[URL="https://docs.google.com/document/d/e/2PACX-1vQb0b1YJBQOu00eot4UN6MncbFtAeLSfgtnK6EasdukCo4gHDMsw47ANKz-jhpArGFnFEXnvzcXZw9-/pub"]Riesel problems[/URL]

Update newest status for R36

Like [URL="https://en.wikipedia.org/wiki/Bunyakovsky_conjecture"]Bunyakovsky conjecture[/URL], it is conjectured that for all integer triples (k, b, c) satisfying these conditions:

1. k>=1, b>=2, c != 0

2. gcd(k, c) = 1, gcd(b, c) = 1

3. there is no finite set {p_1, p_2, p_3, ..., p_u} (all p_i (1<=i<=u) are primes) and finite set {r_1, r_2, r_3, ..., r_s} (all r_i (1<=i<=s) are integers > 1) such that for every integer n>=1:

either

(k*b^n+c)/gcd(k+c, b-1) is divisible by at least one p_i (1<=i<=u)

or

k*b^n and -c are both r_i-th powers for at least one r_i (1<=i<=s)

or

one of k*b^n and c is a 4th power, another is of the form 4*t^4 with integer t

4. the triple (k, b, c) is [B][I]not[/I][/B] in this case: c = 1, b = q^m, k = q^r, where q is an integer not of the form t^s with odd s > 1, and m and r are integers having no common odd prime factor, and the exponent of highest power of 2 dividing r >= the exponent of highest power of 2 dividing m, and the equation 2^x = r (mod m) has no solution

Then there are infinitely many integers n>=1 such that (k*b^n+c)/gcd(k+c, b-1) is prime.

Searched SR66 up to k=10000

Searched SR120 up to k=10000

[QUOTE=sweety439;547407]Like [URL="https://en.wikipedia.org/wiki/Bunyakovsky_conjecture"]Bunyakovsky conjecture[/URL], it is conjectured that for all integer triples (k, b, c) satisfying these conditions:
...
Then there are infinitely many integers n>=1 such that (k*b^n+c)/gcd(k+c, b-1) is prime.[/QUOTE]

[QUOTE]
from the Wikipedia page:

The Bunyakovsky conjecture (or Bouniakowsky conjecture) gives a criterion for a polynomial {\displaystyle f(x)}f(x) in one variable with integer coefficients to give infinitely many prime values...
[/QUOTE]

(k*b^n+c)/gcd(k+c, b-1) is not a polynomial sequence so it isn't at all related to the Bunyakovsky conjecture. Steps 1 and 2 are trivial enough, but steps 3 and 4 are what make the difference. Step 4 isn't even part of the Bunyakovsky conjecture since that would imply a polynomial f(x) is irreducible over the integers. However, it can be proven for exponential sequences.

For polynomial sequences, it's easy to prove Step 3: f(0) = C is the constant of a polynomial, so the infinitude of primes is implied by finding an integer x with gcd(x,C)=1 and gcd(f(x),C)=1.

As for exponential type sequences, we can only assume that there "appear" to be infinitely many primes, and we can't prove if there exists a "covering set" or not. For example, we can't prove there doesn't exist a covering set for the sequence "3*2^n+-1", although it is extremely unlikely it exists.

An exception, however, is divisibility sequences. For example, 2^n-1 does not have a covering set --- and we can prove this by showing that gcd(2^n-1,f)=1 for any prime f<n if n is prime --- and there are infinitely many primes, so no finite set is possible.

Back to your original problem, if you "conjecture" there are infinitely many primes of the form (k*b^n+c)/gcd(k+c, b-1), you are really conjecturing that step 3 is true, alongside from conjecturing that if all 4 steps are true, there are infinitely many primes of that form.

Search SR126 up to k=30000, only search the k not in CRUS (i.e. gcd(k+-1,b-1) is not 1)

[QUOTE=carpetpool;547416](k*b^n+c)/gcd(k+c, b-1) is not a polynomial sequence so it isn't at all related to the Bunyakovsky conjecture. Steps 1 and 2 are trivial enough, but steps 3 and 4 are what make the difference. Step 4 isn't even part of the Bunyakovsky conjecture since that would imply a polynomial f(x) is irreducible over the integers. However, it can be proven for exponential sequences.

