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 April 5th, 2012, 09:39 AM #1 Senior Member   Joined: Jan 2011 Posts: 560 Thanks: 1 Mertens function and prime counting function Hi everybody, I want you to check this conjecture if it is true. I computed a sieved Mertens function on the basis that the composite number are valued to 1 and the prime number valued to zero. So the sum of the new function M'(n) will be : M'(n)=Sigma (mu(n)*c(n)) with c(n)=1 if c(n) is composite and c(n)=0 otherwise (prime) When n >3607 M'(n) is amost equal to pi(n) (+ or - 3,6%) pi(n) is the prime counting function. mu(n) the mobius function I do not know if it holds after n>10000. If the conjecture is true then we can go forward to something more interesting.
 April 5th, 2012, 09:51 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Mertens function and prime counting function The conjecture is implied by (but does not itself imply) the Riemann Hypothesis. So yes, it's true, subject to the RH. It may be known to be true unconditionally; I'd have to see what the best-known bounds on M are.
April 5th, 2012, 10:22 AM   #3
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Re: Mertens function and prime counting function

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 Originally Posted by CRGreathouse The conjecture is implied by (but does not itself imply) the Riemann Hypothesis. So yes, it's true, subject to the RH. It may be known to be true unconditionally; I'd have to see what the best-known bounds on M are.
It seems to me that you did not read carefully what I said above.
The Mertens function proposed do not take onto account the prime numbers.
You compute the Mertens function ONLY for the composite numbers.
Is M'(n) asymptotic to Pi(n)?
That is my question.
It has nothing to do with Riemann hypothesis.
Nothing!

 April 5th, 2012, 11:27 AM #4 Senior Member   Joined: Jan 2011 Posts: 560 Thanks: 1 Re: Mertens function and prime counting function Between n=10000 and n=50000 M'(n)=pi(n)+c c= min -2.98% c= max 2.76% I can not check for n>50000. It maybe holds when n is infinite. I'm not sure. I need confirmation ....
 April 5th, 2012, 11:32 AM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Mertens function and prime counting function I answered your question. It doesn't bother me if you can't see the connection between it and the Riemann hypothesis.
April 5th, 2012, 11:38 AM   #6
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Re: Mertens function and prime counting function

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 Originally Posted by CRGreathouse I answered your question. It doesn't bother me if you can't see the connection between it and the Riemann hypothesis.
WHO ARE YOU?????
YOU ARE A PROGRAMMER NOT A MATHEMATICIAN!!!!
SO STOP TREATING OTHERS WITH DEEP CONTEMPT.

GOOD BYE!!!

 April 5th, 2012, 11:40 AM #7 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Mertens function and prime counting function Do you really want me to spoil it? The proof is so simple that you could surely find it on your own.
 April 5th, 2012, 11:50 AM #8 Senior Member   Joined: Nov 2011 Posts: 595 Thanks: 16 Re: Mertens function and prime counting function Hi Bogauss, When you talk about the behaviour of $\pi(x)$ at infinite, it has always somehow a relation with RH. Ok , so here, si je ne m'abuse, you have $M'(n)=M(n)+\pi(n)$ is that true? So your M'(n) jumpsas M(n) everywhere but at the primes. Now if RH is TRUE then you have that $M(n)=O(\sqrt{n})$ (roughly because this is a bit stronger than RH but almost the same). So since it is known from the PNT that $\pi(n)\~\frac{n}{ln(n)}$ (sorry I don't manage to do a descent equivalent sign in Latex!) and since of course from the above condition (assuming RH) you have $M(n)=o(\frac{n}{ln(n)})$ then you finally get $M'(n)\~\pi(n)$ and the relative precision here of $M'(n)/\pi(n)$ is $M(n)/\pi(n)=O(\frac{ln(n)}{\sqrt{n}})$. For n=50000 it gives 4.8%
April 5th, 2012, 04:09 PM   #9
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Re: Mertens function and prime counting function

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 Originally Posted by Bogauss YOU ARE A PROGRAMMER NOT A MATHEMATICIAN!!!!
No, actually. Unlike you, I have advanced mathematical training. It's what I studied in school, both undergraduate and graduate.

April 5th, 2012, 04:11 PM   #10
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Re: Mertens function and prime counting function

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 Originally Posted by Dougy Now if RH is TRUE then you have that $M(n)=O(\sqrt{n})$ (roughly because this is a bit stronger than RH but almost the same). So since it is known from the PNT that $\pi(n)\~\frac{n}{ln(n)}$ (sorry I don't manage to do a descent equivalent sign in Latex!) and since of course from the above condition (assuming RH) you have $M(n)=o(\frac{n}{ln(n)})$ then you finally get $M'(n)\~\pi(n)$ and the relative precision here of $M'(n)/\pi(n)$ is $M(n)/\pi(n)=O(\frac{ln(n)}{\sqrt{n}})$. For n=50000 it gives 4.8%
That's it (though, as you intimate, $O(\sqrt n)$ should actually be $O(n^{0.5+\varepsilon})$ for some $\varepsilon>0.$

So assuming the RH, this M' function varies as the prime-counting function.

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