the free group with $$b$$ generators. Let $$v$$ be such a word in terms of the generators $$x_{a+1},\dots,x_{n-1},x_n$$. Now, consider the word:

$$\! w = [u,v]$$

When we expand this out, $$u,v,u^{-1},v^{-1}$$ are all pieces of the word. The lengths of both $$u$$ and $$u^{-1}$$ are $$f(a)$$ and the lengths of both $$v$$ and $$v^{-1}$$ are $$f(b)$$. The total length of $$w$$ is thus $$2(f(a) + f(b))$$. Further, $$w$$ solves the picture hanging problem in terms of all the generators $$x_1,x_2,\dots,x_n$$.