Skip to main content
6 events
when toggle format what by license comment
Apr 1, 2021 at 13:46 comment added Timothy Chow I don't know if this is how Carlo Beenakker did it, but one way to arrive at his identity is to search the OEIS for "34511 bernoulli"; this will take you to A096053. The OEIS will also tell you that 34511 divides $3^{17}-1$ (see A074477).
Apr 1, 2021 at 6:18 history edited Jens Reinhold CC BY-SA 4.0
Added "even" somewhere where it is highly relevant
Mar 31, 2021 at 16:54 comment added Carlo Beenakker $-\tfrac{1}{3}9^9 B_{18}(1/3)/B_{18}=1871\cdot 34511$
Mar 31, 2021 at 15:45 comment added Will Sawin Similarly $p$ divides $2^{k-1}-1$ if and only if $k$ is congruent to $1$ modulo the order of $2$ mod $p$, which itself divides $p-1$. For $34511$ this is $595$. Thus once $k$ is a solution, anything congruent to $k$ mod $p-1$ is a solution.
Mar 31, 2021 at 15:35 comment added Will Sawin The $p$-divisibility of the Bernoulli number $B_{2k}$ depends only on $2k \mod p-1$ (slightly simplified) so the reason $2^{23}$ shows up in one case and $2678$ shows up in the other should be because of the congruence $2^{23} \equiv 2678 \mod (34511-1)$.
Mar 31, 2021 at 15:16 history asked Jens Reinhold CC BY-SA 4.0