Finding the integer $\le n$ with largest number of divisors As mentioned in an answer to this question an integer less than $n$ with largest number of divisors can be found exploring the numbers $x$ of the form
  $$ x = 2^{a_1} 3^{a_2} \dots p_k^{a_k} \dots $$
(where $p_k$ is the $k$-th prime number) with the conditions 
  $$ x \le n < 2x \quad \text{and}\quad a_1 \ge a_2 \ge \dots \ge a_k \ge \dots$$
to determine the complexity of this algorithm I would like to know the asymptotic number of tuples $(a_1, a_2, \dots)$ verifying these conditions as $n\to \infty$. I suspect that this number is $\gg_l \log^l n$ for every $l$ and $\ll_\epsilon n^\epsilon$ for every $\epsilon > 0$, but I don't know how to prove it. 
Thanks for your help.
 A: For a given $n$, let $N(n)$ be the number of admissible exponent sequences, that is, the number of sequences $a_1 \geqslant a_2 \geqslant \ldots$ on nonnegative integers such that
$$\frac{n}{2} < \prod_{\nu = 1}^{\infty} p_{\nu}^{a_{\nu}} \leqslant n\,. \tag{1}$$
Then indeed we have $N(n) \ll_{\epsilon} n^{\epsilon}$ for every $\epsilon > 0$ and $N(n) \gg_l \log^l n$ for every $l$.
This is easy to prove if we write the product in $(1)$ as a product of primorials. The non-increasingness of the exponent sequence makes this possible. Define $k = k(n)$ as the index such that
$$p_k\# \leqslant n < p_{k+1}\#\,. \tag{2}$$
Then by the prime number theorem we have $k \sim \frac{\log n}{\log \log n}$. If
$$\prod_{\nu = 1}^{k} (p_{\nu}\#)^{b_{\nu}} \leqslant n\,,$$
then a fortiori we have
$$(p_{\nu}\#)^{b_{\nu}} \leqslant n \iff b_{\nu} \leqslant \frac{\log n}{\log (p_{\nu}\#)} = \frac{\log n}{\vartheta(p_{\nu})}$$
for each $\nu$. Hence
$$N(n) \leqslant \prod_{\nu = 1}^{k}\biggl(1 + \biggl\lfloor \frac{\log n}{\vartheta(p_{\nu})}\biggr\rfloor\biggr) < 2^k\prod_{\nu = 1}^{k} \frac{\log n}{\vartheta (p_{\nu})}\,.$$
Now
$$2^k = \exp (k\log 2) = \exp \frac{(1 + o(1))(\log 2)\log n}{\log \log n} = n^{o(1)}$$
and
$$(\log n)^k = \exp (k\log \log n) = n^{1 + o(1)}$$
are straightforward. To estimate the denominator, we use
\begin{align}
\sum_{\nu = 1}^{k} \log \vartheta(p_{\nu}) &= \sum_{\nu = 1}^{k} \bigl(\log p_{\nu} + O(1)\bigr) \\
&= \vartheta(p_k) + O(k) \\
&= \log n + O(\log \log n) + O(k) \\
&= \log n + O\biggl(\frac{\log n}{\log \log n}\biggr)
\end{align}
per the Chebyshev bounds and $(2)$, thus
$$\prod_{\nu = 1}^{k} \vartheta(p_{\nu}) = n^{1 + O\bigl(\frac{1}{\log \log n}\bigr)}\,.$$
Putting everything together gives $N(n) = n^{o(1)}$, whence $N(n) \ll_{\epsilon} n^{\epsilon}$ for all $\epsilon > 0$.
For the lower bound, fix an arbitrary $l > 0$ and assume $n$ is so large that $k > l+1$. Then we count the products
$$\prod_{\nu = 2}^{l+1} (p_{\nu}\#)^{b_{\nu}} \leqslant n\,.\tag{3}$$
For every such product there is a unique $b_1$ such that
$$\frac{n}{2} < \prod_{\nu = 1}^{l+1} (p_{\nu}\#)^{b_{\nu}} \leqslant n\,,$$
hence the number of products $(3)$ is a lower bound for $N(n)$. These products correspond to the lattice points in the simplex $S \subset \mathbb{R}^l$ with vertices $0$ and
$$\frac{\log n}{\vartheta(p_{\nu})}\cdot e_{\nu-1},\qquad 2 \leqslant \nu \leqslant l+1,$$
whose ($l$-dimensional) volume is
$$\frac{1}{l!}\prod_{\nu = 2}^{l+1} \frac{\log n}{\vartheta(p_{\nu})} = C_l \cdot \log^l n\,.$$
As $n \to \infty$ the number of lattice points in $S$ is asymptotically equal to the volume of $S$, hence $N(n) \gg_l \log^l n$ for every $l$.
A: As the question is posed, it is the largest m that 2^m is smaller than n. m is the answer.
In the discussion of the question, there is no mention the devisors are different from one another. If there is such requirement than the solution is the largest 2*3*5*7... all primes that maintain a value less than n.
