Timeline for Number of prime divisors with multiplicity in a sum of Gaussian binomial coefficients
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2019 at 21:05 | answer | added | Kasper Andersen | timeline score: 2 | |
| Apr 25, 2019 at 19:00 | comment | added | Mare | @MaxAlekseyev I edited it. | |
| Apr 25, 2019 at 18:59 | history | edited | Mare | CC BY-SA 4.0 |
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| Apr 25, 2019 at 7:38 | history | edited | Mare | CC BY-SA 4.0 |
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| Apr 25, 2019 at 7:02 | comment | added | Mare | @T.Amdeberhan I forgot adding the condition $a_t^q+1 \geq q^{t+1}$, which for $t=2$ is $a_2^q+1 \geq q^3$ and makes it look like it does not work for $t=2$. | |
| Apr 25, 2019 at 6:55 | comment | added | Mare | @T.Amdeberhan I modified the question, sorry it was late when I asked it. | |
| Apr 25, 2019 at 6:54 | history | edited | Mare | CC BY-SA 4.0 |
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| Apr 25, 2019 at 6:53 | history | undeleted | Mare | ||
| Apr 25, 2019 at 6:53 | history | deleted | Mare | via Vote | |
| Apr 25, 2019 at 5:38 | history | edited | Greg Martin | CC BY-SA 4.0 |
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| Apr 25, 2019 at 1:18 | comment | added | T. Amdeberhan | $a_2^q+1=q+4$, so you can choose $q$ so that $q+4$ has more than $3$ factors. I don't understand. | |
| Apr 25, 2019 at 0:38 | history | edited | Mare | CC BY-SA 4.0 |
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| Apr 25, 2019 at 0:17 | history | asked | Mare | CC BY-SA 4.0 |