For the discussion of math. Duh.

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### Re: Primes, sequences and infinite

lorb wrote:It seems you are looking for a closed formula for this?

Exactly!
But I want to reach it using other sequence easy to find.
I still do not know how to simplify the second sequence the negative.
All the results will be the first step to have another formula for the prime counting function.
To compute the numbers of prime up to n is not easy.
I know that I`m using a list of primes that is obvious.
I`m not fool.

lorb
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### Re: Primes, sequences and infinite

Oh well. While you are at it, would you mind to proof the Riemann hypothesis too? It's related, and shouldn't be too hard compared to what you are aiming for now.
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### Re: Primes, sequences and infinite

lorb wrote:Oh well. While you are at it, would you mind to proof the Riemann hypothesis too? It's related, and shouldn't be too hard compared to what you are aiming for now.

Do not worry. Selling roasted almonds to toothless people is much harder than solving Riemann hypothesis.

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### Re: Primes, sequences and infinite

A(p)=S(p)*(p-r)
where r and S(p) are real numbers.
Now I need to compute the values of r.

PsiCubed
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### Re: Primes, sequences and infinite

lorb wrote:Oh well. While you are at it, would you mind to proof the Riemann hypothesis too? It's related, and shouldn't be too hard compared to what you are aiming for now.

Depending on what is meant by "closed expression".

If you're willing to use the Reimann zeta function itself in your expression, finding a closed formula for pi(x) is certainly doable. Ramanujan already solved that problem a century ago.

But somehow, I don't think this is what Goahead had in mind...

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### Re: Primes, sequences and infinite

After some computation I reached the conclusion that pi(n) the counting function does need only the first prime numbers to be computed with big precision.
Out of n (assuming that n is very large) how many first primes I need to know to have something precise? I still do not know. It needs lot of tests before reaching to the final conclusion.
I have found a close formula for A(p).
I have found the the number 1/e (e=2.7....) play a role when p goes to infinite.
So my idea is to put in place some kind of "zigzag" function.
You compute for what is yet known (example n=10^9) to have pi(10^12 for example) and you assume that your formula for 10^12 is accurate and you go forward (10^15) etc....

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### Re: Primes, sequences and infinite

Goahead52, why do you believe there is a (simple) way to express A(p) in terms of S(p)?

Sure, they're both sums of reciprocals of primes, but so are an infinite number of other things, so I'm unsure why you expect a closed-form expression in this case.
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lorb
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### Re: Primes, sequences and infinite

PsiCubed wrote:
lorb wrote:Oh well. While you are at it, would you mind to proof the Riemann hypothesis too? It's related, and shouldn't be too hard compared to what you are aiming for now.

Depending on what is meant by "closed expression".

If you're willing to use the Reimann zeta function itself in your expression, finding a closed formula for pi(x) is certainly doable. Ramanujan already solved that problem a century ago.

But somehow, I don't think this is what Goahead had in mind...

I was going by this:
Goahead52 wrote:I`m not looking for an approximation [...] The exact closed formula.
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PsiCubed
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### Re: Primes, sequences and infinite

The formula I'm talking about is exact. It uses the relation between the positions of the Riemann Zeros and the distribution of the primes (and it isnt Ramanujan's. I was mistaken. But the formula does exist).

But on second thought, it doesn't count as a "closed formula". You need to actually find the position of some Riemann Zeros for it to work, and there's no closed formula for doing that. I guess that if we're allowed to do that, we could just count all the primes up to x directly and get it over with.

And speaking of Riemann:

I wonder... would a clean and closed formula for pi(x) automatically settle the Reimann Hypothesis? I know that if we assume RH to be true, it leads to some very tight bounds for pi(x), but does it also work in reverse?

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### Re: Primes, sequences and infinite

The main idea behind those computation is to find an easily computable sequence U(n) such as :
- when n exceed what is yet known (exact values of pi(n) obtained by sieving or any other mean) U(n) will be exactly equal to pi(n) and will stay equal to the infinite.
- we know exactly for which value of the n are identical

We need forcibly a proof of that equality otherwise it will be unuseful.
I`m saying this because because I discovered a sequence identical to another one found in OEIS but the method of the first sequence is TOTALLY different. The 2 were identical element by element. So what I`m looking is an equality after n reach some level.

