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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.

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3 might have information. – Qiaochu Yuan Jun 19 '11 at 22:51
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

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.

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"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

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