Timeline for Is there a simple expression for $\sum_{k =1}^{(p-3)/2} \frac{1\cdot 3\cdots (2k-1)}{2\cdot 4 \cdots 2k\cdot(2k+1)} \bmod p$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 10, 2022 at 22:15 | comment | added | Zhi-Wei Sun | Note that the two congruences stated above already have the modulus $p^2$. | |
| Aug 10, 2022 at 4:53 | vote | accept | matt stokes | ||
| Aug 10, 2022 at 3:45 | comment | added | David E Speyer | Note that a prime such that $\tfrac{2^{p-1}-1}{p} \equiv 0 \bmod p$ is called a Wieferich prime en.wikipedia.org/wiki/Wieferich_prime . Only two such are known, 1093 and 3511, and there are very few results about them. | |
| Aug 10, 2022 at 1:38 | comment | added | Zhi-Wei Sun | The result indicates that it is natural to let $k$ start from $0$ rather than from $1$. Please note the infinite series for $\pi/2$. | |
| Aug 10, 2022 at 1:00 | comment | added | Zhi-Wei Sun | Of course, $$\sum_{k=1}^{(p-3)/2}\frac{\binom{2k}k}{(2k+1)4^k}=\sum_{k=0}^{(p-3)/2}\frac{\binom{2k}k}{(2k+1)4^k}-1.$$ | |
| Aug 9, 2022 at 22:22 | history | answered | Zhi-Wei Sun | CC BY-SA 4.0 |