A Sierpinski/Riesel-like problem

[sweety439]

k,n
1,3
2,2
3,16
4,1

With conjectured k=5, this conjecture is proven.

[sweety439]

k,n
1,7
2,16
3,3
4,5
5,30
6,2
7,1

With conjectured k=8, this conjecture is proven.

[sweety439]

k,n
1,7
2,4
3,1

With conjectured k=4, this conjecture is proven.

[sweety439]

k,n
1,3
2,2
3,2
4,1

With conjectured k=5, this conjecture is proven.

[sweety439]

k,n
1,13
2,1
5,1
6,1
7,6
8,22

With conjectured k=9, k=4 proven composite by partial algebraic factors, k=3 remains.

[sweety439]

k,n
1,3
2,2
3,1

With conjectured k=4, this conjecture is proven.

[sweety439]

k,n
1,3
2,8
3,6

With conjectured k=5, k=4 remains.

[sweety439]

k,n
2,1
3,1
5,2

With conjectured k=6, k=4 proven composite by partial algebraic factors, k=1 remains.

[sweety439]

k,n
1,19
2,2
3,16

With conjectured k=4, this conjecture is proven.

[sweety439]

Reserve SR1024. (SR256 and R512 are already proven, and S512 is already tested to at least n=1M for all remain k's)

[sweety439]

Originally Posted by sweety439

Reserve SR1024. (SR256 and R512 are already proven, and S512 is already tested to at least n=1M for all remain k's)

Found these (probable) primes:

(44*1024^1933+1)/3
(13*1024^1167-1)/3
(43*1024^2290-1)/3

These k's are still remain:

S1024: 4, 16, 29, 38, 56
R1024: 29, 31, 56, 61

Note that both sides have k=29 and k=56 remain, but the divisors (i.e. gcd(k+-1,b-1)) for two sides for these k's aren't the same.