For polynomial sequences, it's easy to prove Step 3: f(0) = C is the constant of a polynomial, so the infinitude of primes is implied by finding an integer x with gcd(x,C)=1 and gcd(f(x),C)=1.

As for exponential type sequences, we can only assume that there "appear" to be infinitely many primes, and we can't prove if there exists a "covering set" or not. For example, we can't prove there doesn't exist a covering set for the sequence "3*2^n+-1", although it is extremely unlikely it exists.

An exception, however, is divisibility sequences. For example, 2^n-1 does not have a covering set --- and we can prove this by showing that gcd(2^n-1,f)=1 for any prime f<n if n is prime --- and there are infinitely many primes, so no finite set is possible.

Back to your original problem, if you "conjecture" there are infinitely many primes of the form (k*b^n+c)/gcd(k+c, b-1), you are really conjecturing that step 3 is true, alongside from conjecturing that if all 4 steps are true, there are infinitely many primes of that form.[/QUOTE]

I know that (k*b^n+c)/gcd(k+c, b-1) is not a polynomial sequence, I just make a conjecture like Bunyakovsky conjecture when the n is in exponent, and I know that we can't prove there doesn't exist a covering set for the sequence "k*2^n+-1" for most k, but I also conjectured....

* k*2^n+1 has no covering set for all k<78557
* k*3^n+1 has no covering set for all k<125050976086 and gcd(k+1,3-1)=1
* (k*3^n+1)/gcd(k+1,3-1) has no covering set for all k<11047
* k*4^n+1 has no covering set for all k<66741 and gcd(k+1,4-1)=1
* (k*4^n+1)/gcd(k+1,4-1) has no covering set for all k<419
* k*5^n+1 has no covering set for all k<159986 and gcd(k+1,5-1)=1
* (k*5^n+1)/gcd(k+1,5-1) has no covering set for all k<7
* k*6^n+1 has no covering set for all k<174308 and gcd(k+1,6-1)=1
* (k*6^n+1)/gcd(k+1,6-1) has no covering set for all k<174308
* k*7^n+1 has no covering set for all k<1112646039348 and gcd(k+1,7-1)=1
* (k*7^n+1)/gcd(k+1,7-1) has no covering set for all k<209
* k*8^n+1 has no covering set for all k<47 and gcd(k+1,8-1)=1
* (k*8^n+1)/gcd(k+1,8-1) has no covering set for all k<47
* k*9^n+1 has no covering set for all k<2344 and gcd(k+1,9-1)=1
* (k*9^n+1)/gcd(k+1,9-1) has no covering set for all k<31
* k*10^n+1 has no covering set for all k<9175 and gcd(k+1,10-1)=1
* (k*10^n+1)/gcd(k+1,10-1) has no covering set for all k<989
* k*11^n+1 has no covering set for all k<1490 and gcd(k+1,11-1)=1
* (k*11^n+1)/gcd(k+1,11-1) has no covering set for all k<5
* k*12^n+1 has no covering set for all k<521 and gcd(k+1,12-1)=1
* (k*12^n+1)/gcd(k+1,12-1) has no covering set for all k<521
etc.

Conjecture 1 (the strong Sierpinski conjecture): For b>=2, k>=1, if there [B][I]is an n[/I][/B] such that:

(1) k*b^n is neither a perfect odd power (i.e. k*b^n is not of the form m^r with odd r>1) nor of the form 4*m^4.
(2) gcd((k*b^n+1)/gcd(k+1,b-1),(b^(9*2^s)-1)/(b-1)) = 1 for all s, i.e. for all s, every prime factor of (k*b^n+1)/gcd(k+1,b-1) does not divide (b^(9*2^s)-1)/(b-1). (i.e. ord_p(b) is not of the form 2^r (r>=0 if p = 2 or p = 3, r>=1 if p>=5), 3*2^r (r>=0) or 9*2^r (r>=0) for every prime factor p of (k*b^n+1)/gcd(k+1,b-1)).
(3) this k is not excluded from this Sierpinski base b by the post [URL="http://mersenneforum.org/showpost.php?p=459405&postcount=265"]#265[/URL]. (the first 6 Sierpinski bases with k's excluded by the post [URL="http://mersenneforum.org/showpost.php?p=459405&postcount=265"]#265[/URL] are 128, 2187, 16384, 32768, 78125 and 131072)

Then there are infinitely many primes of the form (k*b^n+1)/gcd(k+1,b-1).

