6・110 For positive integers $a, n$, define
$$
F_{n}(a)=q+r,
$$

where $q, r$ are non-negative integers, $a=q^{n+r}$, and $0 \leqslant r<n$.
Find the largest positive integer $A$ such that there exist positive integers $n_{1}, n_{2}, n_{3}, n_{4}, n_{5}, n_{6}$, for any positive integer $a \leqslant A$, we have
$$
F_{n_{6}}\left(F_{n_{5}}\left(F_{n_{4}}\left(F_{n_{3}}\left(F_{n_{2}}\left(F_{n_{1}}(a)\right)\right)\right)\right)\right)=1 .
$$

Prove your conclusion.
[Solution] Let the maximum positive integer $B$ that satisfies the condition "there exist positive integers $n_{1}, n_{2}, \cdots, n_{k}$, such that for any positive integer $a \leqslant B$,
$$
F_{n_{k}}\left(F_{n_{k-1}}\left(\cdots\left(F_{n_{1}}(a)\right) \cdots\right)\right)=1 "
$$

be denoted as $x_{k}$.
Clearly, the maximum positive integer $A=x_{6}$.
(1) First, we prove that $x_{1}=2$.
Indeed, $F_{2}(1)=1, F_{2}(2)=1$, so $x_{1} \geqslant 2$.
However, when $n_{1} \geqslant 3$, $F_{n_{1}}(2)=2, F_{2}(3)=2, f_{1}(2)=2$, so $x_{1}<3$.
Thus, we have $x_{1}=2$.
(2) Suppose $x_{k}$ has been determined, and $x_{k}$ is even, clearly $x_{k} \geqslant x_{1}=2$. Clearly, the necessary condition for $x_{k+1}$ is: there exists $n_{1}$, such that for any $a \leqslant x_{k+1}$, we have $F_{n_{1}}(a) \leqslant x_{k}$.
Let $x_{k+1}=q n_{1}+r$, then
$$
\begin{array}{l}
F_{n_{1}}\left(x_{k+1}\right)=q+r \leqslant x_{k}, \\
x_{k+1}=q n_{1}+r \leqslant q n_{1}+x_{k}-q=x_{k}+q\left(n_{1}-1\right) .
\end{array}
$$

If we take $n_{1}=2$, then $x_{k+1}=2 x_{k} \geqslant x_{k}+2$. Thus, we have
$$
\begin{array}{l}
q\left(n_{1}-1\right) \geqslant 2, \\
q \geqslant 1, n_{1} \geqslant 2 .
\end{array}
$$

Therefore,
$$
0<(q-1) n_{1}+n_{1}-1=q n_{1}-1<x_{k+1} .
$$

Thus,
$$
F_{n_{1}}\left((q-1) n_{1}+n_{1}-1\right)=q+n_{1}-2 \leqslant x_{k} .
$$

Hence,
$$
\begin{aligned}
q\left(n_{1}-1\right) & \leqslant\left[\left(\frac{q+n_{1}-1}{2}\right)^{2}\right] \leqslant\left[\left(\frac{x_{k}+1}{2}\right)^{2}\right] \\
& =\left[\frac{x_{k}^{2}}{4}+\frac{x_{k}}{2}+\frac{1}{4}\right] .
\end{aligned}
$$

Since $x_{k}$ is even, we have
$$
q\left(n_{1}-1\right) \leqslant \frac{x_{k}^{2}}{4}+\frac{x_{k}}{2},
$$

and from $x_{k+1} \leqslant x_{k}+q\left(n_{1}-1\right)$, we get
$$
x_{k+1} \leqslant x_{k}+\frac{x_{k}^{2}}{4}+\frac{x_{k}}{2}=\frac{x_{k}\left(x_{k}+6\right)}{4} \text {. }
$$

On the other hand, if we take $n_{1}=\frac{x_{k}}{2}+2$, then
$$
\frac{x_{k}\left(x_{k}+6\right)}{4}=\frac{x_{k}}{2} \cdot n_{1}+\frac{x_{k}}{2} \text {. }
$$

For each $a \leqslant \frac{x_{k}\left(x_{k}+6\right)}{4}$, let $a=q n_{1}+\dot{r}$, then:
$$
q=\frac{x_{k}}{2}, r \leqslant \frac{x_{k}}{2}
$$

or
$$
q \leqslant \frac{x_{k}}{2}-1, r \leqslant n_{1}-1=\frac{x_{k}}{2}+1 .
$$

In both cases, we have $q+r \leqslant x_{k}$, i.e., $F_{n_{l}}(a) \leqslant x_{k}$.
In summary, we have
$$
x_{k+1}=\frac{x_{k}\left(x_{k}+6\right)}{4} .
$$

Since $x_{k}$ is even, $x_{k}+6$ is also even, and one of $x_{k}$ and $x_{k}+6$ is a multiple of 4, thus $x_{k}\left(x_{k}+6\right)$ is a multiple of 8, hence $x_{k+1}$ is even.
Thus, by the principle of mathematical induction, we have proved the recurrence relation:
$$
x_{1}=2, x_{k+1}=\frac{x_{k}\left(x_{k}+6\right)}{4},(k=1,2, \cdots \cdots)
$$

From this relation, it is not difficult to compute successively:
$$
x_{2}=4, x_{3}=10, x_{4}=40, x_{5}=460, x_{6}=53590 \text {. }
$$

Therefore, the maximum positive integer $A=53590$.