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-   -   k=1 thru k=12 (https://www.mersenneforum.org/showthread.php?t=10354)

henryzz 2008-06-01 12:38

k=1 thru k=12
 
3 Attachment(s)
I don't really know where to post this but does anyone know if anyone has tested riesel or sierp numbers with k=1? It seems that something overlooked to me
of because gimps is doing this for base2 but what testing has been done on k=1 other bases?

[B]Admin edit: All primes and remaining k's/bases/search depths for k=1 thru 12 and bases<=1030 are attached to this post.[/B]

henryzz 2008-06-01 15:50

forget reisel it always has a factor b+1
there also seems to to algebraic factors of sierp except if n is a power of 2 but i cant quite place my finger on it

gd_barnes 2008-06-02 07:16

Riesel bases always have a trivial factor of b-1 rather than b+1. Technically k=1 IS considered in the Riesel base 2 conjecture because you cannot have a trivial factor of 1 since it is not considered prime. But k=1 has a prime at n=2 and hence is quickly eliminated.

For Sierp, k=1 always make Generalized Fermat #'s (GFNs). GFNs are forms that can reduce to b^n+1, hence k's where k=b^q and q>=0 are also not considered.

We do not consider GFNs in testing because n must be 2^q to make a prime, resulting in few possibilities of primes. Most mathematicians agree that the number of primes of such forms is finite.

See the project definition for more details about exclusions and inclusions of k-values in the 'come join us' thread.


Gary

robert44444uk 2008-06-02 13:16

For the infinity of bases, the smallest Sierpinski k may take any integer value except 2^x-1, x=integer.

There are generating functions to discover instances of certain values such as k=2,5,65 which do not appear for small bases.

This is down to the work of Chris Caldwell and his last year students.

for example k=2 for b=19590496078830101320305728

gd_barnes 2008-11-02 22:43

[quote=robert44444uk;134998]For the infinity of bases, the smallest Sierpinski k may take any integer value except 2^x-1, x=integer.

There are generating functions to discover instances of certain values such as k=2,5,65 which do not appear for small bases.

This is down to the work of Chris Caldwell and his last year students.

for example k=2 for b=19590496078830101320305728[/quote]


Is this the lowest base where k=2 is the Sierpinski number?

robert44444uk 2008-11-03 03:06

[QUOTE=gd_barnes;147615]Is this the lowest base where k=2 is the Sierpinski number?[/QUOTE]

This k was generated from looking at (x^2)^n-1 factorisations -covering set is 3,5,17,257,641,65537,6700417 which I think is 32-cover. I do not think anyone has claimed it is the smallest k, it just comes from the smallest-cover.

gd_barnes 2008-11-03 04:49

[quote=robert44444uk;147641]This k was generated from looking at (x^2)^n-1 factorisations -covering set is 3,5,17,257,641,65537,6700417 which I think is 32-cover. I do not think anyone has claimed it is the smallest k, it just comes from the smallest-cover.[/quote]


OK, very good. I asked because I'm undertaking an effort on 2 slow cores to see which small bases do not yield an easy prime for k=2. I started with the Riesel side and am testing bases 2 to 1024.

Here are the 20 Riesel bases <= 1024 remaining that do NOT have a prime of the form 2*b^n-1 at n=10K:
[code]
b
107
170
278
303
383
515
522
578
581
590
647
662
698
704
845
938
969
989
992
1019
[/code]

Here are the primes for n>=1000 found for the effort:
[code]
b (n)
785 (9670)
233 (8620)
618 (8610)
627 (7176)
872 (6036)
716 (4870)
298 (4202)
572 (3804)
380 (3786)
254 (2866)
669 (2787)
551 (2718)
276 (2484)
382 (2324)
968 (1750)
550 (1380)
434 (1166)
1013 (1116)
734 (1082)
215 (1072)
[/code]


I'm going to take it up to n=10K and then work on the Sierp side to the same depth. The hard part about the effort is that each base has to be sieved individually. AFAIK sr(x)sieve will not sieve more than one base at a time.

Obviously PROVING that the lowest base that has a Sierp k=2 would not be possible using the brute force approach such as this but it would be quite possible for higher values of k.

If anyone else has any input or info. for searches done like this with a fixed k and variable base, please post it here.

I will edit this post with additional primes found and update the search limit as I progress.

[B]Admin edit: Effort has now been completed to n=10K. 20 bases remain.[/B]


Gary

henryzz 2008-11-03 07:55

wouldnt a pfgw script work

gd_barnes 2008-11-03 11:07

[quote=henryzz;147663]wouldnt a pfgw script work[/quote]

How might one sieve using PFGW? I'm not referring to factoring like would be done with the -f100 or -f10000 option.

Sieving is the issue when attempting to search this way.

henryzz 2008-11-03 14:46

[quote=gd_barnes;147677]How might one sieve using PFGW? I'm not referring to factoring like would be done with the -f100 or -f10000 option.

Sieving is the issue when attempting to search this way.[/quote]
yes u would have to skip sieving and do trial factoring instead

robert44444uk 2008-11-04 04:31

Somebody should also be looking at the theory - by checking higher (x^2)^2-1 factorisations, to see whether a smaller k is feasible, by running through bigcover.exe

gd_barnes 2008-11-06 04:14

[quote=robert44444uk;147641]This k was generated from looking at (x^2)^n-1 factorisations -covering set is 3,5,17,257,641,65537,6700417 which I think is 32-cover. I do not think anyone has claimed it is the smallest k, it just comes from the smallest-cover.[/quote]

Wouldn't this be a 64-cover?

3 covers 0mod2; 1mod2 remains
5 covers 3mod4; 1mod4 remains
17 covers 5mod8; 1mod8 remains
257 covers 9mod16; 1mod16 remains
65537 covers 17mod32; 1mod32 remains
641 covers 33mod64; 1mod64 remains
6700417 covers 1mod64; all covered

Hence factors repeat every 64 n.

I see that all factors except 641 and 6700417 were generated by the form 2^(2^q)+1 where q>=0. Two questions:

1. How do these factors of the form 2^(2^q)+1 relate to the (x^2)^n-1 factorizations that you are talking about?
2. How were 641 and 6700417 arrived at? Did they just happen to "pop out" as a result of looking for bases with covers that contained factors of the form in which you are talking about.

I know that 641 is a commonly occurring factor that pops out in GFN factorizations. I'm sure that has something to do with why it occurred here.

Here is what I need to understand what you mean: Put each factor here in the (x^2)^n-1 form that you described. That will help further research on this topic. For me, examples help the best.


Thanks,
Gary

robert44444uk 2008-11-06 07:24

Sorry for misleading if it is 64-cover - should have checked!

There is unpublished work in this area, and for this reason I cannot really go into details on why these factorisations are key as it involves an unpublished theorem.

But applying this to 128-cover, 256-cover etc has not been carried out to my knowledge.

I will look into this a bit more this weekend

gd_barnes 2008-11-06 07:51

[quote=robert44444uk;147779]Somebody should also be looking at the theory - by checking higher (x^2)^2-1 factorisations, to see whether a smaller k is feasible, by running through bigcover.exe[/quote]

[quote=robert44444uk;148064]Sorry for misleading if it is 64-cover - should have checked!

There is unpublished work in this area, and for this reason I cannot really go into details on why these factorisations are key as it involves an unpublished theorem.

But applying this to 128-cover, 256-cover etc has not been carried out to my knowledge.

I will look into this a bit more this weekend[/quote]


It's definitely a 64-cover.

I'm confused now because you state one thing in one post and then imply the opposite in a follow-up post. At first, you say someone should be looking into the theory of (x^2)^2-1 factorizations but then you imply that we should not because there is unpublished work on the topic.

Two more questions:

1. What exactly is it that we can look into? A specific example or 2 would help.

2. I have a program called covering.exe. Where is the bigcover.exe program that you are referring to?


Thanks,
Gary

kar_bon 2008-11-06 09:48

don't know if your realized that the 5. Fermat number is F5 = 641 * 6700417 !

see here [url]http://www.prothsearch.net/fermat.html[/url]

gd_barnes 2008-11-06 10:09

[quote=kar_bon;148077]don't know if your realized that the 5. Fermat number is F5 = 641 * 6700417 !

see here [URL]http://www.prothsearch.net/fermat.html[/URL][/quote]

Very interesting! I thought those factors looked familiar from somewhere. I had seen 641 many times when messing around with GFNs but I didn't specifically remember 6700417.

If you or Robert can tell me how that fact relates to what he is doing here, perhaps I can help find a smaller base with k=2 that has a covering set.


Gary

gd_barnes 2008-11-06 10:11

Renamed thread to "k=1 [B]and k=2[/B]" since discussion in this thread is related to both.

robert44444uk 2008-11-06 12:39

[QUOTE=gd_barnes;148066]It's definitely a 64-cover.

I'm confused now because you state one thing in one post and then imply the opposite in a follow-up post. At first, you say someone should be looking into the theory of (x^2)^2-1 factorizations but then you imply that we should not because there is unpublished work on the topic.

Two more questions:

1. What exactly is it that we can look into? A specific example or 2 would help.

2. I have a program called covering.exe. Where is the bigcover.exe program that you are referring to?


Thanks,
Gary[/QUOTE]

Gary

Clumsy of me, the theory maintains that Fermat and GFN factors are key (not as stated), to find a specific, but not necessarily efficient b for any value of k, except k=1 or k=Mersenne.

The theory will be published in due course, but the key point is that if you find a fermat or GFN number with two primitive primes (fermat number wise) then a cover is possible. For k=2 we only look at Fermat numbers.

So if F6 has 2 new prime factors not in F1..F5 then that will provide a new cover. Each new Fermat number introduces at least 1 new odd prime as a factor.

Ex W. Keller

F5 = 641 . 6700417
F6 = 274177 . 67280421310721
F7 = 59649589127497217 . 5704689200685129054721
F8 = 1238926361552897 . P62
F9 = 2424833 . 7455602825647884208337395736200454918783366342657 . P99
F10 = 45592577 . 6487031809 . 4659775785220018543264560743076778192897 . P252
F11 = 319489 . 974849 . 167988556341760475137 . 3560841906445833920513 . P564

Will these produce a smaller b? The theorem only dealt with the fact that k=2=Sierpinski is possible, and did not look for the lowest b.

Bigcovering.exe is the big integer equivalent to covering.exe available from Robert Gerbicz's site.

kar_bon 2008-11-06 12:40

[QUOTE=robert44444uk;147641]This k was generated from looking at (x^2)^n-1 factorisations -covering set is 3,5,17,257,641,65537,6700417 which I think is 32-cover. I do not think anyone has claimed it is the smallest k, it just comes from the smallest-cover.[/QUOTE]

BTW
the set are the first Fermat numbers F0 to F6 with their factors!!

PS
ok, Robert saw this!

gd_barnes 2008-11-11 08:04

I have completed the "Riesel k=2 conjecture" up to n=10K for all bases <= 1024. 20 bases remain. See [URL="http://www.mersenneforum.org/showpost.php?p=147653&postcount=7"]post 7[/URL].

