All those unit fractions add to 1?

Consider $$S(n)=\{x \mid x=(a_1 ,a_2,a_3 \cdots a_n) \text{ where } \sum_{r=1}^{n}\frac{1}{a_r} =1 \}$$

Now let $|S(n)|$ denote the cardinaly (order) of set $S(n)$.

Thus:

$S(1)= \{(1)\} \implies |S(1)|=1$

$S(2)= \{(2,2)\} \implies |S(2)|=1$

$S(3)= \{(3,3,3) ,(2,3,6) , (2,4,4)\} \implies |S(3)|=3$

And similarly $|S(4)|= 14$.

Now define $$S^*(n)=\{x \mid \ x=(a_1 ,a_2,a_3 \cdots a_n) \text{ where } \sum_{r=1}^{n}\frac{1}{a_r} =1 \quad \text{and } a_i =a_j \Longleftrightarrow i=j \ \}$$ Meaning the elements of each size $n$ should be unique . And let its cardinary be denoted by $|S^*(n)|$.

So I have been able to prove $$|S^{*}(n)| \ge \sum_{k=3}^{n-1}|S(n)|+\frac{n^2 -5n+8}{2}$$ So my question is

Is it possible to come up with a stronger result ? as I believe that there should be more terms on the right.

• First comment: Since you are talking about tuples, you should adjust your definition of the set $S(n)$: $$S(n)=\{x \mid x=(a_1 ,a_2,a_3 \ldots a_n) \text{ where } \sum_{r=1}^{n}\frac{1}{a_r} =1 \text{ and } a_1 \le a_2 \le\ldots\le a_n\}$$ (otherwise e.g. your $S(3)$ contains $(2, 3, 6)$ as well as $(2, 6, 3)$ etc.) – jpvee May 22 '14 at 6:24
• Second comment: Your inequality holds only for $n\ge 2$ since $|S^\ast(1)|=1$ and $|S^\ast(2)|=0$ whereas the corresponding right hand sides are $0+2=2$ and $0+1=1$. – jpvee May 22 '14 at 6:28
• Third comment: The first 8 values of $|S(n)|$ can be found here; the first 8 values of $|S^\ast(n)|$ can be found here. In particular, they show that your lower bound is in fact quite low compared to the actual value of $|S^\ast(n)|$. – jpvee May 22 '14 at 6:31