[quote=MrOzzy;150940] prove the first, second, third, ... conjectured k for one specific base with a lot of small conjectured k (Riesel base 68 for example has a conjectured k at k=22, 43, 142, 185, 783, 1394, 3051)
[/quote]

This is a very interesting idea that I have toyed around with at different times but never stuck with it very long. Riesel and Sierp base 8 would be interesting bases to attack to prove the 2nd/3rd/etc. conjectured k's since their 1st one is so low and was already easily proven. Also, since base 8 is a power of 2, LLRing would be fast.


Gary

Riesel update
 

Hi Gary,

I am comparing my results with your excellent pages. Here is the difference:

Riesel base 49: I searched until n = 116,000, sieved until n = 200,000.
Riesel base 36: I reached the end of my reservation at n = 25,000. These primes are not yet listed on the pages:
107819*36^24637-1
94152*36^24621-1
114403*36^24366-1
61040*36^24332-1
43215*36^23692-1
100937*36^23147-1
80733*36^22504-1

Cheers, Willem.

I have been working on many of the easier Riesel bases 51-100. Below is a general synopsis of what I have done and found. I will show more details a little later on. All bases were searched to n=10K or until proven unless otherwise stated below.

I am listing any reservations after the k's remaining.

[code]
base k's remaining
53 [128 k's remaining at n=10K; shown on web pages]
54 proven
55 68, 2330, 3060, 3158, 3240, 3578, 3942, 4640, 4878,
5286, 5354, 5690, & 6098 at n=25K
56 proven
57 proven
59 proven
60 [81 k's remaining at n=10K; shown on web pages]
61 198, 1520, 1644, 6168, 9642, 10572, & 10968 at n=25K
62 proven
65 proven
67 242, 1274, 1886, 2228, 2846, & 2906 at n=25K
68 proven by Flatlander; final prime at n=25395 !
69 proven
70 729, 1776, 2202, & 5468 at n=25K
72 4 at n=54K; sieve file available to n=100K
73 proven
74 proven
76 proven
77 proven
80 10, 31, 106, 170, & 214 at n=10K
81 proven
83 proven
84 proven
86 proven
87 172, 186, 384, 472, 536, 562, 672, 714, 848, 862, 1004, 1112,
1132, 1154, 1418, & 1628 at n=20K; sieve file available to n=100K
89 proven
90 proven
92 proven
93 424 & 452 at n=50K; sieve file available to n=100K
94 29 at n=51K
98 proven
99 proven
100 653 at n=60K; sieve file available to n=150K
[/code]Unless shown as reserved, all of the above bases are now unreserved. Anyone can feel free to take them further at this point. [B]The web pages are now fully updated with the details of this effort.[/B] I will also continue updating this post for a while until work on these particular bases comes to a near stand still.


Gary

[quote=gd_barnes;151849]I have been working on many of the easier Riesel bases 51-100. Below is a general synopsis of what I have done and found. I will show more details a little later on. All bases were searched to n=10K or until proven.

The only exception to the n=10K search was base 55, which has now been searched to n=15K. I am going to reserve base 55 up to n=25K.

[code]
base k's remaining
53 [128 k's remaining; to be shown later]
54 proven
55 [14 k's remaining; to be shown later]
56 proven
57 proven
59 proven
62 proven
65 proven
68 5 & 7
69 proven
72 4, 79, 116, & 291
73 36
74 proven
76 proven
77 proven
80 10, 31, 106, 170, & 214
81 proven
83 proven
84 proven
86 proven
87 172, 186, 384, 472, 536, 562, 672, 714, 758, 848, 862, 898, 958,
1004, 1112, 1132, 1154, 1418, & 1628
89 proven
90 proven
92 proven
93 424 & 452
94 29
98 proven
99 proven
100 74 & 653
[/code]
All of the above bases are now unreserved except for base 55. Anyone can feel free to take them further at this point. Within a couple of weeks, I'll get the web pages updated with the detailed info.


Gary[/quote]
Hmm...I think I'll reserve k=36 on Riesel base 73 for at least up to n=20K. With only one k left, that one looks like I might have a pretty decent chance of proving my first conjecture. :grin:

[quote=mdettweiler;151852]Hmm...I think I'll reserve k=36 on Riesel base 73 for at least up to n=20K. With only one k left, that one looks like I might have a pretty decent chance of proving my first conjecture. :grin:[/quote]
Hmm...I just started sieving on this a few minutes ago, and oddly enough, after only 6 CPU minutes of sieving having been done, there were only 20 remaining k/n pairs in the sieve! :shock: Here's the srsieve output:[code]srsieve 0.6.9 -- A sieve for integer sequences in n of the form k*b^n+c.
srsieve started: 10000 <= n <= 20000, 3 <= p <= 4000000000003
Split 1 base 73 sequence into 2 base 73^12 subsequences.
WARNING: 36*73^n-1 has algebraic factors.
WARNING: 36*73^n-1 has algebraic factors.
p=409039997, 6801123 p/sec, 9973 terms eliminated, 28 remain
p=846160019, 7272243 p/sec, 9977 terms eliminated, 24 remain
p=1297080067, 7507826 p/sec, 9980 terms eliminated, 21 remain
p=1754920133, 7615436 p/sec, 9980 terms eliminated, 21 remain
p=2217960077, 7713347 p/sec, 9981 terms eliminated, 20 remain
p=2685040051, 7774818 p/sec, 9981 terms eliminated, 20 remain
p=3156040061, 7841635 p/sec, 9981 terms eliminated, 20 remain
Sieving 3 <= p <= 3163531061 eliminated 9981 terms, 20 remain.
Wrote 20 terms for 1 sequence to abc format file `sr_73.pfgw'.
srsieve stopped: at p=3163531061 because SIGINT was received.
[/code]
Either the sieve somehow got completely messed up, or this is one heck of a low-weight k. :smile:

Okay, that was easy. Riesel base 73 is now complete to n=20K. :grin: Results are attached.

