Timeline for Two conjectural series for $\pi$ involving the central trinomial coefficients
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 26, 2020 at 2:20 | answer | added | Max Alekseyev | timeline score: 4 | |
| Dec 13, 2019 at 22:38 | comment | added | user44191 | That's exactly why I chose those - they separate out the $\log 3$ and $\pi \sqrt{3}$ terms, if I haven't made an arithmetic mistake. | |
| Dec 13, 2019 at 22:33 | comment | added | Zhi-Wei Sun | @user44191 Note that $105k^2-99k+24=(105k-44)(k-1)+10(5k-2)$. So $(1)$ and $(2)$ imply that $$\sum_{k=2}^\infty\frac{(105k^2-99k+24)T_{k-1}}{k^2\binom{2k}k^2(k-1)3^{k-1}}=\frac{9-6\log3}{4}.$$ | |
| Dec 13, 2019 at 17:34 | comment | added | user44191 | The forms of the right-hand sides suggest looking at $\sum_{k = 2}^\infty T_{k - 1} \frac{105k^2 - 109k + 28}{k^2 {{2k}\choose{k}}^2 (k - 1) 3^{k - 1}}$ and $\sum_{k = 2}^\infty T_{k - 1} \frac{105k^2 - 99k + 24}{k^2 {{2k}\choose{k}}^2 (k - 1) 3^{k - 1}}$; have you looked at them? | |
| Dec 13, 2019 at 16:06 | comment | added | L. Milla | How did you find these identities? | |
| Dec 13, 2019 at 14:15 | review | Close votes | |||
| Dec 20, 2019 at 3:01 | |||||
| Dec 13, 2019 at 13:35 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |