Brier any base

[robert44444uk]

Originally Posted by masser

Before this gets out of hand...

http://www.primepuzzles.net/problems/prob_049.htm

GIYF

Masser - all of prob 49 is base 2 though, right?

Here is looking at base b =/= 2

[robert44444uk]

Originally Posted by Siemelink

That site is down for the moment, so allow me to dream. I would try to find the briers like this:
1) generate a series of Riesels R1, R2,..
2) generate a series of Sierpinskis S1, S2,
3) for every Rx loop through Sy
4a) if the cover set of Rx equals the cover set of Sy and Rx = Sy it is a Brier
4b) if the cover sets are not equal there must be a Brier where Rx + a*product(Cover of Rx) = Sy + b*product(Cover of Sy)

Cheers, Willem.

The approach for lowest Brier b=2 is best explained here.

http://www.primepuzzles.net/problems/prob_029.htm

and download the file at the end of the page.

I am looking forward to b=3,7,15,71,280 most. Conjectures of other lowest b-Briers will be relatively straightforward.

What will be interesting is some of the theory. We now know that all b, with a few known exceptions, have S or R covers <=12. I wonder what rule is for Briers?

[gd_barnes]

Originally Posted by masser

Before this gets out of hand...

http://www.primepuzzles.net/problems/prob_049.htm

GIYF

A great and very interesting page! For all who weren't able see it before, it confirms the smallest known Brier number base 2 that was shown previously by Robert at the beginning of this thread.

A little off topic but what had me the most interested is that it deals with a TWIN prime conjecture too! First off, I have a web page of all twin primes for n<48K for all k<100K. It is here. If you compare my page to the puzzle page, it confirms the k's remaining below the lowest conjectered twin-prime k but it also extends the search range. All of the k's have been tested to n=745K! Obviously if any had been found, they would be greater than the largest currently known twin at n=195K.

Another interesting thing that I discovered: These are the 9 k's that are in RPS's 9 k's drive that you supplied the sieved files to them for and that are shown as such on Karsten's www.rieselprime.org k<300 page. NOW I know why you had those k's reserved a long time ago. Very interesting effort!

To demonstrate the limit to myself of barren twin primes, I simply took all the known Riesels for the 9 k's as shown on Karsten's page above the n=140K limit shown on the puzzle page here and tested them for Sierp primes.

Taking the above a step further, the largest known twin for k<300 is:
291*2^1553+/-1
The largest known twin for k<1000 is:
459*2^8529+/-1

With no twins between n=1553 and n=657000 for k<300 (the lowest search limit remaining for all Riesels k<300), I think it's safe to say that THIS twin-prime conjecture will never be proven with currently known math theory!


Gary

[kar_bon]

i begun to build a page for www.rieselprime.org with the Twin and Sophie Germain problem:

a Twin or Sophie Germain can only be produced by a k divisible by 3!

so there is a question:
is there a k (lowest) for which no Twin and/or SG exist?

so the first k without any Twin/SG is k=183!

see menu 'Related' for a first look: http://www.rieselprime.org/Related/RieselTwinSG.htm

PS:
Brier numbers are another Related for the PrimeDatabase but not yet created!

[robert44444uk]

Just getting my head around these Brier numbers is boggling my bird brain.

My approach, looking at Base 12, was to investigate primes p with small modulo order (<=target cover) base 12, represented as Mp(12), and investigate the qualities of k=xmod[Mp(12)] of each p under the power series k.12^n+1. If for a given k, x=0 for some n, then p can contribute towards Sierpinski (S) cover, and if x=2, for some n, then p can contribute towards Riesel (R) cover. Where the power series gives both k.12^n+1=0mod[Mp(12)] and k.12^n+1=2mod[Mp(12)] for two different n, then the prime can contribute to both S- and R-cover.

Base 12 has the following p with small modulo order:

p   Mp(12)
13	2
157	3
5	4
29	4
7	6
19	6
20593 12

All p except 157 can give both S and R cover, but 157 does not give cover for both at the same time.

For example, k=2mod5 covers n=1mod[M5(12)] or n=1mod4 for S, and n=3mod4 for R, but k=145mod157 provides cover for n=1mod3 for S but does not provide cover at all for R. k=12mod157 provides cover for n=1mod3 for R but does not provide cover at all for S.

As 13, 5 and 29 provide 4-cover for either S or R, we must ask whether these can provide for both at the same time.

S:
k=1mod13 contributes to 4-cover at n=1,3
k=2mod5 contributes at n=2
k=28mod29 contributes at n=4

R:
k=1mod13 contributes to 4-cover at n=2,4
k=2mod5 contributes at n=3
k=28mod29 contributes at n=2 :( already covered by p=13

So 4-cover is not possible. 6-cover cannot be obtained as 157 does not fully contribute, and the two primes ordp(12)=6 and the one prime ordp(12)=2 are insufficient where 157 does not contribute.

Looking then at 12-cover, then we must bring in primes p with Mp(12)=4 and check all CRM possibilities for the permutations of the following mods:

n 	S(k)	 R(k)
		
1	1mod13	12mod13
2	12mod13	1mod13
			
		
1	13mod157	144mod157
2	145mod157	12mod157
3	156mod157	1mod157
					
1	2mod5	4mod5
2	1mod5	3mod5
3	3mod5	2mod5
4	4mod5	1mod5
				
1	12mod29	17mod29
2	1mod29	28mod29
3	17mod29	12mod29
4	28mod29	1mod29
				
1	4mod7	3mod7
2	5mod7	2mod7
3	1mod7	6mod7
4	3mod7	4mod7
5	2mod7	5mod7
6	6mod7	1mod7
				
1	11mod19	8mod19
2	12mod19	7mod19
3	1mod19	18mod19
4	8mod19	11mod19
5	7mod19	12mod19
6	18mod19	1mod19

20593 ignored

The resulting CRM solutions might not produce a Brier, as the overlaps have to cover all S and R positions, and as we have seen from 4-cover this is not guaranteed.

One solution is provided by 1mod13, 145mod157, 3mod7, 18mod19, 2mod5, 17mod29. cover is provided on the S- side through 13 for n=1,3,5,7,9,11: 157 at n=2,5,8,11: 7 for n=4,10: and 19 for n=6,12. On the R- side cover is provided by 13 for n=2,4,6,8,10,12: 7 for n=1,7: 19 for n=3,9: 5 for n=3,7,11 and 29 for n=1,5,9.

Solving CRM provides Briers of the form 13376702mod39360685, of which the smallest is of course 13376702.

But this might not be the smallest!

[robert44444uk]

Originally Posted by robert44444uk

All p except 157 can give both S and R cover, but 157 does not give cover for both at the same time.

I may be wrong on this, but I think that for primes p, and base b, if Mp(b)=odd then p will not contribute to both R and S, and, on the flip side if Mp(b)=even then p will contribute to both.