Heck, if it's this easy, I'll just go on ahead and take it up to n=50K! :smile:

[quote=mdettweiler;151857]Hmm...I just started sieving on this a few minutes ago, and oddly enough, after only 6 CPU minutes of sieving having been done, there were only 20 remaining k/n pairs in the sieve! :shock: Here's the srsieve output:[code]srsieve 0.6.9 -- A sieve for integer sequences in n of the form k*b^n+c.
srsieve started: 10000 <= n <= 20000, 3 <= p <= 4000000000003
Split 1 base 73 sequence into 2 base 73^12 subsequences.
WARNING: 36*73^n-1 has algebraic factors.
WARNING: 36*73^n-1 has algebraic factors.
p=409039997, 6801123 p/sec, 9973 terms eliminated, 28 remain
p=846160019, 7272243 p/sec, 9977 terms eliminated, 24 remain
p=1297080067, 7507826 p/sec, 9980 terms eliminated, 21 remain
p=1754920133, 7615436 p/sec, 9980 terms eliminated, 21 remain
p=2217960077, 7713347 p/sec, 9981 terms eliminated, 20 remain
p=2685040051, 7774818 p/sec, 9981 terms eliminated, 20 remain
p=3156040061, 7841635 p/sec, 9981 terms eliminated, 20 remain
Sieving 3 <= p <= 3163531061 eliminated 9981 terms, 20 remain.
Wrote 20 terms for 1 sequence to abc format file `sr_73.pfgw'.
srsieve stopped: at p=3163531061 because SIGINT was received.
[/code]
Either the sieve somehow got completely messed up, or this is one heck of a low-weight k. :smile:[/quote]

Oops sorry. You can't prove your first conjecture so easily. That one is proven! I goofed. k=36 on Riesel base 73 has algebraic factors as shown by srsieve.

Sheesh, and I just now checked. That is an OBVIOUS one since it's a perfect square:
odd n: factor of 37
even n have algebraic factors: Where n=2q, it factors to (6*73^q-1)*(6*73^q+1)

Note: Not all k's that are perfect squares have algebraic factors to make a full covering set. Only on the Riesel side where all odd n have a trivial "numeric" factor like this one.

I eliminated a lot of algebraic factors on several of the bases but obviously missed this one.

I'll edit my original post. Sorry about that. Any time there are an abnormally low # of candidates remaining after sieving, it almost always means algebraic factors.


Gary

[quote=mdettweiler;151862]Okay, that was easy. Riesel base 73 is now complete to n=20K. :grin: Results are attached.

Heck, if it's this easy, I'll just go on ahead and take it up to n=50K! :smile:[/quote]


It looks like you posted this just ahead of my post. Don't waste your time! Like I said, any time there is an abnormally low # of candidates remaining upon sieving, there are likely algebraic factors, hence eliminating the k from consideration. The base is proven.

Edit: It looks like bases 109 and 110 are ripe for a proof.

An interesting note about base 110: k=36 on that base has exactly the same numeric and algebraic factors as does k=36 for base 73. That is a factor of 37 for odd n and where the base = b and n=2q for even n: (6*b^q-1)*(6*b^q+1). That's strange that I found this one and not the one for base 73. Brain fart I guess.


Gary

[quote=gd_barnes;151864]Oops sorry. You can't prove your first conjecture so easily. That one is proven! I goofed. k=36 on Riesel base 73 has algebraic factors as shown by srsieve.

Sheesh, and I just now checked. That is an OBVIOUS one since it's a perfect square:
odd n: factor of 37
even n have algebraic factors: Where n=2q, it factors to (6*73^q-1)*(6*73^q+1)

Note: Not all k's that are perfect squares have algebraic factors to make a full covering set. Only on the Riesel side where all odd n have a trivial "numeric" factor like this one.

I eliminated a lot of algebraic factors on several of the bases but obviously missed this one.

I'll edit my original post. Sorry about that. Any time there are an abnormally low # of candidates remaining after sieving, it almost always means algebraic factors.


Gary[/quote]
Whoops--okay, I'll be sure to keep on the lookout for those low-#-of-candidates k's in the future. :smile:

One quick question though: how do you know that all odd n have a factor of 37? It's probably something blindingly obvious but I'm missing it. :smile:

[quote=mdettweiler;151866]Whoops--okay, I'll be sure to keep on the lookout for those low-#-of-candidates k's in the future. :smile:

One quick question though: how do you know that all odd n have a factor of 37? It's probably something blindingly obvious but I'm missing it. :smile:[/quote]


The web page that I've expoused many times here:

[URL]http://www.alpertron.com.ar/ECM.HTM[/URL]


Just plug in your form for n=1, 3, 5, 7, 9, etc. You'll see it has a factor of 37 every time. It takes any normal mathematical symbols. I use the page extensively and almost exclusively for determining such things. You just have to see the patterns in the way the factors occur. The factor patterns are one of the most interesting things to me about these conjectures.


Gary

[quote=gd_barnes;151868]The web page that I've expoused many times here:

[URL]http://www.alpertron.com.ar/ECM.HTM[/URL]


Just plug in your form for n=1, 3, 5, 7, 9, etc. You'll see it has a factor of 37 every time. It takes any normal mathematical symbols. I use the page extensively and almost exclusively for determining such things. You just have to see the patterns in the way the factors occur. The factor patterns are one of the most interesting things to me about these conjectures.


