Dear Ali Hameed Yassir ,

I think you are talking about a ( Primality tests), this is a different issue.

In fact, there are hundreds of formulas that test numbers if they are prime or not. For example:

(1) n is prime if and only if [(n-1)! +1]/n is an integer number.

(2)If (2n-1 -1)/n is not an integer, then n is a composite number.

(3) n is prime if and only if all coefficients of the polynomial

[(x+y)n - xn - yn ]/xy(x+y) are divisible by n.

(4) Miller - Rabin test

(5) Elliptic curve primality test.

(6) AKS primality test.

(7) Fermat's test

(8) Lucas Lehmer test.

And many more.

The advantage of each test is given by its complexity ).

( The time needed to perform the test)

Hope your algorithm has good complexity.

Best regards

Dear Ali Hameed Yassir , I can certainly help you, but I could not understand your result. Can you please give me some details?

Dear Ali Hameed Yassir ,

I prefer to change your post into:

{I think that I have proved a formula that may generate prime numbers.

Can somebody help?}

Great mathematicians through hundred of years failed to find a formula that generates primes in a reasonable period.

One may prove some interesting relation, but you can't call it a discovery

or (breakthrough) before you publish the results in an indexed journal.

Or you can publish the results on the arXiv (https://arxiv.org/archive/math) to save your copyright.

In the meanwhile, you are welcome for any advice.

Best regards

There are many journals on this topic, ``Experimental Mathematics'', https://www.tandfonline.com/loi/uexm20 for instance; or ``Journal of Number Theory'', https://www.journals.elsevier.com/journal-of-number-theory

Thanks for your responding

About the formula works in tow steps

To check the primarity of given number.

It works

Last night i try with numbers of 16 digits.

Dear Ali Hameed Yassir ,

I think you are talking about a ( Primality tests), this is a different issue.

In fact, there are hundreds of formulas that test numbers if they are prime or not. For example:

(1) n is prime if and only if [(n-1)! +1]/n is an integer number.

(2)If (2n-1 -1)/n is not an integer, then n is a composite number.

(3) n is prime if and only if all coefficients of the polynomial

[(x+y)n - xn - yn ]/xy(x+y) are divisible by n.

(4) Miller - Rabin test

(5) Elliptic curve primality test.

(6) AKS primality test.

(7) Fermat's test

(8) Lucas Lehmer test.

And many more.

The advantage of each test is given by its complexity ).

( The time needed to perform the test)

Hope your algorithm has good complexity.

Best regards

Dear Dr. Issam Kadoura

I thank you

Your answer is the correct and the best.

I hope to get your attention please.

Best Regards

Dear Ali Hameed Yassir ,

Even finding a new method of primality testing is welcome. If this is working for 16 digits numbers that does not mean that the method is right. You need to give a proof for your method. The methods mentioned above by Issam Kaddoura are all proved by mathematicians.

Dear Dr. Tapas Chatterjee

Thanks so much for your replay.

I 'll tacke your advice in mind.

16 digit till now

But it works with

Infinite digits

Best regards

Dear Ali Hameed Yassir

This will be a great discovery if your algorithm works for big numbers. Wish you the best. Maged Z. Youssef

Dear Dr. Maged

I thank you for your comments.

انا اشكركم دكتورنا الكبير لو كان هنالك حاسوب يسع خانات اكثر او مراتب اكثر لكن هذا الموجود.

الحمد لله الصيغة او الطريقة تعمل بدقة وسرعه وساحولها الى كود برمجي قريبا ان شاء الله

Dear Ali Hameed Yassir ,

Could you please take a look at my paper

https://www.researchgate.net/.../221665691_On_formula_to_compute_primes_and_the_...

It presents a new primality test as well as it generates all prime numbers in order.

A formula for the nth prime number is available too and the next prime for any given number.

Also, there is a program available so you can compare the algorithms Vs. yours to decide which is faster.

I hope you find it useful in your research.

Best regards

Dear Ali Yaser,

Regarding your experimental calculation, even though it may be work correctly for 16 digit or even more, it wan’t be worth looking at until you formulate it in an algebraic or analytic formula. For instance the famous unproven hypothesis of Riemann was shown to be correct for huge number ( exceeding 10 to power 100) of complex zeros of the Riemann‘s zeta function, but this means almost nothing toward proving this unattainable hypothesis which is by the way is crucial in determining prime numbers.

