k=1 thru k=12

[sweety439]

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]

Originally Posted by sweety439

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.

Originally Posted by sweety439

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.

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]

Originally Posted by henryzz

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.

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

Check all covering combinations to find smallest b.

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]

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]

Originally Posted by sweety439

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.

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: https://numberfields.asu.edu/NumberF...scription.html
And his dissertation: https://numberfields.asu.edu/NumberF...ssertation.pdf .

[henryzz]

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]

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.

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

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.

Admin edit: Moved all attachments to 1st post of this thread.

[LaurV]

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]

Originally Posted by LaurV

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.

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

[sweety439]

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]

Originally Posted by sweety439

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)

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.