Gary[/quote]
Ah, I see. I assumed that there was some kind of fancy mathematical formula that proved that all odd n had a factor of 37 or something like that. :smile:

At this time I haven't been able to get the ECM factoring applet you linked to to work under Linux on my computer (I've used it before under Windows), but I guess the important thing is that now I understand how you came to conclude that all odd n for k=36 have a factor of 37. :smile:

[quote=mdettweiler;151869]Ah, I see. I assumed that there was some kind of fancy mathematical formula that proved that all odd n had a factor of 37 or something like that. :smile:

[B]At this time I haven't been able to get the ECM factoring applet you linked to to work under Linux on my computer (I've used it before under Windows)[/B], but I guess the important thing is that now I understand how you came to conclude that all odd n for k=36 have a factor of 37. :smile:[/quote]
I have problems with it too. (Windows.)
But the batch one at the bottom of the page works okay.

I too thought Gary was using fancy stuff. Cheat!

[quote=mdettweiler;151869]Ah, I see. I assumed that there was some kind of fancy mathematical formula that proved that all odd n had a factor of 37 or something like that. :smile:

At this time I haven't been able to get the ECM factoring applet you linked to to work under Linux on my computer (I've used it before under Windows), but I guess the important thing is that now I understand how you came to conclude that all odd n for k=36 have a factor of 37. :smile:[/quote]


Oh, that's easy. It's just faster to use Alptertron's site. lol :smile:

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

[quote=gd_barnes;151995]Oh, that's easy. It's just faster to use Alptertron's site. lol :smile:

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[/quote]
Ah, I see now. :smile:

As for Alpertron's site on Linux machines: the main problem on my machine in particular is that I think something might be broken on the backend that's causing errors whenever I try to set up the Java browser plugin in Firefox. As you said, I did have a Windows virtual machine set up that I could use, but it seems that an upgrade of the virtualization software broke that too... :ermm: (I blame Windows for requiring a reinstall due to the virtual "IDE controller" having changed ever-so-slightly...but I guess the VirtualBox people could have put in some sort of compatibility mode pretty easily if they wanted to. :smile:)

However, I just found that under Linux I can use PARI/GP (a handy-dandy freeware math application that can do all sorts of stuff) to factor these numbers at least as easily as with Alpertron's applet. :smile:

Just for fun, I also tested Riesel base 94, k=29 to 20,000. Results attached. (No sieve file. No reservation. Stopping there.)

[quote=Flatlander;152017]
Just for fun, I also tested Riesel base 94, k=29 to 20,000. Results attached. (No sieve file. No reservation. Stopping there.)[/quote]


Thanks for the nice sieve file. I'm compiling a list of things to do while on a business trip for doing when I get back next Tuesday. I'll make sure I get the sieve file posted to the web pages after getting them updated for all of these bases.

I think I've stuck a chord here...easier to prove bases. I may continue doing this for the Sierp side as well as for some bases > 125. Proving bases is fun! :-)


Gary

How many of the 19 ks for Reisel base 87 would you 'expect' to fall before n=20,000?

[quote=Flatlander;152151]How many of the 19 ks for Reisel base 87 would you 'expect' to fall before n=20,000?[/quote]

In other words, I've already started testing this base to n=20,000. :whistle:

[QUOTE=Flatlander;152151]How many of the 19 ks for Reisel base 87 would you 'expect' to fall before n=20,000?[/QUOTE]

I'd guess 3.

Willem.

[quote=gd_barnes;150987]...
There is a thread that has all of the conjectured values for all bases on both sides up to 1024...
Gary[/quote]
Can someone post a link to these values please. :smile:

[url]http://robert.gerbicz.googlepages.com/sierpinski.txt[/url]
[url]http://robert.gerbicz.googlepages.com/riesel.txt[/url]

Hope this helps!

KEP

[quote=Siemelink;152186]I'd guess 3.

Willem.[/quote]


I would concur with that. I'd say it's slightly more likely that you'll get 4 primes than 2 primes so an expection of a bit more than 3. For info. purposes, here are the top 10 primes that I found up to n=10K that gives a general idea of how they have fallen:

1128*87^3171-1
1002*87^3257-1
1398*87^3699-1
1426*87^3778-1
1284*87^4444-1
1186*87^5151-1
102*87^5508-1
1586*87^6195-1
628*87^6371-1
508*87^9016-1

This is another one of those highly misleading bases. It is unlikely to be proven in several hundred CPU years; believe it or not! :-) Except for highly prime bases, any base that is not a power-of-2 that has more than 10 k's remaining at n=10K will be virtually impossible to prove anytime in the near future and for most of them, not likely in most of our lifetimes without new concepts or math on finding primes; even taking into account increases in computer speeds.

So far, we've only had 2 bases proven with conjectures > 1000...Sierp bases 11 at k=1490 and 21 at k=1002. But we have two bases with only one k remaining that would shatter those records: Sierp base 9 and 10. Proving Sierp base 10 would be a great coupe! With a conjecture of k=9175, it would be an outstanding proof! :-)


Gary

I have added and udpated search limits and reservations to the original post about these bases 50-100. I will continue doing so until I get my web pages updated with all of the info.

Let me know if you see any problems.


Gary

Riesel Base 87
 

[quote=KEP;152196][URL]http://robert.gerbicz.googlepages.com/sierpinski.txt[/URL]
[URL]http://robert.gerbicz.googlepages.com/riesel.txt[/URL]

Hope this helps!

KEP[/quote]
Thanks for the links Kenneth! Just what I was looking for. When Gary's got 5 minutes I'm sure he'll add all this to his website. lol

I have tested Riesel Base 87 to 20,000. Three primes were found, good estimates guys!
Verified primes:
758*87^13638-1
898*87^14455-1
958*87^17047-1

(Though Gary suggested there should be 3-and-a-bit primes, and I can't find the bit...)

