A Sierpinski/Riesel-like problem - Page 43

sweety439 · 2017-09-28, 22:41

These k's are still remain for these bases, they are likely tested to at least n=15K.

S5, k=181
S16, k=89
R8, k=239 and 757

sweety439 · 2017-09-28, 22:41

Reserve k=386 and 744 for R9.

sweety439 · 2017-09-28, 22:58

Update the file for the status for the 1st, 2nd and 3rd conjecture for SR10.

All the three conjectures for R10 are proven, but S10 has 5 k's remain for k < 3rd CK: 100, 269, 1343, 2573, 3356 (k = 1000 and 2690 are included in the conjectures but excluded from testing, since these k-values will have the same (probable) prime as k = 100 and 269)

sweety439 · 2017-09-29, 19:47

1037*12^6281-1 is prime!!!

No prime found for R9 k=386, R9 k=744 and R12 k=1132, they are likely tested to at least n=15K.

sweety439 · 2017-10-03, 04:10

Update the current file for Sierpinski bases 5, 8, 9, 11 for all k's <= 1024.

sweety439 · 2017-10-03, 04:10

Update the current file for Riesel bases 5, 8, 9, 11 for all k's <= 1024.

sweety439 · 2017-10-04, 14:05

Quote:

Originally Posted by sweety439

Update the file for the status for the 1st, 2nd and 3rd conjecture for SR10.

All the three conjectures for R10 are proven, but S10 has 5 k's remain for k < 3rd CK: 100, 269, 1343, 2573, 3356 (k = 1000 and 2690 are included in the conjectures but excluded from testing, since these k-values will have the same (probable) prime as k = 100 and 269)

Reserve S10 (for k=1343, 2573 and 3356).

sweety439 · 2017-10-04, 14:07

Quote:

Originally Posted by sweety439

Update the text files for the 1st, 2nd and 3rd conjecture of SR12.

R12 has only 2 k's remain: 1037 and 1132, but S12 has many k's remain. (thus, I did not search S12 very far)

S12 has 9 k's remain:

12, 563, 846, 885, 911, 976, 1041, 1052, 1057. (k = 144 is included in the conjectures but excluded from testing, since this k-value will have the same prime as k = 12)

Since 563*12^4020+1 is prime, k=563 can be removed, I will run other k's (except k=12) after the reservations for S10 were done (see post #469).

sweety439 · 2017-10-04, 14:10

Quote:

Originally Posted by sweety439

Reserve S10 (for k=1343, 2573 and 3356).

(3356*10^4584+1)/9 is (probable) prime!!!

sweety439 · 2017-10-10, 13:21

If k is a rational power of b, then...

* In Riesel case, if the divisor (i.e. gcd(k-1,b-1)) is d, then these numbers are repunit numbers in positive base d+1. (if and only if d+1 is perfect power, then these numbers have algebra factors)

* In Sierpinski case, if the divisor (i.e. gcd(k+1,b-1)) is 1, then these numbers are generalized Fermat numbers in base m, where m is the largest integer such that both k and b are integer powers of m. (if and only if m is perfect odd power, then these numbers have algebra factors)

* In Sierpinski case, if the divisor (i.e. gcd(k+1,b-1)) is 2, then these numbers are half generalized Fermat numbers in base m, where m is the largest integer such that both k and b are integer powers of m. (if and only if m is perfect odd power, then these numbers have algebra factors)

* In Sierpinski case, if the divisor (i.e. gcd(k+1,b-1)) is d and d>=3, then these numbers are repunit numbers in negative base -(d-1). (if and only if d-1 is either perfect odd power or of the form 4*m^4, then these numbers have algebra factors)

sweety439 · 2017-10-10, 13:28

Quote:

Originally Posted by sweety439

Reserve S10 (for k=1343, 2573 and 3356).

No (probable) primes found for k=269, 1343 and 2573, all of the k's are likely tested to at least n=15K.

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2017-09-28, 22:41	#463
sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·13·113 Posts	These k's are still remain for these bases, they are likely tested to at least n=15K. S5, k=181 S16, k=89 R8, k=239 and 757

2017-09-28, 22:41	#464
sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·13·113 Posts	Reserve k=386 and 744 for R9.

2017-09-29, 19:47	#466
sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2×13×113 Posts	1037*12^6281-1 is prime!!! No prime found for R9 k=386, R9 k=744 and R12 k=1132, they are likely tested to at least n=15K.

2017-10-10, 13:21	#472
sweety439 "99(4^34019)99 palind" Nov 2016 (P^81993)SZ base 36 2·13·113 Posts	If k is a rational power of b, then... * In Riesel case, if the divisor (i.e. gcd(k-1,b-1)) is d, then these numbers are repunit numbers in positive base d+1. (if and only if d+1 is perfect power, then these numbers have algebra factors) * In Sierpinski case, if the divisor (i.e. gcd(k+1,b-1)) is 1, then these numbers are generalized Fermat numbers in base m, where m is the largest integer such that both k and b are integer powers of m. (if and only if m is perfect odd power, then these numbers have algebra factors) * In Sierpinski case, if the divisor (i.e. gcd(k+1,b-1)) is 2, then these numbers are half generalized Fermat numbers in base m, where m is the largest integer such that both k and b are integer powers of m. (if and only if m is perfect odd power, then these numbers have algebra factors) * In Sierpinski case, if the divisor (i.e. gcd(k+1,b-1)) is d and d>=3, then these numbers are repunit numbers in negative base -(d-1). (if and only if d-1 is either perfect odd power or of the form 4*m^4, then these numbers have algebra factors)