I have also completed the "Sierpinksi k=2 conjecture" up to n=10K for all bases <= 1024. Unlike the Riesel side where no bases could be removed with trivial factors, all b==(1 mod 3) were removed from the Sierp side with a trivial factor of 3. Despite this, the Sierp side is proving far more challenging with 30 vs. 20 bases remaining for the Riesels.

Here are the 30 Sierpinski bases <= 1024 remaining that do NOT have a prime of the form 2*b^n+1 at n=10K:
[code]
b
101
206
218
236
257
305
365
383
461
467
542
578
626
647
695
752
773
788
801
836
869
878
887
899
908
914
917
932
947
1004
[/code]

Here are the primes for n>=1000 found for the effort:

[code]
b (n)
758 (8309)
954 (8100)
368 (7045)
167 (6547)
287 (5467)
518 (4453)
38 (2729)
395 (2625)
635 (2535)
416 (2517)
353 (2313)
842 (1919)
698 (1885)
497 (1339)
867 (1280)
948 (1242)
104 (1233)
764 (1189)
992 (1179)
812 (1003)
[/code]

Interesting stats:
Only 3 Riesel bases are remaining that are divisible by 3: 303, 522, 969
Only 1 Sierp base is remaining that is divisible by 3: 801
ZERO Riesel bases are remaining where b==(1 mod 3). Base 298 was the last to drop at n=4202.
3 bases are remaining on both Riesel and Sierp: 383, 578, 647

Base conjectures proofs:
Riesel base 254 conjecture of k=4 is proven.
The largest prime was 2*254^2866-1.

Riesel base 669 conjecture of k=66 is proven.
k=4 & 64 were eliminated with algebraic factors.
The largest primes were 44*669^3132-1 and 2*669^2787-1.

Sierp base 38 conjecture of k=14 is proven.
The largest primes were 2*38^2729+1 and 9*38^21+1.

Sierp base 167 conjecture of k=8 is proven.
The largest primes were 2*167^6547+1 and 6*167^25+1.

Below are bases where the only k remaining at n=10K is k=2 in order to prove the base conjecture.
Note: While many lower bases have had all their k's tested to n=10K, a majority of the bases have only had other k-values (besides k=2) tested to n=2500.

Riesel:
[code]
b conjecture
278 14
515 44
704 4
845 46
969 96 (k=4 & 64 eliminated with algebraic factors)
989 4
1019 4
[/code]

Sierpinski:
[code]
b conjecture
101 16
206 22
305 16
461 8
626 10
752 16
869 4
899 4
914 4
917 16
1004 4
[/code]

I'm still doing some testing on other k-values for these "difficult k=2 bases" up to n=10K. I will also be doing some testing up to n=25K for k=2 on some of the lower bases where k=2 is the only k remaining. Because they have been difficult to find a prime for the lowest possible k-value on a base, most of them are the hardest to find primes for many k-values and hence the most difficult to prove.


Gary

Citrix 2009-12-26 23:57

This was also discussed here:

[url]http://www.mersenneforum.org/showthread.php?t=6918[/url]

Anyway, A solution for k=2 Sierpinski is base
12576017419215635147

(This is not the smallest possible, just that I have not put enough effort into it.)

:smile:

gd_barnes 2009-12-27 07:38

[quote=Citrix;199983]This was also discussed here:

[URL]http://www.mersenneforum.org/showthread.php?t=6918[/URL]

Anyway, A solution for k=2 Sierpinski is base
12576017419215635147

(This is not the smallest possible, just that I have not put enough effort into it.)

:smile:[/quote]

Nice work Citrix.

For clarification:

The covering set is:
{3, 5, 17, 257, 641, 65537, 6700417}

The period is n=64.

641 covers all n==(31 mod 64) and 6700417 covers all n==(63 mod 64) with the other 5 factors covering all of the rest.

Is that what you show?

Citrix 2010-01-01 20:14

Here is the smallest number with this covering set.

b=201446503145165177
(18 digits)
:smile:

sweety439 2016-11-23 19:18

k=1 and k=2
 
2 Attachment(s)
According to the CRUS link, some k=2 primes are known:

Riesel:
b (n)
107 (21910)
170 (166428)
278 (43908)
303 ?
383 (20956)
515 (58466)
522 ?
578 ?
581 ?
590 (15526)
647 (21576)
662 (16590)
698 (127558)
704 (62034)
845 (39406)
938 (40422)
969 (24096)
989 (26868)
992 ?
1019 ?

Sierpinski:
b (n)
101 (192275)
206 (46205)
218 ?
236 ?
257 (12183)
305 (16807)
365 ?
383 ?
461 ?
467 (126775)
542 ?
578 (44165)
626 (174203)
647 ?
695 (94625)
752 (26163)
773 ?
788 (72917)
801 ? (the only not-started base with k=2 remaining, what is the final test limit?)
836 ?
869 (49149)
878 ?
887 (27771)
899 (15731)
908 ?
914 ?
917 ?
932 (13643)
947 ?
1004 ?

Although I know exactly which bases have k=2 remaining (since we does not include GFN, no bases have k=1 remaining), I also want to know exactly which bases have k=3, k=4, k=5, k=6, ..., k=12 remaining (even including the not-started base and the bases which have conjectured k less than the k). e.g. I know some bases (like S718 and S912) have k=3 remaining, but how about a not-started base S358? I cannot find a k=3 prime for S358 (this is the first such Sierpinski base). Besides, I tests all Sierpinski bases b<=400 for k=10 and b<=600 for k=12, but I cannot find prime for some bases, see the test files.

gd_barnes 2016-11-23 20:35

Long ago I undertook an effort to search all k=2, 3, 4, 5, 6, and 7 to n=25K for bases <= 1030 regardless of how far the project has searched them. Although I personally searched no further I have continued to maintain lists of them and update them periodically with the search depth of the project and with primes found by the project. In the lists, I specify the bases remaining for each k, the search depth for each k, the # of k's remaining in the project conjecture for each base in question, and a list of bigger primes found for each base that was eliminated. I also specify which bases should not be searched due to trivial and algebraic factors.

I think you will find my lists interesting. I will post them in the next couple of days.

Batalov 2016-11-23 22:30

There existed some unrelated interest from other researchers to k=2.

I came across it when I had an email conversation with David Broadhurst and W.Keller concerning the "Divides Phi(*,2)" UTM Top20 [URL="http://primes.utm.edu/top20/page.php?id=37"]category[/URL].
It turns out that for b = 11 (mod 12), 2*b^n+1 may (frequently) divide a Phi(b^m,2). For other b values and other k values, the divisibility of this sort is more more rare.

There are a few bases b for which no 2*b^n+1 primes are yet known.
Here is an abbreviated quote from Keller (who did this work with Broadhurst & I.Buechel)
[QUOTE]I wanted to compare your upper limits with those I had reached some
years ago before stopping my whole search, and to add information about
a few bases q > 131 (in the same category) which I had also investigated,
particularly q = 383, 467, 647, 947 and 67607. These five bases continued
having no known prime at all.
...
I could verify that I had searched q = 47 as far as to 246k, and
q = 71 up to 261k.
...
Also, much effort had been invested in q = 383 (433k), 467 (347k),
647 (322k), 947 (much less: 105k), 60607 (412k).

Now, examining the files, near the end of the last one corresponding
to q = 467 I casually noted something "irregular" - which on a closer
look appeared to be this surprising entry:

[Thu Dec 07 10:45:01 2006]
2*467^126775+1 is a prime.
[/QUOTE]

gd_barnes 2016-11-24 09:02

1 Attachment(s)
Attached are my files for searches of k=2, 3, 4, 5, 6, and 7 for bases <= 1030 undertaken about 5 years ago. Here are the steps that I took to create the lists:
1. Use a PFGW sript to search all Riesel and Sierp bases for all six k's to n=5K.
2. Do some analysis to remove bases with trivial and algebraic factors as well as GFN's.
3. Do some sieving and test remaining bases up to n=25K.
4. Use the CRUS project to continually update them.

I have made every effort to keep these updated and synced up with CRUS but I could have missed something so anyone is welcome to double-check them. The file names should be self-explanatory. Here is an explanation of the columns in each file:

1. The k and base remaining.
2. The search depth of the k for that base.
3. The conjecture for that base.
4. The # of CRUS k's remaining for that base.*

*If the CRUS search depth of the base is different than the search depth for the k shown then the CRUS search depth will be shown next to the k's remaining in parenthesis.

*If the conjecture is less than the applicable k (meaning that CRUS has done no searching on the k) then the k's remaining reflect what the CRUS pages show as remaining PLUS the applicable k. In such cases I have searched the k/base to n=25K. This would only apply to k=5, 6, and 7 where the base conjecture is 4 or 6. (There are no conjectures of k=2, 3, or 5.)

*If CRUS has not searched the base then "many" is shown in the k's remaining.

The following k's have bases with trivial or algebraic factors to make a full covering set -or- make GFN's and so are not included in the searches (similar to the CRUS conjectures):
[B]Riesel k=2:[/B]
none
[B]Riesel k=3:
[/B]b==(1 mod 2); factor of 2
[B]Riesel k=4:[/B]
b==(1 mod 3); factor of 3
b==(4 mod 5): odd n, factor of 5; even n, algebraic factors (see CRUS pages)
b=m^2 proven composite by full algebraic factors
[B]Riesel k=5:[/B]
b==(1 mod 2); factor of 2
[B]Riesel k=6:[/B]
b==(1 mod 5); factor of 5
b==(34 mod 35); covering set [5, 7]
b=24, 54, 294, and 864: even n, factor of 5; odd n, algebraic factors (see CRUS pages)
[B]Riesel k=7:[/B]
b==(1 mod 2); factor of 2
b==(1 mod 3); factor of 3
[B]Sierp k=2:[/B]
b==(1 mod 3); factor of 3
b=512 is a GFN with no known prime
[B]Sierp k=3:[/B]
b==(1 mod 2); factor of 2
[B]Sierp k=4:[/B]
b==(1 mod 5); factor of 5
b==(14 mod 15); covering set [3, 5]
b=625 proven composite by full algebraic factors
b=32, 512, and 1024 are GFN's with no known prime
[B]Sierp k=5:[/B]
b==(1 mod 2); factor of 2
b==(1 mod 3); factor of 3
[B]Sierp k=6:[/B]
b==(1 mod 7); factor of 7
b==(34 mod 35); covering set [5, 7]
[B]Sierp k=7:[/B]
b==(1 mod 2); factor of 2

See the files for the k's remaining and their search depth. In a synopsis, here are the number of bases <= 1030 remaining for each k that do not have a prime:
Riesel k=2: 6
Riesel k=3: 2
Riesel k=4: 12
Riesel k=5: 2
Riesel k=6: 12
Riesel k=7: 9
Sierp k=2: 16
Sierp k=3: 2
Sierp k=4: 19
Sierp k=5: 4
Sierp k=6: 9
Sierp k=7: 1*

*Sierp k=7 has only one base remaining, 1004. The conjecture for the base is k=4 and therefore k=7 was only searched by me to n=25K. It would be interesting if someone would like to undertake an effort to search 7*1004^n+1 to n=100K or 200K to see if we can eliminate all bases for k=7. :-)


