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What you observed resonates with the Von Staudt–Clausen theorem.

Indeed, by this theorem, the denominator of $(2k)\zeta(1-2k)=-B_{2k}$ equals $\prod_{p-1\mid 2k} p$. In particular, the numerator is odd, hence the denominator of $\zeta(1-2k)$ is divisible by $2\cdot 2\cdot 3=12$ for $k$ odd, and by $2\cdot 2\cdot 2\cdot 3\cdot 5=120$ for $k$ even.

Added. I misunderstood (misread) the question, and I thought the OP was only interested in the above divisibility relation. I certainly don't know how to prove that the denominator of $\zeta(1-2k)$ is infinitely often $12$ for $k$ odd and infinitely often $120$ for $k$ even. This seems and interesting but difficult question.

What you observed resonates with the Von Staudt–Clausen theorem.

Indeed, by this theorem, the denominator of $(2k)\zeta(1-2k)=-B_{2k}$ equals $\prod_{p-1\mid 2k} p$. In particular, the numerator is odd, hence the denominator of $\zeta(1-2k)$ is divisible by $2\cdot 2\cdot 3=12$ for $k$ odd, and by $2\cdot 2\cdot 2\cdot 3\cdot 5=120$ for $k$ even.

Added. I misunderstood (misread) the question, and I thought the OP was only interested in the above divisibility relation. I certainly don't know how to prove that the denominator of $\zeta(1-2k)$ is infinitely often $12$ for $k$ odd and infinitely often $120$ for $k$ even. This seems an interesting but difficult question.

What you observed resonates with the Von Staudt–Clausen theorem.

Indeed, by this theorem, the denominator of $(2k)\zeta(1-2k)=-B_{2k}$ equals $\prod_{p-1\mid 2k} p$. Let us assume that the numerator is coprime with $2k$, then we get that the denominator of $\zeta(1-2k)$ is divisible by $2\cdot 2\cdot 3=12$ for $k$ odd, and by $2\cdot 2\cdot 2\cdot 3\cdot 5=120$ for $k$ even.

In particular, if all prime factors $p\mid 2k$ satisfy $p-1\mid 2k$, then the above conclusion is true. For example, for $k=2^a3^b$, the conclusion is true.

What you observed resonates with the Von Staudt–Clausen theorem.

Indeed, by this theorem, the denominator of $(2k)\zeta(1-2k)=-B_{2k}$ equals $\prod_{p-1\mid 2k} p$. In particular, the numerator is odd, hence the denominator of $\zeta(1-2k)$ is divisible by $2\cdot 2\cdot 3=12$ for $k$ odd, and by $2\cdot 2\cdot 2\cdot 3\cdot 5=120$ for $k$ even.

Added. I misunderstood (misread) the question, and I thought the OP was only interested in the above divisibility relation. I certainly don't know how to prove that the denominator of $\zeta(1-2k)$ is infinitely often $12$ for $k$ odd and infinitely often $120$ for $k$ even. This seems and interesting but difficult question.

What you observed resonates with the Von Staudt–Clausen theorem.

Indeed, by this theorem, the denominator of $(2k)\zeta(1-2k)=B_{2k}$ equals $\prod_{p-1\mid 2k} p$. Let us assume that the numerator is coprime with $2k$, then we get that the denominator of $\zeta(1-2k)$ is divisible by $2\cdot 2\cdot 3=12$ for $k$ odd, and by $2\cdot 2\cdot 2\cdot 3\cdot 5=120$ for $k$ even.

In particular, if all prime factors $p\mid 2k$ satisfy $p-1\mid 2k$, then the above conclusion is true. For example, for $k=2^a3^b$, the conclusion is true.

What you observed resonates with the Von Staudt–Clausen theorem.

Indeed, by this theorem, the denominator of $(2k)\zeta(1-2k)=-B_{2k}$ equals $\prod_{p-1\mid 2k} p$. Let us assume that the numerator is coprime with $2k$, then we get that the denominator of $\zeta(1-2k)$ is divisible by $2\cdot 2\cdot 3=12$ for $k$ odd, and by $2\cdot 2\cdot 2\cdot 3\cdot 5=120$ for $k$ even.

In particular, if all prime factors $p\mid 2k$ satisfy $p-1\mid 2k$, then the above conclusion is true. For example, for $k=2^a3^b$, the conclusion is true.