I want to find the next $n \in \mathbb{N}$ such that
$$s < n = \prod_{p_i \in \mathbb{P}_B} {p_i}^{a_i}$$
Where $\mathbb{P}_B$ is the set of primes not greater than $B$ I know that we can generate the factors of these numbers recursively with complexity less than
$$ \prod_{p_i \in \mathbb{P}_B} \frac{log(s)}{log(p_i)}$$
If we restrict the search to the values such that $n < z$, we have $\sum a_i \log(p_i) < \log(z)$, or $\sum a_i \log(p_i)/log(z) < 1$.
accordingly to wikipedia the volume of under a standard $n-$simlex $1/(n+1)!$, so we reduce the number of candidates to
$$ \frac{1}{\pi(B)!}\prod_{p_i \in \mathbb{P}_B} \frac{log(z)}{log(p_i)}$$
In particular we know that there is one power of $p_i$ between $n$ and $n p_i$, and the formula becomes
$$ \frac{2^{\pi(B)}}{\pi(B)!}\prod_{p_i \in \mathbb{P}_B} \frac{log(s)}{log(p_i)} = \frac{(2\log(s))^{\pi(B)}}{\pi(B)!}\prod_{p_i \in \mathbb{P}_B} \frac{1}{log(p_i)}$$
What would be the complexity for the factor set algorithm?
Are there standard algorithms for this problem?
