Bases 251-500 reservations/statuses/primes

[KEP]

These sierpinski conjectures have I come up with:

Base: k : Covering set:
----- ------- -------------
251 8 (3,7,43,1471,17,109,13,1609,37,9601,1553,1249) (most likely 4 else n>900000000, covering set will be p<=5, most likely 3,5)**
252 45 (11,23,103,619,5,13,977,43,1471,337,997,4057,17,193,241)
253 65914 (127,31,691,5,37,173,103,619,17,257,13,353, 8369,1201)
254 4 (3,5,17,7,19,487,149,433,31,691,13,401) *
255 110094 (97,673,13,41,61,7,19,487,89,1621,1609,3529)

* Proven
** Most likely proven, but lack of covering set

Notice that Sierpinski base 254 is proven. Also base 251 is msot likely 4 with a covering set of 3 and 5 else n>900M

Is not going to work any further on these conjectures, accept conjecture 255, it is going to be taken to n<=2,500. Base 252 has 2 k's remaining, 1 and 27. I'm going to take these to n<=25,000. (at 11351 there is still no prime). This will most likely be my last contribution...

Thank you.

Kenneth!

[KEP]

@all: Forgot to tell that I'm about to take base 255 Sierpinski to n<=2500. Also at the moment I'm searching through the log files of Riesel base 3 k<=500M for PRP being composites... at the moment 3 has been found

[gd_barnes]

Thanks for the info. guys. It's kind of fun doing those new bases isn't it? After a short review, I'll post it on the web pages.

KEP, I'm not sure I understand your large covering sets. Most will have small covering sets of 6 or less factors. Examples:

For Sierp base 254, k=4 has a covering set of {3, 5}. That is the factors of 3 and 5 knock out all n-values for k=4.

You are correct, it is proven because:
k n-prime
2 1
3 2
k=1 is a GFN and is not considered, although it does have a prime at n=4.

On Sierp base 251, your later analogy is correct. k=4 has a covering set of {3, 5}. As you stated, it is proven:
k n-prime
2 1
k=1 and 3 have trivial factors of 2.

One more thing, you do not need to test k=1 on any Sierp base. k=1 makes the form a Generalized Fermat number (GFN). GFN's can only have a prime when n is a perfect power of 2.

Based on the above, on your base 252, if you wanted to test k=1, you would only need to test n=1, 2, 4, 8, 16, 32, etc. There would be no need to test any other n-values as they would yield composites. But it's not necessary to test at all to prove the conjecture because it is a GFN. So per your analysis, for base 252, only k=27 is considered remaining.


Gary

[KEP]

Originally Posted by gd_barnes

Thanks for the info. guys. It's kind of fun doing those new bases isn't it? After a short review, I'll post it on the web pages.

KEP, I'm not sure I understand your large covering sets. Most will have small covering sets of 6 or less factors. Examples:

For Sierp base 254, k=4 has a covering set of {3, 5}. That is the factors of 3 and 5 knock out all n-values for k=4.

You are correct, it is proven because:
k n-prime
2 1
3 2
k=1 is a GFN and is not considered, although it does have a prime at n=4.

On Sierp base 251, your later analogy is correct. k=4 has a covering set of {3, 5}. As you stated, it is proven:
k n-prime
2 1
k=1 and 3 have trivial factors of 2.

One more thing, you do not need to test k=1 on any Sierp base. k=1 makes the form a Generalized Fermat number (GFN). GFN's can only have a prime when n is a perfect power of 2.

Based on the above, on your base 252, if you wanted to test k=1, you would only need to test n=1, 2, 4, 8, 16, 32, etc. There would be no need to test any other n-values as they would yield composites. But it's not necessary to test at all to prove the conjecture because it is a GFN. So per your analysis, for base 252, only k=27 is considered remaining.



Gary

The covering sets was copied directly from the output presented by the covering.exe program. Maybe I missunderstood and copied to much, but I decided to copy the entire amount of numbers, to make sure that you got what was needed.

Regarding base 252, it actually appears that k=1 is also very easily sieved. At n<=17426 I've completed 826 tests, and only 8 of these were for k=1 the remaining were for k=27.

Glad that you could use this work. Actually it's kind of need to have a thread were people can tell which bases they try to conjecture. However if I remember correctly, there is no way to verify PRPs for bases>255 (unless they are powers of 2?), so maybe we should encourage people to only work on bases <=255 for both Riesel or Sierpinski.

Take care.

Kenneth!

[gd_barnes]

Originally Posted by KEP

The covering sets was copied directly from the output presented by the covering.exe program. Maybe I missunderstood and copied to much, but I decided to copy the entire amount of numbers, to make sure that you got what was needed.

Regarding base 252, it actually appears that k=1 is also very easily sieved. At n<=17426 I've completed 826 tests, and only 8 of these were for k=1 the remaining were for k=27.

Glad that you could use this work. Actually it's kind of need to have a thread were people can tell which bases they try to conjecture. However if I remember correctly, there is no way to verify PRPs for bases>255 (unless they are powers of 2?), so maybe we should encourage people to only work on bases <=255 for both Riesel or Sierpinski.

Take care.

Kenneth!

The "exponent" that you are entering is too large on the covering software. On each base, start with an exponent of 4 and if no covering set found, go to an exponent of 6, 8, 12, 16, 24, 36, 48, 60, 72, 96, 120, and 144 until you find a conjectured k. (Actually keep going after you find a conjectured k because you may find a slightly lower k with a larger covering set; although this is unlikely.)

Besides being the lowest conjectured k-value, the covering set needs to be the smallest number of factors that make the k-value always composite.