Thanks Doctor for your replay,

I drive a formula that work for all number for infinite number of digits.

Do you know doctor that i have not any idea about riemann or any other hypotheses.

I hope to have the knowledge about that theories to connect my work with them.

Best regards

OK brother Ali,

I am a professor of Algebra in Al-Imam Mohammad Ibn Saud University in Riyadh, Saudi Arabia. So to help you with your discovery - If it is a serious one - , I would like to know where do you live and to communicate with me on my e-mail: [email protected]. Regards ..

RG is a popular platform to share knowledge and to provide popular qualitative answers that may help the progress of research activities for all.

(We live in the world ) where one can publish his work in many trusted journals.

At least, the arXiv is available for all.

Sounds great mate, looking forward to see your research paper on this big achievement.

Regards,

Thanks doctor

Dear Ali Hameed Yassir

The best way to approve or disprove your new formula is to write a good article about it with some examples and submit it to a peer reviewed journal.

Also you can pre-publish your article in RG so you can have some opinions from other mathematicians in number theory.

I wish you the best luck

The quick and best way to make some progress about your new formula is to give example to RG followers.

Please find below a new method to estimate or compute the next prime after a given prime.

First compute the integer Q=sqrt (N) then compute R=Q(Q+1).

Compare the result to N. The difference D is positive (nearrest to N) then compute E= (Q-1)Q ,

the difference F with N is negative (nearrest to N).

The next prime after N could be Q(Q+1)-D or (Q-1)Q+F. If you do not get a prime number then decrease D or increase F. After (logN)/2 tests you will reach the next prime. Choose carefully the new values for D and F in order to avoid even number.

For a prime p=100003

sqrt(100003)= 316 , R=316 x 317=100172 , D=+169 and

E=315 x 316=99540 ,F= - 463

D={165,163,161,159,157,153}

F={467,469,471,473,477,479}

Number of test is 6=(logN)/2

next prime=99540 + 479=100019

or

next prime=100172 - 153=100019

Best regards

Dear Mohamed Azzedine

To show your method correctly, could you please answer the following:

1. You said that

sqrt(100003)= 316 , R=316 x 317=100172 , D=+169

How did you determine D? Is it by trial and error?

2. Similarly, for the number F. How did you determine F?

2. If you test Q(Q+1) - D and you obtain a prime number, how did you decide that this prime is indeed the next?

3.What about the numbers between N and Q(Q+1) - D?

Why did you consider them composites?

4. According to you, you need only log(N)/2 trials, which is not the case.

It seems that you need more than that number.

Best regards

Dear Issam Kaddoura

1/ D is equal to R -N=100172 - 100003=+169

2/ F is equal to E -N= - 463

3/We start with 100172 -169=100003 (given prime number) and we decrease D.

All the even values of D is discarded (difference between 2 even numbers) until D=153 then we get the next prime number 100172-153=100019.

4/ The same method is applied to F with ascending order. We start with 99540+463=100003 (given prime number) until F=479; then we get 99540+479=100019 wich is the next prime number.

5/The number of test is log(N)/2 you can check it.

The method is powerful and new. It could be used for various and different problems related to prime numbers.

Best regards

M. Azzedine

Dear Mohamed Azzedine

Frankly, I didn't see any method yet.

You did some calculations around the prime 100003 and the next prime 100019.

they are very close and trivially, we test

100003 given prime

100003+4 composite

100003+6 composite

100003+10 composite

100003+12 composite

100003+16 prime

We reach the next prime by applying only 5 steps(the others divides 3 ).

Any trial and error methods reach the next prime in a short time.

(I mean close primes)

The efficiency of any method (complexity) to test the next prime appears when there is a large gap between a prime and the next prime.

Best regards

Dear Issam Kaddoura

Why you start with 4 and not 8 ?

You built the sequence because you know the result.

You can check your method against my method when there is a large gap between a prime and the next prime and then you will see the efficiency .

Best regards

Dear Mohamed Azzedine ,

I didn't show any method. I said the trial and error method can do the job because of the primes 100003 and 100019 are very close.

I told you that your procedure is not clear. You need to show every step and to give proof that the method will produce a NEXT PRIME!

I have a published paper in this concern that used a different approach.

