A Sierpinski/Riesel-like problem

[sweety439]

Originally Posted by sweety439

S103 tested to n=8K (4K-8K)

1 (probable) prime found, 1 remain

R97 tested to n=8K (1K-8K)

1 (probable) prime found, 2 remain

R43 tested to n=12K (5K-12K)

nothing found, 1 remain

S83 tested to n=8K (4K-8K)

nothing found, 2 remain

S73 tested to n=10K (5K-10K)

nothing found, 1 remain

R107 tested to n=8K (4K-8K)

1 (probable) prime found, base proven

R100 tested to n=8K (4K-8K)

1 (probable) prime found, base proven

S67 tested to n=10K (5K-10K)

nothing found, 3 remain

R33 tested to n=12K (6K-12K)

nothing found, 2 remain

New (probable) primes:

S103 k=13: (13*103^7010+1)/2
R97 k=16: (16*97^1627-1)/3
R107 k=3: (3*107^4900-1)/2
R100 k=133: (133*100^5496-1)/33

[sweety439]

Originally Posted by sweety439

I also reserved S33 and R61 to n=12K (S61 is already proven) and found that (407*33^10961+1)/8 is (probable) prime!!! S33 now has only 2 k's remain.

Also reserve S36 k=1814 to n=12K.

[sweety439]

Originally Posted by sweety439

S112 has totally 48 k's remain:

8, 92, 122, 183, 209, 269, 428, 467, 547, 553, 668, 677, 813, 896, 926, 941, 943, 947, 953, 983, 1013, 1131, 1171, 1217, 1286, 1292, 1346, 1412, 1445, 1463, 1470, 1499, 1517, 1573, 1581, 1604, 1613, 1664, 1696, 1712, 1780, 1791, 1807, 1920, 1937, 2082, 2189, 2237

Some primes given by CRUS:

1780*112^62794+1
547*112^8124+1
1920*112^5333+1
2082*112^5308+1
1807*112^3619+1
1470*112^3096+1
1131*112^2768+1

[sweety439]

Originally Posted by sweety439

R112 has totally 37 k's remain:

9, 31, 68, 72, 79, 142, 187, 310, 340, 349, 421, 424, 451, 498, 529, 619, 636, 646, 703, 749, 758, 790, 853, 898, 940, 948, 981, 1008, 1018, 1024, 1051, 1093, 1204, 1254, 1268, 1349, 1353

Some primes given by CRUS:

948*112^173968-1
1268*112^50536-1
758*112^35878-1
1353*112^7751-1
498*112^6038-1
9*112^5717-1

[sweety439]

Originally Posted by sweety439

Some primes given by CRUS:

1780*112^62794+1
547*112^8124+1
1920*112^5333+1
2082*112^5308+1
1807*112^3619+1
1470*112^3096+1
1131*112^2768+1

Due to CRUS, k=1696 for S112 is already tested to n=1M with no prime found.

[sweety439]

Originally Posted by sweety439

S112 has totally 48 k's remain:

8, 92, 122, 183, 209, 269, 428, 467, 547, 553, 668, 677, 813, 896, 926, 941, 943, 947, 953, 983, 1013, 1131, 1171, 1217, 1286, 1292, 1346, 1412, 1445, 1463, 1470, 1499, 1517, 1573, 1581, 1604, 1613, 1664, 1696, 1712, 1780, 1791, 1807, 1920, 1937, 2082, 2189, 2237

Since 896 = 112 * 8, k=896 will have the same (probable) prime as k=8, thus, S112 in fact has only 47 k's remain at n=1K.

[sweety439]

Originally Posted by sweety439

R112 has totally 37 k's remain:

9, 31, 68, 72, 79, 142, 187, 310, 340, 349, 421, 424, 451, 498, 529, 619, 636, 646, 703, 749, 758, 790, 853, 898, 940, 948, 981, 1008, 1018, 1024, 1051, 1093, 1204, 1254, 1268, 1349, 1353

Since 1008 = 112 * 9, k=1008 will have the same (probable) prime as k=9, thus, R112 in fact has only 36 k's remain at n=1K.

[sweety439]

There is no pseudoprime (i.e. probable prime but not prime) if gcd(k+-1,b-1) (+ for Sierpinski, - for Riesel) = 1, because of the N-1/N+1 primality proof.

[sweety439]

Originally Posted by sweety439

These text files are the status for these conjectures: (only for bases 5, 8, 9, 11, 13, 14 and 16) (the prime (41*9^2446+1)/2 (S9, k=41) is converted by S3, k=41) (the prime 186*16^5229+1 (S16, k=186) is converted by S4, k=186)

These bases and k's are remain:

base    k
S13     29
S16     89, 215, 459, 515

Interestingly, there are no k's remain in the Riesel conjectures.

Thus, for all these Sierpinski/Riesel bases except S13 and S16, all of the 1st, 2nd and 3rd conjectures are proven. (the 1st and 2nd conjectures for S13, and the 1st conjecture for S16 are also proven, but the 3rd conjecture for S13, and the 2nd and 3rd conjecture for S16 are not proven)

The 3rd conjecture for S13 is also proven, see post #347 for the (probable) prime (29*13^10574+1)/6.

Reserve S16 (for all remain k's < 3rd CK).

[sweety439]

Originally Posted by sweety439

The 3rd conjecture for S13 is also proven, see post #347 for the (probable) prime (29*13^10574+1)/6.

Reserve S16 (for all remain k's < 3rd CK).

Found 3 (probable) primes:

(215*16^3373+1)/3
(459*16^3701+1)/5
(515*16^940+1)/3

Thus, the 2nd and 3rd conjecture for S16 both have only k=89 remain.

[sweety439]

Originally Posted by sweety439

Found 3 (probable) primes:

(215*16^3373+1)/3
(459*16^3701+1)/5
(515*16^940+1)/3

Thus, the 2nd and 3rd conjecture for S16 both have only k=89 remain.

Update newest file for the Sierpinski side for the 1st, 2nd and 3rd conjectures for bases 5, 8, 9, 11, 13, 14 and 16.