A Sierpinski/Riesel-like problem

[Batalov]

Originally Posted by sweety439

For the original Sierpinski/Riesel problem, it is finding and proving ...
Now, I extend to the k's such that gcd(k+-1, b-1) is not 1.

The strong (extended) Sierpinski problem ...

Staying away from words that have an established meaning in the community is probably a good idea before you actually can talk the language of the community. How do you know that you are not actually meaning 'weak' when you are saying 'strong'?

Or rather, how do you know that things you are talking about are altogether related? Because they aren't!

If one proved Goldbach, the would have also proven weak Goldbach.

If one proved weak Goldbach (and they did!), then nothing happened to Goldbach.

If you (well, let's imagine) proved "The strong Sierpinski problem", then ... <care to fill the blanks?> ... Nothing would happen to the "normal" Sierpinski problem, of course!

Origin: https://xkcd.com/1310/

[sweety439]

Originally Posted by Batalov

Staying away from words that have an established meaning in the community is probably a good idea before you actually can talk the language of the community. How do you know that you are not actually meaning 'weak' when you are saying 'strong'?

Or rather, how do you know that things you are talking about are altogether related? Because they aren't!

If one proved Goldbach, the would have also proven weak Goldbach.

If one proved weak Goldbach (and they did!), then nothing happened to Goldbach.

If you (well, let's imagine) proved "The strong Sierpinski problem", then ... <care to fill the blanks?> ... Nothing would happen to the "normal" Sierpinski problem, of course!

Origin: https://xkcd.com/1310/

Since this is the extending of the original Sierpinski/Riesel problem.

If you found all (probable) primes of the form (k*b^n+-1)/gcd(k+-1, b-1) for all k, then of course, you also found all primes of the form k*b^n+-1 for all k such that gcd(k+-1, b-1) = 1, and the latter is the original Sierpinski/Riesel conjecture.

However, although the conjecture smallest k does need to be the same, if we also include the k's > conjectured smallest k, then this conjecture covers the original conjecture.

For example, for R10, the conjecture smallest Riesel k is 10176, but the conjecture smallest strong Riesel k is 334. Of course, this conjecture does not cover the original conjecture. However, if we also include the k's > conjectured smallest k (334), if we include all k's < 10176 which are not proven composite for all exponent n, then this conjecture covers the original R10 conjecture.

[sweety439]

This is only my extending of the Sierpinski/Riesel problem. Now, I know that it does not apply to CRUS since the correspond primes are often only probable primes, i.e. not proved primes.

Thanks. Now, I know this should be in "and now for something completely different".

[sweety439]

Originally Posted by sweety439

Can you find the smallest conjectured strong Sierpinski number in base 3 and 6?

base 3: conjectured smallest strong Sierpinski k=11047, cover set: {2, 5, 7, 13, 73}, period=12

For the number (11047*3^n+1)/2:

If n=0 (mod 2), then this number is divisible by 2.
If n=3 (mod 4), then this number is divisible by 5.
If n=1 (mod 12), then this number is divisible by 73.
If n=5 (mod 12), then this number is divisible by 13.
If n=9 (mod 12), then this number is divisible by 7.

[sweety439]

base 6: conjectured smallest strong Sierpinski k=174308, cover set: {7, 13, 31, 37, 97}, period=12.

This k is the same as the conjectured smallest original base 6 Sierpinski k. Thus, the strong Sierpinski conjecture base 6 covers the original Sierpinski conjecture in the same base.

Update the text file for all conjectured smallest strong Sierpinski/Riesel number base 2 to 12.

[sweety439]

Due to the CRUS, if we only consider the k's with cover set (i.e. not with full or partial algebra factors), then the conjectured k's for S8, R4, R9 and R12 should be larger, they are in this text file.

Of course, there are k's can be proven composite with full or partial algebra factors, e.g. square k's in R4 and R9, cube k's in S8, and k=25, 27, 64, 300, 324 in R12, these k's are excluded from the conjectures.

The conjectured k should be:

S8: 1 --> 47
R4: 9 --> 361
R9: 1 --> 41
R12: 25 --> 376

[sweety439]

Update the extend S8, R4, R9, R12 files. (with the CRUS definition, i.e. exclude the k's can be proven composite with full or partial algebra factors)

All of the four conjectures are proven.

[sweety439]

For base 2, the strong Sierpinski/Riesel problem is completely the same as the original Sierpinski/Riesel problem, since gcd(k+-1,2-1) = 1 for all k.

[sweety439]

I continued to search bases 13 to 24.

This is the text file for the conjectured smallest strong Sierpinski/Riesel k.

I will find the conjectured smallest strong Sierpinski/Riesel k for base 15 and 24. (The two bases are harder)

[sweety439]

I continued to search bases 25 to 64.

This is the text file for the conjectured smallest strong Sierpinski/Riesel k.

For these bases, I still found no k with a cover set. (but of course, there are infinitely many such k's)

S15, S24, S40, S42, S52, S60, S63.

R15, R24, R30, R36, R40, R42, R48, R52, R60.

All such bases have been tested to at least k=1000 (k=1000 is the test limit for all bases b>=25 except S28, R28 and S36)

Edit: For R28, I found k=3769, cover: {5, 29, 157} period=4.

I will find (probable) primes for these problems in next few weeks.

[sweety439]

Add the conjectured k's I found for S63, R28, R30 and R48.

Now, I still found no k with a cover set only for these bases <= 64.

S15, S24, S40, S42, S52, S60.

R15, R24, R36, R40, R42, R52, R60.