Counting odd smooth numbers

Question

Let $P(n)$ be the largest prime factor of $n$, and let $\Psi(x,B) = |\{ n \mid n \leq x \wedge P(n) \leq B\}|$. (This is a well-studied function in analytic number theory.) Define $\Psi'(x,B) = | \{ n \mid n \leq x \wedge P(n) \leq B \wedge \mbox{$n$ odd}\}|$. Is there a good estimate for $\Psi'(x,B)$, or for the ratio $\Psi'(x,B)/\Psi(x,B)$?

The answer to this post shows how to prove that $\Psi(x, B) \sim \frac{1}{\pi(B)!} \cdot \prod_{p \leq B} \frac{\log x}{\log B}$. If I repeat the argument to try to estimate $\Psi'(x,B)$ I get $\Psi'(x,B) \sim \frac{1}{(\pi(B)-1)!} \cdot \prod_{2 < p \leq B} \frac{\log x}{\log B}$. But then $\Psi'(x,B)/\Psi(x,B) = \frac{\pi(B)}{\log x} \sim \frac{B}{(\log B) \cdot (\log x)}$, which can exceed 1. So that wasn't very useful.

score 0 · Answer 1 · answered Mar 07 '24 at 22:22

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Error terms matter.

Both of the asymptotic formulas in the OP have error terms, which means that the approximations shown might not be very accurate until $x$ is quite large in terms of $B$.

The ratio $\frac B{(\log B)(\log x)}$ can be greater than $1$ for small $x$, but when $x$ is large in terms of $B$ it will be less than $1$. So there is no inconsistency.

answered Mar 07 '24 at 22:22

Greg Martin

78,820

B will have to be very small relative to x for this to be below 1, so this bound is just not useful. Is a better bound known? – user432944 Mar 08 '24 at 03:17
From your derivation, it seems that this is actually the correct asymptotic bound.... – Greg Martin Mar 08 '24 at 03:48
But the bound is greater than 1 for most “interesting” cases (e.g., B=x^{1/u} with u fixed) – user432944 Mar 08 '24 at 11:10
The asymptotic formula in the OP definitely does not hold for $B$ that large. Look into the Dickman–de Bruijn function for the correct asymptotic formula in that range. – Greg Martin Mar 08 '24 at 18:55

user432944 · Answer 2 · 2024-03-11T11:03:49.587

Let $\Psi_{\rm odd}(x,B) = |\{n \mid n \leq x \wedge P(n) \leq B \wedge \mbox{$n$ is even}\}|$, so $\Psi(x,B) = \Psi_{\rm odd}(x,B) + \Psi_{\rm even}(x,B)$. Note that $\Psi_{\rm even}(x,B) = \Psi(x/2,B)+1$ since there is a natural one-to-one correspondence given by dividing by 2. (The “+1” is due to the edge case of 2 itself.) This allows for calculating the asymptotic behavior of $\Psi_{\rm odd}(x,B)/\Psi(x,B)$.

Counting odd smooth numbers

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