Timeline for $2$-adic valuations: a tale of two $q$-series
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 26, 2021 at 13:39 | vote | accept | T. Amdeberhan | ||
| Sep 25, 2021 at 18:41 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
some lead
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| Sep 25, 2021 at 18:32 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
more accurate
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| Sep 25, 2021 at 17:06 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 386 characters in body
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| Sep 25, 2021 at 11:32 | comment | added | Max Alekseyev | For an odd $m$, we have $2\sigma(m)\equiv 2\tau(m)\pmod4$. | |
| Sep 25, 2021 at 5:13 | comment | added | მამუკა ჯიბლაძე | Sorry, one more question: in the last display, should not it be $\sigma$ rather than $\tau$? | |
| Sep 24, 2021 at 18:14 | comment | added | მამუკა ჯიბლაძე | @GHfromMO Max I see, thanks | |
| Sep 24, 2021 at 16:57 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added some lead ; added 7 characters in body
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| Sep 24, 2021 at 16:48 | comment | added | Max Alekseyev | @მამუკაჯიბლაძე: Perhaps, the easiest way is to think of the coefficient 4 as $x$ satisfying $x^2\equiv 0$. Then $$\prod_i (1+xc_i) \equiv 1 + x\sum_i c_i.$$ | |
| Sep 24, 2021 at 16:18 | comment | added | GH from MO | მამუკა ჯიბლაძე: $4$ times $4$ is divisible by $8$. Of course the congruence is meant in the ring $\mathbb{Z}[[q]]$. | |
| Sep 24, 2021 at 16:06 | comment | added | მამუკა ჯიბლაძე | How do you pass from the first congruence to the second? | |
| Sep 24, 2021 at 16:03 | history | edited | Max Alekseyev | CC BY-SA 4.0 |
added 12 characters in body
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| Sep 24, 2021 at 15:26 | history | answered | Max Alekseyev | CC BY-SA 4.0 |