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?

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 $X\prod_{p\leq x}\left(11/p^2\right)$ 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 inclusionexclusion 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. 


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. 

