1.44 Let $x \in N$. If a sequence of natural numbers $1=x_{0}, x_{1}, \cdots, x_{1}=x$ satisfies
$$
x_{i-1}<x_{i}, x_{i-1} \mid x_{i}, i=1,2, \cdots, l \text {, }
$$

then $\left\{x_{0}, x_{1}, x_{2}, \cdots, x_{l}\right\}$ is called a factor chain of $x$, and $l$ is the length of this factor chain. $L(x)$ and $R(x)$ represent the length of the longest factor chain of $x$ and the number of the longest factor chains of $x$, respectively. For $x=5^{k} \times 31^{m} \times 1990^{n}(k, m, n \in N)$, find $L(x)$ and $R(x)$.
[Solution] First, let's derive the general expressions for $L(y)$ and $R(y)$. Suppose the prime factorization of $y \in N$ is
$$
y=p_{1}^{a} \cdot p_{2}^{a_{2}} \cdots \cdots \cdot p_{s}^{a s},
$$

where $p_{1}, p_{2}, \cdots, p_{s}$ are distinct prime numbers. Since $y$ has only a finite number of factor chains, there must exist a longest one. If the quotient of two consecutive numbers in a factor chain is a composite number, then an additional number can be inserted between these two numbers to extend the factor chain. Therefore, in the longest factor chain, the quotient of any two consecutive numbers must be a prime number. Starting from 1 and successively multiplying by $p_{1}, p_{2}, \cdots, p_{s}$, and multiplying $p_{i}$ a total of $\alpha_{i}$ times, $i=1,2, \cdots, s$, we eventually obtain $y$ as shown in (1), after a total of $\alpha_{1}+\alpha_{2}+\cdots+\alpha_{s}$ multiplications. Thus, the length of the longest factor chain is
$$
L(y)=\alpha_{1}+\alpha_{2}+\cdots+\alpha_{s} .
$$

It is easy to see that any permutation of the elements
$$
p_{1}, \cdots, p_{1}, p_{2}, \cdots, p_{2}, \cdots, p_{s}, \cdots, p_{s}
$$

corresponds to a longest factor chain of (1). That is, when the first 1, the first 2, the first 3, ..., the $\left(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{s}\right)$ numbers in a permutation are successively multiplied, the corresponding longest factor chain is obtained. Conversely, in any longest factor chain of (1),

the quotients of consecutive terms from front to back form a permutation corresponding to that longest factor chain. Clearly, this correspondence is a bijection. According to the formula for permutations with repetition, we have
$$
R(y)=\frac{\left(\alpha_{1}+\alpha_{2}+\cdots+\alpha_{s}\right)!}{\alpha_{1}!\alpha_{2}!\cdots \alpha_{s}!} .
$$

For the given $x$, we write
$$
x=5^{k} \times 31^{m} \times 1990^{n}=2^{n} \times 5^{k+n} \times 31^{m} \times 199^{n} .
$$

Using formulas (2) and (4), we get
$$
L(x)=k+m+3 n, R(x)=\frac{(k+m+3 n)!}{(k+n)!m!(n!)^{2}} .
$$