How do you calculate the width of the Poset Lattice of Divisors? Let $n = 10800 = 2^43^35^2$
I can find a set of eleven divisors of $n$ such that none divides another:
$$\begin{array}{ccccc}
  &   &   & 2 3^3 & 3^35\\
  &   & 2^23^2 & 23^25 & 3^25^2 \\
  & 2^33  & 2^235 & 235^2 &   \\
 2^4 & 2^35  & 2^25^2 &   &   
\end{array}$$
I have written them in this way to show some structure. The factors of two decrease to the right so no term may divide any term to the right. The factors of three decrease and the factors of five increase going down. So the only chance for a term to divide another is to the left somehow, but since the total number of prime divisors is constantly four this is impossible.
Question number 1) Is it possible to find a larger set of such divisors with no element dividing another?
Question number 2) (the answer implies an answer to #1) 
Given a prime decomposition of $n$ how do I calculate the size of a largest set of divisors such that no element divides another?
$$n = p_1^{\alpha_1}\cdots p_k^{\alpha_k}$$
Clearly this must be a function of the $\alpha_i$ only, but I don't see how do to it for the general case.
 A: I think I see at least how to calculate this at all. here is my answer.
Start with the tuple $$( \alpha_1, \cdots, \alpha_k )$$
Then form a set of the $k$ possible tuples by subtracting one from each entry.
Note the size of this set. 
Continue making new sets by forming all possible tuples by subtracting one from each possible entry--every element of the list should have the same tuple sum at each stage. (These are sets: do not count duplicate elements!)
The maximum length of this list is your answer.
With $n = 10800$ we proceed as follows:
Step 0:$$\{(4,3,2)\}$$
Step 1:$$\{(3,3,2),(4,2,2),(4,3,1)\}$$
Step 2:$$\{(2,3,2),(3,2,2),(3,3,1),(4,1,2),(4,2,1),(4,3,0)\}$$
Step 3:$$\{(1,3,2),(2,2,2),(2,3,1),(3,1,2),(3,2,1),(3,3,0),(4,0,2),(4,1,1),(4,2,0)\}$$
Step 4:$$\{(0,3,2),(1,2,2),(1,3,1),(2,1,2),(2,2,1),(2,3,0),(3,0,2)(3,1,1),(3,2,0),(4,0,1),(4,1,0)\}$$
At step five you have the same number of elements (11) and then it decreases after that until you get down to $(0,0,0)$ at step 10.
This algorithm will work in the general case, although I'm still very interested if there is a better, number theoretic way.
