How many tuples of {$a, b, c, ...$} satisfy $abc... \leq n$? Let $n$ and $k$ be positive integers.
Let $a, b, c, ...$ be $k$ positive integers such that $abc... \leq n$.
How many tuples of {$a, b, c, ...$} satisfy the inequality?
Note that the tuples {$a=1, b=2$} and {$a=2, b=1$} are two different tuples.
For $k = 1$, the answer is $n$.
So for $k = 2$, the answer is $\sum_{i=1}^n floor(\frac{n}{i})$ also $\sum_{i=1}^n d(i)$, where $d$ = number of divisors. http://oeis.org/A006218
 A: Even for $k=2$ this is a hard problem; the first few terms (for $n=1\ldots 10$) would be
$$ \{ 1,3,5,8,10,14,16,20,23,27\}
$$
While no closed form exists in terms of familiar functions, the large-$n$ behavior is a well-studied problem (the Dirichlet divisor problem) with the result that it goes like
$$
n(\log n + 2 \gamma -1) + O(\sqrt{n})
$$
A: Let's consider the case in which the product is equal to $q$.
If $q$ has the prime factorization
$$
q = p_{\,1} ^{\,k_{\,1} } p_{\,2} ^{\,k_{\,2} }  \cdots p_{\,h} ^{\,k_{\,h} } 
$$
then a divisor $a_j$ of $q$ shall have a prime factorization where each exponent ($x_{j,l}$) is not greater than the corresponding exponent of $p_l$ for $q$.
So we have that
$$
a_{\,1} a_{\,2}  \cdots a_{\,m}  = q\quad  \Rightarrow \quad \left\{ \matrix{
  0 \le x_{\,u,v}  \in Z \hfill \cr 
  x_{\,1,1}  + x_{\,1,2}  +  \cdots  + x_{\,1,m}  = k_{\,1}  \hfill \cr 
  x_{\,2,1}  + x_{\,2,2}  +  \cdots  + x_{\,2,m}  = k_{\,2}  \hfill \cr 
  \quad \quad  \vdots  \hfill \cr 
  x_{\,h,1}  + x_{\,h,2}  +  \cdots  + x_{\,h,m}  = k_{\,h}  \hfill \cr}  \right.
$$
The number of solutions to each line is the number of [weak compositions][1] of $k_j$ into $m$ parts, i.e.
$$
\left( \matrix{
  k_{\,j}  + m - 1 \cr 
  m - 1 \cr}  \right)
$$
I do not see how we can formulate the product of the binomials above in a compact way.
