/sci/ - Science & Math » Thread #11489149

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Anonymous Sat 21 Mar 13:40:03 2020 No.11489149 View Reply Original Report

Quoted By: >>11489199 >>11489251 >>11489320

discuss

Anonymous

Anonymous Sat 21 Mar 2020 13:40:44 No.11489151 Report

Quoted By: >>11489199 >>11489320

made a mistake lol, its "p < n"

Anonymous

Anonymous Sat 21 Mar 2020 13:45:23 No.11489159 Report

Quoted By: >>11489173

equivalent to 1/2.log n

Anonymous

Anonymous Sat 21 Mar 2020 13:48:33 No.11489163 Report

Quoted By:

It goes to infinity.

Anonymous

Anonymous Sat 21 Mar 2020 13:51:11 No.11489173 Report

Quoted By: >>11489185

>>11489159
How do you get that?

Anonymous

Anonymous Sat 21 Mar 2020 13:56:31 No.11489185 Report

Quoted By: >>11489245

>>11489173
Prime number theorem. Not sure where he gets the 1/2 from though.

Anonymous

Anonymous Sat 21 Mar 2020 14:04:19 No.11489199 Report

Quoted By:

>>11489149
It not only diverges, it's constantly infinite.
>>11489151
Ah, k.

Anonymous

Anonymous Sat 21 Mar 2020 14:22:13 No.11489245 Report

Quoted By: >>11489247

>>11489185
i got the 1/2 from assuming the primes are normally distributed lol so the average value of p/n is like 1/2 idk

Anonymous

Anonymous Sat 21 Mar 2020 14:23:15 No.11489247 Report

Quoted By:

>>11489245
just to be clear i meant 1/(2logn)

Anonymous

Anonymous Sat 21 Mar 2020 14:23:48 No.11489251 Report

Quoted By: >>11489253

>>11489149
Number of primes less than n ~ n/log(n)

Sum of first n/log(n) primes ~ sum from k=1 to n/log(n) of k log(k)

= log(H(n/log(n)))/n where H is the hyperfactorial function.

Anonymous

Anonymous Sat 21 Mar 2020 14:26:34 No.11489253 Report

Quoted By: >>11489291

>>11489251
even more retarded than my take congrats

Anonymous

Anonymous Sat 21 Mar 2020 14:50:29 No.11489291 Report

Quoted By:

>>11489253
https://www.wolframalpha.com/input/?i=li%28n%5E2%29%2Fn

Anonymous

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Anonymous Sat 21 Mar 2020 15:02:18 No.11489320 Report

Quoted By:

>>11489149
>>11489151
I don't think it converges.
We set $f(n)= \Sigma _{p ~ prime, ~ p<n}\frac{p}{n}$ .
Then, for a prime p, $f(p+1)-f(p)$ should go to one when $p$ goes to infinity.

Anonymous

Anonymous Sat 21 Mar 2020 15:33:48 No.11489410 Report

Quoted By:

Sum p/n is greater than (# prime between n/2 and n)/2 which goes to infinity by prime number theorem

Anonymous

Anonymous Sat 21 Mar 2020 15:35:25 No.11489411 Report

Quoted By:

4, right?

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