# 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.

Can you explain further what you mean by " $\gg \log^k n$ for every $k$"? If this means $(\log n)^k$, is there some upper limit on $k$, perhaps based on primorial numbers? – Henry Jun 19 '11 at 23:33
@Henry: I don't understand what you are asking. Although I will guess that the problem is that the OP should not use the same $k$ to denote both the $k^{th}$ prime and the growth condition concerning $\log n$. Specifically, the last line should read "I suspect that this number is $\gg_l \log^l n$ for every $l$, and $\dots$" – Eric Naslund Jun 20 '11 at 2:39
"number of divisors" means "number of distinct divisors" ---- what sense would it make to say a number has 17 divisors, only 11 of which are distinct? And $2\times3\times5\times7=210$ has 16 divisors, but the smaller number $2^3\times3\times5$ also has 16 divisors, and $2^2\times3^2\times5=180$ has 18 divisors. – Gerry Myerson Feb 16 '15 at 6:35