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Old 2017-10-16, 20:16   #496
sweety439
 
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

B7A16 Posts
Default Riesel base 137

Code:
k,n
1,11
2,2
3,27
4,1
5,12
6,1
7,1
8,2
9,algebra
10,5
11,?
12,2
13,?
14,4
15,?
16,231
With CK=17, k=9 proven composite by partial algebraic factors, k=11, 13 and 15 remain.
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Old 2017-10-16, 20:18   #497
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2·13·113 Posts
Default Riesel base 140

See CRUS, compare with the prime (1*140^79-1)/139.

With CK=46, this base is proven.
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Old 2017-10-16, 20:18   #498
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"99(4^34019)99 palind"
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55728 Posts
Default Riesel base 142

Code:
k,n
1,?
2,1
3,26
4,3
5,1
6,3
7,1
8,7
9,1
10,2
11,14
With CK=12, k=1 remains.
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Old 2017-10-16, 20:20   #499
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2×13×113 Posts
Default Riesel base 144

Code:
k,n
1,algebra
2,24
3,1
4,algebra
5,1
6,1
7,5
8,1
9,algebra
10,1
11,1
12,1
13,1
14,4
15,10
16,algebra
17,1
18,1
19,4
20,1
21,1
22,1
23,134
24,2
25,algebra
26,5
27,2
28,2
29,4
30,519
31,1
32,3
33,1
34,8
35,1
36,algebra
37,3
38,1
39,964
40,1
41,1
42,1
43,2
44,6
45,3
46,97
47,2
48,1
49,algebra
50,2
51,3
52,1
53,1
54,8
55,1
56,1
57,20
58,35
With CK=59, k=1, 4, 9, 16, 25, 36 and 49 proven composite by full algebraic factors, this base is proven.
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Old 2017-10-16, 20:25   #500
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2·13·113 Posts
Default Riesel base 147

Code:
k,n
1,3
2,1
3,2
4,1
5,1
6,1
7,14
8,2
9,1
10,14
11,?
12,112
13,31
14,3
15,46
16,1
17,1
18,2
19,140
20,1
21,1
22,48
23,4
24,1
25,5
26,1
27,2
28,2
29,1
30,1
31,10
32,1
33,619
34,43
35,4
36,algebra
37,1
38,131
39,12
40,1
41,9
42,1
43,20
44,3
45,1
46,1
47,8
48,96
49,?
50,1
51,?
52,1
53,3
54,1
55,?
56,1
57,13
58,?
59,?
60,1
61,1
62,29
63,?
64,169
65,5
66,3
67,2
68,7
69,13
70,1
71,114
72,2
With CK=73, k=36 proven composite by partial algebraic factors, k=11, 49, 51, 55, 58, 59 and 63 remain.
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Old 2017-10-24, 07:26   #501
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Nov 2016
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1011011110102 Posts
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k-values that are a multiple of base (b) and where (k+-1)/gcd(k+-1,b-1) (+ for Sierpinski, - for Riesel) is not prime are included in the conjectures but excluded from testing, since such k-values will have the same prime as k / b.

However, k-values that are a multiple of base (b) and where (k+-1)/gcd(k+-1,b-1) (+ for Sierpinski, - for Riesel) is prime are included from testing since the exponent n must be >=1 (n can be 1, but cannot be 0 or -1 or -2 or ...), and the same prime n=1 for k / b would be n=0 for this k but n must be >=1 hence it is not allowed so this k must continue to be searched. (of course, k-values that are not a multiple of base (b) are included from testing)

Thus, for S3, k = 42, 45, 57, 60, 66 and 72 are included from testing since although 42, 45, 57, 60, 66 and 72 are multiples of 3, but 42+1, (45+1)/2, (57+1)/2, 60+1, 66+1 and 72+1 are primes. However, k = 48, 51, 54, 63, 69 and 75 are excluded from testing since 48, 51, 54, 63, 69 and 75 are multiples of 3, and 48+1, (51+1)/2, 54+1, (63+1)/2, (69+1)/2 and (75+1)/2 are not primes. Besides, for R3, k = 42, 48, 54, 60, 63, 72 and 75 are included from testing since although 42, 48, 54, 60, 63, 72 and 75 are multiples of 3, but 42-1, 48-1, 54-1, 60-1, (63-1)/2, 72-1 and (75-1)/2 are primes. However, k = 45, 51, 57, 66 and 69 are excluded from testing since 45, 51, 57, 66 and 69 are multiples of 3, and (45-1)/2, (51-1)/2, (57-1)/2, 66-1 and (69-1)/2 are not primes.

Note: Since 1 is not prime, thus for R3, k = 3 is excluded from testing. ((3-1)/2 = 1) However, since 2 is prime, thus for S3, k = 3 is included from testing. ((3+1)/2 = 2)

Last fiddled with by sweety439 on 2017-10-24 at 08:08
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Old 2017-10-24, 08:11   #502
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Quote:
Originally Posted by sweety439 View Post
k-values that are a multiple of base (b) and where (k+-1)/gcd(k+-1,b-1) (+ for Sierpinski, - for Riesel) is not prime are included in the conjectures but excluded from testing, since such k-values will have the same prime as k / b.

