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I was very surprised when I first saw that the product of all primes $p$ such that $p-1|2n,$$p-1\mid 2n,$ is the denominator of Bernoulli number $B_{2n}$.
I was very surprised when I first saw that the product of all primes $p$ such that $p-1|2n,$ is the denominator of Bernoulli number $B_{2n}$.
I was very surprised when I first saw that the product of all primes $p$ such that $p-1\mid 2n,$ is the denominator of Bernoulli number $B_{2n}$.
