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April 5th, 2012, 10:39 AM   #1
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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.
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April 5th, 2012, 10:51 AM   #2
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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.
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April 5th, 2012, 11:22 AM   #3
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Re: Mertens function and prime counting function

Quote:
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!
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April 5th, 2012, 12:27 PM   #4
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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 ....
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April 5th, 2012, 12:32 PM   #5
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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.
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April 5th, 2012, 12:38 PM   #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.
Are your claims a proof????????????????????
WHO ARE YOU?????
IS YOUR ANSWER A GOD'S ANSWER???????????????????
YOU ARE A PROGRAMMER NOT A MATHEMATICIAN!!!!
YOU KNOW NOTHING ABOUT MATHEMATICS
SO STOP TREATING OTHERS WITH DEEP CONTEMPT.

GOOD BYE!!!
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April 5th, 2012, 12:40 PM   #7
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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.
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April 5th, 2012, 12:50 PM   #8
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Re: Mertens function and prime counting function

Hi Bogauss,
When you talk about the behaviour of at infinite, it has always somehow a relation with RH. Ok , so here, si je ne m'abuse, you have is that true? So your M'(n) jumpsas M(n) everywhere but at the primes. Now if RH is TRUE then you have that (roughly because this is a bit stronger than RH but almost the same). So since it is known from the PNT that (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 then you finally get and the relative precision here of is . For n=50000 it gives 4.8%
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April 5th, 2012, 05:09 PM   #9
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Re: Mertens function and prime counting function

Quote:
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.
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April 5th, 2012, 05:11 PM   #10
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Re: Mertens function and prime counting function

Quote:
Originally Posted by Dougy
Now if RH is TRUE then you have that (roughly because this is a bit stronger than RH but almost the same). So since it is known from the PNT that (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 then you finally get and the relative precision here of is . For n=50000 it gives 4.8%
That's it (though, as you intimate, should actually be for some

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