Brier any base
 

Not wishing to disturb anyone from what they are doing, but maybe it is time to explore Brier numbers any base.

Briers are k such that k*2^n+/-1 are never prime

Without thinking at all deeply, (it is late and the wine is having its wicked way) I have to assume they exist for b>2

[QUOTE=robert44444uk;148110]Briers are k such that k*2^n+/-1 are never prime[/QUOTE]
Ehh, maybe I don't understand the question.. but aren't that the numbers, seventeen or bust (for example) is looking for?
SoB is looking for primes - so the remaining k could be "briers"?

[QUOTE=Xentar;148121]Ehh, maybe I don't understand the question.. but aren't that the numbers, seventeen or bust (for example) is looking for?
SoB is looking for primes - so the remaining k could be "briers"?[/QUOTE]

I presume he is referring to k where both k*2^n+1 and k*2^n-1 are composite for all n.

For base 20:
the lowest Riesel is 8.
the lowest Sierpinski is 8.

If a Brier number for a base is a number that is both a Riesel and a Sierpinski then the lowest Brier number for base 20 is 8.

Willem.

[URL]http://mathworld.wolfram.com/BrierNumber.html[/URL]
this definition seems to be a little different

Anyway, the concept is clearer now I have slept and shaken off the wine.

So the Mathworld looks odd because they have inverted not only Sierpinski and Riesel but also k and n. Written by someone doubly dyslexic. It is only conjectured that k=878503122374924101526292469 is smallest for n=2.

Siemelink has already spotted that Brier any base can give up results easily where the lowest k for either is the same, then this is the lowest Brier as well. Same with many of those where low k=4 i.e. b=14mod15

[QUOTE=Xentar;148121]Ehh, maybe I don't understand the question.. but aren't that the numbers, seventeen or bust (for example) is looking for?
SoB is looking for primes - so the remaining k could be "briers"?[/QUOTE]

SoB is looking for numbers k such that k*2^n+1 never prime, not k*2^n+[U][B]&[/B][/U]-1

[quote=robert44444uk;148173]SoB is looking for numbers k such that k*2^n+1 never prime, not k*2^n+[U][B]&[/B][/U]-1[/quote]

The "&" sign clears it up. I didn't know what you meant either.

Mathworld needs to correct their k and n. There's even a link on that page to a different page on their own site where the k and n are correct. They need to be consistent with the rest of the math world as well as themselves. lol

This would be an interesting exercise. Willem has already come up with one solution. I think I'll mess around a little with this myself.


Gary

am i right in thinking that this modified script will prove brier conjuctures
[code]SCRIPT
DIM base, 2
DIM min_k, 1
DIM max_k, 100000
DIM max_n, 1000
OPENFILEAPP k_file,pl_remain.txt
OPENFILEAPP p_file,pl_prime.txt
DIMS tmpstr
DIM n
DIM k_step, 1
DIM k
IF (base % 2 == 1) THEN SET k_step, 2
IF (base % 2 == 1) && (min_k % 2 == 1) THEN SET min_k, min_k + 1
SET k,min_k - k_step

LABEL next_k
SET k, k + k_step
IF (k > max_k) THEN GOTO END
SET n, 0

LABEL next_n
SET n, n + 1
PRP k*base^n-1

IF (ISPRIME) THEN GOTO Riesel_Prime_found
PRP k*base^n+1
IF (ISPRIME) THEN GOTO Sierpinski_Prime_found
IF (n < max_n) THEN GOTO next_n

SETS tmpstr,%d*%d^n+-1;k;base;
WRITE k_file,tmpstr
GOTO next_k

LABEL Riesel_Prime_found
SETS tmpstr,%d*%d^%d-1;k;base;n;
WRITE p_file,tmpstr
GOTO next_k
LABEL Sierpinski_Prime_found
SETS tmpstr,%d*%d^%d+1;k;base;n;
WRITE p_file,tmpstr
GOTO next_k
END
[/code]
it is a modification of the script posted in the base 101 thread

Link
 

Before this gets out of hand...

[url]http://www.primepuzzles.net/problems/prob_049.htm[/url]

GIYF

[QUOTE=masser;148200]Before this gets out of hand...

[/QUOTE]

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.

[QUOTE=masser;148200]Before this gets out of hand...

[url]http://www.primepuzzles.net/problems/prob_049.htm[/url]

GIYF[/QUOTE]

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

Here is looking at base b =/= 2

[QUOTE=Siemelink;148254]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.[/QUOTE]

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

[url]http://www.primepuzzles.net/problems/prob_029.htm[/url]

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?

[quote=masser;148200]Before this gets out of hand...

[URL]http://www.primepuzzles.net/problems/prob_049.htm[/URL]

GIYF[/quote]


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 [URL="http://gbarnes017.googlepages.com/twins100K.htm"]here[/URL]. 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 [URL="http://www.rieselprime.org"]www.rieselprime.org[/URL] k<300 page. NOW I know why you had those k's reserved a long time ago. Very interesting effort! :smile:

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

i begun to build a page for [url]www.rieselprime.org[/url] 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: [url]http://www.rieselprime.org/Related/RieselTwinSG.htm[/url]

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

Brier base 12
 

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:
[CODE]
p Mp(12)
13 2
157 3
5 4
29 4
7 6
19 6
20593 12
[/CODE]
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:

[CODE]
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

[/CODE]

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!

[QUOTE=robert44444uk;150157]
All p except 157 can give both S and R cover, but 157 does not give cover for both at the same time.
[/QUOTE]

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.