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Old 2018-01-10, 19:15   #562
sweety439
 
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

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S8 has 4 k's remain: 256, 467, 1028, 1132.
S9 has 6 k's remain: 1039, 1627, 1801, 2007, 2036, 2287.
S11 has 2 k's remain: 195 and 237.
S13 has no k's remain.
R8 has 2 k's remain: 239 and 247.
R9 has no k's remain.
R11 has 5 k's remain: 201, 243, 851, 855, 856.
R13 has no k's remain.
Reserve them.
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Old 2018-01-10, 19:17   #563
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Reserve them.
Found these (probable) primes:

(1627*9^2939+1)/4
(2007*9^3942+1)/8
(243*11^2384-1)/2
(856*11^2105-1)/5

Continue reserving...
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Old 2018-01-14, 10:43   #564
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Originally Posted by sweety439 View Post
Found these (probable) primes:

(1627*9^2939+1)/4
(2007*9^3942+1)/8
(243*11^2384-1)/2
(856*11^2105-1)/5

Continue reserving...
Those are all proven by Edwin Hall.
Reserving (751*4^6615-1)/3, should take ~12K sec.
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Old 2018-01-14, 16:54   #565
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Canidate (751*4^6615-1)/3, proven.
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Old 2018-01-15, 06:20   #566
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"99(4^34019)99 palind"
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Canidate (751*4^6615-1)/3, proven.
Thanks!!!

You fully proved the 2nd and the 3rd conjecture for R4!!!

I think you can prove the primality for the probable primes in the post #552 first. Some bases only need one primality proving, e.g. R17, it only needs the primality proving for the probable prime (29*17^4904-1)/4.
Last fiddled with by sweety439 on 2018-01-15 at 06:24
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Old 2018-02-04, 08:42   #567
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Reserve S93 and S117.
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Old 2018-02-04, 08:53   #568
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Reserve S93 and S117.
Found 2 (probable) primes:

(11*117^1164+1)/4
(75*117^1428+1)/4

Current likely at n=2K, S93 has no (probable) primes found.

Continue to find...
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Old 2018-02-04, 13:48   #569
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Quote:
Originally Posted by sweety439 View Post
Found 2 (probable) primes:

(11*117^1164+1)/4
(75*117^1428+1)/4

Current likely at n=2K, S93 has no (probable) primes found.

Continue to find...
Found 2 (probable) primes:

(19*93^4362+1)/4
(43*93^2994+1)/4

S93 k=67, S93 k=87 and S117 k=59 are still remain (also the half GFN's, i.e. S93 k=93 and S117 k=117).
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Old 2018-02-04, 14:13   #570
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"99(4^34019)99 palind"
Nov 2016
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Quote:
Originally Posted by sweety439 View Post
Found 2 (probable) primes:

(19*93^4362+1)/4
(43*93^2994+1)/4

S93 k=67, S93 k=87 and S117 k=59 are still remain (also the half GFN's, i.e. S93 k=93 and S117 k=117).
No primes found for these k's, they are likely tested to n=8K.
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Old 2018-02-04, 14:14   #571
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Reserve R93 and R117.
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Old 2018-02-11, 06:50   #572
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Reserve R93 and R117.
No (probable) primes found for R93 and R117, they are likely tested to n=8K.

Reserve R85 and R115.
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