I have also sieved a file to p=560bn, n=100k. I estimate that this is sufficient to test further to about n=28,000. (Hopefully by then another prime will have been found making further sieving faster.)

Unreserving this er, conjecture.


edit:
My estimate above is based on the fact that tests at n=20,000 took one minute and the sieve was rejecting candidates about every two minutes; 28^2 is about twice 20^2.

edit2:
The exact sieve depth is 560973747863

Taking Riesel Base 68 to n=20,000.

[QUOTE=gd_barnes;151995]Oh, that's easy. It's just faster to use Alptertron's site. lol :smile:

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[/QUOTE]

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.

[QUOTE=robert44444uk;152579]

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

[/QUOTE]

Instead of the above,

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

Riesel base 68 is proven. :smile:

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.

[quote=Flatlander;152607]Riesel base 68 is proven. :smile:
[/quote]

Congrats!

How does it feel, proving a conjecture? :)

[quote=michaf;152652]Congrats!

How does it feel, proving a conjecture? :)[/quote]
Thanks. :smile:
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.

[quote=Flatlander;152607]Riesel base 68 is proven. :smile:

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.[/quote]


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! :smile:


Gary

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.

[quote=gd_barnes;151995]Oh, that's easy. It's just faster to use Alptertron's site. lol :smile:

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[/quote]

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

[quote=MyDogBuster;152797]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.[/quote]


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... :smile:

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

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

I'm using that file.

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

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.

[quote=MyDogBuster;152860]I'm using that file.[/quote]

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.

[quote]
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.[/quote]

lol, just what does this machine like?

[QUOTE]lol, just what does this machine like?
[/QUOTE]

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.

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
[/code]

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... :grin:

[quote=kar_bon;153029]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... :grin:[/quote]


Thanks for letting me know. You were doing too many posts for base 35! I'm bound to miss some of them. :smile: 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

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

[quote=Flatlander;152607]Riesel base 68 is proven. :smile:

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

I'll push Riesel base 101 higher [B]and take Riesel base 93 to at least n=20k.[/B][/quote]
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.

[QUOTE=gd_barnes;151995]Oh, that's easy. It's just faster to use Alptertron's site. lol :smile:

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[/QUOTE]

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.

[quote=Flatlander;153110]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.[/quote]


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

[quote=robert44444uk;153136]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.[/quote]


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. :smile:


Gary

[QUOTE=gd_barnes;153140]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."
[/QUOTE]

Understood

[QUOTE=gd_barnes;153140]
I'm not sure what you mean by a "prime free power series" or a "CRT solution".
[/QUOTE]

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

CRT= Chinese Remainder Theorem


[QUOTE=gd_barnes;153140]
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.
[/QUOTE]

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.

[QUOTE=gd_barnes;153140]
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.
[/QUOTE]

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

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

[QUOTE=robert44444uk;153163]
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[/QUOTE]

So x is 6mod37 or 31mod37, and the first non trivial partial algebraic factorisation after x=6 is provided by x=228=6mod37 and k= 228^2=51984, well above the lowest Riesel value.

The lowest 31mod37 value that is non trivial is x=438, k=191844

[quote=gd_barnes;153139]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[/quote]
Sorry I missed your post. Yes, those three reservations are correct.

Status:
Base 72 primes for:
116*72^13887-1
291*72^26322-1
79*72^28009-1
k=4 tested to >41000. (I assume it doesn't have falgebraic actors? :wink:)

Base 93 tested to >46000, no primes.

Base 100 prime for:
74*100^44709-1 :showoff:
k=653 tested to >55,000

I'll take base 100 to 60k, bases 93 and 72 to 50k. (edit: at least)

[quote=Flatlander;153441]Sorry I missed your post. Yes, those three reservations are correct.

Status:
Base 72 primes for:
116*72^13887-1
291*72^26322-1
79*72^28009-1
k=4 tested to >41000. (I assume it doesn't have falgebraic actors? :wink:)

Base 93 tested to >46000, no primes.

Base 100 prime for:
74*100^44709-1 :showoff:
k=653 tested to >55,000

I'll take base 100 to 60k, bases 93 and 72 to 50k. (edit: at least)[/quote]


Impressive amount of work Chris, especially on base 72. To get that to one k remaining is excellent.

I finally finishing reviewing Riesel base 3 early this morning and getting the web pages updated accordingly. After doing some other things this afternoon, I should have time to finish updating the pages for all of these Riesel bases up to 125. In the mean time, I'm still updating that one post here for the work done.


Gary

[quote=Flatlander;153441]Sorry I missed your post. Yes, those three reservations are correct.

Status:
Base 72 primes for:
116*72^13887-1
291*72^26322-1
79*72^28009-1
k=4 tested to >41000. (I assume it doesn't have falgebraic actors? :wink:)

Base 93 tested to >46000, no primes.

Base 100 prime for:
74*100^44709-1 :showoff:
k=653 tested to >55,000

I'll take base 100 to 60k, bases 93 and 72 to 50k. (edit: at least)[/quote]


4*72^n-1 is very low weight but eventually should yield a prime. Any Riesel k that is a perfect square will always be lower weight than average because the even-n have algebraic factors.

For this one, many of the odd-n's are also knocked out by factors of 7 and 17 as follows:
n==(1 mod 3) have a factor of 7
n==(3 mod 4) have a factor of 17

This leaves only: n==(5, 9, 17, 21) mod 24 that don't have algebraic factors nor a factor of 7 or 17.

I'm seeing plenty of random factors in k's with those mods so there should be a prime at some point.