Here is a list of all primes found for n>5000:
[code]
base (n-value)
Riesel k=2:
107 (21910)
170 (166428)
233 (8620)
278 (43908)
383 (20956)
515 (58466)
590 (15526)
618 (8610)
627 (7176)
647 (21576)
662 (16590)
698 (127558)
704 (62034)
785 (9670)
845 (39406)
872 (6036)
938 (40422)
969 (24096)
989 (26868)

Riesel k=3:
432 (16002)

Riesel k=4:
72 (1119849)
212 (34413)
218 (23049)
270 (89661)
422 (21737)
480 (93609)
527 (46073)
537 (7287)
566 (23873)
582 (5841)
686 (16583)
758 (15573)
783 (12507)
786 (8001)
800 (33837)
947 (10055)
965 (8755)
998 (8427)

Riesel k=5:
14 (19698)
68 (13574)
196 (9849)
254 (15450)
800 (20508)
986 (5580)

Riesel k=6:
258 (212134)
272 (148426)
307 (26262)
354 (25561)
433 (283918)
635 (36162)
678 (40858)
692 (45446)
719 (20551)
783 (5022)
857 (23082)
867 (61410)
972 (36702)
999 (8887)
1025 (8958)

Riesel k=7:
68 (25395)
332 (15221)
338 (42867)
362 (146341)
458 (9823)
488 (33163)
566 (164827)
866 (7227)
980 (50877)
986 (12505)
1016 (23335)

Sierp k=2:
101 (192275)
167 (6547)
206 (46205)
257 (12183)
287 (5467)
305 (16807)
368 (7045)
467 (126775)
578 (44165)
626 (174203)
695 (94625)
752 (26163)
758 (8309)
788 (72917)
869 (49149)
887 (27771)
899 (15731)
932 (13643)
954 (8100)

Sierp k=3:
358 (9560)
996 (6549)

Sierp k=4:
77 (6098)
83 (5870)
107 (32586)
227 (13346)
257 (160422)
264 (9647)
355 (10989)
410 (144078)
440 (56086)
452 (14154)
470 (5218)
482 (30690)
542 (15982)
579 (67775)
608 (20706)
635 (11722)
650 (96222)
679 (69449)
737 (269302)
740 (58042)
818 (7726)
934 (101403)
962 (84234)

Sierp k=5:
350 (20391)
536 (8789)
554 (10629)
662 (13389)
926 (40035)

Sierp k=6:
108 (16317)
129 (16796)
409 (369832)
587 (24119)
614 (7480)
643 (164915)
669 (5450)
704 (5282)
762 (11151)
789 (27296)
919 (5092)
929 (9148)
969 (5888)
977 (6404)
986 (21633)
1018 (9943)

Sierp k=7:
398 (17472)
632 (8446)
836 (5700)
[/code]What I initially found interesting about these efforts is that there are few bases remaining for most k's and very few remaining for almost all odd k's (mainly since odd bases have a trivial factor of 2 for such k's).

Theoretically someone could reverse this project and have k-values with conjectured bases. Such bases would be conjectured to be the lowest base with a full covering set of numeric factors for each k. Of course this would be extraordinarily difficult to determine for k=2 and 3. But then you would have k=4, which is trivial since base=14 would be the conjecture and would be easily proven on both sides to be the lowest such base since all lower bases have a prime for k=4 (or a trivial set of factors). But it would be an interesting mathematical excercise to attempt to determine such conjectures for some of the more difficult k's. The problem with such a project is that the bases for each k would get very large very quickly and so would yield fewer primes and take much longer to search. (Imagine searching base=12345 for a small k!) Hence the project would quickly become less interesting than CRUS.

I did not desire to search beyond k=7 because, (see Riesel k=6 and Sierp k=4 above), as the k's get higher there becomes more and more bases with trivial factors and complex arrangements of covering sets and algebraic factors. k=8 would have gotten difficult on both sides with many combinations of algebraic factors. Therefore k=7 was a natural stopping point.

Nice effort Sweety. Your bases remaining and primes shown exactly match what I show for k=2. The attached files and this post should give you additional info. for furthering such efforts. One problem that I see for your k=10 and 12 files: You are including n=0 in the searches. CRUS requires that primes be n>=1. The reason for this is that n=0 makes the k become trivial for all bases and hence not interesting. For example k=14 would be eliminated for all Riesel bases because 13 is prime and k=18 would be eliminated for all bases on both sides since both 17 and 19 are prime.

I have updated the thread title to show k=1 thru 12.

sweety439 2016-11-24 12:08

I don't include n=0 in the searches. In these files, n=0 means no such prime exists, e.g. in the k=12 file, since 12*14^n+1 is always divisible by 13, no primes are of the form 12*14^n+1, and 12*142^n+1 is always divisible by 11 or 13, no primes are of the form 12*142^n+1, and 12*296^n+1 is always divisible by 7, 11, 13, or 19, no primes are of the form 12*296^n+1. Thus, in the file, the n of k=14, 27, 40, ..., 142, ..., 296, etc. are 0.

sweety439 2016-11-24 12:17

Thank you, gd_barnes. However, your file still miss some primes, such as 4*737^269302+1 and 4*72^1119849-1. Besides, you only tested 2<=k<=7 for bases 2<=b<=1030, but I want the list of the primes with n>1000 (you only listed n>5000) for 2<=k<=12, 2<=b<=1728 = 12^3 and remain bases for 2<=k<=12 and 2<=b<=1728 = 12^3. I am curious that why you stopped at b=1030.
I only want the k's <= 12, I will not have you give the list of the primes and remain bases for k=13, 14, 15, 16, ..., you don't need to search beyond k=12.

sweety439 2016-11-24 12:55

1 Attachment(s)
I want a list like this.

sweety439 2016-11-24 14:56

See [URL]http://mersenneforum.org/showthread.php?t=15188[/URL] for bases 1031<=b<=2048.

For Sierp k=8, all perfect cube bases and all bases =(1 mod 3), =(20 mod 21) should not be searched.
For Riesel k=8, all perfect cube bases and all bases =(1 mod 7), =(20 mod 21) should not be searched.
For Sierp k=9, all bases =(1 mod 2) and all bases =(1 mod 5) should not be searched.
For Riesel k=9, all perfect square bases and all bases =(1 mod 2) and all bases =(4 mod 5) should not be searched.
For Sierp k=10, all bases =(1 mod 11) and all bases =(32 mod 33) should not be searched.
For Riesel k=10, all bases =(1 mod 3) and all bases =(32 mod 33) should not be searched.
For Sierp k=11, all bases =(1 mod 2) and all bases =(1 mod 3) and all bases =(14 mod 15) should not be searched.
For Riesel k=11, all bases =(1 mod 2) and all bases =(1 mod 5) and all bases =(14 mod 15) should not be searched.
For Sierp k=12, all bases =(1 mod 13) and all bases =(142 mod 143) should not be searched.
For Riesel k=12, all bases =(1 mod 11) and all bases =(142 mod 143) should not be searched.

Also, for Riesel k=4, all bases =(4 mod 5) can be proven composite for partial algebraic factors and should not be searched.

gd_barnes 2016-11-24 17:46

[QUOTE=sweety439;447688]Thank you, gd_barnes. However, your file still miss some primes, such as 4*737^269302+1 and 4*72^1119849-1. Besides, you only tested 2<=k<=7 for bases 2<=b<=1030, but I want the list of the primes with n>1000 (you only listed n>5000) for 2<=k<=12, 2<=b<=1728 = 12^3 and remain bases for 2<=k<=12 and 2<=b<=1728 = 12^3. I am curious that why you stopped at b=1030.
I only want the k's <= 12, I will not have you give the list of the primes and remain bases for k=13, 14, 15, 16, ..., you don't need to search beyond k=12.[/QUOTE]

I did see those primes and you can see that those bases are not remaining in the bases remaining files. But you are correct, I did miss listing them in my primes files. Thanks for pointing it out. I will add them. After I did the initial search 5 years ago I have just manually updated the primes files and bases remaining files so it doesn't surprise me that I missed posting some primes.

You are asking a lot. I have all of the primes but digging up the primes for n=1K-5K would take a while to organize them. I did not search beyond b=1030 because that is as far as the CRUS project decided to go. We had to stop at some point. The project is already huge. I did not search beyond k=7 because...well...I didn't want to. It was an interesting effort but it wasn't that interesting and I grew tired of determining bases with trivial and algebraic factors. I'll leave it up to you to search b>1030 and k>7.

sweety439 2016-11-24 18:18

I will search all 1<=k<=12 and 2<=b<=1728 to 400K, but not now.

gd_barnes 2016-11-25 07:26

[QUOTE=sweety439;447687]I don't include n=0 in the searches. In these files, n=0 means no such prime exists, e.g. in the k=12 file, since 12*14^n+1 is always divisible by 13, no primes are of the form 12*14^n+1, and 12*142^n+1 is always divisible by 11 or 13, no primes are of the form 12*142^n+1, and 12*296^n+1 is always divisible by 7, 11, 13, or 19, no primes are of the form 12*296^n+1. Thus, in the file, the n of k=14, 27, 40, ..., 142, ..., 296, etc. are 0.[/QUOTE]

[QUOTE=sweety439;447688]Thank you, gd_barnes. However, your file still miss some primes, such as 4*737^269302+1 and 4*72^1119849-1. Besides, you only tested 2<=k<=7 for bases 2<=b<=1030, but I want the list of the primes with n>1000 (you only listed n>5000) for 2<=k<=12, 2<=b<=1728 = 12^3 and remain bases for 2<=k<=12 and 2<=b<=1728 = 12^3. I am curious that why you stopped at b=1030.
I only want the k's <= 12, I will not have you give the list of the primes and remain bases for k=13, 14, 15, 16, ..., you don't need to search beyond k=12.[/QUOTE]

[QUOTE=sweety439;447697]See [URL]http://mersenneforum.org/showthread.php?t=15188[/URL] for bases 1031<=b<=2048.

For Sierp k=8, all perfect cube bases and all bases =(1 mod 3), =(20 mod 21) should not be searched.
For Riesel k=8, all perfect cube bases and all bases =(1 mod 7), =(20 mod 21) should not be searched.
For Sierp k=9, all bases =(1 mod 2) and all bases =(1 mod 5) should not be searched.
For Riesel k=9, all perfect square bases and all bases =(1 mod 2) and all bases =(4 mod 5) should not be searched.
For Sierp k=10, all bases =(1 mod 11) and all bases =(32 mod 33) should not be searched.
For Riesel k=10, all bases =(1 mod 3) and all bases =(32 mod 33) should not be searched.
For Sierp k=11, all bases =(1 mod 2) and all bases =(1 mod 3) and all bases =(14 mod 15) should not be searched.
For Riesel k=11, all bases =(1 mod 2) and all bases =(1 mod 5) and all bases =(14 mod 15) should not be searched.
For Sierp k=12, all bases =(1 mod 13) and all bases =(142 mod 143) should not be searched.
For Riesel k=12, all bases =(1 mod 11) and all bases =(142 mod 143) should not be searched.

Also, for Riesel k=4, all bases =(4 mod 5) can be proven composite for partial algebraic factors and should not be searched.[/QUOTE]

Nice research effort Sweety. It now makes sense what you mean for the bases where n=0. I have corrected my post #27 to show the partial algebraic factors for Riesel k=4 and added the primes for Riesel and Sierp k=4. Fortunately these were just posting omissions on my part. They did not affect which bases were remaining for each k so the previously attached files are still correct.

