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jrsousa2 2020-09-12 19:39

The prime counting function conjecture
Don't swallow me alive for this, but I was wondering if somebody can be kind enough to test this conjecture with a large enough say [TEX]x>50[/TEX] using a powerful computer:

[TEX]\pi(x) \sim -16\sum _{h=1}^{\infty}\frac{x^{2h+1}}{2h+1}\sum _{i=1}^h \log\zeta(2i)\sum _{v=i}^{h}\frac{(-1)^{h-v}(4\pi )^{2h-2v}}{\zeta(2v-2i)(2h+2-2v)!} \text{, if }x\text{ is sufficiently large.}[/TEX]

If you can let me know, and am willing to provide you with the typed out formula to be used in your math software.
Notice this formula assumes 1 is not a prime (as it should.)

The reasoning behind it can be found [URL=""]here[/URL].

Batalov 2020-09-12 19:55

The reasoning behind posting here on this forum can be found here:
[QUOTE=jrsousa2;516410]If you know how one deletes their account here, let me know. It will be my pleasure to leave this forum and its forumers, only non-sense and trolls and BS from you idiots.
This site is rather abusive too, disabling us to delete topics and posts.
I'm fed up and want out. You people suck, as does this forum.

As they say, when you hate a place, shrug the dust of your shoes, leave and never go back. Why did I come back.[/QUOTE]
Very consistent.

Dr Sardonicus 2020-09-12 21:33

I question the use of the term "conjecture," which, as explained [url=]here[/url], implies the statement is important and worthy of study. The term "claim" appears to be more appropriate. Offhand, I see no reason to consider it, even if correct, as anything other than a curiosity.

The claim is also poorly stated. The phrase "if x is sufficiently large" is entirely superfluous. The definition of asymptotic equality involves the limit as x increases without bound, or tends to plus infinity.

Your request indicates to me that your series is difficult to evaluate. There are already expressions proven to be asymptotically equal to the prime-counting function, such as li(x) and x/log(x), which are easy to evaluate.

CRGreathouse 2020-09-13 04:04

[QUOTE=jrsousa2;556822][TEX]\pi(x) \sim -16\sum _{h=1}^{\infty}\frac{x^{2h+1}}{2h+1}\sum _{i=1}^h \log\zeta(2i)\sum _{v=i}^{h}\frac{(-1)^{h-v}(4\pi )^{2h-2v}}{\zeta(2v-2i)(2h+2-2v)!} \text{, if }x\text{ is sufficiently large.}[/TEX][/QUOTE]

What are you taking the log of? If it's just the zeta, I don't get convergence. If it's the whole inner sum, what branch are you taking?

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