Sum Equals Product Sum Equal Product
There are many articles in re this issue on the Web, although many are restricted to special cases (e.g., John Cook’s sum of tangents = product of tangents), and even some involving mixed numbers. 
There is one excellent article in re: sum equal product written by two Polish mathematicians which is replete with wonderful proofs that I am not sophisticated enough to understand. (Kurlandchick & Nowicki, Nov `98)
Actually, my question is quite straightforward. Is there a formula or some kind of algorithm that can show which sums of positive, unequal integers equal their products?
Falco
 A: Let the positive, unequal integers be $a_1 < a_2 < \cdots < a_k$ with $a_1+a_2+\cdots+a_k = a_1a_2\cdots a_k = n$.  Obviously if $k = 1$ all positive integers work; suppose from now on that $k > 1$.  Note that $a_k \ge k$.  Then we have $a_1 + a_2 + \cdots + a_k < ka_k$, while $a_1a_2\cdots a_k \ge (k-1)!a_k$.  So we must have $ka_k > (k-1)!a_k$, which means that $k > (k-1)!$.  This means $k \le 3$.  If $k = 2$, then we have $a_1 + a_2 = a_1a_2$, or $\frac{1}{a_1}+\frac{1}{a_2} = 1$.  This can easily be seen to have only the solution $(2,2)$, which doesn't satisfy our hypothesis that the $a_i$ be unequal.
The case $k = 3$ is then the only tricky case.  If $a_1 > 1$, then we have $a_1+a_2+a_3 < 3a_3$ and $a_1a_2a_3 \ge 6a_3$.  So $a_1 = 1$.  We then must find solutions to $1 + a_2 + a_3 = a_2a_3$.  This can be rewritten as $(a_2-1)(a_3-1) = 2$, which as $a_2 < a_3$ are integers immediately gives $a_2 = 2, a_3 = 3$.
In summary, all possible solutions are $\{n\}$ for all positive integers $n$ and $\{1,2,3\}$.
History: this is definitely a classic problem; see for example 2006 USA Mathematical Olympiad problem #4 (pdf, problem is on second page), which is your problem with several restrictions removed.
