Timeline for How to prove the identity $L(2,(\frac{\cdot}3))=\frac2{15}\sum\limits_{k=1}^\infty\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}$?
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19 events
| when toggle format | what | by | license | comment | |
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| Dec 9 at 2:04 | answer | added | Zacky | timeline score: 4 | |
| Sep 25 at 1:58 | vote | accept | Zhi-Wei Sun | ||
| Sep 25 at 1:01 | answer | added | CarP24 | timeline score: 0 | |
| Dec 21, 2022 at 1:19 | answer | added | pisco | timeline score: 11 | |
| Jun 18, 2018 at 9:55 | comment | added | Nemo | Alternative form $$ \int_0^{\pi/3}\frac{\left(2-\sqrt{3} \sin y\right) (y-\sin y\cos y)}{\sin ^3y \sqrt{3-2 \sqrt{3} \sin y}}dy=\frac{5}{4}L(2,\chi) $$ | |
| Jun 17, 2018 at 19:39 | comment | added | j.c. | The paper of Kh. & T. Hessami Pilehrood, cited in the comments above can be found here: combinatorics.org/ojs/index.php/eljc/article/view/v18i2p35 . In their notation, $K$ is the constant of interest in this question. The result is proved on page 10 after Corr. 4, using the following identity involving Hurwitz zeta functions: $9K = \zeta(2,1/3)-\zeta(2,2/3)$. | |
| Jun 17, 2018 at 19:18 | comment | added | Sylvain JULIEN | A naive question: have you tried to introduce additional continuous parameters in your RHS before differentiating with respect to these new variables? | |
| Jun 17, 2018 at 15:14 | comment | added | Johannes Trost | $R=\frac{1}{12}(-4 \ _4F_3(1,1,1,1;\frac{4}{3},\frac{3}{2},\frac{5}{3};-\frac{1}{4}) + 15 \ _4F_3(1,1,1,2;\frac{4}{3},\frac{3}{2},\frac{5}{3};-\frac{1}{4}))$ | |
| Jun 17, 2018 at 15:09 | comment | added | Johannes Trost | @Will Sawin: Indeed, Mathematica gives for the sum (1): $\frac{8}{15} \ _4F_3(\frac{1}{2},1,1,2;\frac{5}{4},\frac{3}{2},\frac{7}{4};\frac{3}{4})$. | |
| Jun 17, 2018 at 14:33 | comment | added | Zhi-Wei Sun | Let $$R=\sum_{k=1}^{8000}\frac{(15k-4)(-27)^{k-1}}{k^3\binom{2k}k^2\binom{3k}k}\ \text{and}\ \ S=\frac{2}{15}\sum_{k=1}^{8000}\frac{48^k}{k(2k-1)\binom{4k}{2k}\binom{2k}k}.$$ Then we may use Mathematica to check that $|R/S-1|<10^{-1000}$. | |
| Jun 17, 2018 at 14:32 | comment | added | Will Sawin | Both sides should be periods and it might be possible to directly compare the motives and show they are isomorphic. | |
| Jun 17, 2018 at 14:23 | comment | added | Zhi-Wei Sun | In 2010 I conjectured that $$L\left(2,\left(\frac{\cdot}3\right)\right)=\sum_{k=1}^\infty\frac{(15k-4)(-27)^{k-1}}{k^3\binom{2k}k^2\binom{3k}k}$$ which was confirmed by Kh. Hessami Pilehrood and T. Hessami Pilehrood [Electron. J. Combin. 18(2012), #P35]. Using this, we can check (1) numerically. | |
| Jun 17, 2018 at 14:20 | comment | added | Will Sawin | The ratio between the $k+1$st and $k$th coefficient is $\frac{ 48 (2k+2)^2 (2k+1)^2 (k+1)^2 k (2k-1) } { (k+1) (2k+1) (4k+1)(4k+2) (4k+3) (4k+4) (2k+1)(2k+2)} =\frac{ 24 (k+1) k (2k-1) } { (4k+1)(4k+2) (4k+3) } = \frac{3 (k+1) (k) (k-1/2)} { 4(k+1/4) (k+1/2) (k+3/4)} $ making this a special value at $3/4$ of a rank $3$ hypergeometric function - something like $\frac{2}{15} {}_4 F_3 ( 1,0,-1/2,1 ; 1/4,1/2,3/4; 3/4)$. | |
| Jun 17, 2018 at 12:43 | comment | added | Neil Strickland | Although it is only peripherally relevant to the current question, I would strongly urge you to distribute your conjectures in machine-readable form as well as in LaTeX. You could add Mathematica files to your home page, or as supplementary files for your arxiv submissions, for example. | |
| Jun 17, 2018 at 11:13 | comment | added | Gil Kalai | Fascinating question! A tempting possibility: a polymath project... | |
| Jun 17, 2018 at 10:35 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
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| Jun 17, 2018 at 10:30 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
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| Jun 17, 2018 at 3:05 | history | edited | Zhi-Wei Sun | CC BY-SA 4.0 |
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| Jun 17, 2018 at 3:00 | history | asked | Zhi-Wei Sun | CC BY-SA 4.0 |