I will post the sequence OEIS and my sequence once I retrieve it on my archives.

lorb
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### Re: Primes, sequences and infinite

PsiCubed wrote:And speaking of Riemann:

I wonder... would a clean and closed formula for pi(x) automatically settle the Reimann Hypothesis? I know that if we assume RH to be true, it leads to some very tight bounds for pi(x), but does it also work in reverse?

Doesn't even have to be a "clean and closed" formula, just one that is close enough to pi(x). This source says that if you can obtain an approximation of pi(x) with an error term that never exceeds x, that would be equivalent to a proof for RH.

Edit: so with the work of Goahead coming along I'm confident RH will be solved soon
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### Re: Primes, sequences and infinite

Goahead52 wrote:The main idea behind those computation is to find an easily computable sequence U(n) such as :
- when n exceed what is yet known (exact values of pi(n) obtained by sieving or any other mean) U(n) will be exactly equal to pi(n) and will stay equal to the infinite.
- we know exactly for which value of the n are identical

We need forcibly a proof of that equality otherwise it will be unuseful.
I`m saying this because because I discovered a sequence identical to another one found in OEIS but the method of the first sequence is TOTALLY different. The 2 were identical element by element. So what I`m looking is an equality after n reach some level.

I will post the sequence OEIS and my sequence once I retrieve it on my archives.
Yeah, just post the sequences and we can discuss why the methods to generate them look so different.

That could be an interesting thread, while your attempt to prove Riemann won't be.
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PsiCubed
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### Re: Primes, sequences and infinite

lorb wrote:
PsiCubed wrote:And speaking of Riemann:

I wonder... would a clean and closed formula for pi(x) automatically settle the Reimann Hypothesis? I know that if we assume RH to be true, it leads to some very tight bounds for pi(x), but does it also work in reverse?

Doesn't even have to be a "clean and closed" formula, just one that is close enough to pi(x). This source says that if you can obtain an approximation of pi(x) with an error term that never exceeds x, that would be equivalent to a proof for RH.

You mean "never exceeds x^(1/2+ε) for a large enough x, given any ε>0 we choose", but the general message is clear.

Thanks. That was very interesting.

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### Re: Primes, sequences and infinite

gmalivuk wrote:
Goahead52 wrote:The main idea behind those computation is to find an easily computable sequence U(n) such as :
- when n exceed what is yet known (exact values of pi(n) obtained by sieving or any other mean) U(n) will be exactly equal to pi(n) and will stay equal to the infinite.
- we know exactly for which value of the n are identical

We need forcibly a proof of that equality otherwise it will be unuseful.
I`m saying this because because I discovered a sequence identical to another one found in OEIS but the method of the first sequence is TOTALLY different. The 2 were identical element by element. So what I`m looking is an equality after n reach some level.

I will post the sequence OEIS and my sequence once I retrieve it on my archives.
Yeah, just post the sequences and we can discuss why the methods to generate them look so different.

That could be an interesting thread, while your attempt to prove Riemann won't be.

Here is a quote of my finding long time ago (it is me the author of the quote :

Here is the sequence based on Mobius function
Quote from OEIS

A049697 a(n)=T(n,n+1), array T as in A049695. 0

1, 3, 6, 10, 16, 22, 30, 40, 50, 60, 74, 88, 104, 122, 136, 152, 176, 198, 222, 248, 268, 290, 322, 352, 380, 412, 442, 472, 512, 548, 586, 632, 668, 704, 744, 780, 828, 882, 924, 964, 1020, 1072, 1126 (list; graph; refs; listen; history; internal format)

OFFSET 0,2

FORMULA a(n-1)=sum_{k=1...n}mu(k)floor(n/k)ceiling(n/k); a(n) is asymptotic to (6/Pi^2)*n^2 - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 27 2005

PROG (PARI) a(n)=sum(k=1, n++, n\k*moebius(k)*ceil(n/k)) \\ Charles R Greathouse IV, Mar 02 2012

I obtained the same sequence using Euler totient

A(n)=sigma(phi(i)+phi(n-i)) with i =1 to n

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### Re: Primes, sequences and infinite

I have another one about Combinatorics but I did not find it for now.
My archives are not classified. Only my memory will allow me to find it.
Some are on my second computer.

