The "Hugo"-est Prime Number in the world
 

The Largest Prime Number in the world
We can find bigger prime numbers with this method instead of using Mersenne method.
[URL="https://www.researchgate.net/publication/348730147_The_Largest_Prime_Number_in_the_world"]https://www.researchgate.net/publication/348730147_The_Largest_Prime_Number_in_the_world[/URL]

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.

[QUOTE=Hugo1177;569985]The Largest Prime Number in the world
We can find bigger prime numbers with this method instead of using Mersenne method.
[URL="https://www.researchgate.net/publication/348730147_The_Largest_Prime_Number_in_the_world"]https://www.researchgate.net/publication/348730147_The_Largest_Prime_Number_in_the_world[/URL][/QUOTE]

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.

Generating Large Prime Numbers
 

Generating Large Prime Numbers
We can find bigger prime numbers with this method.
[URL="https://www.researchgate.net/publication/349080188_Generating_Large_Prime_Numbers"]https://www.researchgate.net/publication/349080188_Generating_Large_Prime_Numbers[/URL]

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

We'll wait.

[QUOTE=retina;571045]Please go ahead and find a "bigger prime number" than M82589933.

We'll wait.[/QUOTE]
On risque d'attendre longtemps, je propose:
M82589934!::smile::smile:

[QUOTE=Hugo1177;569985]The Largest Prime Number in the world
We can find bigger prime numbers with this method instead of using Mersenne method.
[URL="https://www.researchgate.net/publication/348730147_The_Largest_Prime_Number_in_the_world"]https://www.researchgate.net/publication/348730147_The_Largest_Prime_Number_in_the_world[/URL][/QUOTE]

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

[QUOTE=Hugo1177;571041]Generating Large Prime Numbers
We can find bigger prime numbers with this method.
[URL="https://www.researchgate.net/publication/349080188_Generating_Large_Prime_Numbers"]https://www.researchgate.net/publication/349080188_Generating_Large_Prime_Numbers[/URL][/QUOTE]


2161 is the next term of your interesting sequance::smile:
{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.

[QUOTE=Hugo1177;571041]Generating Large Prime Numbers
We can find bigger prime numbers with this method.
[URL="https://www.researchgate.net/publication/349080188_Generating_Large_Prime_Numbers"]https://www.researchgate.net/publication/349080188_Generating_Large_Prime_Numbers[/URL][/QUOTE]

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

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

[QUOTE=Kebbaj;571055]On risque d'attendre longtemps, je propose:
M82589934!::smile::smile:[/QUOTE]82 589934 = 2 × 3 × 7 × 1 966427
so M82589934 or M(82589934!) mod 3 = 0.
Not a long wait.

[url]https://www.mersenneforum.org/showpost.php?p=512813&postcount=4[/url]

[QUOTE=kriesel;571077]82 589934 = 2 × 3 × 7 × 1 966427
so M82589934 or M(82589934!) mod 3 = 0.
Not a long wait.

[url]https://www.mersenneforum.org/showpost.php?p=512813&postcount=4[/url][/QUOTE]
M8258994, it was just for fun.
thank you!

On the other hand, could you tell if [COLOR="Red"]25463[/COLOR] 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,[COLOR="red"].....[/COLOR]

[QUOTE=Kebbaj;571073]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 * ...[/QUOTE]

Did Prof. Caldwell tell you that was a requirement? There are several primes in that database with only an ECPP proof. There's [URL="https://primes.utm.edu/top20/page.php?id=27"]an entire top 20 page of them[/URL].

I also ran a third or so of a t30 (ECM) on the cofactor of that number (which is nowhere near small for factorization, where 200 digits is huge for non-special forms), with no dice.

Other closed form of the Hugo sequence and observations
 

There is a much simpler closed form for the Hugo Numbers than what has been posted by Hugo and the others, they are equal to:

(Sum from k=0 to k=(p-1)/2 of (Binomial(p, 2*k)*13^k))/2^(n-1)

It can also be shown that the Hugo Numbers are, in the limit, a geometric series, with first term 1/2(1+sqrt(13)) and common ratio 1/2(7+sqrt(13)).

There are also Mersenne-like primality characteristics of this sequence that I have observed (Hugo(y*x) = Hugo(y) * Hugo(x) * cofactor unless y*x is a perfect power, divisors of Hugo(n) with N prime of specific forms either 2*k*p+1 or 2*k*p-1)

[QUOTE=retina;571045]Please go ahead and find a "bigger prime number" than M82589933.

We'll wait.[/QUOTE]

Okay, let me try

2^82589933+80875047
2^82589933+148316925
2^82589933+205790751
2^82589933+326805255
2^82589933+382879239
2^82589933+432870647
2^82589933+516518021
2^82589933+768187217
2^82589933+785549457
2^82589933+861000575
2^82589933+919430651
2^82589933+937150935
2^82589933+966499157
2^82589933+974862225
2^82589933+981502331
2^82589933+984925185
2^82589933+1016742659
2^82589933+1114725015
2^82589933+1189890905
2^82589933+1315732101
2^82589933+1337194461
2^82589933+1386199751
2^82589933+1402635029
2^82589933+1602826059
2^82589933+1628081609
2^82589933+1674437519
2^82589933+1711934037
2^82589933+1724772015
2^82589933+1988025401
2^82589933+2081143619
2^82589933+2101268115
2^82589933+2161866755
2^82589933+2182473311
2^82589933+2223069261
2^82589933+2248242831

Sorry, I really couldn't resist. It was an unfinished project in my drawer.
Please don't waste any of your precious time. :wink:

[QUOTE=Happy5214;571089]Did Prof. Caldwell tell you that was a requirement? There are several primes in that database with only an ECPP proof. There's [URL="https://primes.utm.edu/top20/page.php?id=27"]an entire top 20 page of them[/URL].

I also ran a third or so of a t30 (ECM) on the cofactor of that number (which is nowhere near small for factorization, where 200 digits is huge for non-special forms), with no dice.[/QUOTE]


Hugo compare his numbers with Mersenne primes which are very large,
if you wanted to have fun proving a number of 1 million digits with ECPP,
have fun alone.
:smile:

[QUOTE=mart_r;571112]2^82589933+80875047
<snip more random numbers>
2^82589933+2248242831[/QUOTE]Go ahead and prove those are prime. We'll wait.

[QUOTE=Kebbaj;571114]Hugo compare his numbers with Mersenne primes which are very large,
if you wanted to have fun proving a number of 1 million digits with ECPP,
have fun alone.
:smile:[/QUOTE]

Sorry, I had a migraine last night and got my wires crossed with the number in that post, which I was trying to factor [i]p[/i]-1 for (which is clearly not top 5k). :redface: Yeah, no chance to prove a 1M+ digit number with ECPP.

[QUOTE=retina;571117]Go ahead and prove those are prime. We'll wait.[/QUOTE]

I'll see to it. As soon as Elon Musk's hyper-space sub-particle Auger flux quantum computer is available.