Bases 33-100 reservations/statuses/primes

[Flatlander]

Taking Riesel Base 68 to n=20,000.

[robert44444uk]

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

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

[robert44444uk]

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

Instead of the above,

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

[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.

[michaf]

Originally Posted by Flatlander

Riesel base 68 is proven.

Congrats!

How does it feel, proving a conjecture? :)

[Flatlander]

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.

[gd_barnes]

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

[MyDogBuster]

Okay, time to get my feet wet.

Reserving Riesel Base 45 k's 372,1264 & 1312 from 10K to 100K
also Riesel Base 45 k=24 from 50K to 100K

Starting out small incase my research into how to do this is flawed.

[gd_barnes]

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.

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

[gd_barnes]

Originally Posted by MyDogBuster

Okay, time to get my feet wet.

Reserving Riesel Base 45 k's 372,1264 & 1312 from 10K to 100K
also Riesel Base 45 k=24 from 50K to 100K

Starting out small incase my research into how to do this is flawed.

This is actually quite a bit of work for such a high base unless you find a couple of primes quickly. Figured I better let you know...

Check with Max on testing using Phrot if you haven't used it already. On my Linux machines, I'm getting a 40% speed boost on my Sierp base 12 effort that is at n>210K right now.


Gary

[MyDogBuster]

Be sure and use the sieve file attached to the reservations web page. It includes both k's remaining on this base.

I'm using that file.

Check with Max on testing using Phrot if you haven't used it already.

I'm using Phrot 0.51. I put this stuff on my lone AMD quad. That quad hates 3.7.1c, Sieve, LLRNET and anything else I put on it. Maybe it will like Phrot.