However, k-values that are a multiple of base (b) and where (k+-1)/gcd(k+-1,b-1) (+ for Sierpinski, - for Riesel) is prime are included from testing since the exponent n must be >=1 (n can be 1, but cannot be 0 or -1 or -2 or ...), and the same prime n=1 for k / b would be n=0 for this k but n must be >=1 hence it is not allowed so this k must continue to be searched. (of course, k-values that are not a multiple of base (b) are included from testing)

Thus, for S3, k = 42, 45, 57, 60, 66 and 72 are included from testing since although 42, 45, 57, 60, 66 and 72 are multiples of 3, but 42+1, (45+1)/2, (57+1)/2, 60+1, 66+1 and 72+1 are primes. However, k = 48, 51, 54, 63, 69 and 75 are excluded from testing since 48, 51, 54, 63, 69 and 75 are multiples of 3, and 48+1, (51+1)/2, 54+1, (63+1)/2, (69+1)/2 and (75+1)/2 are not primes. Besides, for R3, k = 42, 48, 54, 60, 63, 72 and 75 are included from testing since although 42, 48, 54, 60, 63, 72 and 75 are multiples of 3, but 42-1, 48-1, 54-1, 60-1, (63-1)/2, 72-1 and (75-1)/2 are primes. However, k = 45, 51, 57, 66 and 69 are excluded from testing since 45, 51, 57, 66 and 69 are multiples of 3, and (45-1)/2, (51-1)/2, (57-1)/2, 66-1 and (69-1)/2 are not primes.

Note: Since 1 is not prime, thus for R3, k = 3 is excluded from testing. ((3-1)/2 = 1) However, since 2 is prime, thus for S3, k = 3 is included from testing. ((3+1)/2 = 2)
Since 1 is not prime, thus in general, for a Riesel base b>=2, k = b is excluded from testing since (b-1)/gcd(b-1,b-1) = (b-1)/(b-1) = 1, thus k = b would have the same prime as k = 1.

Last fiddled with by sweety439 on 2017-10-24 at 08:40
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Old 2017-11-03, 08:41   #503
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Sieve S10 k=269 to n=100K. (Unfortunately, I cannot sieve R43 k=13, because the cmd.exe says "ERROR: 13*43^n-1: every term is divisible by 2". For both sides (Sierpinski and Riesel), if and only if b and k are both odd, then srsieve cannot sieve it, thus srsieve can only sieve the case which b or k (or both) is even)

For S10 k=269, since the divisor (gcd(k+1,b-1)) is 9, and the only prime factor of 9 is 3, thus we do not sieve the prime 3, and all numbers of the form (269*10^n+1)/(269+1,10-1) are not divisible by 2 or 5, thus, we start with the prime 7. (of course, there are n's such that (269*10^n+1)/(269+1,10-1) is still divisible by 3, it is divisible by 3 if and only if n = 0 (mod 3), we should remove these n's from sieve file)

Last fiddled with by sweety439 on 2017-11-03 at 08:44
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Old 2017-11-03, 08:41   #504
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Quote:
Originally Posted by sweety439 View Post
Sieve S10 to n=100K. (Unfortunately, I cannot sieve R43, because the cmd.exe says "ERROR: 13*43^n-1: every term is divisible by 2". For both sides (Sierpinski and Riesel), if and only if b and k are both odd, then srsieve cannot sieve it, thus srsieve can only sieve the case which b or k (or both) is even)
The result is

Code:
Read 1 sequence from input file `K-list.txt'.
srsieve started: 20000 <= n <= 100000, 7 <= p <= 10000000000
Split 1 base 10 sequence into 15 base 10^30 subsequences.
p=228940087, 3823697 p/sec, 73130 terms eliminated, 6871 remain
p=470340011, 4023332 p/sec, 73394 terms eliminated, 6607 remain
p=742500067, 4536152 p/sec, 73531 terms eliminated, 6470 remain
p=1017820007, 4588053 p/sec, 73634 terms eliminated, 6367 remain
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Old 2017-11-03, 08:52   #505
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"99(4^34019)99 palind"
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(P^81993)SZ base 36

2×13×113 Posts
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Quote:
Originally Posted by sweety439 View Post
Sieve S10 k=269 to n=100K. (Unfortunately, I cannot sieve R43 k=13, because the cmd.exe says "ERROR: 13*43^n-1: every term is divisible by 2". For both sides (Sierpinski and Riesel), if and only if b and k are both odd, then srsieve cannot sieve it, thus srsieve can only sieve the case which b or k (or both) is even)

For S10 k=269, since the divisor (gcd(k+1,b-1)) is 9, and the only prime factor of 9 is 3, thus we do not sieve the prime 3, and all numbers of the form (269*10^n+1)/(269+1,10-1) are not divisible by 2 or 5, thus, we start with the prime 7. (of course, there are n's such that (269*10^n+1)/(269+1,10-1) is still divisible by 3, it is divisible by 3 if and only if n = 0 (mod 3), we should remove these n's from sieve file)
Also sieve S36 k=1814 to n=100K. (similarly, we should start with the prime 7, since the divisor (gcd(k+1,b-1)) is 5, and all numbers of the form (1814*36^n+1)/gcd(1814+1,36-1) are not divisible by 2 or 3)

Last fiddled with by sweety439 on 2017-11-03 at 08:54
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Old 2017-11-03, 11:50   #506
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"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

55728 Posts
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Sieved to p=1e10.

Update the sieve files, I will sieve them using pfgw.
Attached Files
File Type: txt t16_b10_k269.txt (61.3 KB, 120 views)
File Type: txt t16_b36_k1814.txt (45.3 KB, 117 views)

Last fiddled with by sweety439 on 2017-11-03 at 11:51
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