Presentation and doubts with the application and competition of 100000000 prime digits

[apzyx]

Originally Posted by Prime95

Welcome aboard. Prime95 can do a *probable* prime test on numbers of the for k*b^n+c where k,b,n,c are all "small". There are other programs that may or may not be able to prove those numbers prime depending on the values of k,b,n,c.

There are no programs that can prove an arbitrary 100,000,000 digit number prime.

Is there an easy way to use the API for the k*b^n+c expression? I'm reading how to do it but it's crazy parameters, isn't there something like the Assignments section with these parameters on the web www.mersenne.org?

[paulunderwood]

Originally Posted by apzyx

Is there an easy way to use the API for the k*b^n+c expression? I'm reading how to do it but it's crazy parameters, isn't there something like the Assignments section with these parameters on the web www.mersenne.org?

There is no API for 100,000,000 digit k*b^n+c candidates. There is only an interface for 2^p-1.

For k*b^n+c you have to write your own. What k,b,n and c are you trying to test?

[kruoli]

You can manually enter something like PRP=N/A,7,17,112108343,17 into worktodo.txt. This will test \(7\cdot{}17^{112{,}108{,}343}+17\). So the format is PRP=N/A,k,b,n,c.

[paulunderwood]

Originally Posted by kruoli

You can manually enter something like PRP=N/A,7,17,112108343,17 into worktodo.txt. This will test \(7\cdot{}17^{112{,}108{,}343}+17\). So the format is PRP=N/A,k,b,n,c.

I can say that number is not prime by doing the calculations in my head.

c must be -1 or +1 for any sizable number to be provable.

The real point here is that testing should be done by using the 2^p-1 server. Trial factoring will have been done and sometimes P-1.

The chance of claiming the prize money for a 100,000,000 digit prime is very small. It s far better for a new comer to join the herd and crunch a wavefront number -- "World Record Prime" -- and hope for a lesser monetary prize yet plenty of fame.

[kruoli]

Of course it is divisible by 17. It was only an example.

Yes, this way we can only prove compositeness, and yes, for reaching the goal of a 100 million digit prime, I would also suggest using the PrimeNet assignment system and according candidates.

Yes, I would also recommend our "normal" wavefront work, but we can only suggest things and do not force anyone.

[apzyx]

Today this possible prime number can not be validated in prime95, it has an n of more than one integer, there I leave it for the future or for someone who knows how to check this numbers =)

The possible prime is: (2^3486784401) + 3486784401

I have a theory that:

2^3 + 3 = 11 is prime
2^(3^4) + (3^4) = 2^81 + 81 = 2417851639229258349412433 is prime

2^((3^4)^5) + ((3^4)^5) = (2^3486784401) + 3486784401 = ?

or

2^(3^(4^5)) + (3^(4^5)) = 2^373391848741020043532959754184866588225409776783734007750636931722079\
04061726525122999368893880397722046876506543147515810872705459216085858\
13513369828091873141917485942625809388070199519564042855718180410466812\
88797402925517668012340617298396574731619152386723046235125934896058590\
58828465479354050593620237654780744273058214452705898875625145281779341\
33521419207446230275187291854328623757370639854853194764169262638199728\
87006907013899256524297198527698749274196276811060702333710356481 + 37339184874102004353295975418486658822540977678373400775063693172207904\
06172652512299936889388039772204687650654314751581087270545921608585813\
51336982809187314191748594262580938807019951956404285571818041046681288\
79740292551766801234061729839657473161915238672304623512593489605859058\
82846547935405059362023765478074427305821445270589887562514528177934133\
52141920744623027518729185432862375737063985485319476416926263819972887\
006907013899256524297198527698749274196276811060702333710356481 = ?

This theory is complicated to prove hahahahah

Pérez Sánchez, Pablo
apzyx@yahoo.com
03/05/2022

[mathwiz]

Originally Posted by apzyx

Today this possible prime number can not be validated in prime95, it has an n of more than one integer, there I leave it for the future or for someone who knows how to check this numbers =)

The possible prime is: (2^3486784401) + 3486784401

I have a theory that:

2^3 + 3 = 11 is prime
2^(3^4) + (3^4) = 2^81 + 81 = 2417851639229258349412433 is prime

2^((3^4)^5) + ((3^4)^5) = (2^3486784401) + 3486784401 = ?

Your number is roughly a billion decimal digits. And not of a form that can be efficiently tested, like 2^n-1.

Your theory has no mathematical basis, other than that the progression is true for 2 terms.

As such, I doubt you'll find much interest here in trying to test it.

[charybdis]

Originally Posted by apzyx

The possible prime is: (2^3486784401) + 3486784401

Doesn't look very prime to me, it has two 7-digit factors:

gp > forprime(p=2,10^7,if(Mod(2,p)^3486784401+3486784401==0,print(p)))
1454521
4965049

[LaurV]

Originally Posted by kriesel

At the time I had posted that, his post count was shown...

[offtopic]
You were not seeing things, that's a long time known joke from Mike, for the new users before their first posts being approved or so, the numbers of posts being shown is an arbitrary large number. After that, the counting resumes properly. I think Unc knows that, but she is playing with you
[/offtopic]

[LaurV]

Originally Posted by kruoli

Of course it is divisible by 17. It was only an example.

Oh... I though is even, as you add two odds

[kruoli]

Originally Posted by LaurV

Oh... I though is even, as you add two odds

Man sieht den Wald vor lauter Bäumen nicht.