4a.a * Let $x$ be a natural number. If a sequence of natural numbers $x_{0}=1, x_{1}, x_{2}, \cdots, x_{t-1}$, $x_{t}=x$ satisfies $x_{i-1} \mid x_{i}, x_{t-1}<x_{i}, i=1,2, \cdots, t$, then $\left\{x_{0}, x_{1}, \cdots, x_{t}\right\}$ is called a divisor chain of $x$, and $t$ is the length of this divisor chain. $T(x)$ and $R(x)$ represent the length of the longest divisor chain of $x$ and the number of the longest divisor chains of $x$, respectively.
For $x=5^{k} \times 31^{m} \times 1990^{n}(k, m, n \in \mathbf{N})$, try to find $T(x)$ and $R(x)$.
Let the prime factorization of $x$ be $x=p_{1}^{a_{1}} p_{d}^{\alpha} \cdots p_{n}^{\alpha_{n}}$, where $p_{1}, p_{2}, \cdots, p_{n}$ are distinct prime numbers, and $\alpha_{1}, \alpha_{2}, \cdots, a_{n}$ are positive integers. Since each term in the factor chain has at least one more prime factor than the previous term, $T(x) \leqslant \alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$. Arranging $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$ prime factors (where $\alpha_{1}$ are $p_{1}$, $\alpha_{2}$ are $p_{2}$, ..., $\alpha_{n}$ are $p_{n}$) in any order, each arrangement generates a factor chain of length $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$ ($x_{1}$ is the first term of the arrangement, $x_{2}$ is $x_{1}$ multiplied by the second term, $x_{3}$ is $x_{2}$ multiplied by the third term, ..., etc.), thus $T(x)=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}$, and $R(x)$ is the number of arrangements $\frac{\left(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{n}\right)!}{\alpha_{1}!\alpha_{2}!\cdots \alpha_{n}!}$. For $x=5^{k} \times 31^{m} \times 1990^{n}=2^{n} \times 5^{k+n} \times 31^{m} \times 199^{n}$, we have
$$
T(x)=3 n+k+m, R(x)=\frac{(3 n+k+m)!}{(n!)^{2} \cdot m!\cdot(k+n)!} .
$$