Here is an interesting side note: To show how low weight this form is, there are very few n's that are yielding just 2 prime factors. I only found 3 up to n=100:
4*72^1-1 = 7*41
4*72^15-1 = 17*1,704,505,929,743,291,922,743,496,463
4*72^57-1 = 35,406,097*(99-digit prime)

n=57 is the only n<=100 that contains only 2 prime factors that don't include a 7 or 17.

I earlier did a test on all bases <= 1024 for primes at k=2. The lowest base I found so far without a prime for k=2 was b=170 on the Riesel side and b=101 on the Sierp side (tested to n=25K). I wonder if Riesel base 72 is the lowest base that does not have a prime for k=4? I may check that eventually. Of course we don't count the bases that have a proven full covering set nor a trivial factor for k=4.

Oh, BTW, if there are any even n's left after you sieve, you can manually remove them. If the smallest factor for them is > than your sieve limit, sr(x)sieve will not remove them.


Gary

Reserving Riesel base 60 up to n=10K.

Even with a conjecture nearly 4 times as high as base 53 and k's with a trivial factor only every 59 k's, by my calculations and some preliminary testing, there should be less k's remaining at n=10K.

My estimate is 100 k's remaining vs. 128 on base 53 at n=10K.

Riesel base 93 tested to 50k. No primes, unreserving.

Results and sieve to 100k attached. Tests were taking 410sec at 50k, I'm not sure of the efficiency of the sieve. Sieve depth - 114449279449. (Shown in header.)

Riesel base 100 tested to 60k. One prime already shown. Unreserving.

Results and sieve 60k-150k to p=423804689167 attached.

Riesel base 72 tested to 54k. Three primes already shown. Unreserving.

Results and sieve 54k-100k, p=227328339361 attached.

Sorry about the multiple posts, I thought I might mix things up otherwise. :smile:

Riesel base 55 is complete to n=25K; one prime already reported since n=15K. I'm unreserving this base.


Gary

I've started testing Riesel base 61 using the script in the "Starting ... 101" thread.

Conjectured k is 13484.

I'm ignoring ks for:
k == 1 mod 2
k == 1 mod 3
k == 1 mod 5

No doubt someone will tell me if this has been tested or I'm doing it wrong. :smile:

Riesel Base 35
 

new PRP's:
[code]
154468 8443 155912 8466 64424 8472 259750 8479 205112 8482 109876 8511
162410 8512 160340 8534 103728 8539 34264 8551 155492 8552 117608 8562
151988 8564 80900 8576 15484 8585 145246 8611 140240 8614 71360 8616
264626 8646 70064 8652 203798 8656 25190 8688 233194 8707 84422 8734
136046 8738 14534 8744 163336 8759 272242 8763 238234 8797 34646 8810
171122 8828 48926 8860 143492 8864 204218 8878 267722 8904 24896 8908
286844 8924 112162 8955 260852 8962 281570 8994 137918 8998 233918 9006
184926 9013 2172 9031 248340 9040 36994 9057 42758 9058 219622 9061
133048 9065 232764 9065 183626 9074 74042 9098 155482 9135 89060 9152
136666 9157 194726 9162 270052 9171 152494 9221 69002 9222
[/code]

currently at n=9231 with 4.26M candidates upto n=100k left

[quote=Flatlander;153940]I've started testing Riesel base 61 using the script in the "Starting ... 101" thread.

Conjectured k is 13484.

I'm ignoring ks for:
k == 1 mod 2
k == 1 mod 3
k == 1 mod 5

No doubt someone will tell me if this has been tested or I'm doing it wrong. :smile:[/quote]No-one has yelled at me so I'll carry on.
22 ks left at n=5000. They include k=12078 which is a multiple of the base. Do I bin it?

[QUOTE=Flatlander;154040]No-one has yelled at me so I'll carry on.
22 ks left at n=5000. They include k=12078 which is a multiple of the base. Do I bin it?

[/QUOTE]

for k's that are multiples of the base:
if k-1 = prime you have to keep it.
if k-1 is not prime you can discard it.

Willem.
run - start - calc - 12077 - He, no prime button?

[quote=Siemelink;154043]for k's that are multiples of the base:
if k-1 = prime you have to keep it.
if k-1 is not prime you can discard it.

Willem.
run - start - calc - 12077 - He, no prime button?[/quote]
Ah yes, like base 3! Thanks.

12077 = 13 * 929

i am going to take the remaining k for base 94 a bit further

[quote=Flatlander;154040]No-one has yelled at me so I'll carry on.
22 ks left at n=5000. They include k=12078 which is a multiple of the base. Do I bin it?
[/quote]

Wow, nice work Chris. I can see someone is having a wee bit of fun with this stuff now. lol I found once I got started running more-and-more bases that it was quite addictive and I didn't want to stop. :smile:

Yes, as you and Willem determined, you can bin k=12078 on base 61. (I kind of like the use of term "bin" there. I've never seen it used as a verb before.)

I'm somewhat amazed that you chose two bases that are so prime. For such high conjectures and bases, you're fortunate. Frequently you'll end up with as many as 100 or more k's remaining for higher conjectures and bases such as this. I think that choosing two of them with a lot of trivial factors helped. For example, on my base 60, there are 99 k's remaining at n=5K. Of course with a conjecture of 20558 and trivial factors for only (1 mod 59), there are a few more k's to test. :-) Still, having so few k's remaining for base 61 is good.


Gary

[B]Testing Riesel base 67.[/B]

Conjectured k is 3144.

I'm ignoring ks for:
k == 1 mod 2
k == 1 mod 3
k == 1 mod 11
Also for m.o.b. where k-1 is not prime.

