What is the number of ways you can build a sequence of positive integers with each number divisible by all the previous numbers? I am looking for a solution to count the number of ways I can create a sequence of $n$ numbers, repetition allowed, where each number is divisible by all the previous numbers in the sequence, and the sequence is bounded.
For example, what is the number of ways you can create a sequence of 6 numbers $a_1$, $a_2$, $a_3$, $a_4$, $a_5$, $a_6$, all positive numbers, where $a_1 = 1$ and $a_6$ = 2000, and for each $2 \leq n \leq 6$, we have that $a_n$ is divisible by $a_{n-1}$?
I was thinking about using Stirling numbers of the second kind, but I am not getting anywhere with this idea.
 A: Suppose you have $a_n$ and $a_1$ then dividing everything by $a_1$ we see that this is equivalent to going from $1$ to a number $m.$ Lets say that $m=p_1^{\alpha _1}\cdots p_k^{\alpha _k}$ and just extract the exponents (the primes are not important) and you get $(\alpha _1,\cdots ,\alpha _k).$ Now you want to construct $k$ increasing chains from $0$ to $\alpha _k$ and so this is equivalent of having $n-1$ numbers $a_{k,1},\cdots a_{k,n-1}$ such that their sum is $\alpha _k.$ Using stars and bars you probably can get that this number would be $$\prod _{i=1}^k\binom{\alpha _i+n-2}{n-2}.$$
Added later: I came across this paper in which this problem, Thm 3.10 in the paper, can be embedded. So a natural generalization of this problem is in the context of mixed Stirling numbers.
A: If you are given $a_1$ and $a_6$ you should divide them and factor the ratio.  In this case  the ratio is $2000=2^4\cdot 5^3$.  If you allow successive numbers to be equal, you just need to distribute the four factors of $2$ and three factors of $5$ into the five bins of gaps between the numbers.  Stars and bars tells you how many ways to do this-there are ${4+5-1 \choose 5-1}={8 \choose 4}=70$ ways to distribute the $2$s and ${7 \choose 4}=35$ ways to distribute the $5$s, so $70\cdot 35=2450$ ways overall.  If you don't allow successive numbers to be equal, there are many fewer.  You have $7$ prime factors which you have to distribute into $5$ factors.
