Let $a_1 \le a_2 \le ... \le a_n$ be positive integers. I'm looking for a closed formula of the number $f(a_1,...,a_n)$ of all (ordered) tuples $(x_1,...,x_n)$ of positive integers statisfying $x_1 + ... + x_i \le a_i\quad\quad(i=1,...,n).$ $f$ has the recursion $f(a_1,...,a_n) = \sum_{k=1}^{a_1}f(a_2-k,...,a_n-k)$ that also yields the sum expression $f(a_1,...,a_n) = \sum_{x_1=1}^{a_1}\sum_{x_2=1}^{a_2-x_1}...\sum_{x_n=1}^{a_n-x_1-...-x_{n-1}}1$ The only case I know of is for $a_1=...=a_n$ when one obtains $\binom{a_n}{n}$ what is also an upper bound for $f(a_1,...,a_n)$.
Edit: Using the transformation $y_i = x_1 + ... + x_i$ one finds that the searched $f$ equals the number of ordered tuples $1 \le y_1 < y_2 < ... < y_n \le a_n$, satisfying the additional condition $y_i \le a_i$ for all $i$.
It's classical that the number of ordered tuples $1 \le y_1 <...