See:

m-hikari.com › ams-2012 › ams-73-76-2012 › kaddouraAMS73-76-2012

by I Kaddoura - ‎2012 - ‎Cited by 11 -

Dear Issam Kaddoura

I answered to your questions and shown all the steps but it seems you decided to not understand. OK.

I told you try an example with two primes not close to each other and then you will see the efficiency of my method.

The work continues.

Best regards

M.Azzedine

You said

(*) I answered to your questions and shown all the steps but it seems you decided to not understand. OK.

Your reply is ( not accepted ...) !!!

I am trying to help you, so you can improve your method, if any!!

I told you, there is nothing to revise or to correct, your statements about

Q(Q+1), Q(Q -1), F, and D mean nothing. You need to show theorems and proofs.

You can support your claims by references, articles, or any documents.

At least you can suggest a mathematical statement as a conjecture.

Unfortunately, you did nothing, and you ask me to follow and calculate that nothing!!

Please, be scientific when you show answers.

📷Dear Issam Kaddoura

Please find here after the formula for the estimation of next prime (or next prime) after a given prime P.

The succeding prime after a given prime P can always be found in between the prime P and

N + (P- N + (log P)) with N=n(n+1),

n=sqrt (P)

example: P=100003; n=sqrt (100003)=315; N=315*316=99540;

Estimation of next prime =99540 + (100003 -99540 +6)=100019

All the variables are integers.

Best regards

M. Azzedne

Dear Mohamed Azzedine

Your claim is that:

For a given a prime P, then

P < Next prime < N + (P - N) + log(P)

In other words:

P < Next prime < P + log(P)

Which is a wrong statement.

Counterexample: For p =13, we know that the next prime is p =17

applying your claim:

13 < 17 < 13+log(13) = 15. 565

which is wrong.

Best regards

Dear Issam Kaddoura

I am claiming the following.

The estimation of the Next prime is N + (P - N) + log(P)

and not what you write.( P < Next prime < P + log(P) )

Please follow carefully my post.

Your counter example is false. P=13, sqrt(13)= 3,6 rounded up to (4) .According to my original claim the estimation of the next prime is

4*5 +(13-4*5) +(2,565)=15,565 rounded up to the integer 16. In one calculation we get a good estimation of the next prime 17.

We can get a better estimation if we decrease D=7 to D=5 (remember my first post) , the new estimation will be 4*5 +(-5) +2,565)=17,565 (rounded down to 17)

After one test we get the exact next prime as described in my first post . The number of test is (log P)/2 .

Best regards

M.Azzedine

Dear Mohamed Azzedine

You said that in your previous claim : (3 days ago)

[[[[The succeding prime after a given prime P can always be found in between the prime P and N + (P- N + (log P)) with N=n(n+1)]]]]

which means: ( P < Next prime < P + log(P) )

In all cases, your new answer is still unclear.

In fact, your new claim is worse than the previous one!!!

[[[ The estimation of the Next prime is N + (P - N) + log(P) ]]]

this estimation is wrong.

Anyway, have you any statement with clear proof?

You show nothing yet.

Best regards

Dear Issam Kaddoura

The estimation is not wrong. If P=13 the estimation gives 16 which is even then the next prime is 17. The second estimation gives 17,526 which gives 17 after rounding down. In one line we get a good estimation of the next prime for a given prime P. It is a powerful method, quick, efficient and new.

Best regards

M.Azzedine

Dear Mohamed Azzedine

I hope so, but let me ask you the following:

Can you show me where is the ( powerful method, quick, efficient and new.)

method?

You should show something like this:

Theorem: xxxxxxxxxxxxxxx.

Proof: xxxxxxxxxxxxxxxxxxx.

Then we can judge that method.

Best regards

,

Not sure where your work is so I can read it, but I've found a formula for the prime counting function. That is a done deal.

This formula uses a primality test function, which yields 0 whenever n is composite and 1 when n is prime. I think it's the real deal, by the looks of it. It tells us that the prime numbers are given by an even function which is expressible as a Taylor series which depends solely on Pi, Bernoulli numbers, logs and factorials.

