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2008-12-09, 03:27   #122
robert44444uk

Jun 2003
Oxford, UK

2×953 Posts

Quote:
 Originally Posted by gd_barnes Oh, that's easy. It's just faster to use Alptertron's site. lol 36*73^1 = 2628 == (1 mod 37) 36*73^2 = 191844 == (36 mod 37) 36*73^3 = 14004612 == (1 mod 37) 36*73^4 = 1022336676 == (36 mod 37) [etc.] Now, when you subtract 1 from the above, you get (0 mod 37) for the odd n's. :-) Gary
I am not good at modular arithmetic, but does it more generally follow that:

show for integer x that (x-1)*(2x-1)^n-1 is 0modx for n odd

(x-1)*(2x-1)^n-1==-1modx*(-1mod2x)^n-1

== -1modx*(-1modx)^n-1

== -1modx^(n+1)-1

if n even

== -1modx-1=-2modx

if n odd

== 1modx-1=0modx

If this is the case then there is always Riesel algebraic factor partial cover provided through odd n for any odd base b when k=(b+1)/2 -1

Last fiddled with by robert44444uk on 2008-12-09 at 03:56

2008-12-09, 04:34   #123
robert44444uk

Jun 2003
Oxford, UK

190610 Posts

Quote:
 Originally Posted by robert44444uk If this is the case then there is always Riesel algebraic factor partial cover provided through odd n for any odd base b when k=(b+1)/2 -1

There is always Riesel factor x provided through odd n for any odd base b when k== -1modx

Last fiddled with by robert44444uk on 2008-12-09 at 04:36

2008-12-09, 14:02   #124
Flatlander
I quite division it

"Chris"
Feb 2005
England

31·67 Posts

Riesel base 68 is proven.

Confirmed primes:
5*68^13574-1
7*68^25395-1

I'll push Riesel base 101 higher and take Riesel base 93 to at least n=20k.
Attached Files
 rbase68-results.zip (10.9 KB, 67 views)

2008-12-09, 22:00   #125
michaf

Jan 2005

1110111112 Posts

Quote:
 Originally Posted by Flatlander Riesel base 68 is proven.
Congrats!

How does it feel, proving a conjecture? :)

2008-12-09, 22:06   #126
Flatlander
I quite division it

"Chris"
Feb 2005
England

31×67 Posts

Quote:
 Originally Posted by michaf Congrats! How does it feel, proving a conjecture? :)
Thanks.
Well I really just finished it off using software others have written but it feels good! Like finding my first top 5000 prime.

I'd like to try one from scratch soon but I need to do some more research.

2008-12-10, 04:24   #127
gd_barnes

May 2007
Kansas; USA

280116 Posts

Quote:
 Originally Posted by Flatlander Riesel base 68 is proven. Confirmed primes: 5*68^13574-1 7*68^25395-1 I'll push Riesel base 101 higher and take Riesel base 93 to at least n=20k.

VERY nice! To find 2 primes out of 2 remaining k for n=10K-25.4K is a most excellent find! We have our first proof where the final prime was n>25K in a long time!

Gary

2008-12-11, 08:12   #128
gd_barnes

May 2007
Kansas; USA

72×11×19 Posts

Quote:
 Originally Posted by gd_barnes Oh, that's easy. It's just faster to use Alptertron's site. lol 36*73^1 = 2628 == (1 mod 37) 36*73^2 = 191844 == (36 mod 37) 36*73^3 = 14004612 == (1 mod 37) 36*73^4 = 1022336676 == (36 mod 37) [etc.] Now, when you subtract 1 from the above, you get (0 mod 37) for the odd n's. :-) I don't believe it's a true proof in the classical sense of the word, but it demonstrates conclusively that all odd n are divisible by 37. BTW, I'm pretty sure I got Alpertron's site to work on one of my Linux machines. I'll check it when I get home. But...I thought you had a Windows machine also. Gary
First of all, Robert, thanks for what appears to be a nice generalized proof about "numeric" factors that apply to many bases that can combine with algebraic factors to make a full covering set for particular k's.

To all:

I understand a little bit about what Robert posted and should be able to grasp it all after studying it in detail a little later when I have more time. For the time being, the above that I stated is not really a proof at all, it essentially just restates the fact that all odd-n are == (0 mod 37) in a different way. When I thought about it, it was kind of a stupid post by me. I'm glad the higher-math folks didn't attack me. They could have easily. lol

For the time being, here is a better analysis of how the modulo arithmetic works on this that provides what I believe to be a "generalized" proof of this specific base if you can really call it generalized for only one base. I'm calling it that because it proves ALL n for this base whereas my poor previous attempt at a proof did nothing but "prove" that the first 4 n's had a repeating pattern that may or may not repeat for higher n-values.

