A Sierpinski/Riesel-like problem

[sweety439]

Originally Posted by sweety439

These are the conjectured smallest Sierpinski/Riesel numbers for bases 2<=b<=1024, searched up to k=10^6. (NA if this k > 10^6)

Note: only searched for exponent n<=1024 and for primes p<=30000.

Also searched the power-of-2 base, up to 65536 = 2^16.

[sweety439]

Originally Posted by sweety439

Update the file for n <= 21001.

Continue to reserve...

Top 10 k's with largest first primes: k (n) (only sorted by n) for S6 for the k's = 4 mod 5:

162259 (20759)
10879 (20532)
102824 (19096)
104259 (18250)
160364 (17460)
98979 (15567)
13929 (15439)
82139 (14234)
21244 (13335)
71849 (13216)

[sweety439]

Originally Posted by sweety439

Update the file for n <= 21001.

Continue to reserve...

S6 has now only 45 k's remain:

1296, 1814, 9589, 12179, 13215, 14505, 17464, 17984, 22139, 23864, 29014, 43429, 45634, 49874, 50252, 57189, 62614, 67894, 73814, 76441, 80389, 87284, 87289, 87800, 93589, 97131, 100899, 101529, 112783, 117454, 122704, 124125, 124874, 127688, 132614, 135199, 139959, 144509, 145984, 151719, 152209, 152249, 166753, 168610, 170199

[sweety439]

Originally Posted by sweety439

S6 has now only 45 k's remain:

1296, 1814, 9589, 12179, 13215, 14505, 17464, 17984, 22139, 23864, 29014, 43429, 45634, 49874, 50252, 57189, 62614, 67894, 73814, 76441, 80389, 87284, 87289, 87800, 93589, 97131, 100899, 101529, 112783, 117454, 122704, 124125, 124874, 127688, 132614, 135199, 139959, 144509, 145984, 151719, 152209, 152249, 166753, 168610, 170199

Test limit:

k = 1296 at n=268.4M
k = 1814 at n=12K
other k = 4 mod 5 at n=21K
other k at n=2M

[sweety439]

If there is no n such that k*b^n is either a perfect odd power or of the form 4*m^4, then (k*b^n+1)/gcd(k+1,b-1) has no algebra factors.

If there is no n such that k*b^n is a perfect power, then (k*b^n-1)/gcd(k-1,b-1) has no algebra factors.

[sweety439]

Originally Posted by sweety439

If there is no n such that k*b^n is either a perfect odd power or of the form 4*m^4, then (k*b^n+1)/gcd(k+1,b-1) has no algebra factors.

If there is no n such that k*b^n is a perfect power, then (k*b^n-1)/gcd(k-1,b-1) has no algebra factors.

Conjecture: If (k*b^n+-1)/gcd(k+-1,b-1) has no algebra factors and this k < (conjectured k in the post #528 and #529), then there are infinitely many integers n>=1 such that (k*b^n+-1)/gcd(k+-1,b-1) is prime.

[sweety439]

Originally Posted by sweety439

If there is no n such that k*b^n is either a perfect odd power or of the form 4*m^4, then (k*b^n+1)/gcd(k+1,b-1) has no algebra factors.

If there is no n such that k*b^n is a perfect power, then (k*b^n-1)/gcd(k-1,b-1) has no algebra factors.

Conjecture: If (k*b^n+-1)/gcd(k+-1,b-1) has no algebra factors and (k*b^n+-1)/gcd(k+-1,b-1) has no covering set, then there are infinitely many integers n>=1 such that (k*b^n+-1)/gcd(k+-1,b-1) is prime.

[sweety439]

Originally Posted by sweety439

Conjecture: If (k*b^n+-1)/gcd(k+-1,b-1) has no algebra factors and (k*b^n+-1)/gcd(k+-1,b-1) has no covering set, then there are infinitely many integers n>=1 such that (k*b^n+-1)/gcd(k+-1,b-1) is prime.

Conjecture: All covering sets of (k*b^n+-1)/gcd(k+-1,b-1) are finite.

[sweety439]

S6 currently at n=25177.

Upload the file for the (probable) primes found for n>2200.

[sweety439]

I will sieve S6 to n=100K, it can be sieved with R6.

Thus, I will reserve SR6 for 25K<=n<=100K.

[sweety439]

Update the sieve file for S6.