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Let $B_n$ be $n$-th Bernoulli number. And let $N_n$ be the numerator of $B_n$.

For example,

$|N_0| = 1,$ $|N_2| = 1,$ $|N_4| = 1,$ $|N_6| = 1,$ $|N_8| = 1,$ $|N_{10}| = 5.$

Is that $|N_{2k}| > 1$ for $k > 4$ true? If it is true, please tell me the simple proof?

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    A simple glance [here](http://oeis.org/A000367) suggests it is true. – 2017-01-22
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    Which link or reference of the OEIS A000367 page suggests? – 2017-01-22
  • 0
    The OEIS A000367 page gives you the answer. – 2017-01-22

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$|N_{2k}| > 1$ for $4 < k \le 7$.

Let $M_{2k}$ be the numerator of $\frac{N_{2k}}{2k}$.

$|M_{2k}| = \prod_{i}p_i $, where $p_i$ is a irregular prime for $k > 7$.

So $|N_{2k}| \ge |M_{2k}| = \prod_{i}p_i > 1$.