Riesel/Sierp #'s for bases 3, 7, and 15

[gd_barnes]

Once again, nice work Willem. It's unfortunate that we weren't aware of this 11-digit Riesel value for base 3 that was discovered in 2004!

So unofficially, the Riesel value for base 3 is now 63,064,644,938. I will do my checking like I did before and if it looks good, I'll change the web page.

Assuming it is correct, this now makes the Riesel base 3 conjecture potentially proveable in our lifetimes now!

Masser, if you know of any more obscure web pages, discussion threads, etc. that have lower Riesel/Sierpinski numbers for bases 3, 7, or 15 then are currently shown on our web pages here, let us know.


Thanks,
Gary

[Siemelink]

This cover set {5,7,13,17,19,37,41,193,757} produces Riesels with a sequence length 144 (I think). My program has checked all the values under that. It needs another efficiency gain before I can run that length quickly.
Or, if I cap the primes at 758 I can find the smallest Riesel for this cover set.

Having fun, Willem.

[robert44444uk]

You might also put your program to work to look at Sierpinski base 3, where a lot of work has been done, but you might be able to do better than the mooted k
3574321403229074.

also other Sierpinskis
base 7, k value 162643669672445
base 15, 91218919470156
base 71, 5917678826

Chris Caldwell would be interested in your results if you beat these.

Also there is a real challenge to see if you can use partial covering sets created by algebra, as x^n+1 often factors algebraically. This is especially relevent to x^3+1 which should provide algebraic factors, but I have been stumped in my efforts to make this work.

[gd_barnes]

Originally Posted by robert44444uk

You might also put your program to work to look at Sierpinski base 3, where a lot of work has been done, but you might be able to do better than the mooted k
3574321403229074.

also other Sierpinskis
base 7, k value 162643669672445
base 15, 91218919470156
base 71, 5917678826

Chris Caldwell would be interested in your results if you beat these

Also there is a real challenge to see if you can use partial covering sets created by algebra, as x^n+1 often factors algebraically. This is especially relevent to x^3+1 which should provide algebraic factors, but I have been stumped in my efforts to make this work.

We already have these as the Sierp numbers as shown on the web pages with the exception of base 71 that we are not working on.

On algebraic factors, the Riesel side generally has many more of them as k^(2q)-1 factors to (k^q-1)*(k^q+1) for all even n so that odd n only needs a trivial factor -or- (m^2)^(q^2)-1 factors to (m^q-1)*(m^q+1) for all n. For a most unusual example, see the discussion in the "algebraic factors issues for base 24" thread.

We do not consider any Riesel or Sierp value to be one that contains algebraic factors to make a full covering set so finding a k-value that fits that criteria does not lower the conjecture from the project's perspective. Otherwise on the Riesel side, the "conjectures" would be very low and uninteresting and mostly proven, i.e. Riesel base 9 would be k=4, Riesel base 12 would be k=25, Riesel base 16 would be k=9, etc.

I think if Prof. Caldwell takes a look at the Riesel side, he will conclude the same thing. I had responded to the paper that he sent me not long after the project was started suggesting this.

For that reason, our Sierp base 16 conjecture differs from Prof. Caldwell's paper. We consider it to be k=66741 with 28 k's remaining after a search to n=132K whereas his paper considers the "conjecture" to be proven at k=2500, which has algebraic factors to make a full covering set as shown on our powers-of-2 web page. k=40000 also has a similar condition.

That said, it is a good thing to find k-values with algebraic factors that make a full covering set so that we can eliminate them from our testing. Riesel base 24 turned out to be quite the challenge in that regard as did Riesel base 12, the latter of which is now proven and BOTH of which have 2 different sets of algebraic factors that make full covering sets to eliminate various k-values. BTW, Riesel base 24 would have a proven "conjecture" of k=6 if we chose to include algebraic factors making a full covering set as the Riesel number, which of course would be most 'boring'.

We do know this for Sierp base 3: We have found primes for ALL even k<2930054!! So if there are any algebraic factors for the base, they are rare indeed! It seems unlikely at this point but you never really know and always have to be on the lookout for them.


Gary

[gd_barnes]

Originally Posted by gd_barnes

Once again, nice work Willem. It's unfortunate that we weren't aware of this 11-digit Riesel value for base 3 that was discovered in 2004!

So unofficially, the Riesel value for base 3 is now 63,064,644,938. I will do my checking like I did before and if it looks good, I'll change the web page.

Assuming it is correct, this now makes the Riesel base 3 conjecture potentially proveable in our lifetimes now!

Masser, if you know of any more obscure web pages, discussion threads, etc. that have lower Riesel/Sierpinski numbers for bases 3, 7, or 15 then are currently shown on our web pages here, let us know.


