Set of solutions to a sum I want to know the set of integer solutions to the following sum: 
$$\sum\limits_{n=1}^{k}\frac{n}{a_n}=1$$
For instance, the series 
$a_n=\frac{1}{nk}$ satisfies this for all integer $k$. So far, this is the only solution that I have come up with for a general $k$. 
Here are some solutions that I have for specific values of $k$: 
If $k=2$, $a_1=a_2=\frac{1}{3}$  works. 
If $k=3$, $a_1=a_2=\frac{1}{4}, a_3=\frac{1}{12}$ works. 
If $k=4$, $a_1=a_2=\frac{1}{5}, a_3=\frac{1}{15}, a_4=\frac{1}{20}$ works. 
Are there any generalizations or ways to generate new solutions?  

Yet another generic solution:
since $$\sum_{n=1}^{k-2} \frac{n}{2^{n+1}}=1-\frac{k}{2^{k-1}},$$
we can write the fraction $\dfrac{k}{2^{k-1}}$ in the form
$$\frac{k}{2^{k-1}}=\frac{k-1}{2^{k-1}}+\frac{1}{2^{k-1}} = \color{red}{\frac{k-1}{2^{k-1}}}+\color{darkviolet}{\frac{k}{k\cdot 2^{k-1}}}.$$
Then for given $k$ ($k\ge 3$) we have solution
$$(a_1,a_2,\ldots,a_{k-2},a_{k-1},a_k) = \left(4,8,\ldots, 2^{k-1}, \color{red}{2^{k-1}}, \color{darkviolet}{k\cdot 2^{k-1}}\right).$$
Examples:
$k=3$: $(4,\color{red}{4},\color{darkviolet}{12})$;
$k=4$: $(4,8,\color{red}{8},\color{darkviolet}{32})$;
$k=5$: $(4,8,16,\color{red}{16},\color{darkviolet}{80})$;
$k=6$: $(4,8,16,32,\color{red}{32},\color{darkviolet}{192})$;
$\cdots$ .

In general, the set of solutions is rather wide:
$k=3$:
(2,5,30)
(2,6,18)
(2,7,14)
(2,8,12)
(2,10,10)
(2,12,9)
(2,16,8)
(2,28,7)
(3,4,18)
(3,6,9)
(3,12,6)
(3,30,5)
(4,3,36)
(4,4,12)
(4,8,6)
(5,4,10)
(5,10,5)
(5,40,4)
(6,3,18)
(6,4,9)
(6,6,6)
(6,24,4)
(8,4,8)
(8,16,4)
(10,5,6)
(12,3,12)
(12,12,4)
(14,4,7)
(15,6,5)
(20,10,4)
(30,3,10)
(36,9,4)

