mersenneforum.org complement of si
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 2017-02-03, 09:17 #1 PAT291   Feb 2017 Belgium 2 Posts complement of si when s0 = 2, s1 = 2*s0*s0-1, s2= 2*s1*s1-1 and so on everybody will recognize lucas-lehmer test M5=31 and divides s3 M7=127 and divides s5 lots of people are looking if Mp is prime or not? but what with the complement? s3/M5 = 607 i'm pretty sure that s3/M5, s5/M7, s11/M13, s15/M17, s17/M19, s29/M31 and s59/M61 are prime. in the last case, this means a number with more than 1.000.000.000 numbers (for those numbers i have proof) (for the moment i'am looking for a prime number with just a little over 1.000.000.000 numbers with similar methodology, creating such a number will take several months or even more time. maybe i can find proof more easily than creating the number :-) ... in my rare spare time ...)
2017-02-03, 16:05   #2
retina
Undefined

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Jun 2006
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32·647 Posts

Quote:
 Originally Posted by PAT291 (for the moment i'am looking for a prime number with just a little over 1.000.000.000 numbers ...
Unless I misunderstood something it looks like you already got it.
Quote:
 Originally Posted by PAT291 s3/M5, s5/M7, s11/M13, s15/M17, s17/M19, s29/M31 and s59/M61 are prime. in the last case, this means a number with more than 1.000.000.000 numbers (for those numbers i have proof)
BTW: Those little things that make up a number are called digits.

2017-02-03, 16:59   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

8,369 Posts

Quote:
 Originally Posted by PAT291 when s0 = 2, s1 = 2*s0*s0-1, s2= 2*s1*s1-1 and so on everybody will recognize lucas-lehmer test M5=31 and divides s3 M7=127 and divides s5 lots of people are looking if Mp is prime or not? but what with the complement? s3/M5 = 607 i'm pretty sure that s3/M5, s5/M7, s11/M13, s15/M17, s17/M19, s29/M31 and s59/M61 are prime. in the last case, this means a number with more than 1.000.000.000 numbers (for those numbers i have proof) (for the moment i'am looking for a prime number with just a little over 1.000.000.000 numbers with similar methodology, creating such a number will take several months or even more time. maybe i can find proof more easily than creating the number :-) ... in my rare spare time ...)
the one thing I can say is that you can find out a few modular classes they are in all mersennes above 7 are 8x+7 from and all values in the sequence after 7 are 8k+1 so we can prove a remainder on division by 8 for all the divisions at very least.

2017-02-04, 07:30   #4
axn

Jun 2003

4,723 Posts

Quote:
 Originally Posted by PAT291 when s0 = 2, s1 = 2*s0*s0-1, s2= 2*s1*s1-1 and so on everybody will recognize lucas-lehmer test
Interesting... I do NOT recognize it as LL test, and yet, it seems to work!

Quote:
 Originally Posted by PAT291 s3/M5, s5/M7, s11/M13, s15/M17, s17/M19, s29/M31 and s59/M61 are prime.
This is true for s3 & s5. It is not true for s11, s15 & s17. They are all composites (1168, 18737 and 74961 digits respectively). These are things you could have easily checked.
It is highly unlikely that you will find any more primes in this sequence.

2017-02-04, 12:07   #5
science_man_88

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Jul 2009
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Quote:
 Originally Posted by axn Interesting... I do NOT recognize it as LL test, and yet, it seems to work! This is true for s3 & s5. It is not true for s11, s15 & s17. They are all composites (1168, 18737 and 74961 digits respectively). These are things you could have easily checked. It is highly unlikely that you will find any more primes in this sequence.
it's a reduced form. if you realize all of the normal LL test numbers are even, and you know you mersenne numbers won't divide into 2 then you have x^2-2 -> 4*y^2-2 now both parts so the factor fo 2 we can divide it out and get 2*y^2-1 with a start of 2. ( and for the even valued seeds we can do the same 10 creates a start of 5 etc.)

