Timeline for On a double sum involving binomial coefficients
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 12, 2023 at 15:40 | comment | added | Iosif Pinelis | Thank you once again! | |
| Jul 12, 2023 at 15:39 | vote | accept | Iosif Pinelis | ||
| Jul 12, 2023 at 11:44 | history | edited | Julius | CC BY-SA 4.0 |
Added all cases
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| Jul 12, 2023 at 11:07 | history | edited | Julius | CC BY-SA 4.0 |
I added the case $n=4m$ and changed all the $l$'s to $m$'s to remove ambiguity. I also made some simplifications to the sums. The case where $n$ is odd I will finish later. Fixed a typo.
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| Jul 12, 2023 at 11:00 | comment | added | Julius | @IosifPinelis I have changed the indices from $l$ to $m$ as requested. I also simplified the sum of binomial coefficients to say $p$ instead of $2p$, but this doesn't actually change the sum itself. WolframAlpha says apparently that this sum is equal to $\frac{1}{2}{4m+2 \choose 2m+1}$, interestingly. | |
| Jul 12, 2023 at 10:58 | history | edited | Julius | CC BY-SA 4.0 |
I added the case $n=4m$ and changed all the $l$'s to $m$'s to remove ambiguity. I also made some simplifications to the sums. The case where $n$ is odd I will finish later.
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| Jul 11, 2023 at 19:56 | comment | added | Iosif Pinelis | Also, $\displaystyle{\sum_{p=0}^l {2l+1 \choose 2p}^2}=\dfrac{2^{4 l+1} \Gamma \left(\frac{1}{2} (4 l+1)+1\right)}{\sqrt{\pi } (2 l+1)!}$. | |
| Jul 11, 2023 at 19:51 | comment | added | Iosif Pinelis | Thank you for your answer. Can you write $n=4m+2$ and $n=4m$ instead of $n=4l+2$ and $n=4l$, since $l$ was already defined to be $\lfloor n/2 \rfloor$? Also, please do finish all the (four?) cases, and it would be good to have a (table?) summary of all the cases. | |
| S Jul 11, 2023 at 19:38 | review | First answers | |||
| Jul 11, 2023 at 20:49 | |||||
| S Jul 11, 2023 at 19:38 | history | answered | Julius | CC BY-SA 4.0 |