Conjecture 2 (the strong Riesel conjecture): For b>=2, k>=1, if there [B][I]is an n[/I][/B] such that:

(1) k*b^n is not a perfect power (i.e. k*b^n is not of the form m^r with r>1).
(2) gcd((k*b^n-1)/gcd(k-1,b-1),(b^(9*2^s)-1)/(b-1)) = 1 for all s, i.e. for all s, every prime factor of (k*b^n-1)/gcd(k-1,b-1) does not divide (b^(9*2^s)-1)/(b-1). (i.e. ord_p(b) is not of the form 2^r (r>=0 if p = 2 or p = 3, r>=1 if p>=5), 3*2^r (r>=0) or 9*2^r (r>=0) for every prime factor p of (k*b^n-1)/gcd(k-1,b-1)).

Then there are infinitely many primes of the form (k*b^n-1)/gcd(k-1,b-1).

[QUOTE=sweety439;547498]Conjecture 1 (the strong Sierpinski conjecture): For b>=2, k>=1, if there [B][I]is an n[/I][/B] such that:

(1) k*b^n is neither a perfect odd power (i.e. k*b^n is not of the form m^r with odd r>1) nor of the form 4*m^4.
(2) gcd((k*b^n+1)/gcd(k+1,b-1),(b^(9*2^s)-1)/(b-1)) = 1 for all s, i.e. for all s, every prime factor of (k*b^n+1)/gcd(k+1,b-1) does not divide (b^(9*2^s)-1)/(b-1). (i.e. ord_p(b) is not of the form 2^r (r>=0 if p = 2 or p = 3, r>=1 if p>=5), 3*2^r (r>=0) or 9*2^r (r>=0) for every prime factor p of (k*b^n+1)/gcd(k+1,b-1)).
(3) this k is not excluded from this Sierpinski base b by the post [URL="http://mersenneforum.org/showpost.php?p=459405&postcount=265"]#265[/URL]. (the first 6 Sierpinski bases with k's excluded by the post [URL="http://mersenneforum.org/showpost.php?p=459405&postcount=265"]#265[/URL] are 128, 2187, 16384, 32768, 78125 and 131072)

Then there are infinitely many primes of the form (k*b^n+1)/gcd(k+1,b-1).

Conjecture 2 (the strong Riesel conjecture): For b>=2, k>=1, if there [B][I]is an n[/I][/B] such that:

(1) k*b^n is not a perfect power (i.e. k*b^n is not of the form m^r with r>1).
(2) gcd((k*b^n-1)/gcd(k-1,b-1),(b^(9*2^s)-1)/(b-1)) = 1 for all s, i.e. for all s, every prime factor of (k*b^n-1)/gcd(k-1,b-1) does not divide (b^(9*2^s)-1)/(b-1). (i.e. ord_p(b) is not of the form 2^r (r>=0 if p = 2 or p = 3, r>=1 if p>=5), 3*2^r (r>=0) or 9*2^r (r>=0) for every prime factor p of (k*b^n-1)/gcd(k-1,b-1)).

Then there are infinitely many primes of the form (k*b^n-1)/gcd(k-1,b-1).[/QUOTE]

I conjectured that all periods of covering set divides 9*2^r for some r

SR124 done to n=10000

I know that for R124, some k's proven composite by partial algebra factors:

* All k where k = m^2 and m = = 2 or 3 mod 5
* All k where k = 31*m^2 and m = = 1 or 4 mod 5

but since there are too many such k, I didn't change the text file.

Now, all SR conjectures are completed to n>=1000 and (k<CK or k<=10000)

zip files attached.