The "Hugo"-est Prime Number in the world

[Hugo1177]

The Largest Prime Number in the world
We can find bigger prime numbers with this method instead of using Mersenne method.
https://www.researchgate.net/publica...r_in_the_world

[axn]

You say that 2/(29+13*sqrt(5)) = 1/29. Basically an irrational number = a rational number. That is nonsense.

You haven't mentioned how you obtain your "Hugo" number for a given prime index. For example, how did you get Hugo(31) = 169746333457? Please explain the algorithm.

You haven't mentioned how you test/prove a "Hugo" number prime. Please explain the algorithm and its runtime complexity.

Finally, you're comparing the n'th "Hugo" prime to n'th Mersenne prime. That is meaningless. In fact, it merely shows that "Hugo" primes are rarer than Mersenne primes. That's actually a bad thing.

A class of numbers is suitable for prime finding if there is a fast algorithm for proving them and they have higher than average probability (compared to normal numbers of same size) of being prime. You have not discussed either of these aspects.

[Viliam Furik]

Originally Posted by Hugo1177

The Largest Prime Number in the world
We can find bigger prime numbers with this method instead of using Mersenne method.
https://www.researchgate.net/publica...r_in_the_world

Alright, say you get a number (potential prime, according to your method) with a million digits. How quickly can you prove it's prime? Unless you can do it in under 10 minutes, nobody here will be interested.

[Hugo1177]

Generating Large Prime Numbers
We can find bigger prime numbers with this method.
https://www.researchgate.net/publica..._Prime_Numbers

[retina]

Please go ahead and find a "bigger prime number" than M82589933.

We'll wait.

[Kebbaj]

Originally Posted by retina

Please go ahead and find a "bigger prime number" than M82589933.

We'll wait.

On risque d'attendre longtemps, je propose:
M82589934!:

[Kebbaj]

Originally Posted by Hugo1177

The Largest Prime Number in the world
We can find bigger prime numbers with this method instead of using Mersenne method.
https://www.researchgate.net/publica...r_in_the_world

not bad,
but we can better calculate 2 ^p-1 and test it?

[Kebbaj]

Originally Posted by Hugo1177

Generating Large Prime Numbers
We can find bigger prime numbers with this method.
https://www.researchgate.net/publica..._Prime_Numbers

2161 is the next term of your interesting sequance:
{2, 3, 5, 7, 13, 19, 31, 43, 61, 67, 107, 127, 521, 631, 1307, 1619, 2161}

The problem is the proof! /

For exemple the term : 2161 which gives a prime number p = 669520952365187012801734773019730340904793566535457142176946739102321502083823099233857142333677031001451069065917971920110393630255782323617275383
212943748551754288384925199630904669367417248891744299566974703517081390275022885309341549249834446417882777235101767661916361693393317355973961109
361740840788086806308743519390960721768560920934486478379127638410123621994632671617516986749436025270882460170612028677207126354071432082804057852
021567294421457263622220997051953206302416515543266086449620395149849240412230511874765922927106048987389425173585248222630010387862113520387585603
833357934162639402504305302892121580407987900770723163143549061197642212314715010925653640972117561263171417335716694379646482116883767036278222571
382751325272483972072844735317113952796000534401

K.caldwel will not accept it, if you do not know how to factoring p-1 or p + 1 in the helper.

factoring p-1 is difficult for this small number:
2 * 7 * 3 * 1 * 5 * 2 * 7 * 1 * 11 * 1 * 13 * 1 * 17 * 1 * 19 * 1 * 31 * 1 * 61 * 1 * 163 * 1 * 181 * 1 * 433 * 1 * 2161 ** 8641 * 1 * 151201 * 84313972619 * 1163620706029 * ...

we do not talking if it is a large number.

[Kebbaj]

Originally Posted by Hugo1177

Generating Large Prime Numbers
We can find bigger prime numbers with this method.
https://www.researchgate.net/publica..._Prime_Numbers

You can make this interesting sequance in OEIS https://oeis.org

If you want the mathematica code for publication :
d = {}; For[i = 1, i <= 326, i++, a = Prime[i];
sol = Solve[x^(2/a) - 3 x^(4/a) + 1 == 0, x];
p = ToExpression[
StringCases[ToString[Numerator[RootReduce[(x /. sol[[1]])^2]]],
WordCharacter ..][[1]]]; If[PrimeQ[p], AppendTo[d, a]]]; d

[kriesel]

Originally Posted by Kebbaj

On risque d'attendre longtemps, je propose:
M82589934!:

82 589934 = 2 × 3 × 7 × 1 966427
so M82589934 or M(82589934!) mod 3 = 0.
Not a long wait.

https://www.mersenneforum.org/showpo...13&postcount=4

[Kebbaj]

Originally Posted by kriesel

82 589934 = 2 × 3 × 7 × 1 966427
so M82589934 or M(82589934!) mod 3 = 0.
Not a long wait.

https://www.mersenneforum.org/showpo...13&postcount=4

M8258994, it was just for fun.
thank you!

On the other hand, could you tell if 25463 is the next one on Hugo's sequance?
2,3,5,7,13,19,31,43,61,67,107,127,521,631,1307,1619,2161,.....