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This was asked in mathstackexchange (see here) but was not satisfactorily answered beyond my basic observations.

Let $\tau(k)$ be the number of divisors of the positive integer $k$. How does $f(n)=\prod_{k\leq n} \tau(k)$ or a reasonable function of it, such as $\log f(n)$ grow with increasing $n$?

Gerry Myerson commented that it's tabulated at OEIS but with no information on growth rate.

I noted that the upper bound (using the arithmetic geometric mean inequality and the sum of divisors of integers up to $n$) below holds $$f(n)\leq (\log n)^n \left(1+\frac{2 \gamma -1}{n}\right).$$

This is a plot of $\log U$ and $\log f(n)$ where $U$ is the upperbound in (1).

enter image description here

This is a plot $U/f(n)$ where $U$ is the upperbound in (1).

enter image description here

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    $\begingroup$ Well, $f(n)=\log\tau(n)$ is an additive function; that is, $f(n) = \sum_{p^k\parallel n} f(p^k)$. Now insert this expression for $f$ into $\sum_{n \le x}f(n)$ and reverse the order of summation. I believe in this example, one finds that the main term in the asymptotic is $x \sum_{p \le x} f(p)/p$, and this is $\sim (\log 2) x\log\log{x}$, as $x\to \infty$ (Alternatively, note that $\tau(n)$ is between $2^{\omega(n)}$ and $2^{\Omega(n)}$, and use the known results --- as found in Hardy and Wright, for example --- on the partial sums of $\omega$ and $\Omega$.) $\endgroup$ – so-called friend Don Nov 8 '15 at 3:10
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    $\begingroup$ As above, $\log{\tau(n)} = \sum_{p^k || n}\log{(k+1)} = \sum_{p^k\vert n}\log{\left(1+\frac{1}{k}\right)}$. Thus $$\sum_{n\leq x} \log{\tau(n)} = \sum_{p^k\leq x} \log{\left(1 + \frac{1}{k}\right)}\left\lfloor\frac{x}{p^k}\right\rfloor.$$ $\endgroup$ – alpoge Nov 8 '15 at 5:03
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    $\begingroup$ Since $\sum_{p\leq x}\left\lfloor\frac{x}{p}\right\rfloor=\sum_{n\leq x}\omega(n)=x\log\log{x}+\sum_{i=0}^K \frac{a_i x}{(\log{x})^i}+O\left(\frac{x}{(\log{x})^{K+1}}\right)$ for some constants $a_i$, and $\sum_{p^k\leq x, k > 1}\frac{\log{\left(1+\frac{1}{k}\right)}}{p^k}=\text{const.} + O(x^{-1/2})$, it follows that the asymptotic is $(\log{2})\log\log{x} + \widetilde{\text{const.}} x + (\log{2})\sum_{i=1}^K \frac{a_i x}{(\log{x})^i} + O\left(\frac{x}{(\log{x})^{K+1}}\right)$. Hope I haven't made a mistake! $\endgroup$ – alpoge Nov 8 '15 at 5:06
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    $\begingroup$ Indeed! That's why I left it as floor(x/p) and used results on the average order of omega(n) instead (namely an expansion I found in work of Diaconis). $\endgroup$ – alpoge Nov 8 '15 at 7:39
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    $\begingroup$ @GHfromMO: Oh brilliant! Funny how that happens! (Hopefully I've done this "@" construction correctly since nothing special shows up on my phone.) $\endgroup$ – alpoge Nov 8 '15 at 7:47
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Let $r_n(\lambda)=\#\{m:m\le n,\tau(m)\le 2^{e(n)}\}$ where $e(n)=\log\log n+\lambda (2\log\log n)^{1/2}.$ It is known (see Note on the distribution of values of the arithmetic function $d\left( m \right)$) that $$\lim_{n\to\infty}\frac{r_n(\lambda)}{n}=\frac1{\sqrt\pi}\int_{-\infty}^{\lambda}e^{-u^2}du.$$ The main term $\log\log n$ in $e(n)$ gives asymptotic $$\prod_{k\leq n} \tau(k)\approx 2^{n\log\log n}.$$

Probably next term in $e(n)$ assigned with rate of convergence (or some kind of Chebyshev's inequality for $\tau(n)$) may give more precise result.

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I deleted my original answer, because I realized (thanks to user alpoge, see the comments below the original post) that I have already answered the same question earlier, with a much better analysis. Read my response here.

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