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-   -   The "one billion minus 999,994,000" digits prime number (https://www.mersenneforum.org/showthread.php?t=20568)

R.D. Silverman 2015-10-23 22:46

[QUOTE=a1call;413527]That thread is too long for my time availability.
Was it based on a proof based on number ranges or was it something else?
Are there any similar proofs that fail on large numbers due to element divergence?

To make it clear.

P[SUB]7[/SUB]#/2-2[SUP]n[/SUP] can be proven to be prime for all results less than 49. No primality testing is required.
Of course this will become useless for larger numbers when the two sides fail to diverge.[/QUOTE]

Idiot

a1call 2015-10-23 23:18

[QUOTE=alpertron;413523]There are too many people who claim to be able to generate primes of any size, because of the prizes. A decade ago, other people claimed that they could factorize a number of 1000 digits, or so, thus trying to get the money from the RSA challenge.

So if you think that your algorithm always generate huge primes, you should post the algorithm in order to be analyzed. There are some people in this forum that can check very quickly whether you are losing your time or you have something that hundreds of number theorists have not found in centuries.[/QUOTE]

As mentioned I am hoping to submit my theorems for publication at some point, so I am not going to disclose them here in the open.
However, I am perfectly willing to share the theorems along with their proofs to any large-integer-computation-savvy entities that would be willing to assist me.

alpertron 2015-10-23 23:23

[QUOTE=a1call;413527]That thread is too long for my time availability.
Was it based on a proof based on number ranges or was it something else?
Are there any similar proofs that fail on large numbers due to element divergence?

To make it clear.

|P[SUB]7[/SUB]#/2-2[SUP]n[/SUP]| can be proven to be prime for all results less than 49. No primality testing is required.
Of course this will become useless for larger numbers when the two sides fail to diverge.[/QUOTE]

This type of sequences f(n) where f(n) is a polynomial or n is located also in the exponent cannot generate prime numbers for all n.

For polynomials this is very easy to see: suppose that f(n) = p where n is an integer, p a prime number and f() is not constant. Then f(n+kp) is always multiple of p. At most deg(f) values of k will make f(n+kp) equal to p, so there are infinite values for the integer k for which the value of that polynomial f(n+kp) is not prime.

If you want someone to program your algorithm, this person should be completely sure that the algorithm is sound. He will not necessarily believe you, so you will need to publish it somewhere with peer review before the coding start. In general this problem does not exist, because the mathematician can also program the algorithms.

a1call 2015-10-23 23:46

Hello [B]alpertron[/B],
I am aware of the proof for impossibility of regular as well as irregular polynomials generating prime numbers for all values of n.
However there are no proofs that other formulas will fail as well. An example would be Mills' formula, which is generally accepted to generate primes if the constant can be calculated.
To be clear my theorem has nothing to do with Mills' formula. My theorem is very similar to the example I posted above and you quoted(but can be made to diverge for large n).
It obviously does work for n= 6 and n= 7 and it can be proven that it would work for any results less than 49 which happen to be 41 and 23.
A quick proof would be that the two sides do not have a common divisor and thus their sum can not divide any of their prime factors which are all the positive primes less than or equal to 7. Thus any sum that is less than 7[SUP]2[/SUP] has to be prime. No primality testing is required.

alpertron 2015-10-24 00:10

With respect to Mills, it is clear that the constant was computed from prime numbers, so it cannot be used to generate prime numbers.

It is like saying that the sequence N(10[SUP]n[/SUP]) generates always prime numbers (where N(k) means the next prime after k). The sequence exists because of [URL="https://en.wikipedia.org/wiki/Bertrand%27s_postulate"]Bertrand's postulate[/URL]. But it is not easily computable, especially in the billion-digit range.

Your example does not generalize to higher primorials: for example 11*7*5*3-2[SUP]n[/SUP] = 1155 - 2[SUP]n[/SUP]. You get 1155-1024 > 121, so you get no primes using your method.

a1call 2015-10-24 00:30

[QUOTE=alpertron;413536]With respect to Mills, it is clear that the constant was computed from prime numbers, so it cannot be used to generate prime numbers.

