Let $u_1=1$ and $u_{n+1}=u_n(u_n+1)$, so that $\sum_{k\geq 1}\frac{1}{u_k+1}$ and $u_k=\lfloor c_0^{2^k}+1\rfloor$ for $k\geq 1$, where $$c_0=\lim u_n^{1/2^n}=1.264085\cdots.$$ Let $a_1<a_2<\cdots$ be any other sequence with $\sum \frac{1}{a_k}=1$. Is it true that $$\liminf a_n^{1/2^n}<c_0=1.264085\cdots?$$

Let $u_1=1$ and $u_{n+1}=u_n(u_n+1)$, so that $\sum_{k\geq 1}\frac{1}{u_k+1}$ and $u_k=\lfloor c_0^{2^k}+1\rfloor$ for $k\geq 1$, where $$c_0=\lim u_n^{1/2^n}=1.264085\cdots.$$ Let $a_1<a_2<\cdots$ be any other sequence with $\sum \frac{1}{a_k}=1$. Is it true that $$\liminf a_n^{1/2^n}<c_0=1.264085\cdots?$$

This is true, and was proved independently by Kamio [Ka25] and Li and Tang [LiTa25].

An earlier interpretation of this question on this site defined $u_1=2$ and $u_{n+1}=u_n^2-u_n+1$ (Sylvester's sequence), which is the same sequence shifted by $1$; we use the phrasing above as more faithful to [ErGr80]. The constant $c_0$ is called the Vardi constant.