$\{b_n\}_{n\geq0}$ is a number sequence satisfying the following condition: \begin{equation} b_{m}=\sum_{r=0}^m\sum_{h=0}^r\left(\frac{m!}{(m-r)!(r-h)!h!}\right)^2b_{m+h-r}b_{r},~\forall m\in\mathbb{N}. \end{equation} If $\lim_{n\to\infty}b_n=0$, can we conclude that only finitely many of $\{b_n\}_{n\geq0}$ are nonzero? A sequence that satisfies the above condition is \begin{equation} \{b_0,b_1,b_2,b_3,\ldots\}=\{0,1,-\frac{1}{2} \left(7+\sqrt{33}\right),\frac{1}{2} \left(46+9 \sqrt{33}-\sqrt{2089+360 \sqrt{33}}\right),\ldots\}. \end{equation}
If $\{b_n\}_{n\geq0}$ satisfies the following less complicated condition: \begin{equation} b_{m}=\sum_{r=0}^m\sum_{h=0}^r\left(\frac{m!}{(m-r)!(r-h)!h!}\right)b_{m+h-r}b_{r},~\forall m\in\mathbb{N}. \end{equation} The question is easy, because we can conclude that \begin{equation} \left(\sum_{p=0}^m\binom{m}{p}b_p\right)^2=\left(\sum_{p=0}^m\binom{m}{p}b_p\right),~\forall m\in\mathbb{N}. \end{equation} So \begin{equation}\label{bm01} b_m=-\sum_{p=0}^{m-1}\binom{m}{p}b_p\text{ or }1-\sum_{p=0}^{m-1}\binom{m}{p}b_p,~\forall m\in\mathbb{N}. \end{equation} And note that $b_0=0$ or $1$, so we have $b_m\in\mathbb{Z}$, $\forall m\in\mathbb{N}$.
Moreover, what if we replace the square $2$ to any other positive integers?
