[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.