# Question about infinitude of $m$-irregular primes

Let $$p>2$$ be a prime, $$\zeta$$ a primitive $$p$$th root of unity, and $$m >0$$ a square-free integer such that $$m \not\equiv 3 \mod 4$$ and $$\gcd(m,p)=1$$. Let $$\chi$$ be the imaginary quadratic character for $$\mathbb{Q}(\sqrt{-m})$$, and $$\omega_p$$ be the Teichmuller character. If $$h$$ is the class number for $$\mathbb{Q}(\zeta + \zeta^{-1}, \sqrt{-m})$$ then we say $$p$$ is $$m$$-irregular if and only if $$p \mid h$$.

It is known that $$p$$ is $$m$$-irregular if and only if $$p$$ is irregular ($$p$$ divides the class number of $$\mathbb{Q}(\zeta)$$), and for $$n = 0, \dots, p-2$$, none of the numerators of $$B_{n+1,\chi}$$ are divisible by $$p$$, where \begin{align*} B_{n+1,\chi} = (4m)^n \sum_{a = 1}^{n+1} \chi(a) B_{n+1}\left(\frac{a}{4m}\right). \end{align*}

I feel like I should be able to prove there are infinitely many $$m$$-irregular by mimicking the proof of the infinitude of regular primes. However, I have reasons to believe that there are finitely many primes such that $$p\mid B_{1,\chi\omega_p^{-2}}$$, which is equivalent to $$p$$ dividing one of the $$B_{n+1, \chi}$$ with $$0\leq n \leq p-2$$.

Here is my question: Is there any reason that there should be finitely many primes satisfying $$p\mid B_{1,\chi\omega_p^{-2}}$$, but somehow there still manages to be infinitely many primes which divide the $$B_{n+1, \chi}$$ $$n = 0, \dots, p-2$$? In my mind if $$p$$ is $$m$$-irregular it will divide the $$B_{n,\chi}$$ randomly as $$p$$ varies.