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Old 2021-02-07, 15:28   #12
Happy5214
 
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Quote:
Originally Posted by Kebbaj View Post
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 * ...
Did Prof. Caldwell tell you that was a requirement? There are several primes in that database with only an ECPP proof. There's an entire top 20 page of them.

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.
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Old 2021-02-07, 15:55   #13
NHoodMath
 
Jan 2017

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Default 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)
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Old 2021-02-07, 20:54   #14
mart_r
 
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you know...around...

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Quote:
Originally Posted by retina View Post
Please go ahead and find a "bigger prime number" than M82589933.

We'll wait.
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.
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Old 2021-02-07, 21:21   #15
Kebbaj
 
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Quote:
Originally Posted by Happy5214 View Post
Did Prof. Caldwell tell you that was a requirement? There are several primes in that database with only an ECPP proof. There's an entire top 20 page of them.

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.

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.

Last fiddled with by Kebbaj on 2021-02-07 at 21:46
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Old 2021-02-07, 22:21   #16
retina
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Quote:
Originally Posted by mart_r View Post
2^82589933+80875047
<snip more random numbers>
2^82589933+2248242831
Go ahead and prove those are prime. We'll wait.

Last fiddled with by retina on 2021-02-07 at 22:22
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Old 2021-02-08, 04:33   #17
Happy5214
 
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Quote:
Originally Posted by Kebbaj View Post
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.
Sorry, I had a migraine last night and got my wires crossed with the number in that post, which I was trying to factor p-1 for (which is clearly not top 5k). Yeah, no chance to prove a 1M+ digit number with ECPP.
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Old 2021-02-08, 18:03   #18
mart_r
 
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Quote:
Originally Posted by retina View Post
Go ahead and prove those are prime. We'll wait.
I'll see to it. As soon as Elon Musk's hyper-space sub-particle Auger flux quantum computer is available.
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