I found all of my files of primes where n<=5000 for these efforts. They are not sorted quite like you want but you should be able to get them into your desired format. They will be attached to the next post.

gd_barnes 2016-11-25 07:55

1 Attachment(s)
Attached are all primes for k=2 thru k=7 for bases<=1030 where n<=5000. These are sorted by n-value. Combining these primes with all previously posted primes for n>5000 in post #27 should give all known primes for these k and base ranges. :smile:

sweety439 2016-11-28 12:14

There is another problem, the reversed Sierpinski/Riesel problem.

For fixed k, find the smallest base b such that all numbers of the form k*b^n+1 (k*b^n-1) are composite.

If k is of the form 2^n-1 (2^n+1), except k=1 (k=9), it is conjectured for every nontrivial base b, there is a prime of the form k*b^n+1 (k*b^n-1).

However, for all other k's, there is a base b such that all numbers of the form k*b^n+1 (k*b^n-1) are composite.

For the Sierpinski (k*b^n+1) cases: (for k up to 24, only tested bases b<=1030)

S = conjectured smallest base b such that k is a Sierpinski number.

k S remaining bases b with no known primes
1 8 proven
2 201446503145165177 (?) {218, 236, 365, 383, 461, 512, 542, 647, 773, 801, 836, 878, 908, 914, 917, 947, 1004, ...}
3 none {718, 912, ...}
4 14 proven
5 140324348 {308, 326, 512, 824, ...}
6 34 proven
7 none {1004, ...}
8 8 proven
9 177744 {244, 592, 724, 884, 974, 1004, ...}
10 32 proven
11 14 proven
12 142 {12}
13 20 proven
14 38 proven
15 none {398, 650, 734, 874, 876, 1014, ...}
16 38 {32}
17 278 {68, 218}
18 322 {18, 74, 227, 239, 293}
19 14 proven
20 56 proven
21 54 proven
22 68 {22}
23 32 proven
24 114 {79}

sweety439 2016-11-28 17:51

1 Attachment(s)
This is I found primes for Sierpinski k=10 with base b<=1030. I didn't search very far, so there are many bases remaining.

I take Sierpinski k=10 because of [URL]https://oeis.org/A088622[/URL] [URL]https://oeis.org/A088782[/URL] and [URL]https://oeis.org/A088783[/URL].

The smallest prime obtained as the concatenation of a power of b followed by a 1, is in fact the smallest prime of the form 10*b^n+1. Thus, this is the reverse Sierpinski problem with k=10.

A088783 lists the bases for which 10 is a Sierpinski number or a trivial k (gcd(10+1, b-1) is not 1), but only up to 166, since when this sequence was created, 10 is still a remaining k for S173. Recently, a prime 10*173^264234+1 was found, next term of this sequence should be 177. However, the correct value of the term after 177 is still unknown, since 10 is a remaining k for S185 (at n=1M).

gd_barnes 2016-11-28 20:43

[QUOTE=sweety439;447963]There is another problem, the reversed Sierpinski/Riesel problem.

For fixed k, find the smallest base b such that all numbers of the form k*b^n+1 (k*b^n-1) are composite.

If k is of the form 2^n-1 (2^n+1), except k=1 (k=9), it is conjectured for every nontrivial base b, there is a prime of the form k*b^n+1 (k*b^n-1).

However, for all other k's, there is a base b such that all numbers of the form k*b^n+1 (k*b^n-1) are composite.

For the Sierpinski (k*b^n+1) cases: (for k up to 24, only tested bases b<=1030)

S = conjectured smallest base b such that k is a Sierpinski number.

k S remaining bases b with no known primes
1 8 proven
2 201446503145165177 (?) {218, 236, 365, 383, 461, 512, 542, 647, 773, 801, 836, 878, 908, 914, 917, 947, 1004, ...}
3 none {718, 912, ...}
4 14 proven
5 140324348 {308, 326, 512, 824, ...}
6 34 proven
7 none {1004, ...}
8 8 proven
9 177744 {244, 592, 724, 884, 974, 1004, ...}
10 32 proven
11 14 proven
12 142 {12}
13 20 proven
14 38 proven
15 none {398, 650, 734, 874, 876, 1014, ...}
16 38 {32}
17 278 {68, 218}
18 322 {18, 74, 227, 239, 293}
19 14 proven
20 56 proven
21 54 proven
22 68 {22}
23 32 proven
24 114 {79}[/QUOTE]

Interesting list. One thing that I would suggest that CRUS does: Do not include GFN's in your conjectures or bases remaining since the only primes found for them would have to be of the form b^(2^m)+1 where m>=0. Excluding GFNs would change your list as follows:

1. k=1 is not applicable because all odd b have a trivial factor of 2 and all even b are GFNs.
2. For k=2, b=512 is removed since 2*512^n+1=2^(9n+1)+1.
3. For k=12, b=12 is removed since 12*12^n+1=12^(n+1)+1; k=12 is proven.
4. For k=16, b=32 is removed since 16*32^n+1=2^(5n+4)+1; k=16 is proven.
5. For k=18, b=18 is removed since 18*18^n+1=18^(n+1)+1.
6. For k=22, b=22 is removed since 22*12^n+1=22^(n+1)+1; k=22 is proven.

One question: How did you determine the conjectures for k=2, k=5, and k=9?

gd_barnes 2016-11-29 05:28

[QUOTE=sweety439;447990]This is I found primes for Sierpinski k=10 with base b<=1030. I didn't search very far, so there are many bases remaining.

I take Sierpinski k=10 because of [URL]https://oeis.org/A088622[/URL] [URL]https://oeis.org/A088782[/URL] and [URL]https://oeis.org/A088783[/URL].

The smallest prime obtained as the concatenation of a power of b followed by a 1, is in fact the smallest prime of the form 10*b^n+1. Thus, this is the reverse Sierpinski problem with k=10.

A088783 lists the bases for which 10 is a Sierpinski number or a trivial k (gcd(10+1, b-1) is not 1), but only up to 166, since when this sequence was created, 10 is still a remaining k for S173. Recently, a prime 10*173^264234+1 was found, next term of this sequence should be 177. However, the correct value of the term after 177 is still unknown, since 10 is a remaining k for S185 (at n=1M).[/QUOTE]

Here are some CRUS primes to help fill in your k=10 file:

10*460^751+1
10*708^17562+1
10*830^436+1
10*927^4752+1
10*954^1506+1
10*1012^426+1

sweety439 2016-11-29 12:41

There is an OEIS sequence for the reverse-Sierpinski problem [URL]https://oeis.org/A263500[/URL].
However, no OEIS sequence for the reverse-Riesel problem, this is my research of this problem:

R = conjectured smallest base b such that k is a Riesel number.

k R remaining bases b with no known primes
1 none (for all base b>2, 1 is a trivial k, and for base b=2, there is a prime with n=2: 1*2^2-1)
2 none
3 none
4 9 proven
5 none
6 24 proven
7 ? (I found no information of this term, can someone find it?)
8 20 proven
9 4 proven
10 32 proven
11 14 proven
12 142 {65, 98}
13 20 proven
14 8 proven
15 ? (I found no information of this term, can someone find it?)
16 9 proven
17 none
18 50 proven
19 14 proven
20 56 proven
21 54 proven
22 68 {38, 62}
23 32 proven
24 114 {64}

There are also OEIS sequence for Sierpinski/Riesel problem:

Sierpinski problem: [URL]https://oeis.org/A123159[/URL]
Riesel problem: [URL]https://oeis.org/A273987[/URL]

sweety439 2016-11-29 12:55

Thanks for some k=10 prime.

About one month ago, I take my effort to find k=10 prime for b=269, 278, 282, 284, 356.

There are some bases b<=1030 remain for k=10:

185 (1M)
338 (100K)
417 (400K)
432 ?
449 ?
537 ?
614 ?
668 ?
671 ?
726 ?
728 ?
743 (200K)
744 (100K)
773 (200K)
786 ?
827 ?
863 ?
869 ?
885 ?
929 ?
935 (200K)
959 ?
977 (100K)
986 ?
1000 (GFN, searched up to (2^25-1)/3-1)
1004 ?

These bases are either not started or have a conjectured k<10.

sweety439 2016-11-29 18:46

For the remaining bases for the reverse-Riesel problem for k=2, 3, 5, 7, 15 and 17:

k R remaining bases b with no known primes
2 none {303, 522, 578, 581, 992, 1019, ...}
3 none {588, 972, ...}
5 none {338, 998, ...}
7 ? {308, 392, 398, 518, 548, 638, 662, 848, 878, ...}
15 ? {454, 552, 734, 856, ...}
17 none {98, 556, 650, 662, 734, ...}

gd_barnes 2016-11-29 18:54

[QUOTE=sweety439;448031]There is an OEIS sequence for the reverse-Sierpinski problem [URL]https://oeis.org/A263500[/URL].
However, no OEIS sequence for the reverse-Riesel problem, this is my research of this problem:

R = conjectured smallest base b such that k is a Riesel number.

k R remaining bases b with no known primes
1 none (for all base b>2, 1 is a trivial k, and for base b=2, there is a prime with n=2: 1*2^2-1)
2 none
3 none
4 9 proven
5 none
6 24 proven
7 ? (I found no information of this term, can someone find it?)
8 20 proven
9 4 proven
10 32 proven
11 14 proven
12 142 {65, 98}
13 20 proven
14 8 proven
15 ? (I found no information of this term, can someone find it?)
16 9 proven
17 none
18 50 proven
19 14 proven
20 56 proven
21 54 proven
22 68 {38, 62}
23 32 proven
24 114 {64}

There are also OEIS sequence for Sierpinski/Riesel problem:

Sierpinski problem: [URL]https://oeis.org/A123159[/URL]
Riesel problem: [URL]https://oeis.org/A273987[/URL][/QUOTE]

I think it is a mistake to include partial or complete algebraic factors for a conjectured k-value especially for Riesel bases. See all of the Riesel bases where k=4 or k=9 would become the conjecture that become uninteresting. On the Sierp side the Sierp OEIS page has a conjecture of k=1 for bases 8 and 32. This is not in the spirit of a good conjecture project. In the case of the reverse conjectures many Riesel k's will have base=4 or 9 as the conjecture leading to many quick proofs that demonstrate little.

