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.



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