Estimate about primes Can anyone give an estimate (upper bound or lower bound) for the number of divisors $d\mid P_r$ such that $\frac{\sqrt{P_r}}{2}< d < \sqrt{P_r}$, where $P_r$ is the product of the $r$ smallest primes?
 A: Gerhard did not mess up. $\binom{r}{r/2}$ (or more precisely, the central binomial coefficient, but say $r$ is even) is an upper bound. If we shuffle the primes and multiply them together in random order, then the probability that we hit a given product of $k$ primes is $1/\binom{r}{k}$ (since it requires the first $k$ primes  to be a specified set), and therefore at least $$\frac1{\binom{r}{r/2}}.$$ It follows that the probability that some partial product is in the given interval is at least $N/\binom{r}{r/2}$, where $N$ is the number of divisors in the given interval (since we never stay in such a short interval).
Being a probability, this number is at most 1, which gives Gerhard's bound. My guess is that in reality that probability is more like $\log 2/\log r$, since if instead we add up the logarithms of the primes, we move in steps of size roughly $\log (r\log r) \sim \log r$, and we look for the probability of hitting an interval of length $\log 2$. This would suggest that $$N\approx \frac{\log 2}{\log r}\cdot \binom{r}{r/2}.$$
Edit: I'm fairly convinced that $N$ is of order $\binom{r}{r/2}/\log r$, but I'm not so sure about the constant $\log 2$, though it seems it gives asymptotically an upper bound.
A: A bit late to the discussion, but let me give the best known upper bound.
As noted by user22202, this is a question about squarefree smooth integers. Namely, letting $\Psi_{\mu^2}(x,y)$ be the number of $y$-smooth numbers up to $x$, you ask about $$\Psi_{\mu^2}(\sqrt{P_r},r)-\Psi_{\mu^2}(\sqrt{P_r}/2,r).$$
Granville, in

"On positive integers <=x with prime factors <= t log x" (Number Theory and Applications (ed R.A. Mollin) (Kluwer, NATO ASI, 1989), pages 403-422

which can be accessed through the author's website, established the asymptotics of $\log \Psi_{\mu^2}(x,y)$ in the full range of parameters (Theorem 2). In particular, for $x=\sqrt{P_r}$ and $y=r$ we have $y \sim 2\log x$ and his theorem yields
$$\Psi_{\mu^2}(\sqrt{P_r},r) \sim 2^{\frac{r}{\log r}(1+o(1)}.$$
This in particular upper bounds your quantity.
I am not aware of any lower bound, but I would expect this is also a lower bound for your quantity (but I do not know how to show it).
This is significantly smaller than $\asymp 2^{r}/(\log r \sqrt{r})$ which was given in Johan Wästlund's answer.

Finally, let me mention that the famous saddle point method is irrelevant here it seems. When studying $y$-smooth squarefree integers up to $x$, it is helpful exactly in the range $$(\star) \,\sum_{p \le y} \log p > 2 \log x.$$
So your parameters are just outside this range. This range appears in the paper

Régis de la Bretèche, Gérald Tenenbaum, "Sur les lois locales de la répartition du k-ième diviseur d'un entier", Proc. London Math. Soc. (3) 84 (2002), no. 2, 289–323.

It arises for the following reason. For $\Re s >0$ let $$F(s,y) = \prod_{p\le y}(1+p^{-s})=\sum_{n \text{ squarefree and }y\text{-smooth}} n^{-s}.$$
One can express you quantity as a Perron-type integral involving the integrand $F(s,y) (x^s-(x/2)^s)/s$ (for $(x,y) = (\sqrt{P_r},r)$ in your case) where one integrates along $\Re s = c$ for some $c>0$. It is beneficial to choose $c=\alpha>0$ where $\alpha$ minimizes $s\mapsto F(s,y)x^s$. However, differentiating $F(s,y)x^s$, one is led to the equation
$$ \log x= \sum_{p \le y} \frac{\log p}{1+p^{\alpha}}$$
which solvable exactly in the range $(\star)$.
A: Let $x$ be such that $\pi(x) \sim \frac{x}{\log x} \sim r$, so that $P_r$ is about $e^x$. Let $\Psi(x,y)$ be the de Bruijn, function, the number of $y$-smooth numbers less than or equal to $x$. So a crude upper bound for the quantity in question is about
$$
\Psi(e^{x/2},x) - \Psi(e^{x/2}/2,x).
$$
Out of these numbers (call the set of these numbers $X$) we want to get the squarefree ones. The proportion of squarefree numbers in all the integers is $1/\zeta(2)$ so it is tempting to use $|X|/\zeta(2)$ or $|X|\prod_{p\leq x}\left(1-1/p^2\right)$ as a crude first approximation, though I'm not sure how far off it is. The thing is that one is dealing here with only $x$-smooth numbers. Perhaps for each $p \leq z \ll x$ for suitable $z$ one can estimate the size of the subset of $X$ that is in $p^2\mathbb{Z}$, then use a simple inclusion-exclusion sieve to estimate the size of the subset of $X$ of numbers that, if they are divisible by prime squares $p^2$, must have $z \leq p\leq x$. Or, use more powerful sieve methods. (Guessing that sieving for squarefree numbers should be very similar to sieving for primes.)
Good references:
A. Granville, Smooth numbers: computational number theory and beyond, Algorithmic Number Theory, MSRI Publications, Volume 44, 2008.
T. Tao's blog, 254B, Notes 7: Sieving and expanders.
