Timeline for Does the average primeness of natural numbers tend to zero?
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Dec 1, 2021 at 5:51 | comment | added | swami | @NilotpalKantiSinha - This may be interesting for you. Have a look at this question. There is a definition of "compositeness" of a number. Here, the value approaches zero as numbers become more prime-like, and approaches infinity as they become more composite. | |
| Apr 8, 2019 at 19:40 | comment | added | lcv | I was mislead by the notation but in the linked answer it is estimated that the integrated version of $s_n$ goes like $n^2 (\log n)^\alpha$ where $\alpha$ is actually a (horrible) negative number. This matches @AsymptotiacK answer. | |
| Apr 8, 2019 at 14:58 | history | became hot network question | |||
| Apr 8, 2019 at 14:17 | comment | added | user44143 | I'm using Mathematica on a $750 Windows machine: f[n_] := 2 StandardDeviation[Divisors[n]] Sqrt[1 - 1/Length[Divisors[n]]]/(n - 1); uniques[k_] := (Table[f[n], {n, 2, k}] // Union // Length) + PrimePi[k]; uniques[10000000] | |
| Apr 8, 2019 at 14:14 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 4.0 |
edited title
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| Apr 8, 2019 at 14:09 | answer | added | Alexander Kalmynin | timeline score: 23 | |
| Apr 8, 2019 at 14:09 | comment | added | Nilotpal Kanti Sinha | @MattF. Can you share your code? I would implement a similar code at my end. I have written one in Sagemath which is not very efficient for checking $f(m) = f(n)$ and has only verified till 12000 in half a day's run. Also what hardware are you using? | |
| Apr 8, 2019 at 14:07 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 4.0 |
edited title
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| Apr 8, 2019 at 14:03 | comment | added | user74900 | @Nilos thank you, it is indeed clearer. | |
| Apr 8, 2019 at 14:02 | comment | added | Nilotpal Kanti Sinha | @AknazarKazhymurat I have reworded that line. Hope it is clearer now? | |
| Apr 8, 2019 at 14:01 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 4.0 |
added 35 characters in body
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| Apr 8, 2019 at 14:00 | comment | added | user44143 | I have verified that $f$ is injective over composites less than 10,000,000. | |
| Apr 8, 2019 at 13:56 | comment | added | user74900 | "...I wanted to have a continuous function...". In what topology is $f$ continuous? If you put discrete topology on natural numbers, then any function is continuous so you probably have something else in mind. | |
| S Apr 8, 2019 at 11:11 | history | suggested | LeechLattice |
There's nothing to do with statistics.
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| Apr 8, 2019 at 10:30 | review | Suggested edits | |||
| S Apr 8, 2019 at 11:11 | |||||
| Apr 8, 2019 at 10:28 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 4.0 |
edited title
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| Apr 8, 2019 at 10:27 | comment | added | Nilotpal Kanti Sinha | @lcv No $s_n$ doesn't grow faster than $n$. What are you looking at? | |
| Apr 8, 2019 at 10:18 | comment | added | lcv | From the linked question it seems that $s_n$ grows faster than $n$ so that $f(n)$ doesn't go to zero. | |
| Apr 8, 2019 at 10:00 | history | edited | Nilotpal Kanti Sinha | CC BY-SA 4.0 |
added 41 characters in body
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| Apr 8, 2019 at 9:51 | history | asked | Nilotpal Kanti Sinha | CC BY-SA 4.0 |