8 ks left at n=5000:
[code]128*67^n-1
242*67^n-1
[I][[U][B]900[/B][/U]*67^n-1] Deleting.[/I]
1274*67^n-1
1886*67^n-1
2228*67^n-1
2244*67^n-1
2846*67^n-1
2906*67^n-1[/code]Testing to n=25k.

(Bases 61 and 211 are also being tested to n=25k. One more prime found so far for base 211, six more primes found for base 61.)

base 94 is at 30k now with no primes

[B]Riesel base 67 tested to 25k. Unreserving.[/B]

Primes > 5k:
2244*67^6600-1
[I][2244*67^9651-1[/I]]
128*67^10592-1

ks left:
242*67^n-1
1274*67^n-1
1886*67^n-1
2228*67^n-1
2846*67^n-1
2906*67^n-1

[B]Riesel base 61 tested to 25k. Unreserving.[/B]

Primes > 5k: :smile:
6354*61^7292-1
3578*61^7325-1
12182*61^8260-1
5114*61^8432-1
11192*61^9763-1
[I][11192*61^10579-1][/I]
768*61^10705-1
13274*61^11434-1
7470*61^12232-1
6914*61^12261-1
12450*61^12564-1
6230*61^12684-1
11150*61^13014-1
8958*61^17644-1
404*61^18637-1


Ks left:
198*61^n-1
1520*61^n-1
1644*61^n-1
6168*61^n-1
9642*61^n-1
10572*61^n-1
10968*61^n-1
(k=12078 was removed.)

I'll email the files (316KB).

Reserving all Riesel base 45 k's to 100K

[quote=MyDogBuster;154647]Reserving all Riesel base 45 k's to 100K[/quote]


A few CPU years work but not too bad on a quad or two. :smile:

[QUOTE]A few CPU years work but not too bad on a quad or two. :smile:
[/QUOTE]

Keeps the useless AMD busy LMAO.

base 94 is at n=40k with no primes
i will unreserve when i get to 50k

[quote=gd_barnes;154684]OK, now slow down just a little bit. You're causing me to take extra time and you know how :censored: I get when I have to take extra admin time. lol

You didn't send me a list of k's remaining for base 143. I could figure it out but...

I only see primes in one Email up to n=1000. Then in another Email there are primes for n>8700. So I need the primes for n=1000-8700.

So if you can send me both the 29 k's remaining and those primes, then I can post everything on the pages.

Edit: Well, I got on tonight thinking that I could get the pages updated in an hour...not a chance. This is a lot of work that everyone has done and I haven't even looked at any of the base 3 stuff yet. Oh well, I got in some updates for bases <= 32. I'll get to all of these late Tuesday and Wednesday during the day.


Gary[/quote]
Oops, sorry Gary. I'll fish them out. (Or start again.)

[quote=Flatlander;154714]Oops, sorry Gary. I'll fish them out. (Or start again.)[/quote]
Got them. :smile:

[B]Testing Riesel base 70.[/B]

Conjectured k is 6176.

I'm ignoring ks for:
k == 1 mod 3
k == 1 mod 23
Also for m.o.b. where k-1 is not prime.

Just 3 ks left at 5k::smile:
1776*70^n-1
2202*70^n-1
5468*70^n-1

Testing to 25k.

When I tested 125*2^n-1, I was told that it had algebraic factors. Do I need to watch out for cubes, powers of 5 etc. ? (Increasingly unlikely, I know.)

[quote=Flatlander;154755][B]Testing Riesel base 70.[/B]

Conjectured k is 6176.

I'm ignoring ks for:
k == 1 mod 3
k == 1 mod 23
Also for m.o.b. where k-1 is not prime.

Just 3 ks left at 5k::smile:
1776*70^n-1
2202*70^n-1
5468*70^n-1

Testing to 25k.

When I tested 125*2^n-1, I was told that it had algebraic factors. Do I need to watch out for cubes, powers of 5 etc. ? (Increasingly unlikely, I know.)[/quote]
I need to restart this base, I had a constant wrong in the script. (Other bases are fine.)

[quote=Flatlander;154755][B]...[/B]

When I tested 125*2^n-1, I was told that it had algebraic factors. Do I need to watch out for cubes, powers of 5 etc. ? (Increasingly unlikely, I know.)[/quote]

Riesel base 70 hase k=729 left. (9^3) Can I bin it?

[quote=Flatlander;154945]Riesel base 70 hase k=729 left. (9^3) Can I bin it?[/quote]
Srsieve says:
[quote]WARNING: 729*70^n-1 has algebraic factors.
WARNING: 729*70^n-1 has algebraic factors.
WARNING: 729*70^n-1 has algebraic factors.[/quote]Is that because it's a square, a cube and a sixth power?

After the holidays and NPLB have kept me busy the last few days, I finally have time to get back in and review and update everything for CRUS.

Any needed files will be posted for drives and the web pages will be fully updated over the next 8 hours. (yes, it will likely take that long!) :smile:


Gary

Riesel base 60 has reached n=8500. 84 k's still remain. I'll show the details in the web pages in the next few hours.

[quote=Flatlander;154956]Srsieve says:
Is that because it's a square, a cube and a sixth power?[/quote]


Srsieve will display that message on any k that is any perfect power in most conditions for a Riesel form. For Sierpinski forms, they are less common. I believe that only perfect cubes and higher can contain algebraic factors on Sierpinski forms, although I haven't checked that completely. See more explanation below.

It does not mean you can remove the k's from testing. Srsieve makes that statement when there are SOME n-values remaining that can be manually eliminated because they contain algebraic factors. Notice use of the word "SOME". It doesn't mean that all n-values can be elminated; only some of them. For the k to be removed, all n-values would have to be eliminated.

