Consider the Diophantine equation $$\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=1$$ I'll give you the solutions sets $(a_1,a_2,\ldots,a_n)$ for $n=1$ up to $4$, because beyond $n=4$ the number of solutions seem to explode. (These can be found quite easily by considering the smallest value of the $n$ variables and seeing just how large/small it can be, and then doing the same for the next smallest variable, and so on.) I will not consider permutations of each set as a separate solution set. $$\begin{align} n=1:&~~~(1)\\ n=2:&~~~(2,2)\\ n=3:&~~~(3,3,3)~(2,3,6)~(2,4,4)\\ n=4:&~~~(4,4,4,4)~(3,3,4,12)~(3,3,6,6)~(3,4,4,6)~(2,3,7,42)\\&~~~(2,3,8,24) ~(2,3,9,18)~(2,3,10,15)~(2,3,12,12)~(2,4,5,20)\\ &~~~(2,4,6,12)~(2,4,8,8)~(2,5,5,10)~(2,6,6,6) \end{align} $$ Denote by $a_n$ the number of solution sets we have for the Diophantine equation above for $n$ variables. So we have $$a_n=1,~~a_2=1,~~a_3=3,~~a_4=14$$ (Thanks for the correction @player3236). And I suppose you could include $a_0=0$ if you really wanted, although I'm not sure how much sense that makes. My question is:
Is there a formula that tells us the number of solutions sets for the Diophantine equation above? If so, how can I find and prove it?
Thank you for your help.
hard, more
, which says that we don't know the exact values for $n \ge 9$. $\endgroup$