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Old 2008-12-09, 03:27   #122
robert44444uk
 
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Quote:
Originally Posted by gd_barnes View Post
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

Is this a reasonable conjecture, I am really uncertain about this.

Last fiddled with by robert44444uk on 2008-12-09 at 03:56
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Old 2008-12-09, 04:34   #123
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Quote:
Originally Posted by robert44444uk View Post

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
Instead of the above,

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
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Old 2008-12-09, 14:02   #124
Flatlander
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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
File Type: zip rbase68-results.zip (10.9 KB, 67 views)
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Old 2008-12-09, 22:00   #125
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Quote:
Originally Posted by Flatlander View Post
Riesel base 68 is proven.
Congrats!

How does it feel, proving a conjecture? :)
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Old 2008-12-09, 22:06   #126
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Quote:
Originally Posted by michaf View Post
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.
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Old 2008-12-10, 04:24   #127
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Quote:
Originally Posted by Flatlander View Post
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
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Old 2008-12-11, 08:12   #128
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Quote:
Originally Posted by gd_barnes View Post
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.

First, we'll start with the obvious:

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
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Old 2008-12-12, 09:26   #129
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Default 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
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Old 2008-12-12, 11:07   #130
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Quote:
Originally Posted by kar_bon View Post
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

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Old 2008-12-12, 11:12   #131
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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
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Old 2008-12-12, 22:51   #132
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Quote:
Originally Posted by Flatlander View Post
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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