Just my two cents.

sweety439 2016-11-29 18:57

According to [URL]https://web.archive.org/web/20160507115134/http://www.prothsearch.net/riesel2.html[/URL], 3*2^18123-1 is prime, that is, 24*64^3020-1 is prime. Thus, base 64 can be removed for the reverse-Riesel problem with k=24, and the reverse-Riesel problem with k=24 is proven. (This is the smallest prime of the form 24*64^n-1 with n>1, since 18123 is the smallest number greater than 3 and = (3 mod 6) in the k=3 list.

sweety439 2016-12-01 19:12

S = conjectured smallest base b such that k is a Sierpinski number.

k S cover set/algebraic factors
1 8 1*8^n+1 = (1*2^n+1) * (1*4^n-1*2^n+1) period=1
2 201446503145165177 (?) {3, 5, 17, 257, 641, 65537, 6700417} period=64
3 none
4 14 {3, 5} period=2
5 140324348 {3, 13, 17, 313, 11489} period=16
6 34 {5, 7} period=2
7 none
8 8 8*8^n+1 = (2*2^n+1) * (4*4^n-2*2^n+1) period=1
9 177744 {5, 17, 41, 193} period=8
10 32 {3, 11} period=2
11 14 {3, 5} period=2
12 142 {11, 13} period=2
13 20 {3, 7} period=2
14 38 {3, 13} period=2
15 none
16 38 {3, 5, 17} period=4
17 278 {3, 5, 29} period=4
18 322 {17, 19} period=2
19 14 {3, 5} period=2
20 56 {3, 19} period=2
21 54 {5, 11} period=2
22 68 {3, 23} period=2
23 32 {3, 11} period=2
24 114 {5, 23} period=2

R = conjectured smallest base b such that k is a Riesel number.

k R cover set/algebraic factors
1 none
2 none
3 none
4 9 4*9^n-1 = (2*3^n-1) * (2*3^n+1) period=1
5 none
6 24 {5} for even n, 6*24^n-1 = (12*24^((n-1)/2)-1) * (12*24^((n-1)/2)+1) for odd n, period=2
7 ? {3, 5, 17, 1201, 169553} period=16
8 20 {3, 7} period=2
9 4 9*4^n-1 = (3*2^n-1) * (3*2^n+1) period=1
10 32 {3, 11} period=2
11 14 {3, 5} period=2
12 142 {11, 13} period=2
13 20 {3, 7} period=2
14 8 {3, 5, 13} period=4
15 ? {7, 17, 113, 1489} period=8
16 9 16*9^n-1 = (4*3^n-1) * (4*3^n+1) period=1
17 none
18 50 {17} for even n, 18*50^n-1 = (30*50^((n-1)/2)-1) * (30*50^((n-1)/2)+1) for odd n, period=2
19 14 {3, 5} period=2
20 56 {3, 19} period=2
21 54 {5, 11} period=2
22 68 {3, 23} period=2
23 32 {3, 11} period=2
24 114 {5, 23} period=2

sweety439 2016-12-01 19:26

The term 201446503145165177 for Sierpinski problem with k=2 is the smallest known nontrivial base b (i.e. gcd(b-1, 2+1)=1) such that 2*b^n+1 is composite for all n>=1, it is not known whether there is a smaller such base b. Thus, the list lists "201446503145165177 (?)", with a question mark.

There is a pdf file [URL]https://www.rose-hulman.edu/mathjournal/archives/2008/vol9-n2/paper4/v9n2-4pd.pdf[/URL] for the research of the reverse-Sierpinski problem, but not the reverse-Riesel problem, can someone find the smallest nontrivial base b such that 7*b^n-1 (or 15*b^n-1) is composite for all n>=1? (I have found a possible cover set, like the cover set of 5*b^n+1 and 9*b^n+1, they are taken by the odd prime factors of the generated Fermat number with these bases, but the smallest nontrivial base b such that 5*b^n+1 (or 9*b^n+1) is composite for all n>=1 is given by the OEIS sequence [URL]https://oeis.org/A263500[/URL])

I want to find the smallest base b with this cover set.

henryzz 2016-12-02 00:59

[QUOTE=sweety439;448165]The term 201446503145165177 for Sierpinski problem with k=2 is the smallest known nontrivial base b (i.e. gcd(b-1, 2+1)=1) such that 2*b^n+1 is composite for all n>=1, it is not known whether there is a smaller such base b. Thus, the list lists "201446503145165177 (?)", with a question mark.

There is a pdf file [URL]https://www.rose-hulman.edu/mathjournal/archives/2008/vol9-n2/paper4/v9n2-4pd.pdf[/URL] for the research of the reverse-Sierpinski problem, but not the reverse-Riesel problem, can someone find the smallest nontrivial base b such that 7*b^n-1 (or 15*b^n-1) is composite for all n>=1? (I have found a possible cover set, like the cover set of 5*b^n+1 and 9*b^n+1, they are taken by the odd prime factors of the generated Fermat number with these bases, but the smallest nontrivial base b such that 5*b^n+1 (or 9*b^n+1) is composite for all n>=1 is given by the OEIS sequence [URL]https://oeis.org/A263500[/URL])

I want to find the smallest base b with this cover set.[/QUOTE]
I am pretty certain that the argument that proves that there isn't a Sierpinski base if k is a Mersenne number can be twisted for Riesel ks. Similarly there isn't a Riesel base if k is a number of the form 2^n+1.

For k=7 a Riesel b is 33559116638. I do not claim that this is minimal.

Combine the following statements to form a covering set and input into your favourite chinese remainder theorem calculator.
[CODE]
If b = 2 mod 3 then p | 7*b^n-1 if n = 0 mod 2
If b = 2 mod 5 then p | 7*b^n-1 if n = 3 mod 4
If b = 3 mod 5 then p | 7*b^n-1 if n = 1 mod 4
If b = 3 mod 17 then p | 7*b^n-1 if n = 5 mod 16
If b = 5 mod 17 then p | 7*b^n-1 if n = 1 mod 16
If b = 6 mod 17 then p | 7*b^n-1 if n = 11 mod 16
If b = 7 mod 17 then p | 7*b^n-1 if n = 15 mod 16
If b = 10 mod 17 then p | 7*b^n-1 if n = 7 mod 16
If b = 11 mod 17 then p | 7*b^n-1 if n = 3 mod 16
If b = 12 mod 17 then p | 7*b^n-1 if n = 9 mod 16
If b = 14 mod 17 then p | 7*b^n-1 if n = 13 mod 16
If b = 7 mod 1201 then p | 7*b^n-1 if n = 7 mod 8
If b = 343 mod 1201 then p | 7*b^n-1 if n = 5 mod 8
If b = 858 mod 1201 then p | 7*b^n-1 if n = 1 mod 8
If b = 1194 mod 1201 then p | 7*b^n-1 if n = 3 mod 8
If b = 7 mod 169553 then p | 7*b^n-1 if n = 15 mod 16
If b = 343 mod 169553 then p | 7*b^n-1 if n = 5 mod 16
If b = 16807 mod 169553 then p | 7*b^n-1 if n = 3 mod 16
If b = 24222 mod 169553 then p | 7*b^n-1 if n = 1 mod 16
If b = 145331 mod 169553 then p | 7*b^n-1 if n = 9 mod 16
If b = 152746 mod 169553 then p | 7*b^n-1 if n = 11 mod 16
If b = 169210 mod 169553 then p | 7*b^n-1 if n = 13 mod 16
If b = 169546 mod 169553 then p | 7*b^n-1 if n = 7 mod 16[/CODE]Check all covering combinations to find smallest b.

sweety439 2016-12-02 15:28

There is no base b such that k is a Sierpinski number if and only if k is of the form 2^n-1 for positive integer n but k != 1. Similarly, there is no base b such that k is a Sierpinski number if and only if k is of the form 2^n-1 for positive integer n but k != 9 (or k=1, since k=1 is a trivial k for all the Riesel bases b except 2, but R2 has an easy prime for k=1: 1*2^2-1). According to the Catalan conjecture (which was proven), 1 and 9 are the only two perfect powers next to a power of 2. If this k is of these forms, then k+1 (or k-1) is a power of 2 and the only prime factor of it is 2, since the n=0 case would cause the number to be a power of 2. Thus, 2 must be in the cover set, so both of k and b must be odd and both of b-1 and k+-1 are divisible by 2, and this k is a trivial k. The unique possible situation is: there is algebraic factors of k*b^n+-1. Thus, there exists an n such that k*b^n is a perfect power, and since the n=0 case would cause the number to be a power of 2, n=0 must be in the algebraic n's. Thus, n must be a perfect power, due to the Catalan conjecture, for the Sierpinski case, n must be 1, and for the Riesel case, n must be 9.

Similarly, if there exists an n such that all of the prime factor of b^n+k are prime factors of b or prime factors of k, then the sequence of the smallest prime factor of k*b^n+1 is unbounded above, since b^n+k is the dual of k*b^n+1. Besides, if there exists an n such that all of the prime factor of |b^n-k| are prime factors of b or prime factors of k, then the sequence of the smallest prime factor of k*b^n-1 is unbounded above, since |b^n-k| is the dual of k*b^n-1.

If k*b^n+1 (or k*b^n-1) is composite for all n>=1 and with a cover set, then (b^n+k)/gcd(b^n, k) (or (|b^n-k|)/gcd(b^n, k)) is also composite for all n>=1 and with the same cover set. We can find a dual prime to prove the dual Sierpinski/Riesel conjecture base b, or the mixed Sierpinski/Riesel conjecture base b.

In fact, we want to find the smallest positive integer k such that gcd(k+-1, b-1)=1 (+ for Sierpinski, - for Riesel) but the sequence of the smallest prime factor of k*b^n+-1 (+ for Sierpinski, - for Riesel) is bounded above, it is conjectured that these k's are the same as the CRUS k's (the smallest k such a full cover set). However, it is not proven, e.g. is the sequence of the smallest prime factor of 5*2^n+1 unbounded above? Even this case is not proven.

sweety439 2016-12-02 15:46

33559116638 may be the smallest base b with this cover set: {3, 5, 17, 1201, 169553} for 7*b^n-1. Like this number, 140324348, it may be the smallest b with this cover set: {13, 17, 313, 11489} for 5*b^n+1.

For the k=15 case, I want to find the smallest base with this cover set: {7, 17, 113, 1489}, this is my research:

If b = 6 mod 7 then 7 | 15*b^n-1 if n = 0 mod 2
If b = 98 mod 113 then 113 | 15*b^n-1 if n = 1 mod 4
If b = 15 mod 113 then 113 | 15*b^n-1 if n = 3 mod 4
If b = 8 mod 17 then 17 | 15*b^n-1 if n = 1 mod 8
If b = 2 mod 17 then 17 | 15*b^n-1 if n = 3 mod 8
If b = 9 mod 17 then 17 | 15*b^n-1 if n = 5 mod 8
If b = 15 mod 17 then 17 | 15*b^n-1 if n = 7 mod 8
If b = 1092 mod 1489 then 1489 | 15*b^n-1 if n = 1 mod 8
If b = 200 mod 1489 then 1489 | 15*b^n-1 if n = 3 mod 8
If b = 397 mod 1489 then 1489 | 15*b^n-1 if n = 5 mod 8
If b = 15 mod 1489 then 1489 | 15*b^n-1 if n = 7 mod 8

I found the smallest base is 1099279, but it is also a trivial base (i.e. gcd(15-1, 1099279-1) is not 1). Thus, it is the next smallest base, 8241218.

sweety439 2016-12-02 15:49

Thanks! I completed my lists for k up to 24:

S = conjectured smallest base b such that k is a Sierpinski number.