Situations in which srsieve will make that statement:

1. For ANY k on ANY base that is a perfect square, when sieving, all n-values that are divisible by 2 can be manually removed.
2. For any k on any base that is a perfect cube, all n-values that are divisible by 3 can be manually removed.
3. (etc.) for k's that are perfect 5th powers, 7th powers, 11th powers, or any prime power where p=power, any n's divisible by p can be manually removed.

Taken to an extreme, for k=2048, which is 2^11, you could manually eliminate all n-values that are divisible by 11.

To put the above in a different way: You can only eliminate the k if manually removing all of these n-values leaves you with no n's remaining, which would have been the case had you attempted to sieve 900*67^n-1. Had you sieved it, you would have ended up with a sieve file with very few EVEN n-values remaining and zero ODD n-values remaining. Once you manually removed the n's divisible by 2, you would have had nothing left. That means that the k-value can be removed from conjecture testing because it has partial algebraic factors that combine with a numeric factor to make a full "covering set" of factors.

Analysis on both k=125 and 729 shows that they should remain because there is no factor or factors that eliminate the n's that the algebraic factors do not. Sorry; you can't get away with removing them this time! lol

When sieving, as per the above, on k=125, you can manually remove all n-values divisible by 3. On k=729, that is one of the few that you can eliminate n-values that are divisible by 2 -or- that are divisible by 3. In effect, you're only left with n-values that are n==(1 or 5 mod 6). If you manually remove those n's, you'll stop getting that message from srsieve.

k=729 would normally be extremely low weight except for the fact that it is divisible by 3, which eliminates any possibility of a factor of 3 for all n-values. That increases the weight to something like a k that is a perfect square.

A k-value that would be extremely low-weight is one that is a perfect square and cube but is not divisible by 2, 3, or 5. I think the lowest one of that nature would be 7^6=117649, which would only be an issue on very few bases.


Gary

[quote=gd_barnes;155243]Srsieve will display that message on any k that is any perfect power in most conditions for a Riesel form. For Sierpinski forms, they are less common. I believe that only perfect cubes and higher can contain algebraic factors on Sierpinski forms, although I haven't checked that completely. See more explanation below.

It does not mean you can remove the k's from testing. Srsieve makes that statement when there are SOME n-values remaining that can be manually eliminated because they contain algebraic factors. Notice use of the word "SOME". It doesn't mean that all n-values can be elminated; only some of them. For the k to be removed, all n-values would have to be eliminated.

Situations in which srsieve will make that statement:

1. For ANY k on ANY base that is a perfect square, when sieving, all n-values that are divisible by 2 can be manually removed.
2. For any k on any base that is a perfect cube, all n-values that are divisible by 3 can be manually removed.
3. (etc.) for k's that are perfect 5th powers, 7th powers, 11th powers, or any prime power where p=power, any n's divisible by p can be manually removed.

Taken to an extreme, for k=2048, which is 2^11, you could manually eliminate all n-values that are divisible by 11.

To put the above in a different way: You can only eliminate the k if manually removing all of these n-values leaves you with no n's remaining, which would have been the case had you attempted to sieve 900*67^n-1. Had you sieved it, you would have ended up with a sieve file with very few EVEN n-values remaining and zero ODD n-values remaining. Once you manually removed the n's divisible by 2, you would have had nothing left. That means that the k-value can be removed from conjecture testing because it has partial algebraic factors that combine with a numeric factor to make a full "covering set" of factors.

Analysis on both k=125 and 729 shows that they should remain because there is no factor or factors that eliminate the n's that the algebraic factors do not. Sorry; you can't get away with removing them this time! lol

When sieving, as per the above, on k=125, you can manually remove all n-values divisible by 3. On k=729, that is one of the few that you can eliminate n-values that are divisible by 2 -or- that are divisible by 3. In effect, you're only left with n-values that are n==(1 or 5 mod 6). If you manually remove those n's, you'll stop getting that message from srsieve.

k=729 would normally be extremely low weight except for the fact that it is divisible by 3, which eliminates any possibility of a factor of 3 for all n-values. That increases the weight to something like a k that is a perfect square.

A k-value that would be extremely low-weight is one that is a perfect square and cube but is not divisible by 2, 3, or 5. I think the lowest one of that nature would be 7^6=117649, which would only be an issue on very few bases.


Gary[/quote]
Was that a 'yes' or a 'no'? lol
Only kidding. It'll make sense when I'm not tired. (Possibly.)

[B]Riesel base 70 tested to 25k. Unreserving.[/B]

Primes >5K:
2621*70^6247-1
5925*70^8850-1
2699*70^15455-1

Ks remaining:
729*70^n-1
1776*70^n-1
2202*70^n-1
5468*70^n-1

Tested with WinPFGW so no results.out file.

btw When I mentioned k = 125 I was refering to 125*2^n-1 I tested for RPS.

[quote=Flatlander;155273]Was that a 'yes' or a 'no'? lol
Only kidding. It'll make sense when I'm not tired. (Possibly.)

[B]Riesel base 70 tested to 25k. Unreserving.[/B]

Primes >5K:
2621*70^6247-1
5925*70^8850-1
2699*70^15455-1

Ks remaining:
729*70^n-1
1776*70^n-1
2202*70^n-1
5468*70^n-1

Tested with WinPFGW so no results.out file.

btw When I mentioned k = 125 I was refering to 125*2^n-1 I tested for RPS.[/quote]


Riesel bases 67 and 70 would be excellent bases to make team efforts out of. They're proof is just challenging enough to make them interesting but not so challenging that they'll die out. I'd expect a final prime on them in the n=100K-500K range somewhere barring some very good or bad luck on them...still very large but not ridiculous.

