In the study of the consequences of the prime number theorem, it is possible to find

an upper bound for $d(n)$, the number of divisors of a number $n$.

The example takes $n$ as the product of primes less than $x$ (I could not understand if it refers to the power one for a prime or greater).

Then it says that for the prime number theorem $$d(n) = 2^{\pi(x)} = 2^{{(1 + o(1)) \frac{x}{\log x}}}$$

Does someone understand the first passage?


  • 1
    $\begingroup$ If $n$ is the product of the primes less than some number $x$, then any divisor of $n$ is the same as a subset of the primes $\{p_1,p_2,\cdots, p_{\pi(x)}\}$. The number of those subsets is $2^{\pi(x)}$. Is that what you are asking? $\endgroup$ – lulu Jun 15 at 13:02
  • $\begingroup$ Your observation is correct if you consider the empty subset as the divisor "one". I thought also that each number have two possibility: included or excluded so the product is a product of $ 2\cdot 2 \cdots$. Thank you anyway what you say is what i needed. $\endgroup$ – Utente Flow Jun 15 at 13:17
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    $\begingroup$ This is actually unrelated to the Prime Number Theorem (which gives an asymptotic estimate for $\pi(x)$). It is just an application of the method of finding $d(n)$ based on the prime factorization of $n$. $\endgroup$ – paw88789 Jun 15 at 14:26
  • 2
    $\begingroup$ In general if $n=p_1^{a_1}\dots p_r^{a_r}$ then the number of divisors $d(n)=(a_1+1)\dots (a_r+1)$. In this case they are considering when $n$ is a product of the (distinct)primes less than $x$. So, $a_i=1$ and $r=\pi(x)$. $\endgroup$ – Julian Mejia Jun 15 at 15:03
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    $\begingroup$ It is customary in math, and often convenient, to define an empty product as 1 and an empty sum as 0, and this also agrees with the intuition that not multiplying by anything has the same effect as multiplying by 1 and that not adding anything has the same effect as adding 0. $\endgroup$ – DanielWainfleet Jun 15 at 21:57

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