k S cover set/algebraic factors
1 8 1*8^n+1 = (1*2^n+1) * (1*4^n-1*2^n+1) period=1
2 201446503145165177 (?) {3, 5, 17, 257, 641, 65537, 6700417} period=64
3 none
4 14 {3, 5} period=2
5 140324348 {3, 13, 17, 313, 11489} period=16
6 34 {5, 7} period=2
7 none
8 8 8*8^n+1 = (2*2^n+1) * (4*4^n-2*2^n+1) period=1
9 177744 {5, 17, 41, 193} period=8
10 32 {3, 11} period=2
11 14 {3, 5} period=2
12 142 {11, 13} period=2
13 20 {3, 7} period=2
14 38 {3, 13} period=2
15 none
16 38 {3, 5, 17} period=4
17 278 {3, 5, 29} period=4
18 322 {17, 19} period=2
19 14 {3, 5} period=2
20 56 {3, 19} period=2
21 54 {5, 11} period=2
22 68 {3, 23} period=2
23 32 {3, 11} period=2
24 114 {5, 23} period=2

R = conjectured smallest base b such that k is a Riesel number.

k R cover set/algebraic factors
1 none
2 none
3 none
4 9 4*9^n-1 = (2*3^n-1) * (2*3^n+1) period=1
5 none
6 24 {5} for even n, 6*24^n-1 = (12*24^((n-1)/2)-1) * (12*24^((n-1)/2)+1) for odd n, period=2
7 33559116638 {3, 5, 17, 1201, 169553} period=16
8 20 {3, 7} period=2
9 4 9*4^n-1 = (3*2^n-1) * (3*2^n+1) period=1
10 32 {3, 11} period=2
11 14 {3, 5} period=2
12 142 {11, 13} period=2
13 20 {3, 7} period=2
14 8 {3, 5, 13} period=4
15 8241218 {7, 17, 113, 1489} period=8
16 9 16*9^n-1 = (4*3^n-1) * (4*3^n+1) period=1
17 none
18 50 {17} for even n, 18*50^n-1 = (30*50^((n-1)/2)-1) * (30*50^((n-1)/2)+1) for odd n, period=2
19 14 {3, 5} period=2
20 56 {3, 19} period=2
21 54 {5, 11} period=2
22 68 {3, 23} period=2
23 32 {3, 11} period=2
24 114 {5, 23} period=2

Note: The Sierpinski case the base b does not exist if and only if k is of the form 2^n-1 and k != 1. The Riesel case the base b does not exist if and only if k is of the form 2^n+1 and k != 9, or k = 1.

sweety439 2016-12-02 16:02

It is only conjectured these are the smallest Sierpinski/Riesel base for these k's, like the original Sierpinski/Riesel problems, these are also only conjectured (e.g. 6694 is the smallest base 22 Sierpinski number, and 4461 is the smallest base 22 Riesel number, both of them are not proven and with 1 k remain). However, it is very hard (as I think) to proven the Sierpinski k=2, k=5, and k=9 case and the Riesel k=7 and k=15 case. Besides, there is no base b such that k=3, 7, 15, ... is a Sierpinski number, and there is no base b such that k=2, 3, 5, 17, ... is a Riesel number (of course for Riesel k=1 case, but k=1 is a trivial k for all base b>=3 and there is a prime for base b=2: 1*2^2-1. Thus, 1 is not a Riesel number to any base b). However, it is also only conjectured, it is unknown that for every nontrivial base b, there is a prime of the form 3*b^n+1, 7*b^n+1, 15*b^n+1, 2*b^n-1, 3*b^n-1, 5*b^n-1, 17*b^n-1, ..., I think since there are infinitely many bases b need to be tested, unlike the 2*b^n+1, 5*b^n+1, 9*b^n+1, 7*b^n-1, 15*b^n-1, ... cases (these cases there are only a finite number of base b need to be tested), the problem may be still unsolved at 2200!

sweety439 2016-12-02 19:46

1 Attachment(s)
These are the Sierpinski k=12 primes up to b=1030.

gd_barnes 2016-12-05 05:39

3 Attachment(s)
I have now searched all k=2 thru k=10 for bases <= 1030. As previously stated all k=2 thru 7 have been searched to n=25K for all bases. The new effort for k=8 thru k=10 has been searched to n=5K for all bases.

Also done with the effort were:
1. Research on the CRUS pages and prime files for k=8 thru 10 to add primes for n>5K.
2. Put primes found for n>5K into separate files for all bases.
3. Check all exclusions, that is bases with trivial factors, algebraic factors, and covering sets, previously posted by Sweety for k=8 thru k=10. All were correct with the exception of k=8 on both sides where I found some additional covering sets. I previously posted all exclusions for k=2 thru k=7. Shown below are exclusions for k=8 thru 10:

[code]
Riesel k=8:
b==(1 mod 7) has a factor of 7
b==(20 mod 21) has a covering set of [3, 7]
*b==(83, 307 mod 455) has a covering set of [5, 7, 13]
(This includes bases 83, 307, 538, 762, 993.)
b=m^3 proven composite by full algebraic factors

Riesel k=9:
b==(1 mod 2) has a factor of 2
b==(4 mod 5): odd n has a factor of 5; even n has algebraic factors
b=m^2 proven composite by full algebraic factors

Riesel k=10:
b==(1 mod 3) has a factor of 3
b==(32 mod 33) has a covering set of [3, 11]

Sierp k=8:
b==(1 mod 3) has a factor of 3
b==(20 mod 21) has a covering set of [3, 7]
*b==(47 or 83 mod 195) has a covering set of [3, 5, 13]
(This includes bases 47, 83, 242, 278, 437, 473, 632, 668, 827, 863, 1022.)
*base 467 has a covering set of [3, 5, 7, 19, 37]
*base 722 has a covering set of [3, 5, 13, 73, 109]
b=m^3 proven composite by full algebraic factors
base 128 is a GFN with no possible prime

Sierp k=9:
b==(1 mod 2) has a factor of 2
b==(1 mod 5) has a factor of 5

Sierp k=10:
b==(1 mod 11) has a factor of 11
b==(32 mod 33) has a covering set of [3, 11]
base 1000 is a GFN with no known prime
[/code]*Covering set not previously found.

Attached are the following:
1. Primes for all n<=5000.
2. Primes for all n>5000.
3. Remaining bases and their search limits for each k.

Sometime in the next month I will search all bases for k=11 and k=12 to n=5K. I also hope to eventually search all k=8 thru 12 to n=25K like I did for k=2 thru 7 many years ago. Doing so requires a lot of personal effort because nearly all k/base combos must be sieved separately.

sweety439 2016-12-05 12:51

Thanks very much!

sweety439 2016-12-05 14:03

The files are not right: 10*611^n-1 (R611, k=10) is already tested to n=200K, not just 5K.

Besides, why the "number of remaining k's" column of 7*1004^n+1 lists 2k, but that of 10*1004^n+1 lists 1k? S1004 has conjecture only k=4, and with only k=2 (for k<4) remaining, but k=7 and k=10 are also remaining with k's > conjecture k.

sweety439 2016-12-05 16:57

Besides, do you miss any primes (that were found recently)? At beginning, you missed two primes: 4*737^269302+1 and 4*72^1119849-1. I don't know whether you miss other primes.

gd_barnes 2016-12-05 18:56

[QUOTE=sweety439;448446]The files are not right: 10*611^n-1 (R611, k=10) is already tested to n=200K, not just 5K.

Besides, why the "number of remaining k's" column of 7*1004^n+1 lists 2k, but that of 10*1004^n+1 lists 1k? S1004 has conjecture only k=4, and with only k=2 (for k<4) remaining, but k=7 and k=10 are also remaining with k's > conjecture k.[/QUOTE]

[QUOTE=sweety439;448467]Besides, do you miss any primes (that were found recently)? At beginning, you missed two primes: 4*737^269302+1 and 4*72^1119849-1. I don't know whether you miss other primes.[/QUOTE]

You are correct. R611 should show n=200K not n=5K. I missed the testing on the CRUS page. 7*1004^n+1 and 10*1004^n+1 should show 3k remaining not 2k or 1k...because k=2, 7, and 10 remain for that same base. I failed to cross reference them since I did k=7 years ago and k=10 recently.

There should be no primes missing and the bases remaining lists should be correct because I have been updating them with primes found by the CRUS project. As you found the only problems were with search depth and k's remaining. The latter of those is mostly irrelevant for this particular effort but I show them as a curiosity.

I will correct the lists in my records so that they are correct the next time that they are posted.

Thanks for checking everything.

sweety439 2016-12-06 12:10

[QUOTE=henryzz;448181]I am pretty certain that the argument that proves that there isn't a Sierpinski base if k is a Mersenne number can be twisted for Riesel ks. Similarly there isn't a Riesel base if k is a number of the form 2^n+1.

For k=7 a Riesel b is 33559116638. I do not claim that this is minimal.

Combine the following statements to form a covering set and input into your favourite chinese remainder theorem calculator.
[CODE]
If b = 2 mod 3 then p | 7*b^n-1 if n = 0 mod 2
If b = 2 mod 5 then p | 7*b^n-1 if n = 3 mod 4
If b = 3 mod 5 then p | 7*b^n-1 if n = 1 mod 4
If b = 3 mod 17 then p | 7*b^n-1 if n = 5 mod 16
If b = 5 mod 17 then p | 7*b^n-1 if n = 1 mod 16
If b = 6 mod 17 then p | 7*b^n-1 if n = 11 mod 16
If b = 7 mod 17 then p | 7*b^n-1 if n = 15 mod 16
If b = 10 mod 17 then p | 7*b^n-1 if n = 7 mod 16
If b = 11 mod 17 then p | 7*b^n-1 if n = 3 mod 16
If b = 12 mod 17 then p | 7*b^n-1 if n = 9 mod 16
If b = 14 mod 17 then p | 7*b^n-1 if n = 13 mod 16
If b = 7 mod 1201 then p | 7*b^n-1 if n = 7 mod 8
If b = 343 mod 1201 then p | 7*b^n-1 if n = 5 mod 8
If b = 858 mod 1201 then p | 7*b^n-1 if n = 1 mod 8
If b = 1194 mod 1201 then p | 7*b^n-1 if n = 3 mod 8
If b = 7 mod 169553 then p | 7*b^n-1 if n = 15 mod 16
If b = 343 mod 169553 then p | 7*b^n-1 if n = 5 mod 16
If b = 16807 mod 169553 then p | 7*b^n-1 if n = 3 mod 16
If b = 24222 mod 169553 then p | 7*b^n-1 if n = 1 mod 16
If b = 145331 mod 169553 then p | 7*b^n-1 if n = 9 mod 16
If b = 152746 mod 169553 then p | 7*b^n-1 if n = 11 mod 16
If b = 169210 mod 169553 then p | 7*b^n-1 if n = 13 mod 16
If b = 169546 mod 169553 then p | 7*b^n-1 if n = 7 mod 16[/CODE]Check all covering combinations to find smallest b.[/QUOTE]

For Riesel k=7, there are 8 situations:

(1) b%3=2, b%5=3, b%1201=1194, b%17=10, b%169553=7
(2) b%3=2, b%5=3, b%1201=1194, b%17=7, b%169553=169546
(3) b%3=2, b%5=3, b%1201=7, b%17=11, b%169553=152746
(4) b%3=2, b%5=3, b%1201=7, b%17=6, b%169553=16807
(5) b%3=2, b%5=2, b%1201=858, b%17=3, b%169553=169210
(6) b%3=2, b%5=2, b%1201=858, b%17=14, b%169553=343
(7) b%3=2, b%5=2, b%1201=343, b%17=5, b%169553=145331
(8) b%3=2, b%5=2, b%1201=343, b%17=12, b%169553=24222

Which situation can give the smallest b, by the chinese remainder theorem? (note that the b should not congruent to 1 mod 2 or 1 mod 3, or k=7 become a trivial k) You said that you do not claim that this is minimal.

sweety439 2016-12-06 12:22

Like my research for Riesel k=15, there are 4 situations:

(1)b%7=6, b%113=98, b%17=2, b%1489=15

The smallest such b is 12196414.