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### Re: Primes, sequences and infinite

While working on the "Factorial puzzle" posted on Puzzles here I discovered an identical sequence (OEIS A088179) computed based on the Mobius function.
I will post it on the factorial puzzle post.
Amazing surprise with a theorem I have to prove.

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### New Theorem (prime numbers)?

Let us note F(n)=n*phi(n) where phi () is the Euler totient
n positive integer > 0

Theorem :

If mu(F(n))=-1 then n is prime (m() is the Mobius function)

Check it or find a counterexample.

Here is the list of the first few prime :
2,7,11,23,47,59,83,107,167,179,211,227,263,331,347,359,383,463,467,479,503,547,563,571,587,691,719,839,859,863,887,911,967,983,1019,1123,1187,1231,1283,1291

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### Re: New Theorem (prime numbers)?

I have an explanation not a proof.
If n is composite with mu(n)=-1 then it has to be square-free and product of an odd number of factors (3,5,7...).
If we multiply this number n by its Euler totient we will the product of 3,5,7.... even factors. Hence mu(n*phi(n))=0 (not square free)
On the contrary if n is prime mu(n)=-1 its Euler totient could be equal to 0,-1,1.
If its Euler totient is equal to 1 then mu(n*phi(n))=-1

lorb
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### Re: Primes, sequences and infinite

PsiCubed wrote:
lorb wrote:
PsiCubed wrote:And speaking of Riemann:

I wonder... would a clean and closed formula for pi(x) automatically settle the Reimann Hypothesis? I know that if we assume RH to be true, it leads to some very tight bounds for pi(x), but does it also work in reverse?

Doesn't even have to be a "clean and closed" formula, just one that is close enough to pi(x). This source says that if you can obtain an approximation of pi(x) with an error term that never exceeds x, that would be equivalent to a proof for RH.

You mean "never exceeds x^(1/2+ε) for a large enough x, given any ε>0 we choose", but the general message is clear.

Thanks. That was very interesting.

I actually meant "To restrict the location of zeta zeros through a formula for prime numbers, the prime formula would have to be strong enough to estimate π(n) with an error of order n^(1−ϵ) for a positive ϵ" which is given in another answer. It seems there are different ideas, but the consenus is clear enough: If you come up with a really really really god prime counting function it solves the RH, but going the other way round is probably easier less hard. (solving RH first, using it to obtain better prime counting function)
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### Re: New Theorem (prime numbers)?

I think you can prove a slightly stronger statement, that is mu(n * phi(n)) != 0 implies n is prime.
My maths is a little rusty, but I'll give it a go.

Suppose n is *not* prime.

Case 1: n is not square-free
Then n * phi(n) is not square-free, hence mu(n*phi(n)) = 0

Case 2(a): n is square-free and has two prime factors, p and q, where 2 < p < q (i.e. neither is 2)
phi(ab) = phi(a) * phi(b) if gcd(a,b) = 1
phi(x) = x - 1 for prime x, hence phi(x) is even for x > 2

Consider phi(n) = phi(p) * phi(q) * phi(d)
where d = (n / q) / p
Since p and q are > 2, phi(n) must have 2 as a prime factor with multiplicity 2, hence phi(n) is not square-free, so mu(n * phi(n)) = 0

Case 2(b): n is square-free and has two prime factors, p = 2 and q > p
phi(p) = 1, phi(q) is even. So phi(n) is even.
n has 2 as a prime factor, hence is even.
So n*phi(n) has 2 as a prime factor with multiplicity 2, hence n * phi(n) is not square-free, so mu(n * phi(n)) = 0

So n not prime implies mu(n * phi(n)) = 0.
Hence "mu(n * phi(n)) != 0 implies n is prime".

It is, however, not at all clear why this would be a useful or meaningful result.

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### Re: New Theorem (prime numbers)?

