Use the pumping lemma to show it's not regular
Suppose that $B$ were regular. According to the pumping lemma, there exists an integer $p \geq 1$ such that every word $w \in B$ of length at least $p$ has a decomposition $w = xyz$, with $|xy| \leq p$ and $y \neq \epsilon$, such that $xy^tz \in B$ for all $t \geq 0$.

Pick some Fibonacci number $F_n$ such that (1) $F_n \geq p$ and (2) $F_{n+1} - F_n &gt; p$. Then $1^{F_n} \in B$ has length at least $p$, and so it can be decomposed as $1^{F_n} = xyz$, where $|xy| \leq p$ and $y \neq \epsilon$, and $xy^2z \in B$. But $xy^2z = 1^m$, where $F_n &lt; m \leq F_n + p &lt; F_{n+1}$.