The number of $n \leq x$ composed using only the given primes $p_1, p_2, ... p_k$ as $x \rightarrow \infty$ satisfies $$ \frac {\log^k x} {k! \prod_1^k p_j} + O \left ( \log^{k1} x \right ) . $$ I am looking for the references to this result. Thanks, AndreyF

3$\begingroup$ Are you sure denominator is not $k! \prod \log p_j$? We need to count integer points in the simplex $x_1, \ldots, x_k \geq 0$, $x_1 \log p_1 + \ldots + x_k \log p_k \leq \log x$, and its volume is $\frac{\log^k x}{k! \prod \log p_j}$. $\endgroup$– Mikhail TikhomirovApr 22, 2022 at 20:01
1 Answer
As Mikhail Tikhomirov commented, the denominator should have $\log$ of $p_j$ instead of $p_j$ itself.
For a reference see Theorem III.5.3, and more generally, section 2 (`The geometric method') of Chapter 5 of part III of G. Tenenbaum's book "Introduction to analytic and probabilistic number theory". This section is contained in pages 516517.
If we let $$N_k(x):=\left \left\{ (\nu_1,\ldots,\nu_k) \in \mathbb{N}^k: \sum_{1 \le j \le k} \nu_j a_j \le x\right\} \right$$ for given positive $a_j$, the theorem states that $$\frac{x^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} \le N_k(x)\le \frac{(x+\sum_{1 \le j \le k}a_j)^k}{k!}\prod_{1\le j\le k}\frac{1}{a_j} .$$ Now apply this with $a_j = \log p_j$. This answers your question, because you count solutions to $\prod_{1 \le j \le k} p_j^{\nu_j} \le x$, which (upon taking natural logarithms) we see is a quantity equal to $N_k(x)$. This is directly related to V. Ennola's work on $y$friable (i.e. smooth) numbers, which is mentioned in the aforementioned section. It could be that your result is in Ennola's paper "On numbers with small prime divisors" (Ann. Acad. Sci. Fenn., Ser. A I 440, 16 p. (1969)) but I am not able to get my hands on it.

$\begingroup$ Thanks Mikhail and Ofir for the correction and for the proof. I managed to find Ennola (1969) via Google: it deals with asymptotics for integers $\leq x$ with prime factors $\leq y$, not quite the case I mentioned. I'm trying to locate an original paper, if any, that has this result. $\endgroup$– AndreyFApr 22, 2022 at 23:00