Thank you very much for your comment.
If you except n=1 your statement is stronger than mine. I did not check the value where n*phi(n)=1 because I was looking for something else.
I have to restate and enlarge my claim to the other prime. I will do it until no one has any comment to add.
I checked your statement up to 10000 it is correct and accurate . You are right.

PsiCubed
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### Re: New Theorem (prime numbers)?

Goahead52 wrote:I have to restate and enlarge my claim to the other prime. I will do it until no one has any comment to add.

Ehm, what? I'm not sure what you just said, but it can't be good...

I checked your statement up to 10000 it is correct and accurate . You are right.

This post is getting stranger and stranger.

For one thing, "checking up to 10000" doesn't tell us anything. It isn't even strong evidence. There are plenty of conjectures in number theory which "break down" at much higher numbers.

Also, the guy gave you a proof which is easy enough to follow. So what's the point of "checking"?

PsiCubed
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### Re: Primes, sequences and infinite

lorb wrote:If you come up with a really really really god prime counting function it solves the RH

Couldn't help but chukle at that typo. It sums the difficulty of the problem nicely

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### Re: New Theorem (prime numbers)?

Thank you all.
I`m not in good health not able to focus on anything. So I will stop posting
Very sorry for bothering you.
Good luck to everybody.

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### Factorization of odd semi prime is over!

The factorization of odd semi prime is over and could be reduced to a simple linear congruence equation :

Here is the solution of an amateur :
p and q are odd semi prime

We start by this equation which is true :

pq*(p-1)(q-1)mod(pq-2)=(p-2)(q-2)

This equation is based on some properties of Eulerian Factorial Numbers.
An EFN(n) is of the form = n*phi(n) where phi(n) is the Euler totient.

-------------------------------------
Here is my solution which uses only elementary tools.

p*q*(p-1)*(q-1)mod((p*q)-2)=(p-2)(q-2)

A=p*q

y=p+q

This equation is nothing more than linear congruence that anyone could solve :
I replaced p*q which is known by A and (p+q) wich is unknown by y

A(A-y+1) mod (p*q-2)=A-(2*y)+4

My job is finished.
I do not have enough energy (because of health problems) to simplify and to give a closed formula of y

Now I will go back to my bed.

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### Re: Factorization of odd semi prime is over!

Sorry I forget to mention that q>p.
I have in mind the RSA cryptographic saystem.

For p^2 I have a different equation.
Anyway confirm it or infirm it.

lorb
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### Re: Factorization of odd semi prime is over!

Please be gracious in judging my english. (I am not a native speaker/writer.)
http://decodedarfur.org/

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### Re: Factorization of odd semi prime is over!

lorb wrote:Try those numbers: https://en.wikipedia.org/wiki/RSA_numbers

I know the RSA numbers and challenge so do not send what is known to anyone.
Focus on what I said first.
You will surely find flaws. I know it.
There are 2 main equations.
Are they false or true?

PsiCubed
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### Re: Factorization of odd semi prime is over!

My dear boy,

If anybody here had any idea how to break RSA (or factor huge semiprimes - which is the same thing), he would most certainly not share his ideas on public forum.

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### Re: Factorization of odd semi prime is over!

I had sent 2 equations : are they false or true?
That is all I want to know.
Sharing results is not prohibited by any law.
I`m not kid and I know what it means to send a solution for problem as hard as factoring huge numbers. All the secret services will try to kill you or to abduct you. I know that. But I do not care.
Am I right or wrong? that`s it.
Because I did not reveal all until now.

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### Re: Factorization of odd semi prime is over!

I got it!!!
After simplification I obtained a final (it could be improved) formula :

2x+y=a (where is a is known and more important 2x is equal to phi(n)/2)

Example : n=1783*2879

2x=2564298 and phi(n)=5128596
y=66
a=2564364
We know the value of a ((n+1)/2)-int(sqrt(n)))then it will be easy to find phi(n) by approximation as phi(n)=(p-1)(q-1)
I assume that n is a big odd semiprime
I did not finished yet but I gave the more important not the algorithm with all details.
Many of you are very suspicious I know it.

