Egyptian Fractions with Restrictions

Let $T_o(k)$ denote the number of solutions of $\sum_{i=1}^k\frac 1{x_i}=1$ in odd numbers $1<x_1<x_2<...<x_k$. It is clear that $T_o(2k)=0$. For distinct primes $p_1, p_2,..., p_t$, let $S(p_1, p_2,..., p_t)=\{p_1^{α_1}...p_t^{α_t}\mid α_i\in \mathbb{N}_0, i=1,2,..., t}$. Let $T_k(p_1,..., p_t)$ be the number of solutions $\sum_{i=1}^{k}\frac 1{x_i}=1$ with $1<x_1<x_2<...<x_{k}$ and $x_i\in S(p_1, p_2,..., p_t)$. It is clear that if $T_k(p_1,..., p_t)\not= 0$ for some $k$, then the inverse sum of all elements $s_j>1$ in $S(p_1, p_2,..., p_t)$ is more than 1. In this paper we study $T_o(k)$ and $T_k(p_1,..., p_t)$. Three of our results are: 1) $T_o(2k+1)\ge (\sqrt 2)^{(k+1)(k-4)}$ for all $k\ge 4$; 2) if the inverse sum of all elements $s_j>1$ in $S(p_1, p_2,..., p_t)$ is more than 1, then $T_k(p_1,..., p_t)\not= 0$ for infinitely many $k$ and the set of these $k$ is the union of finitely many arithmetic progressions; 3) there exists two constants $k_0=k_0(p_1,..., p_t)>1$ and $c=c(p_1,..., p_t)>1$ such that for any $k>k_0$ we have either $T_k(p_1,..., p_t)= 0$ or $T_k(p_1,..., p_t)>c^k$.

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