Bases 33-100 reservations/statuses/primes

[gd_barnes]

Originally Posted by MyDogBuster

I'm using that file.

I assume you know that post was intended for Max. He only reserved 1 of 2 k's remaining on Sierp base 23 and since the file contains both k's, I'm wondering if he missed it.

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.

lol, just what does this machine like?

[MyDogBuster]

lol, just what does this machine like?

Not much. The cooling fan runs constantly in 5th gear. I think it knows that everytime I go to a computer store, it's life is in jeopardy. A perfect machine for long term prime searching. Set it and forget it.

[kar_bon]

new PRP's

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...

[gd_barnes]

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

[gd_barnes]

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

[Flatlander]

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.

[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. :-)

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.



Gary

All that we have done is shown that 37moduloorder73 is 2. Hardly remarkable. 37 is going to be part of a 4-cover set, along with a combination of 5,13, or 41 for which modulo order base 73 is in each case 4. The smallest CRT solution using 4-cover is k=408, using 37, 5, 13. The question is, is there a prime free power series for a smaller k, possibly using a very large cover.

As part of the CRT calculation 37 needs to appear in the power series as a factor, which it does when k is 1mod37 or 36mod37.

This is therefore not a algebraic issue, in the sense of what you define it to be. There exist a lot of k that will have 37 contributing to eliminating odd or even n; or 5,13,41 eliminating one in every four n. Most k that are difficult to eliminate have only selected n not covered by primes that are of low modulo order in that base.

In the general case, we are saying that for odd bases b, then the integer x={[b+1]/2} is 2 modulo order base b, and consequently two values of k == 1modx or-1modx provide part cover for n = even or odd respectively.

[gd_barnes]

Originally Posted by Flatlander

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.

Yep, I missed the reservation on bases 93 and 100. Are you still keeping them reserved? If so, how high will you be taking them?

Right now, I've got you reserved for bases 72, 93, and 100.


Gary

[gd_barnes]

Originally Posted by robert44444uk

All that we have done is shown that 37moduloorder73 is 2. Hardly remarkable. 37 is going to be part of a 4-cover set, along with a combination of 5,13, or 41 for which modulo order base 73 is in each case 4. The smallest CRT solution using 4-cover is k=408, using 37, 5, 13. The question is, is there a prime free power series for a smaller k, possibly using a very large cover.

As part of the CRT calculation 37 needs to appear in the power series as a factor, which it does when k is 1mod37 or 36mod37.

This is therefore not a algebraic issue, in the sense of what you define it to be. There exist a lot of k that will have 37 contributing to eliminating odd or even n; or 5,13,41 eliminating one in every four n. Most k that are difficult to eliminate have only selected n not covered by primes that are of low modulo order in that base.

In the general case, we are saying that for odd bases b, then the integer x={[b+1]/2} is 2 modulo order base b, and consequently two values of k == 1modx or-1modx provide part cover for n = even or odd respectively.

A couple of things here:

I realize I have thrown around the term "contains algebraic factors" a little loosely at times. Frequently when I've stated that, I should have been saying:

"Trivial factors that combine with algebraic factors to make a full covering set."

I'm not sure what you mean by a "prime free power series" or a "CRT solution".

We do know this about base 73:
We have searched all k where k is not == (1 mod 2) nor (1 mod 3), which have a trivial factor of 2 and 3 respectively. All remaining k's had a prime (after eliminating k=36) so the "conjecture" of k=408 is now proven. The last k to fall was k=242 with a prime at n=2280.

I did an extensive analysis last night of k's that were perfect squares for Riesel base 73 all the way up to k=50^2. Surprisingly, there were no other k's except for k=36 where trivial factor(s) on odd-n combined with algebraic factors on even-n to make a full covering set. (I think I said it right that time. lol) I suspect there are more k's for this base that have such factors but I didn't go high enough to find them and don't quite have the math skills to generalize it out further using modulo arithmetic or additional "kinds" of algebraic factors for this specific base.

Also about my post that you quoted: It was rather ludicrous and proved nothing. I stated as such in my last post where I actually analyzed the modulo arithmetic involved specifically for 36*73^n-1 about this issue with the following:

"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"

So thanks for not attacking me too badly on it.


Gary

[robert44444uk]

Originally Posted by gd_barnes

A couple of things here:

I realize I have thrown around the term "contains algebraic factors" a little loosely at times. Frequently when I've stated that, I should have been saying:

"Trivial factors that combine with algebraic factors to make a full covering set."

Understood

Originally Posted by gd_barnes

I'm not sure what you mean by a "prime free power series" or a "CRT solution".

Prime free power series = The power series k*b^n-1 are all composite, k,b fixed, n variable

CRT= Chinese Remainder Theorem

Originally Posted by gd_barnes

We do know this about base 73:
We have searched all k where k is not == (1 mod 2) nor (1 mod 3), which have a trivial factor of 2 and 3 respectively. All remaining k's had a prime (after eliminating k=36) so the "conjecture" of k=408 is now proven. The last k to fall was k=242 with a prime at n=2280.

I realise now I have been barking up the wrong tree, concentrating my comments on odd n, whereas the algebraic factors arise out of even n:

36*73^n-1 = 6^2*73^n (even) is square therefore factors algebraically.

Originally Posted by gd_barnes

I did an extensive analysis last night of k's that were perfect squares for Riesel base 73 all the way up to k=50^2. Surprisingly, there were no other k's except for k=36 where trivial factor(s) on odd-n combined with algebraic factors on even-n to make a full covering set. (I think I said it right that time. lol) I suspect there are more k's for this base that have such factors but I didn't go high enough to find them and don't quite have the math skills to generalize it out further using modulo arithmetic or additional "kinds" of algebraic factors for this specific base.

For such a condition to arise, let some y=x^2, x integer
Then y*b^2a-1, a=integer, factorises into (x*b^a-1)(x*b^a+1)
This provides algebraic cover at n=even

We also require y+1 to be a prime that is 2 modulo order b, and this is provided by the primitive prime factors of b^2-1 = primitive prime factors of (b+1), of which, for b=73, there is 1, p=37

For cover at n odd in k*b^n-1, then k==ymodp==36mod37
Or alternatively x^2==-1mod37..... x^2+1=0mod37
The only valid solution is given by x=6 (x=31 provides that 31^2*b^n-1=even, and therefore trivial)

Think thats right

[gd_barnes]

Riesel base 55 is at n=22K; one prime so far:

2022*55^19568-1 is prime


This is somewhat disappointing. I had expected at least 3 primes for n=15K-25K on this very prime base. With only 13 k's remaining on a conjecture of k=6852, it should be kicking out primes pretty quickly at all ranges. Perhaps at least 2 will come in the next n=3K range.


Gary