It is like saying the the sequence N(10[SUP]n[/SUP]) generates always prime numbers (where N(k) means the next prime after k). The sequence exists because of [URL="https://en.wikipedia.org/wiki/Bertrand%27s_postulate"]Bertrand's postulate[/URL]. But it is not easily computable, especially in the billion-digit range.

Your example does not generalize to higher primorials: for example 11*7*5*3-2[SUP]n[/SUP] = 1155 - 2[SUP]n[/SUP]. You get 1155-1024 > 121, so you get no primes using your method.[/QUOTE]

I only mentioned Mills' formula to point out formulas can generate all primes and there is no proof that no formula will do so. Not knowing the constant is irrelevant to the the fact that there is no proof that there can be no formula that can generate all primes.
There are other formulas as well. Regardless my theorem as mentioned has nothing to do with Mills' formula.

Your sum should be 131 and 131 is not less than 11[SUP]2[/SUP] and does not apply to the proof I made. Your examples fails to diverge to less than 121. But I already pointed that out before. The example is not my theorem but is similar. The difference is that my theorem can be made to diverge to [B]less than[/B] the largest prime factor squared.

Uncwilly 2015-10-24 00:31

[QUOTE=a1call;413531]As mentioned I am hoping to submit my theorems for publication at some point, so I am not going to disclose them here in the open.[/QUOTE]
Posting here gives you a time stamp on your work. It will establish that you were the first to bring your method to light. And there are enough reliable people in the field that frequent this board that any claims otherwise can be dismissed.
:popcorn:

a1call 2015-10-24 00:42

[QUOTE=Uncwilly;413538]Posting here gives you a time stamp on your work. It will establish that you were the first to bring your method to light. And there are enough reliable people in the field that frequent this board that any claims otherwise can be dismissed.
:popcorn:[/QUOTE]

That is true and I planned to submit the theorems for publication. Until I came across the EFF prize. The official rules explicitly disqualify any theorems from winning the prize and require a specific prime. I find it idiotic to publish a thereon which will almost immediately enable someone with the right computing power to claim the EFF prize. Still if there are no takers I will go with the submit for publication option.

alpertron 2015-10-24 00:45

[QUOTE=a1call;413537]I only mentioned Mills' formula to point out formulas can generate all primes and there is no proof that no formula will do so. Not knowing the constant is irrelevant to the the fact that there is no proof that there can be no formula that can generate all primes.
There are other formulas as well. Regardless my theorem as mentioned has nothing to do with Mills' formula.

131 so 131 is not less than 11[SUP]2[/SUP] and does not apply to the proof I made. Your examples fails to diverge to less than 121. But I already pointed that out before. The example is not my theorem but is similar. The difference is that my theorem can be made to diverge to [B]less than[/B] the largest prime factor squared[/QUOTE]
You have a strange definition of divergence. But I can say that for larger primorials, it will be more difficult to find "divergence". Since p# is about e[SUP]n[/sup], you are trying to find a difference of two numbers of about n/log(10) digits to be less than n[SUP]2[/SUP]. For big values of n this is very difficult to find.

Since your prime is bounded by n[SUP]2[/SUP], in order to find the billion-digit prime number, you will need to find the difference of two numbers of about 10[SUP]5*10^8[/SUP]/log(10) digits. Forget it.

a1call 2015-10-24 00:54

[QUOTE=alpertron;413542]You have a strange definition of divergence. But I can say that for larger primorials, it will be more difficult to find "divergence". Since p# is about e[SUP]n[/sup], you are trying to find a difference of two numbers of about n/log(10) digits to be less than n[SUP]2[/SUP]. For big values of n this is very difficult to find.[/QUOTE]

My apologies. By divergence I mean divergence of a sum to less than a value (largest prime factor squared in this case).
The example that I provided is not only difficult to diverge, but virtually impossible. That is why I posted it on the open forum, since it will not make anyone claim any prizes.
However, my theorem can result in a formula that will diverge to a value where the associated proof confirms the output number is prime.

a1call 2015-10-24 00:57

My bad [B]alpertron[/B],
I meant [B]converge[/B]. Sorry about that


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