Thanks,
Gary

Originally Posted by Siemelink

This cover set {5,7,13,17,19,37,41,193,757} produces Riesels with a sequence length 144 (I think). My program has checked all the values under that. It needs another efficiency gain before I can run that length quickly.
Or, if I cap the primes at 758 I can find the smallest Riesel for this cover set.

Having fun, Willem.

I agree with this. The new Riesel base 3 value for the project is now 63064644938. Proof:

Factor   n-occurrences   n-remaining
13       n==(1 mod 3)    n==(0,2 mod 3)
5        n==(3 mod 4)    n==(0,2,5,6,8,9 mod 12)
7        n==(0 mod 6)    n==(2,5,8,9 mod 12)
41       n==(5 mod 8)    n==(2,8,9,14,17,20 mod 24)
757      n==(2 mod 9)    n==(8,9,14,17,26,32,33,41,44,50,57,62,68 mod 72)
17       n==(1 mod 16)   n==(8,9,14,26,32,41,44,50,57,62,68,80,86,89,98,104,105,116,122,134,140 mod 144)
193      n==(9 mod 16)   n==(8,14,26,32,44,50,62,68,80,86,98,104,116,122,134,140 mod 144)
19       n==(8 mod 18)   n==(14,32,50,68,86,104,122,140 mod 144)
37       n==(14 mod 18)  (none)

The web page will be updated shortly. I think KEP will be happy to hear this!

I renamed the thread to include base 3.


Gary

[michaf]

I can now test base 3 with a speed of about 1M k's per hour, upto n=1000

Testing higher takes 'too much' time.

But this means that, with the mooted riesel conjecture of k=63,064,644,938
we need about 63000 hours worth of testing them ALL to 1k.

Which is about 2625 days, which is 'only' about 7-8 CPU years.

Testing 10M candidates resulted in a handy 2150 remaining,
so in the end: 6306*2150 = about 14M candidates remaining.

[gd_barnes]

Originally Posted by michaf

I can now test base 3 with a speed of about 1M k's per hour, upto n=1000

Testing higher takes 'too much' time.

But this means that, with the mooted riesel conjecture of k=63,064,644,938
we need about 63000 hours worth of testing them ALL to 1k.

Which is about 2625 days, which is 'only' about 7-8 CPU years.

Testing 10M candidates resulted in a handy 2150 remaining,
so in the end: 6306*2150 = about 14M candidates remaining.

OK, fire away but be sure and coordinate with KEP on this reservation and let me know what k-range you will take. (I'm assuming you won't be taking all of them...yet, anyway, lol.) You'll have quite a bit of manual intervention with so many k's remaining but with a much lower conjecture, I suppose it won't be too bad.


Gary

[michaf]

Originally Posted by gd_barnes

OK, fire away but be sure and coordinate with KEP on this reservation and let me know what k-range you will take. (I'm assuming you won't be taking all of them...yet, anyway, lol.) You'll have quite a bit of manual intervention with so many k's remaining but with a much lower conjecture, I suppose it won't be too bad.

Gary

The manual intervention is indeed too much when dealing with such huge number of numbers...

Does anyone know of the existance of a program to delete all lines which are in one file, from another file?
Or is anyone skilled in perl to do so with a quick few commands?

edit: I also tried testing at once upto 10k, which leads to 1.5M done in about 5 hours, leaving some 30 candidates

[michaf]

Originally Posted by gd_barnes

OK, fire away but be sure and coordinate with KEP on this reservation and let me know what k-range you will take. (I'm assuming you won't be taking all of them...yet, anyway, lol.) You'll have quite a bit of manual intervention with so many k's remaining but with a much lower conjecture, I suppose it won't be too bad.

Gary

Which k's were tested above 100M? I know only of my own 100-110M

[mdettweiler]

Originally Posted by michaf

The manual intervention is indeed too much when dealing with such huge number of numbers...

Does anyone know of the existance of a program to delete all lines which are in one file, from another file?
Or is anyone skilled in perl to do so with a quick few commands?

edit: I also tried testing at once upto 10k, which leads to 1.5M done in about 5 hours, leaving some 30 candidates

Actually, I have just recently been working on a Perl script to do just that for KEP. I've got it working fine on Linux already; I still have yet to make the necessary modifications for it to run on Windows, and then compile it into a .exe file.

If you'd like me to send you the finished Linux version, just PM me your email address. When I get the Windows version finished, too, I'll post both here on the forum.

[michaf]

Originally Posted by Anonymous

Actually, I have just recently been working on a Perl script to do just that for KEP. I've got it working fine on Linux already; I still have yet to make the necessary modifications for it to run on Windows, and then compile it into a .exe file.

If you'd like me to send you the finished Linux version, just PM me your email address. When I get the Windows version finished, too, I'll post both here on the forum.

Just the .pl script would be fine, I can run it here (or at least I think I can)