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 $S(n)$.


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

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

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

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.

  • $\begingroup$ The first sequence is oeis.org/A002966. The OEIS entry has a couple of references but not much information. $\endgroup$ – Douglas Zare Jun 6 '14 at 15:03
  • $\begingroup$ @DouglasZare yes thats the problem , and none talk about the inequity either . $\endgroup$ – Shivam Patel Jun 6 '14 at 15:10

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