Three, 50 points 5) For positive integers $a, n$, define $F_{a}(a)=q+r$, 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.
Three, the maximum positive integer $B$ that satisfies the condition "there exist positive integers $n_{1}, n_{2}, \ldots, n_{k}$, such that for any positive integer $a \leqslant B$,
$$
F_{n_{k}}\left(F_{\pi_{k-1}}\left(\cdots\left(F_{n_{1}}(a)\right) \cdots\right)\right) \doteq 1^{\prime}
$$
is denoted as $x_{k}$. Clearly, the maximum positive integer $A$ we are looking for is $x_{6}$.
(1) First, prove that $x_{1}=2$.
In fact, $F_{2}(1)=F_{2}(2)=1$, so $x_{1} \geq 2$.
Furthermore, when $n_{1} \geqslant 3$, $F_{n_{1}}(2)=2$, and $F_{2}(3)=F_{1}(2)=2$, so $x_{1}0, n_{1}>1$. Thus, $0<(q-1) n_{1}+n_{1}-1$ $=q n_{1}-1<x_{k+1}$. Therefore,
$$
F_{n_{1}}\left((q-1) n_{1} a_{s}-1\right)=q+n_{1}-2 \leqslant x_{k} \text {. }
$$

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

Since $x_{k}$ is even, thus,
$q\left(n_{1}-1\right) \leqslant \frac{x_{k}^{2}}{4}+\frac{x_{k}}{2}$.
Since $x_{k} \geqslant 2$, so,
$$
x_{k}+\frac{x_{k}^{2}}{4}+\frac{x_{k}}{2} \geqslant x_{k}+2 \text {. }
$$

Therefore, we always have
$x_{k+1} \leqslant x_{k}+\frac{x_{k}^{2}}{4}+\frac{x_{k}}{2}=\frac{x_{k}\left(x_{k}+6\right)}{4}$.
On the other hand, if we take $n_{1}=\frac{x_{k}}{2}+2$, since $\frac{x_{k}\left(x_{k}+6\right)}{4}=$ $\frac{x_{k}}{2} \cdot n_{1}+\frac{x_{k}}{2}$, for each $a \leqslant \frac{x_{k}\left(x_{k}+6\right)}{4}$, let $a=q n_{1}+r$, then,
either $q=\frac{x_{k}}{2}, r \leqslant \frac{x_{k}}{2}$;
or $q \leqslant \frac{x_{k}}{2}-1, r \leqslant n_{1}-1=\frac{x_{k}}{2} \div 1$.
In both cases, we have $q+r \leqslant x_{k}$. Therefore,
$$
x_{k+1}=\frac{x_{k}\left(x_{k}+6\right)}{4} \text {. }
$$

Moreover, since $x_{k}$ is even, if $4 \mid x_{k}$, by $2 \mid\left(x_{k}+6\right)$ we get $8 \mid x_{k}\left(x_{k}+6\right)$; if $x_{k} \equiv 2(\bmod 4)$, by $x_{k}+6 \equiv 0$ $(\bmod 4)$ we also get $8 \mid x_{k}\left(x_{k}+6\right)$. Therefore, $x_{k+1}$ is also even.
Thus, the induction proof is completed: $x_{k+1}=\frac{x_{k}\left(x_{k}+6\right)}{4}$.
Starting from $x_{1}=2$, we successively obtain $x_{2}=4, x_{3}=10, x_{4}=40$, $x_{5}=460, x_{6}=\frac{460 \times 446}{4}=53590$.
Therefore, the maximum positive integer $A=53590$.
(Provided by 邆玉懒)