2017-02-04, 13:22   #6
a1call

"Rashid Naimi"
Oct 2015
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1,931 Posts

Quote:
 Originally Posted by axn Interesting... I do NOT recognize it as LL test, and yet, it seems to work! This is true for s3 & s5. It is not true for s11, s15 & s17. They are all composites (1168, 18737 and 74961 digits respectively). These are things you could have easily checked. It is highly unlikely that you will find any more primes in this sequence.
What do "they"/"these" refer to exactly?
S15 is composite?
Or
S15/M17 is composite?

ETA
Coming from axn and including S3 (which is not claimed to be prime), logically it should refer to S15/M17.
I just wished members here would be a bit more descriptive for comprehension of folks like myself.

Last fiddled with by a1call on 2017-02-04 at 13:45

2017-02-05, 05:14   #7
axn

Jun 2003

111638 Posts

Quote:
 Originally Posted by a1call What do "they"/"these" refer to exactly?
I PRP tested s11/M13, s15/M17, and s17/M19, and all three turned out to be composite. I thought it was obvious (from the quoted portion that preceded) that all references were to s3/M5, s5/M7, etc., but I can see now how it cold be confusing. Sorry.

The "these" refer to basic things that people should be doing before making "bold statements".

2017-02-05, 06:03   #8
Batalov

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Mar 2008
Phi(4,2^7658614+1)/2

23×7×163 Posts

Quote:
 Originally Posted by PAT291 when s0 = 2, s1 = 2*s0*s0-1, s2= 2*s1*s1-1 and so on everybody will recognize lucas-lehmer test
Yep, it is the Lucas-Lehmer test with bit rotation-by-1: sj = normal_Sj >> 1. Functionally identical to LL. Doing bit rotation mod Mp is nothing new (Prime95 does it all the time, as you all know).

The cofactor of sp-2/Mp doesn't seem to have any reason to be prime. Sometimes it may be, most of the time it won't.

 2017-02-05, 13:15 #9 PAT291   Feb 2017 Belgium 2 Posts on the basis, I started with some sequence f0 = 7 f1 = 7*2702-f0 ... fi=fi-1*2702-fi-2 ... the strange thing was that 7 + i*24 seemed to be prime when it was a divisor of fi it worked for the first thousands and thousands and more i and I still don't know where it goes wrong the first time fi mod (7+i24) = 0 seems to hold in most cases (primes of the form 7 + i*24) although I know sometimes you get 0 not on i-th place and apparently there are cases when you get 0 and 7 + i*24 is not a prime ... it was to good to be true (like most of the times)
 2017-02-05, 14:05 #10 henryzz Just call me Henry     "David" Sep 2007 Cambridge (GMT/BST) 10110011001112 Posts An alternative formulation for the Lucas-Lehmer test: $$s_i=\sqrt2s_{i-1}+s_{i-2}; s_0=2, s_1=\sqrt2$$ $$M_p=2^p-1$$ is prime iff $$s_{2^{n-1}}\equiv0 (mod M_p)$$ This is an example of a lucas-lehmer sequence which is a generalization of lucas sequences. The normal formula that we know doubles i. $$s_{2i}=s_i^2-2$$ See http://siauliaims.su.lt/pdfai/2009/Ericksen-09.pdf for further reading.
2017-02-15, 02:02   #11
henryzz
Just call me Henry

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Sep 2007
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Quote:
 Originally Posted by henryzz An alternative formulation for the Lucas-Lehmer test: $$s_i=\sqrt2s_{i-1}+s_{i-2}; s_0=2, s_1=\sqrt2$$ $$M_p=2^p-1$$ is prime iff $$s_{2^{n-1}}\equiv0 (mod M_p)$$ This is an example of a lucas-lehmer sequence which is a generalization of lucas sequences. The normal formula that we know doubles i. $$s_{2i}=s_i^2-2$$ See http://siauliaims.su.lt/pdfai/2009/Ericksen-09.pdf for further reading.
It appears that the even iterations can be worked out using:
$$s_i=4s_{i-2}-s_{i-4}; s_0=2, s_2=4$$

Looks like I am reinventing the wheel. https://oeis.org/A003500 is already connected to https://oeis.org/A003010 Not surprising really.
The standard way of referring to this Lucas sequence of the second kind is $$V_n(4,1)$$
It should be possible to work out SNFS polynomials for this sequence. I am unsure whether this will just boil down to x^2-2 and x^4-4x^2+2 for n a power of 2.