The "exponent" is the first value that you enter after typing "covering" to start the program.

To be blunt, you are wasting your time on k=1. It is mathematically proven that n must be a power of 2 because it is a GFN for all Sierp bases. Also, you don't need to test it to prove the conjecture. Just remove it from your sieved file. If you really want to test it for n>10K, all you need to do is test n=16384, 32768, 65536, 131072, 262144, and 524288 and TA-DA; you've now tested it all the way to n=1M (actually n=1048575) because the next test would be at n=1048576. In other words, at these high bases, you don't want to do even one test for high n-values that is a mathematically proven composite.

Getting into finding primes on GFN's is a whole other topic. There are several web pages (links are in the top-5000 site) dedicated to finding the primes and factors of them for various bases < 20 that can be generalized for all bases.

Edit: I wasn't aware that PRP's for bases > 256 that are not power of 2 could not be proven. If that is true then I agree, we need to keep bases that are not powers of 2 at < 256.


Gary

[gd_barnes]

Originally Posted by gd_barnes

On Sierp base 251, your later analogy is correct. k=4 has a covering set of {3, 5}. As you stated, it is proven:
k n-prime
2 1
k=1 and 3 have trivial factors of 2.

My quick analogy was incorrect. Sierp base 251 has trivial k's of k==(1 mod 2) and (4 mod 5) so k=4 could not be the conjecture. KEP, you're original conjecture of k=8 was correct.

I don't know what you meant by "most likely 4 else n>900000000" because k=8 is quickly proven:

k=8 covering set {3 7}
k n-prime
2 1
6 17
k=1, 3, 5, 7 have trivial factors of 2.
k=4 has a trivial factor of 5.

I have to check myself too!


Gary

[KEP]

Originally Posted by gd_barnes

I don't know what you meant by "most likely 4 else n>900000000"

Well I did use NewPGen to test base 251 for k=4, for as many n as NewPGen could by curtan verify (before starting to test -2148....), the limit of n is higher than 900M, for k=4 for base 251, but I decided to stop testing there since p<=5 for the entire range removed all n's. Well I guess it has something to do with the trivial factors

Also I'm sorry that I concluded wrong, concerning the bases. Well there you have it, then I actually learned new stuff today also. Actually I was also kind of puzzled why base 255 should in fact be the limit, since I figured that the coding should be kind of general for any bases as compared to a fixed amount of bases

Tonight, I'll hand over the last work that I have for the conjectures I found, and then I'll hopefully next weekend be able to hand over the remaining work for Riesel base 3. Once I hand over Riesel base 3 work, it will also conclude my stay here. At the moment, I guess that even though I check out, I'll not entirely leave, just like in "hotel california" But heck it has been a need a very educational time for me, and I'm glad that whatever I contributed with was actually needfull.

Regards

Kenneth!

[gd_barnes]

Originally Posted by KEP

Well I did use NewPGen to test base 251 for k=4, for as many n as NewPGen could by curtan verify (before starting to test -2148....), the limit of n is higher than 900M, for k=4 for base 251, but I decided to stop testing there since p<=5 for the entire range removed all n's. Well I guess it has something to do with the trivial factors

Definition of trivial factors for the conjectures: Each and every n-value has the same factor.

Hence k's with trivial factors cannot be considered the conjecture nor can they beconsidered remaining if they are lower than the conjecture. This is because they will always be composite.

As you found, NewPGen or Srsieve would quickly sieve all n-values out.

Here: 4*251^n+1 always has a trivial factor of 5. Here's a demonstration:

4*251^1+1 = 1005 = 3*5*67
4*251^2+1 = 252005 = 5*13*3877
4*251^3+1 = 63253005 = 3*5*17*248051
4*251^4+1 = 15876504005 = 5*41*3061*25301
etc.


If factors and covering sets confuse you, try plugging the above into Alperton's excellent prime factoring web page to get the prime factors of the first few n-values of a base before starting on it. I learn a lot by looking at the patterns of factors that occur in the various forms. It is also how I determine the smallest covering set after determining the lowest conjectured k.

One more thing: If you still plan on testing Sierp base 255 to n=2500, be sure and remove the following k's before reporting primes remaining:
k==(1 mod 2) [odd k values]
k==(126 mod 127) [k's that leave a remainder of 126 after dividing by 127, i.e. 126, 253, 380, etc.]

When starting new bases, it's essential to understand how all of the trivial factors work or you wind up with k's remaining that you shouldn't and you end up searching things that are proven composite for all n-values.

I hope this helps.


Gary

[KEP]

@Gary:

About the base 255 Sierpinski, it actually helped a lot. It has at the moment been taken to n=1968 and is continueing to n<=2500. I will try to see if I remember to remove all k's that you mentioned ... I'm not sure you will get the base 255 Sierpinski primes as well as k's remaining before the coming friday, since I don't think it will finish getting to n<=2500 before I leave this machine for the next ~110 hours

Regards

Kenneth

EDIT: Forgot to supply with the info, that I'm only testing even k's and no odd k's, hence the fact that the odd k's never yield a prime

[gd_barnes]

KEP reported in an Email that he has completed Sierp base 252 k=27 to n=25K. No primes were found and he is unreserving it.

KEP, the only reservations I now have you down for are Sierp base 255 up to n=2500 and Riesel base 3 up to k=500M and n=25K.


Gary

[gd_barnes]

For Sierp bases not already done so in KEP's list that had a conjecture < 100, I searched them to a shallow depth and proved a couple of them as follows:

Sierp base 251; conjecture k=8 proven; highest prime is k=6 at n=17. (Shown previously by me.)

Additional info. for the bases on the web pages.

I'm not working on these further.


Gary