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 $1<x_1<x_2<\mathrm{.}\mathrm{.}\mathrm{.}<x_k$. It is clear that $T_o(2k)=0$. For distinct primes $p_1,p_2,\mathrm{.}\mathrm{.}\mathrm{.},p_t$, let $S(p_1, p_2,…, p_t)={p_1^{\alpha_1}…p_t^{\alpha_t}\mid \alpha_i\in $\mathbb{N}$0, i=1,2,…, t}$\mathrm{.}Let$T_k(p_1,…, p_t)$bethenumberofsolutions$\sum{i=1}^{k}\frac 1{x_i}=1$with$1<x_1<x_2<…<x_{k}$and$x_i\in S(p_1, p_2,…, p_t)$. It is clear that if $T_k(p_1,\mathrm{.}\mathrm{.}\mathrm{.},p_t)≠0$ for some $k$, then the inverse sum of all elements $s_j>1$ in $S(p_1,p_2,\mathrm{.}\mathrm{.}\mathrm{.},p_t)$ is more than 1. In this paper we study $T_o(k)$ and $T_k(p_1,\mathrm{.}\mathrm{.}\mathrm{.},p_t)$. 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 $S(p_1,p_2,\mathrm{.}\mathrm{.}\mathrm{.},p_t)$ is more than 1, then $T_k(p_1,\mathrm{.}\mathrm{.}\mathrm{.},p_t)≠0$ for infinitely many $k$ and the set of these $k$ is the union of finitely many arithmetic progressions;
3. there exists two constants $k_0=k_0(p_1,\mathrm{.}\mathrm{.}\mathrm{.},p_t)>1$ and $c=c(p_1,…, p_t)>1$suchthatforany$k>k_0$wehaveeither$T_k(p_1,…, p_t)= 0$or$T_k(p_1,…, p_t)>c^k$.