Good job brother

there are multiple theorems to the effect that given a sufficiently large prime p, the next prime after p will occur before p+x, for various values of x (bertrand's postulate takes x = p, but there are many theorems with smaller values)

the issue is the extremely irregular distribution of primes; there are primes with difference 2 (101, 103 and 107, 109) and long gaps with no primes at all (for example, n!+2 through n!+n for any n > 2)

that said, the prime distribution functions (discovered, it seems, by gauss in 1792 and finally proven by hadamard and de la vallee poussin in 1896) are approximations over large intervals (they are the basis for the best values of x above), but they are asymptotic formulas not usefully applied locally

all of the methods have to decide: are you guaranteeing a prime, suggesting a prime, of saying that a prime is nearby? the results are quite different

Thanks for your interests, but I would like to suggest you make a good literature review and find out the research question? That is to find out where is the gape? Solve the gape and file your report in the form of regular formal manuscript for publication. You may also read : https://www.elsevier.com/books/elements-of-the-theory-of-numbers/dence/978-0-08-050549-7?countrycode=AU&format=print&campaign_source=google_ads&campaign_medium=paid_search&campaign_name=Australia_shopping&gclid=EAIaIQobChMIxNq-86zX5wIVT6mWCh2iaAfPEAYYASABEgKV_PD_BwE

Dear Ali Hameed Yassir

Any progress? What is the complexity of the algorithm?

What is the largest prime you succeed in testing?

Sharing results may help you to improve the algorithm.

You can post the results at arxiV for copyright.

I hope to hear good news.

3*(n+1th prime) -2*(nth prime) is either a prime or a semi prime

Dear Aziz Lokhandwala

Your claim is not correct.

Counterexample:

3×7919 - 2×7907= 13²47 is neither prime nor semiprime.

Where 7907 is the 999th prime and 7919 is the 1000th prime.

There are many counterexamples.

Best regards

Yes sir Issam your counter example is correct

it is a curious fact that the polynomial x^2 + x + 41 gives a prime value for integers x from x = -40 to x = 39 (the value is clearly 41^2 for x = -41 and x = 40), though it misses some (59 and 67, for example)

there are hundreds of examples of varyingly complicated formulas that seem to give primes for a long time; nowadays you must spend a lot of computer time checking, and most of us do not have the time or the computer power

Dear Christopher Landauer

In the introductory number theory, we have the elementary fact:

"There are no polynomials with integer coefficients that provide only prime numbers."

Polynomials always generate prime and composite numbers as well.

The challenge is:

Can we find an equation ( Not a polynomial) that generates only prime numbers all the time?

Best regards

yes, i do know that; i just used that polynomial example to show that it is easy to stop before finding the counterexample for any proposed formula; i also vaguely remember a set of diophantine equations in a very large number of variables that produces only primes, but it is way to complex to generate primes quicklyj

i don't think there is a formula of any kind that not only generates only primes, but also generates all of them (in some order)

If you have a look at our paper,

On Formula to Compute Primes and the nth Prime - Hikari

m-hikari.com › ams › ams-73-76-2012 › kaddouraAMS73-76-2012

It includes a formula that generates all prime numbers in ascending order. It can compute the nth prime, without any knowledge about the previous ones.

Unfortunately, its complexity is not as efficient as some other methods.

But the formula is useful to produce the primes for cryptographic purposes.

Yes sir I have refered to your paper. It is using the fact that all primes are of form 6k+-1.

In that paper what you have apply is a kind of sieve which or like a scan that in a field if you have n numbers. You are tying its high and low bounds as l and h.Then looking for average of l and h.Now if that average is 1 then x is prime else not prime.

I have referred your paper while making my way of approximating nth prime using geometrical analysis and prime number theorem.

And I need to say that time complexity of your test is very less.

But sir to me its like divide and win type thing. Because you have firstly divided the whole set in two different parts and compare each elements indivually and then merge the two sets to get the net results

Dear Aziz, I hope you find it useful. Some authors used our formula in their papers to cipher and decipher some texts.

Dear Aziz Lokhandwala would you please give more explenation about the method? Thanks

Dear Ali Hameed Yassir would you please give some explenation about your method ,or send us your paper? Thanks

Primality test and easy factorization

This is an algorithm that test if one number is prime and if the number it´s not prime returns the biggest factor of that number.

https://www.researchgate.net/publication/344177203_Primality_test_and_easy_factorization

PHGP - could you explain in your formula why the exponent (4/4) is not just 1? and why you write (4/2347...) later on? the notation is not well explained

also, why the 8th derivative some places and the 10th others?

finally,

it appears that you are assuming that your result fractions are in lowest terms, which assumes that you already know that the originals are prime (or not) - thus this reasoning seems to be completely circular - a clarification would be helpful

Yes you can take (4/4) as 1 obviously. The reason is that I work with that function in other problems and I use 4/n (n like an integer). (4/2347...) is the number to test you can use any integer. I use the 8th derivative because is more simple than 10th. But If you use bigger derivatives the factorization in more specific and if you work with big numbers you need bigger derivatives because the final division could be near of 0. At the end is a division between two similar numbers in magnitude.