(2)b%7=6, b%113=98, b%17=15, b%1489=200

The smallest such b is 11047091.

(3)b%7=6, b%113=15, b%17=8, b%1489=397

The smallest such b is 1099279.

(4)b%7=6, b%113=15, b%17=9, b%1489=1092

The smallest such b is 8241218.



The smallest number of which is 1099279. However, gcd(15-1, 1099279-1) is not 1. Thus, 15 is a trivial k of R1099279 and not a Riesel number to base 1099279. Besides, the second such k of (3) is 1099279+7*113*17*1489=21121862, which is much large. Thus, the (conjectured) smallest b such that 15 is a Riesel number to base b is 8241218 (I have checked that gcd(15-1, 8241218-1)=1)

MisterBitcoin 2016-12-06 16:01

[QUOTE=sweety439;448564]For Riesel k=7, there are 8 situations:

(1) b%3=2, b%5=3, b%1201=1194, b%17=10, b%169553=7
(2) b%3=2, b%5=3, b%1201=1194, b%17=7, b%169553=169546
(3) b%3=2, b%5=3, b%1201=7, b%17=11, b%169553=152746
(4) b%3=2, b%5=3, b%1201=7, b%17=6, b%169553=16807
(5) b%3=2, b%5=2, b%1201=858, b%17=3, b%169553=169210
(6) b%3=2, b%5=2, b%1201=858, b%17=14, b%169553=343
(7) b%3=2, b%5=2, b%1201=343, b%17=5, b%169553=145331
(8) b%3=2, b%5=2, b%1201=343, b%17=12, b%169553=24222

Which situation can give the smallest b, by the chinese remainder theorem? (note that the b should not congruent to 1 mod 2 or 1 mod 3, or k=7 become a trivial k) You said that you do not claim that this is minimal.[/QUOTE]


I´m not that sort of math pro, but this is somethink like NumberFields@Home is doing. (It sound like that...)

I can give you contact to the project admin, Eric. He knows a lot about the chinese reminder theorem an a lot of other math stuff. NF@H was created to prove his Doctor-work
Just let me know if you want his private Mail.
There is the project description: [URL]https://numberfields.asu.edu/NumberFields/ProjectDescription.html[/URL]
And his dissertation: [URL]https://numberfields.asu.edu/NumberFields/Dissertation.pdf[/URL] .

henryzz 2016-12-06 21:16

I am not aware of any method that is better than running the CRT on all of these combinations.
However, it might not be necessary. It may be possible to find the smallest b with a given covering set easily once you know a possible k. This can be done using similar methods to what you did in post #49 for k=15.
This ignores the fact that there could be a different covering set with a larger period that produces a smaller b.

gd_barnes 2017-01-07 20:44

I have now searched k=11 and 12 for all bases <= 1030. Therefore all k=2 thru 12 for all bases <= 1030 have been completed. All k=2 thru 7 have been searched to n=25K for all bases and k=8 thru k=12 have been searched to n=5K for all bases.

Attached are all primes for n<=5K found by my effort, n>5K found by CRUS, and bases remaining for each k. There have been some updates for k=2 thru 10 so all of k=2 thru 12 are included.

Below are all exclusions including bases with trivial factors, algebraic factors, and covering sets for k=11 and 12. Exclusions for k<=10 were previously posted.
[code]
Riesel k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 5) has a factor of 5
b==(14 mod 15) has a covering set of [3, 5]

Riesel k=12:
b==(1 mod 11) has a factor of 11
b==(142 mod 143) has a covering set of [11, 13]
base 307 has a covering set of [5, 11, 29]
base 901 has a covering set of [7, 11, 13, 19]

Sierp k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 3) has a factor of 3
b==(14 mod 15) has a covering set of [3, 5]

Sierp k=12:
b==(1 mod 13) has a factor of 13
b==(142 mod 143) has a covering set of [11, 13]
bases 562, 828, and 900 have a covering set of [7, 13, 19]
base 563 has a covering set of [5, 7, 13, 19, 29]
base 597 has a covering set of [5, 13, 29]
bases 296 and 901 have a covering set of [7, 11, 13, 19]
base 12 is a GFN with no known prime
[/code]I am done with this effort. As the k's get higher, the exclusions get much more complex. Many of the bases for k>=8 are only searched to n=5K. That would be a good starting point for people to do some additional searching if they are interested in this effort.

[B]Admin edit: Moved all attachments to 1st post of this thread.[/B]

LaurV 2017-01-17 02:15

Hey Gary, if you are still interested in this, I got a bit hooked with 2*b^n-1 for small bases (b<=2048), and found (in your area of interest, b<=1030) that 2*303^40174-1 is prime.

gd_barnes 2017-01-20 10:14

[QUOTE=LaurV;451078]Hey Gary, if you are still interested in this, I got a bit hooked with 2*b^n-1 for small bases (b<=2048), and found (in your area of interest, b<=1030) that 2*303^40174-1 is prime.[/QUOTE]

Great thanks! I keep my files updated with all primes found for these from everyone. I will note that one.

sweety439 2017-01-20 16:13

Some additional exclusions for k<=12 and b<=2048:

Riesel k=6: all bases b=6*m^2 and m = 2 or 3 mod 5 can be proven composite by partial algebra factors. (this includes bases 24, 54, 294, 384, 864, 1014, 1734, 1944)

Riesel k=11: all bases b=11*m^2 and m = 2 or 3 mod 5 can be proven composite by partial algebra factors. (this includes bases 44, 99, 539, 704, 1584, 1859)

gd_barnes 2017-01-20 21:40

[QUOTE=sweety439;451278]Some additional exclusions for k<=12 and b<=2048:

Riesel k=6: all bases b=6*m^2 and m = 2 or 3 mod 5 can be proven composite by partial algebra factors. (this includes bases 24, 54, 294, 384, 864, 1014, 1734, 1944)

Riesel k=11: all bases b=11*m^2 and m = 2 or 3 mod 5 can be proven composite by partial algebra factors. (this includes bases 44, 99, 539, 704, 1584, 1859)[/QUOTE]

These are not additional exclusions. This exclusion was already posted for Riesel k=6. For Riesel k=11, those bases all have a numeric covering set of [3, 5], which has already been posted.

sweety439 2017-02-01 13:34

Found an additional exclusion:

Riesel k=12: base 307 has a covering set of [5, 11, 29]

Thus, there are only 6 bases remain for Riesel k=12: 263, 593, 615, 717, 912, 978.

sweety439 2017-02-01 15:43

Also, for the "number of remaining k's" column of the text files for the remain bases, why 7*1004^n+1 and 10*1004^n+1 lists 3k, but 2*1004^n+1 lists 1k? As you say, S1004 should list 3k since k=2, 7 and 10 remain for that same base, like the example for S593, all 4*593^n+1, 8*593^n+1 and 12*593^n+1 list 3k since k=4, 8 and 12 remain for that same base, and the example for S824, both 5*824^n+1 and 8*824^n+1 lists 2k since k=5 and 8 remain for that same base, but why for S230, 12*230^n+1 lists 2k but 4*230^n+1 lists 1k? S230 should list 2k since k=4 and 12 remain for that same base.

gd_barnes 2017-02-01 22:24

My files have been fixed on my machine. Whenever I post them again, you will see the corrections.

LaurV 2017-02-09 01:39

[QUOTE=gd_barnes;451259]Great thanks! I keep my files updated with all primes found for these from everyone. I will note that one.[/QUOTE]
One more, turned out by LLR this morning:

2*522^62288-1 is prime! (169279 decimal digits, P = 29) Time : 768.095 sec.

And out of your interest range, but yet ok for our sweety-tweety (see the thread I posted yesterday about cllr bug),

2*1487^36432-1 is prime! (115574 decimal digits, P = 9) Time : 420.446 sec.

sweety439 2017-02-09 15:49

There is a research for k=2 and some Sierpinski/Riesel bases (including some bases b>1030): [URL]http://mersenneforum.org/showthread.php?t=6918[/URL]

sweety439 2017-03-10 19:10

[QUOTE=gd_barnes;450652]I have now searched k=11 and 12 for all bases <= 1030. Therefore all k=2 thru 12 for all bases <= 1030 have been completed. All k=2 thru 7 have been searched to n=25K for all bases and k=8 thru k=12 have been searched to n=5K for all bases.

Attached are all primes for n<=5K found by my effort, n>5K found by CRUS, and bases remaining for each k. There have been some updates for k=2 thru 10 so all of k=2 thru 12 are included.

Below are all exclusions including bases with trivial factors, algebraic factors, and covering sets for k=11 and 12. Exclusions for k<=10 were previously posted.
[code]
Riesel k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 5) has a factor of 5
b==(14 mod 15) has a covering set of [3, 5]

Riesel k=12:
b==(1 mod 11) has a factor of 11
b==(142 mod 143) has a covering set of [11, 13]
base 307 has a covering set of [5, 11, 29]
base 901 has a covering set of [7, 11, 13, 19]

Sierp k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 3) has a factor of 3
b==(14 mod 15) has a covering set of [3, 5]

Sierp k=12:
b==(1 mod 13) has a factor of 13
b==(142 mod 143) has a covering set of [11, 13]
bases 562, 828, and 900 have a covering set of [7, 13, 19]
base 563 has a covering set of [5, 7, 13, 19, 29]
base 597 has a covering set of [5, 13, 29]
bases 296 and 901 have a covering set of [7, 11, 13, 19]
base 12 is a GFN with no known prime
[/code]I am done with this effort. As the k's get higher, the exclusions get much more complex. Many of the bases for k>=8 are only searched to n=5K. That would be a good starting point for people to do some additional searching if they are interested in this effort.[/QUOTE]

@Gary, your text file for n > 5K for Riesel k=10 is not right, 10*992^5443-1 is not prime, it is divisible by 7.

MisterBitcoin 2017-03-10 19:26

[QUOTE=sweety439;454637]@Gary, your text file for n > 5K for Riesel k=10 is not right, 10*992^5443-1 is not prime, it is divisible by 7.[/QUOTE]

[CODE]Primality testing 10*992^5443-1 [N-1, Brillhart-Lehmer-Selfridge]
Running N-1 test using base 3
Factored: 13
composite
10*992^5443-1 is composite (5.8175s+0.0003s)[/CODE]

gd_barnes 2017-03-10 21:10

OK thanks. The prime is 10*992^5433-1. There was a typo in my file. I have corrected it on my machine.

gd_barnes 2017-03-23 05:31

I have updated the file links in post 62 [URL="http://www.mersenneforum.org/showpost.php?p=450652&postcount=62"]here[/URL] with the most recent corrections and updates.

sweety439 2017-08-16 17:58

1 Attachment(s)
I am now reserving 2*801^n+1, 7*1004^n+1, 10*449^n+1 and 12*312^n+1 and found that 10*449^18506+1 is prime. (2*801^n+1 is currently at n=26600, 7*1004^n+1 is currently at n=28374, and 12*312^n+1 is currently at n=12394, all no prime found)

This is the result text file for 10*449^n+1.

gd_barnes 2017-08-16 22:27

I have updated the files in post #62.

sweety439 2017-08-21 01:58

1 Attachment(s)
12*312^21162+1 is prime.
Result text file attached.

sweety439 2017-08-21 02:04

2*801^n+1 is currently at n=28661, and 7*1004^n+1 is currently at n=31030, both no primes found.

