Sierpinski/ Riesel bases 6 to 18

[Citrix]

I am not sure why you are stuck, it is quite easy to prove that every base will has infinite number of covering sets.

Consider this suppose you have base b. (Sierpinski first)

Then you factorize b^2+1, b^4+1,...b^(2^n)+1
Now all the above will provide atleast one factor. If all of them are prime then there is no covering set. But atleast one number in this sequence will be composite and will have 2 factors which will generate a covering set.

Similarly the Riesel case can also be solved.
Also more covering sets can be generated using other powers like b^3+1, b^9+1...

[ValerieVonck]

Hi everyone!

Am I doing something wrong?
I tested 22*22^n+1 till n=145407 and sieved it to 514M and I am still getting no prime(s)?
In NewPGen I did choose: k*b^n+1 with k fixed??

Is this ok?

Thx

[robert44444uk]

Originally Posted by Citrix

I am not sure why you are stuck, it is quite easy to prove that every base will has infinite number of covering sets.

Consider this suppose you have base b. (Sierpinski first)

Then you factorize b^2+1, b^4+1,...b^(2^n)+1
Now all the above will provide atleast one factor. If all of them are prime then there is no covering set. But atleast one number in this sequence will be composite and will have 2 factors which will generate a covering set.

Similarly the Riesel case can also be solved.
Also more covering sets can be generated using other powers like b^3+1, b^9+1...

The problem comes with b=2^n-1, as the factorisation of b^n-1 does not guarantee a new factor. For example b=2^4-1=15, 15^1-1=14=7*2, 15^2-1=2^5*7 so no new prime factor. If all other values of n^2 for that b only produce 1 new prime factor, then no covering set is there as 1/4+1/8+1/16....<1.

What we have proved that, for all other b, there is a covering set with a repeat of no greater than 12.

Of course, we do not expect that, even in the isolated cases above, that there is no covering set, we have not proved it yet. I have checked up to b=2^24-1 and have found covering sets for them all.

[robert44444uk]

Originally Posted by CedricVonck

Hi everyone!

Am I doing something wrong?
I tested 22*22^n+1 till n=145407 and sieved it to 514M and I am still getting no prime(s)?
In NewPGen I did choose: k*b^n+1 with k fixed??

Is this ok?

Thx

Hi, Cedric. You might want to check Axn1's post. We are not really expecting to find a prime for this value. You might want to take over from where he left off, or look at another base and k.

[robert44444uk]

Base 24 is slightly more complex

Sierpinski 77554 [5,7,79,601]

Riesel 135249 [5,7,79,601]

There is an chance that lower sierpinskis and riesels exist, but remote.

Almost all of the values need to be checked as trivial solutions are only one in every 23.

[ValerieVonck]

Ok unreserving
Base 22
n = 22
n = 484

Reserving:


1343*16^n-1

[michaf]

Originally Posted by CedricVonck

Ok unreserving
Base 22
n = 22
n = 484

How far have you checked for primes, Cedric?

[michaf]

In the meantime, I have found primes:

472*23^2379-1
154*23^2898+1
124*23^3118+1
328*23^5001-1
230*23^6228-1
464*23^7548-1
122*23^14049+1
314*23^17268-1
394*23^20169-1

[Citrix]

There is an error on the base 16 Riesel candidates I posted previously. The list is wrong. I will have to do redo the work.... The sierpinski candidates should be ok.

[robert44444uk]

Hi guys, the mailman has arrived, with a new goody.

25 has a lot of trivial k, as 24 has prime factors 2 and 3.

Annoyingly many of the best candidates for Sierpinskis and Riesels are trivial, and the best I can manage are:

Sierpinski 262638 [7,13,31,601]

Riesel 346802 [7,13,31,601]

Anyone taking this one forward needs to make sure to elimiate the trivials first before prime checking. 4 in every 6k are trivial.

As always, where these candidates are relatively large number there is a chance that some other primes provide cover with a lower k.

[ValerieVonck]

Ok relaasing 1343*16^n-1.

LLR seemed to test this number as a 1343*2^n-1.