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.