Many of the bases for 8<=k<=12 are only searched to n=5K, I will reserve all such bases to n=25K.

sweety439 2017-10-30 14:21

1 Attachment(s)
Update current file.

sweety439 2017-10-30 19:32

The exclusions are:

[CODE]
Riesel k=2:
none

Riesel k=3:
b==(1 mod 2); factor of 2

Riesel k=4:
b==(1 mod 3); factor of 3
b==(4 mod 5): odd n, factor of 5; even n, algebraic factors
b=m^2 proven composite by full algebraic factors

Riesel k=5:
b==(1 mod 2); factor of 2

Riesel k=6:
b==(1 mod 5); factor of 5
b==(34 mod 35); covering set [5, 7]
b=6*m^2 with m==(2, 3 mod 5): even n, factor of 5; odd n, algebraic factors
(This includes bases 24, 54, 294, 384, 864, 1014.)

Riesel k=7:
b==(1 mod 2); factor of 2
b==(1 mod 3); factor of 3

Riesel k=8:
b==(1 mod 7) has a factor of 7
b==(20 mod 21) has a covering set of [3, 7]
b==(83, 307 mod 455) has a covering set of [5, 7, 13]
(This includes bases 83, 307, 538, 762, 993.)
b=m^3 proven composite by full algebraic factors

Riesel k=9:
b==(1 mod 2) has a factor of 2
b==(4 mod 5): odd n has a factor of 5; even n has algebraic factors
b=m^2 proven composite by full algebraic factors

Riesel k=10:
b==(1 mod 3) has a factor of 3
b==(32 mod 33) has a covering set of [3, 11]

Riesel k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 5) has a factor of 5
b==(14 mod 15) has a covering set of [3, 5]

Riesel k=12:
b==(1 mod 11) has a factor of 11
b==(142 mod 143) has a covering set of [11, 13]
base 307 has a covering set of [5, 11, 29]
base 901 has a covering set of [7, 11, 13, 19]

Sierp k=2:
b==(1 mod 3); factor of 3
base 512 is a GFN with no known prime

Sierp k=3:
b==(1 mod 2); factor of 2

Sierp k=4:
b==(1 mod 5); factor of 5
b==(14 mod 15); covering set [3, 5]
base 625 proven composite by full algebraic factors
bases 32, 512, and 1024 are GFN's with no known prime

Sierp k=5:
b==(1 mod 2); factor of 2
b==(1 mod 3); factor of 3

Sierp k=6:
b==(1 mod 7); factor of 7
b==(34 mod 35); covering set [5, 7]

Sierp k=7:
b==(1 mod 2); factor of 2

Sierp k=8:
b==(1 mod 3) has a factor of 3
b==(20 mod 21) has a covering set of [3, 7]
b==(47 or 83 mod 195) has a covering set of [3, 5, 13]
(This includes bases 47, 83, 242, 278, 437, 473, 632, 668, 827, 863, 1022.)
base 467 has a covering set of [3, 5, 7, 19, 37]
base 722 has a covering set of [3, 5, 13, 73, 109]
b=m^3 proven composite by full algebraic factors
base 128 is a GFN with no possible prime

Sierp k=9:
b==(1 mod 2) has a factor of 2
b==(1 mod 5) has a factor of 5

Sierp k=10:
b==(1 mod 11) has a factor of 11
b==(32 mod 33) has a covering set of [3, 11]
base 1000 is a GFN with no known prime

Sierp k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 3) has a factor of 3
b==(14 mod 15) has a covering set of [3, 5]

Sierp k=12:
b==(1 mod 13) has a factor of 13
b==(142 mod 143) has a covering set of [11, 13]
bases 296 and 901 have a covering set of [7, 11, 13, 19]
bases 562, 828, and 900 have a covering set of [7, 13, 19]
base 563 has a covering set of [5, 7, 13, 19, 29]
base 597 has a covering set of [5, 13, 29]
base 12 is a GFN with no known prime
[/CODE]

gd_barnes 2017-10-31 01:45

I have updated the files in post 62.

sweety439 2018-06-05 21:32

Now I reserve all bases for k = 8 to 12 that are at n<25K to n=25K and found that 8*284^5266-1 and 10*1020^6944-1 are primes. (using pfgw)

Current at:

Riesel k=8: n=6144
Riesel k=9: n=7445
Riesel k=10: n=7025
Riesel k=11: n=9679
Riesel k=12: n=8690
Sierp k=8: n=6135
Sierp k=9: n=9541
Sierp k=10: n=5828
Sierp k=11: n=9568
Sierp k=12: n=5631

Only found the above two primes.

sweety439 2018-06-06 17:41

[QUOTE=sweety439;489251]Now I reserve all bases for k = 8 to 12 that are at n<25K to n=25K and found that 8*284^5266-1 and 10*1020^6944-1 are primes. (using pfgw)

Current at:

Riesel k=8: n=6144
Riesel k=9: n=7445
Riesel k=10: n=7025
Riesel k=11: n=9679
Riesel k=12: n=8690
Sierp k=8: n=6135
Sierp k=9: n=9541
Sierp k=10: n=5828
Sierp k=11: n=9568
Sierp k=12: n=5631

Only found the above two primes.[/QUOTE]

12*826^5786+1 is prime.

sweety439 2018-06-07 03:02

8*854^6500-1 is prime.

sweety439 2018-06-07 03:06

Currently at:

Riesel k=8: n=6632
Riesel k=9: n=8346
Riesel k=10: n=7750
Riesel k=11: n=11009
Riesel k=12: n=9840
Sierp k=8: n=6599
Sierp k=9: n=10619
Sierp k=10: n=6196
Sierp k=11: n=11120
Sierp k=12: n=5915

kar_bon 2018-06-08 07:16

8*194^38360-1 is prime

kar_bon 2018-06-09 09:23

4*312^51565-1 is prime!

kar_bon 2018-06-10 07:17

4*513^38031-1 is prime

sweety439 2018-06-20 18:48

8*728^7399+1 is prime.

sweety439 2019-04-11 16:52

[QUOTE=gd_barnes;450652]I have now searched k=11 and 12 for all bases <= 1030. Therefore all k=2 thru 12 for all bases <= 1030 have been completed. All k=2 thru 7 have been searched to n=25K for all bases and k=8 thru k=12 have been searched to n=5K for all bases.

Attached are all primes for n<=5K found by my effort, n>5K found by CRUS, and bases remaining for each k. There have been some updates for k=2 thru 10 so all of k=2 thru 12 are included.

Below are all exclusions including bases with trivial factors, algebraic factors, and covering sets for k=11 and 12. Exclusions for k<=10 were previously posted.
[code]
Riesel k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 5) has a factor of 5
b==(14 mod 15) has a covering set of [3, 5]

Riesel k=12:
b==(1 mod 11) has a factor of 11
b==(142 mod 143) has a covering set of [11, 13]
base 307 has a covering set of [5, 11, 29]
base 901 has a covering set of [7, 11, 13, 19]

Sierp k=11:
b==(1 mod 2) has a factor of 2
b==(1 mod 3) has a factor of 3
b==(14 mod 15) has a covering set of [3, 5]

Sierp k=12:
b==(1 mod 13) has a factor of 13
b==(142 mod 143) has a covering set of [11, 13]
bases 562, 828, and 900 have a covering set of [7, 13, 19]
base 563 has a covering set of [5, 7, 13, 19, 29]
base 597 has a covering set of [5, 13, 29]
bases 296 and 901 have a covering set of [7, 11, 13, 19]
base 12 is a GFN with no known prime
[/code]I am done with this effort. As the k's get higher, the exclusions get much more complex. Many of the bases for k>=8 are only searched to n=5K. That would be a good starting point for people to do some additional searching if they are interested in this effort.[/QUOTE]

Are there any update of these files? e.g. recently the prime 8*410^279991+1 (for Sierpinski k=8) was found.

gd_barnes 2019-04-12 22:41

The files in post #62 have been updated.

gd_barnes 2019-06-07 02:49

I searched all remaining k<=12 and b<=1030 up to n=25K. There were 45 k/base combos for k=8 thru 12 that needed to be searched for n=5K-25K. I found the following 11 primes:

8*997^15814-1
9*990^23031-1
10*599^11775-1
12*593^16063-1
10*537^7117+1
10*827^9894+1
10*929^13064+1
10*1004^10644+1
12*600^11241+1
12*607^7582+1
12*673^7789+1

The files in post #62 have been updated accordingly. :smile:

sweety439 2019-06-07 21:24

[QUOTE=gd_barnes;518762]I searched all remaining k<=12 and b<=1030 up to n=25K. There were 45 k/base combos for k=8 thru 12 that needed to be searched for n=5K-25K. I found the following 11 primes:

8*997^15814-1
9*990^23031-1
10*599^11775-1
12*593^16063-1
10*537^7117+1
10*827^9894+1
10*929^13064+1
10*1004^10644+1
12*600^11241+1
12*607^7582+1
12*673^7789+1

The files in post #62 have been updated accordingly. :smile:[/QUOTE]

So you can add the prime 8*997^15814-1 in the CRUS page, currently R997 is only tested to n=10K.

kar_bon 2019-06-08 12:50

I've included two pages in the Prime-Wiki for these values:

- [url='https://www.rieselprime.de/ziki/Riesel_prime_small_bases_least_n']Riesel type[/url]
- [url='https://www.rieselprime.de/ziki/Proth_prime_small_bases_least_n']Proth type[/url]

I took the data from post #62, compiled as CSV (link for download given) for all 2 ≤ k ≤ 12 and displayed all wanted values. The table columns are sortable.

Dylan14 2019-06-08 18:18

I am presently working on trying to find a prime for 7*1004^n+1, currently past 50k, will take to n = 100k.

Dylan14 2019-06-08 20:33

And we have a prime:


[CODE]7*1004^54848+1 is 3-PRP! (97.6687s+0.0034s)


C:\Users\Dylan\Desktop\prime finding\prime testing\pfgw>pfgw64 -t -q"7*1004^54848+1"
PFGW Version 3.8.3.64BIT.20161203.Win_Dev [GWNUM 28.6]

Primality testing 7*1004^54848+1 [N-1, Brillhart-Lehmer-Selfridge]
Running N-1 test using base 3
7*1004^54848+1 is prime! (93.6171s+0.0033s)[/CODE] With this, all bases within CRUS limits (b <= 1030) have a prime for k = 7 on the Sierpinski side. (*)


(*) if they are not excluded by covering sets.


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