I don´t assume that I alredy know if the number is prime or not. I test with any number that I dont know previously if is prime or not. The explanation that I think why this occur is that the function in the denominator x^(4/4) returns all primes and when I divide it with the function with the test number, if the test number is a prime is a whole factor, and if is a composite numbers dont complete the division

That´s were good questions.

Thank you

Hugo

Unreadable article.Nothing is clear and the notations are not appropriate.

Lack of any theory.

PHGP - yes, you actually do assume that the factors are computed when you do the arithmetic, because otherwise, you cannot prevent the system from returning 1053/2347967901 instead of 351/782655967, which would cause you to decide that 2347967901 is prime, which it is not

(and you want to be more careful about the very first statement about primes, which is false: 2 and 3 are primes NOT of the form 6*n +/- 1)

simple gcd calculations can work this particular case, but nothing in the method prevents some fraction that has a non-prime denominator and a numerator relatively prime to it, which will not reduce as your example (accidently) did - in this case, the denominator will be not a prime and your method can not show that

PHGP - also, when you write

``In the structure of the derivative we choose the first number'',

which derivative are you talking about? or are you talking about the quotient? and what is the ``first number'' anyway? do you mean the highest degree term?

these things are not at all made clear

The derivative is:

(8th derivative x^(4/4)-3x^(2/4)+1) / (8th derivative x^(4/2347967901)-3x^(2/2347967901)+1)

And the structure of the derivative after the operations is:

(-----972670208587507037972081590054148967298435679876606390772097420491454642030653--------- x^(2347967897/4695935802))/(63623515096683345297912974704640 (82889055288145367997349961181951360634 x^(2/2347967901) – 124333583206820800237761746278619343435))

We choose the first number

Best regards

Hugo

Dear

Issam Kaddoura:

If you have anything better post it, otherwise go you and your camels to the school to learn.

Best regards

Hugo

Dear Pedro Hugo García Peláez

Each researcher has an academic profile that presents his educational activities. I don't care about your aggressive words that reflect your personality.

( [Camels] taught humanity how to write and how to talk! It was the origin of humankind civilization on earth. It would be best if you read more about human history.)

My remarks about your work are not personal.

They are academic issues that may motivate you to rewrite your article in better form to show and present the theory behind your computations.

Your work may include valuable ideas, but first, you should make it readable. I am an expert in number theory, but I didn't understand your point of view because of the mess of notations and the lack of precise, explicit theorem.

I wish you good luck in the academic community and hope you may change your language of hate.

I think science and knowledge are the ultimate moral of humanity.

Regards

.

How many science discoveries had you made. Search me in google. And most researchers shut up his mouth if they haven´t anything better to offer.

You can´t generalize, my best friend is from Syria. I don´t extend the comment to all Arabs. I said only in he particular case of Issam Kaddoura

Dear friends,

Please, stop hate.

Dear Marouane Rhafli

I respect your point, but I can't entirely agree with it. Your good work should be your title.

I accept Hugo's accusation, I think he got me wrong, and then he understands what I mean.

So, Kindly I ask you to remove your previous answer, please.

By the way, I like camels and will follow the advice of my colleague Hugo and try to learn more, why not?

Regards

Thank you dear Issam Kaddoura I respect you a lot, yeah it is good idea, let's gather camels and let's learn all together in the desert, fresh air will make us more focused :D

Dear brothers

I will publish near, Insyallah

the new method is not my previous articles ( Prime Numbers New Pattern, Formulas and ASA Method ).

the new method is formula not pattern, its deferent.

Best regard

Dear Ali Hameed Yassir

In the meanwhile, I prefer to post your results on the arXiv, so we can share you

our points of view.

I wish you good luck.

Dear Pedro Hugo García Peláez

Your reply"

How many science discoveries had you made. Search me in google. And most researchers shut up his mouth if they haven´t anything better to offer."

I appreciate it if you can change your language.

You can see the RG platform evaluation(attached).

I live in Canary Islands (Spain) We also have camels here.