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### Re: Factorization of odd semi prime is over!

knowing that n=pq is odd semiprime there is a solution to :
1. approximate phi(n)=2x
2. approximate y
As we know that 2x+y=a (with a known) then we will obtain the definitive solution by convergence. We will go quickly to the solution.
We coud reduce the global equation 2x+y=a to more handable equation 2z+t=b (where z and t are of tiny size in terms of digits).
It needs a collaborative work.
I can not do it with my little knowledge and my little tools.
Someone of you will surely come with more elaborated solution.

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I'm merging these because you have a habot of starting a new thread every time you have a slightly different idea, even when it's clearly about a discussion in an existing thread. We don't need the forum cluttered up with a bunch of your threads that never get past the first page.
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### Re: Factorization of odd semi prime is over!

Goahead52 wrote:knowing that n=pq is odd semiprime there is a solution to :
1. approximate phi(n)=2x
2. approximate y
As we know that 2x+y=a (with a known)

1. How do you approximate phi(n) without factoring n in the first place?
2. What is y and how do you approximate it?
3. What is a?

Cauchy
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### Re: Factorization of odd semi prime is over!

Since pq=A, we get that A(A-y+1) = A - 2y + 4 (mod (A-2)).

Since A is congruent to 2 mod A-2, we find that 2(2-y+1) = 2-2y+4 (mod (A-2)).

But this is an identity. That is to say, it's true for any values of A and y. So its truth doesn't tell us anything about y.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

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### Re: Factorization of odd semi prime is over!

Demki wrote:
Goahead52 wrote:knowing that n=pq is odd semiprime there is a solution to :
1. approximate phi(n)=2x
2. approximate y
As we know that 2x+y=a (with a known)

1. How do you approximate phi(n) without factoring n in the first place?
2. What is y and how do you approximate it?
3. What is a?

a was yet given if you read carefully my post.

1. There is a way to approximate phi(n) if you know that n is odd semiprime. I have the solution principle on how to reach it. But that is another subject.
2. y is the value that you add to int(sqrt(n)) such as (int(sqrt(n))+2y)^2-n=z^2 (if you know y then you factorize n easily). There is a way to approximate y.
Example for illustration: n=35=5*7
y=1
int(sqrt(35))=5
(5+1)^2-35=36-1=6^2-1^2=(6+1)*(6-1)=7*5
n and phi(n) share the same y

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### Re: Factorization of odd semi prime is over!

Cauchy wrote:

Since pq=A, we get that A(A-y+1) = A - 2y + 4 (mod (A-2)).

Since A is congruent to 2 mod A-2, we find that 2(2-y+1) = 2-2y+4 (mod (A-2)).

But this is an identity. That is to say, it's true for any values of A and y. So its truth doesn't tell us anything about y.

Right! But you need to transform that identity to have the required y.
Any equation is an identity except you need to isolate some variables to make it solvable.
How do you find a root of quadratic equation ax^2+bx+c?
By reversing the process

lorb
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### Re: Factorization of odd semi prime is over!

Goahead52 wrote:Any equation is an identity except you need to isolate some variables to make it solvable.

No it's not. Identity is a mathematical term that means:
An identity is a statement resembling an equation which is true for all possible values of the variable(s) it contains.
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### Re: Factorization of odd semi prime is over!

lorb wrote:
Goahead52 wrote:Any equation is an identity except you need to isolate some variables to make it solvable.

No it's not. Identity is a mathematical term that means:
An identity is a statement resembling an equation which is true for all possible values of the variable(s) it contains.

What is this?
A(A-y+1) mod (p*q-2)=A-(2*y)+4

It is an identity.
I do not understand why you are rejecting as on identity?

Ps : wikipedia is not a reference. It contains many errors. I did not read your reference.

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Do not forget that the way to handle the right part and the left part of any equation could lead you to solving it.
How did I handle such equation is another debate?
For those who are asking me about how to know phi(n) without factorizing it?
Hint : many numbers have the same value of phi(n). If you know that some number has properties such as another it will have the same phi(n).
13
21
26
28
36
42

All those numbers have the phi(x)=12 (replace x by each number of the set above.