25. S3 (POL) For an integer \( x \geq 1 \), let \( p(x) \) be the least prime that does not divide \( x \), and define \( q(x) \) to be the product of all primes less than \( p(x) \). In particular, \( p(1)=2 \). For \( x \) such that \( p(x)=2 \), define \( q(x)=1 \). Consider the sequence \( x_{0}, x_{1}, x_{2}, \ldots \) defined by \( x_{0}=1 \) and 
\[ x_{n+1}=\frac{x_{n} p\left(x_{n}\right)}{q\left(x_{n}\right)} \]
for \( n \geq 0 \). Find all \( n \) such that \( x_{n}=1995 \).
25. By the definition of \( q(x) \), it divides \( x \) for all integers \( x > 0 \), so \( f(x) = x p(x) / q(x) \) is a positive integer too. Let \( \{p_{0}, p_{1}, p_{2}, \ldots\} \) be all prime numbers in increasing order. Since it easily follows by induction that all \( x_{n} \)'s are square-free, we can assign to each of them a unique code according to which primes divide it: if \( p_{m} \) is the largest prime dividing \( x_{n} \), the code corresponding to \( x_{n} \) will be \( \ldots 0 s_{m} s_{m-1} \ldots s_{0} \), with \( s_{i}=1 \) if \( p_{i} \mid x_{n} \) and \( s_{i}=0 \) otherwise. Let us investigate how \( f \) acts on these codes. If the code of \( x_{n} \) ends with 0, then \( x_{n} \) is odd, so the code of \( f(x_{n}) = x_{n+1} \) is obtained from that of \( x_{n} \) by replacing \( s_{0}=0 \) by \( s_{0}=1 \). Furthermore, if the code of \( x_{n} \) ends with \( 011 \ldots 1 \), then the code of \( x_{n+1} \) ends with \( 100 \ldots 0 \) instead. Thus if we consider the codes as binary numbers, \( f \) acts on them as an addition of 1. Hence the code of \( x_{n} \) is the binary representation of \( n \) and thus \( x_{n} \) uniquely determines \( n \). Specifically, if \( x_{n} = 1995 = 3 \cdot 5 \cdot 7 \cdot 19 \), then its code is 10001110 and corresponds to \( n=142 \).