I have a combinatoric problem: find the number of positive integer solutions $x_i$ s.t.
$$ \begin{cases} x_1+...+x_n = a\\ px_1 \geq x_2\\ px_2 \geq x_3\\ \vdots \\ px_n\geq x_1 \end{cases} $$
in which $a,p$ are constants. The solution is called $P_n$. My approach is to find a recursion formula by letting $x_n$ runs through all possible values, however it does not work since a value of $x_n$ would create more conditions.
Can you help me? Thanks.
This should get you started.
Given $x_1+x_2+...+x_n=a$, lets do a substitution. Often a good technique in such problems is to rerepresent the terms in a form that's easier to count.
Let $a=u_1+u_2+...+u_a$
Where each $u_i$=1. Then we have $(a-1)$ plus signs in the expression.
Pick a single plus sign. Add up all the terms to the left of the plus sign $x_1$ and everything to the right of the plus sign $x_2$. Then we have $a=x_1+x_2$.
How many ways are there of picking such plus signs? $\frac{(a-1)!}{(a-1-1)!(1!).}$,i.e. the number of ways of picking one plus sign from the (a-1) plus signs there.
How many ways of there of adding 3 numbers to get $a$? Pick 2 plus signs instead of one.
Keep in mind commutativity. In some problems x+y=a and y+x=a are the same solution. In others they are two separate solutions. The above technique does not allow 0 to be one of the addends. One way to address this is to put a "+0" at the end of the u's to represent "adding zero". Additional, but similar modifications may be necessary depending on commutatiivity requirements.
What is p, another integer? If it's greater than one, then you have ordered solutions implying eliminating solutions that are the same up to commutativity. This means if you pick $n$ plus signs, you have to divide by $(n+1)!$ to eliminate redundancies.