I posted the question on MSE here but it did not get any answer.
Consider $$S(n)=\left\{(a_1 ,a_2,a_3, \dots, a_n)\mid a_1\le a_2\le\cdots\le a_n, \; \sum_{r=1}^{n}\frac{1}{a_r} = 1\right\} \subset \mathbb{Z}_+^n$$
Now let $|S(n)|$ denote the cardinality (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 $$S^*(n)=\left\{(a_1 ,a_2,a_3, \dots, a_n) \mid a_1 < a_2 < \cdots < a_n, \; \sum_{r=1}^{n}\frac{1}{a_r} =1\right\}\subset \mathbb{Z}_+^n$$ i.e. the $a_i$ are pairwise distinct. And let its cardinality 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?
I believe that there should be more terms on the right.