73 == (36 mod 37)

Use the above with variables as in:
a == (b mod c)

Using elementary modulo arithmetic, we have:

To square a above, you would multiply (b mod c) by a, which would yield a^2 == (a*b mod c), where a*b reduces to some value in the range of 0 thru c-1.

Therefore:
We have 73^2 == (36*73 mod 37) == (2628 mod 37) == (1 mod 37).

Going one more exponent, you have 73^3 = 73^2*73 and since 73^2 == (1 mod 37), you have:
73^3 == (1*73 mod 37) == (73 mod 37) == (36 mod 37)

Here, you can stop, because you have a repeated mod. The exponents of 1 and 3 are == (36 mod 37). You now know that all 73^n where n is odd will be == (36 mod 37) and where n is even will be == (1 mod 37).

We know that the mods repeat every 2n for 73^n but we don't know if there is a trivial factor for the original form of 36*73^n-1 at this point. Therefore:

73^1 == (36 mod 37); multiplying that by 36 and subtracting 1 gives:
36*73^1-1 == (36*36-1 mod 37) == (1295 mod 37) == (0 mod 37); therefore a factor of 37.

73^2 == (1 mod 37); multiplying that by 36 and subtracting 1 gives:
36*73^2-1 == (36*1-1 mod 37) == (35 mod 37); therefore a remainder of 35 when dividing by 37.

Since 73^n is always == (36 mod 37) when n is odd and applying a multiplier of 36 and subtracting one always gives (0 mod 37), then the above proves that all odd n give a factor of 37. And taking it further for the conjecture: Since all even-n for any k that is a perfect square always yield algebraic factors on the Riesel side, then the k is eliminated from consideration.

After grasping what Robert has done here a little better, I should avoid missing "basic" trivial factors that combine with algebraic factors to make a full covering set such as what I did on base 73.

Gary

Last fiddled with by gd_barnes on 2008-12-11 at 21:18

 2008-12-12, 09:26 #129 kar_bon     Mar 2006 Germany 2×1,427 Posts Riesel Base 35 new PRP's Code: 87064 8031 64760 8046 244466 8054 179312 8064 132574 8065 131290 8091 206962 8091 223562 8146 17468 8160 70526 8164 249296 8174 62008 8195 122156 8212 46120 8239 89338 8245 272306 8268 227564 8278 46288 8279 81772 8305 21346 8327 57848 8338 81532 8353 100504 8365 135062 8366 101648 8368 165608 8378 36214 8431 at n=8444 with 4.55M candidates left to n=100k to Gary: the PRP's from post 149 are missing on your reservation-page for base 35. jerky job.... no time to update the pages... i know that too... Last fiddled with by kar_bon on 2008-12-12 at 09:28
2008-12-12, 11:07   #130
gd_barnes

May 2007
Kansas; USA

72·11·19 Posts

Quote:
 Originally Posted by kar_bon new PRP's at n=8444 with 4.55M candidates left to n=100k to Gary: the PRP's from post 149 are missing on your reservation-page for base 35. jerky job.... no time to update the pages... i know that too...

Thanks for letting me know. You were doing too many posts for base 35! I'm bound to miss some of them. It wasn't a matter of time. I completely missed that post.

I'm sure I would have noticed when I went to post these primes; wondering why there was a large n-range gap in primes.

As I sit, I've been updating the Riesel pages for bases > 50 so I'll get these and post 149 for base 35 added in a little while.

Gary

Last fiddled with by gd_barnes on 2008-12-12 at 11:42

 2008-12-12, 11:12 #131 gd_barnes     May 2007 Kansas; USA 72×11×19 Posts To all: I have my web pages updated for Riesel bases up to 80 now. What takes quite a while is correctly generalizing the algebraic factors. I had already found them and took them into account on the k's remaining that I posted previously. But to properly prove a conjecture, they need to be generalized, especially if we eventually go to proving the 2nd or 3rd conjectured k-value in the future. Quite a few of the bases have them that allowed k's to be removed. The one that took the longest was Riesel base 54. It has 2 different sets of algebraic factors that are very similar to Riesel base 24. Both kinds were needed to eliminate k=4, 6, and 9. I've haven't uploaded the pages yet but will do so shortly. I'm beat now and will do some more updating Friday afternoon. Gary Last fiddled with by gd_barnes on 2008-12-12 at 11:16
2008-12-12, 22:51   #132
Flatlander
I quite division it

"Chris"
Feb 2005
England

1000000111012 Posts

Quote:
 Originally Posted by Flatlander Riesel base 68 is proven. Confirmed primes: 5*68^13574-1 7*68^25395-1 I'll push Riesel base 101 higher and take Riesel base 93 to at least n=20k.
I think you missed my reservation.

Bases 93 and 100 are both at n>37k with no primes. :surprised It must be Max's turn to find one.

I'll also take Riesel base 72 to at least n=20k.

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