We won't do more team drives right now but they are definite possibilities for the future.


Gary

Reserving Sierpinski base 63 :smile: Can anyone tell me which k's can be excluded from testing beyond n=1000?

Here is how I'm going to do it:

1. Strict test all k<=37565866 up to n=1000 (already in progress)
2. Remove the k's that can be excluded before sieving
3. Sieve all remaining candidates for n>1000 to n<=25000
4. PRP test all candidate-pairs to n=25000
5. Verify the PRPs
6. Send primes for n>1000 aswell remaining k's to Gary

I know this sounds like a big task, but as soon as my base 3 reservations clear up, I'm gonna put all six availeable cores on the task.

Regards

KEP

Riesel Base 45

4484*45^16012-1 is prime! (186.3988s+0.0016s)

Only 21k's remaining

Started that k and found the prime on test 156, about 2 hours worth.

[quote=MyDogBuster;155697]Riesel Base 45

4484*45^16012-1 is prime! (186.3988s+0.0016s)

Only 21k's remaining

Started that k and found the prime on test 156, about 2 hours worth.[/quote]


I show that there are 20 k's remaining since you've eliminated 2 of them now.

One question: You said you were testing each k individually up to n=100K. Have you already tested several of them to n=100K? If so, I'll show that on the pages.


Gary

[quote=KEP;155669]Reserving Sierpinski base 63 :smile: Can anyone tell me which k's can be excluded from testing beyond n=1000?

Here is how I'm going to do it:

1. Strict test all k<=37565866 up to n=1000 (already in progress)
2. Remove the k's that can be excluded before sieving
3. Sieve all remaining candidates for n>1000 to n<=25000
4. PRP test all candidate-pairs to n=25000
5. Verify the PRPs
6. Send primes for n>1000 aswell remaining k's to Gary

I know this sounds like a big task, but as soon as my base 3 reservations clear up, I'm gonna put all six availeable cores on the task.

Regards

KEP[/quote]


Well, I can say that you like the really BIG and TOUGH efforts Kenneth! lol :smile: I noticed that you really like the bases that are 2^q-1, which of course are some of the biggest conjectures because they are usually the most prime bases. Your plan of action sounds very good.

You do realize that this is likely 2-5 CPU years worth of work, correct? Regardless, no problem. If you end up testing it to n=10K or 15K and leaving some sieved files behind for us, that's OK. If you're able to get all the way to n=25K (I definitely could not! :smile:), then I would congradulate you on a job very well done!

Exclusions:
1. Odd k's due to a trivial factor of 2.
2. k==(30 mod 31) due to a trivial factor of 31.
3. k's that are multiples of the base where k+1 is not prime.

On odd Sierp bases, GFNs have a trivial factor of 2 so are not an issue.

On #3, I'll give some examples:

k=62, 124, 186, and 248 would be excluded from testing because 63, 125, 187, and 249 are not prime.

k=310 and 372 would be INcluded in testing because 311 and 373 are prime.

The exclusions for multiples of the base are the same on all bases. Check for k-1 being prime on the Riesel side and check for k+1 being prime on the Sierp side. If it's prime, include it; if it's not prime, exclude it.


Gary

Riesel Base 45

6372*45^23067-1 is prime! (327.8198s+0.0024s)

19 left

[QUOTE]
One question: You said you were testing each k individually up to n=100K. Have you already tested several of them to n=100K? If so, I'll show that on the pages.[/QUOTE]

The following k's are tested up to n=100K:

372, 1312, 15432

Do you need the results files on this stuff? Primes & non-primes?

Riesel Base 45

20654*45^18103-1 is prime! (217.9591s+0.0017s)

18 Left

[quote=MyDogBuster;155815]

Do you need the results files on this stuff? Primes & non-primes?[/quote]


Yes on both but to avoid inundating me with Emails, you can send them all when you are done. However you want is fine but I'd prefer more than one k per Email. lol

BTW, one reason that I asked if you could post the k's that you had tested to n=100K is that I can now unreserve just those k's for you since you said that was your testing limit. Technically others can now take them higher. That's the good thing about searching by k-value! :smile:

If you now think you'd like to continue above n=100K on the k's without a prime, let me know and I'll re-reserve them.


Gary

i have now taken riesel base 94 to 51k with no primes
unreserving

[quote=henryzz;156023]i have now taken riesel base 94 to 51k with no primes
unreserving[/quote]


Can you post a results file? Thanks.


Gary

Riesel base 60 is complete to n=10K. 3 primes found since n=8.5K. 81 k's remain. Details on the web pages. It is now unreserved.

Now reserving Sierp base 37 to n=25K. I'm currently at n=5K with 8 k's remaining.


Gary

[quote=gd_barnes;156038]Can you post a results file? Thanks.


Gary[/quote]
attached

Reserving Riesel Base 72

k=4 (last one) reserving from n=54K to n=200K

Riesel Base 45

5128*45^31528-1 is prime! (1471.1138s+0.0036s)

17 left

Riesel Base 37

Reserving all 27 k's from n=20K to n=70K

Just thinning out the ranks

Riesel base 75 tested to 10k. (No reservation.) No more testing.

17 ks remaining.

Riesel base 95 tested to 10k. (No reservation.) No more testing.

24 ks remaining.

I believe k=324 can be removed. Odd n has factors of 7,13,37 or 229 for as far as I could test.
I think other ks can remain but I was having problems with the factoring website.

The reason I am testing without making reservations is because it's difficult to tell which bases are hard until they are part tested. I regularly check this thread to make sure there is no repetition of work. :smile:

I am trying to push Riesel base 39 to at least n=1000.