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 2017-04-17, 17:28 #1 sweety439   Nov 2016 2,819 Posts Semiprime and n-almost prime candidate for the k's with algebra for the Sierpinski/Riesel problem For some k's without covering set, k*b^n+-1 still cannot be prime because of the algebra factors, thus, we can choose the solve another problem, if k*b^n+-1 has r-almost prime candidate but no (r-1)-almost prime candidate, then we want to find an r-almost prime (this r-almost prime should not have small prime factor) of the form k*b^n+-1. For the k's with full algebra factors, we need to find an n such that all the factors are primes. For the k's with partial algebra factors, we need to find an n with algebra factors but all the factors are primes. These are the status for some k's with full/partial algebra factors for Sierpinski/Riesel bases b<=64: S8, k=1: factors to (1*2^n + 1) * (1*4^n - 1*2^n + 1), and the two numbers are both primes for n = 1, 2, 4, ... S8, k=8: factors to (2*2^n + 1) * (4*4^n - 2*2^n + 1), and the two numbers are both primes for n = 1, 3, ... S16, k=2500: factors to (50*4^n + 10*2^n + 1) * (50*4^n - 10*2^n + 1), and the two numbers are both primes for n = 6, ... S16, k=40000: factors to (200*4^n + 20*2^n + 1) * (200*4^n - 20*2^n + 1), and the two numbers are both primes for n = 5, ... S27, k=8: factors to (2*3^n + 1) * (4*9^n - 2*3^n + 1), and the two numbers are both primes for n = 1, 2, 4, ... S27, k=216: factors to (6*3^n + 1) * (36*9^n - 6*3^n + 1), and the two numbers are both primes for n = 1, 3, ... S27, k=512: factors to (8*3^n + 1) * (64*9^n - 8*3^n + 1), and the two numbers are both primes for n = 2, ... S32, k=1: factors to (1*2^n + 1) * (1*16^n - 1*8^n + 1*4^n - 1*2^n + 1), and the two numbers are both prime for n = 1, 4, 8, ... S63, k=3511808: if n=3q, factors to (152*63^q + 1) * (23104*3969^q - 152*63^q + 1), the two numbers are not both prime for all q<=1000. S63, k=27000000: if n=3q, factors to (300*63^q + 1) * (90000*3969^q - 300*63^q + 1), the two numbers are not both prime for all q<=1000. S64, k=1: factors to (1*4^n + 1) * (1*16^n - 1*4^n + 1), and the two numbers are both primes for n = 1, 2, ... R9, k=4: factors to (2*3^n - 1) * (2*3^n + 1), and the two numbers are both primes for n = 1, 2, ... R9, k=16: factors to (4*3^n - 1) * (4*3^n + 1), and the two numbers are both primes for n = 1, 3, 15, ... R9, k=36: factors to (6*3^n - 1) * (6*3^n + 1), and the two numbers are both primes for n = 1, ... R9, k=64: factors to (8*3^n - 1) * (8*3^n + 1), and the two numbers are both primes for n = 2, 10, ... R12, k=25: if n=2q, factors to (5*12^q - 1) * (5*12^q + 1), and the two numbers are both primes for q = 1, ... R12, k=27: if n=2q+1, factors to (18*12^q - 1) * (18*12^q + 1), and the two numbers are both primes for q = 1, 3, ... R12, k=64: if n=2q, factors to (8*12^q - 1) * (8*12^q + 1), and the two numbers are both primes for q = 2, ... R12, k=300: if n=2q+1, factors to (60*12^q - 1) * (60*12^q + 1), the two numbers are not both prime for all q<=1000. R12, k=324: if n=2q, factors to (18*12^q - 1) * (18*12^q + 1), and the two numbers are both primes for q = 2, ... For all the bases above except S63 and R12, this problem is proven for the k's with algebra factors in the original problems. However, S16, S27 and S32 are still not proven since their original problem are not proven, all other bases above are completely proven. (note that I do not exclude GFNs in the conjectures, thus, in my definition, S32 is not proven, and it has only k=4 remain) That is, there is known semiprime of all the forms above but 3511808*63^n+1 (n%3=0), 27000000*63^n+1 (n%3=0), 300*12^n-1 (n%2=1), all the forms above have no prime candidate, but all have semiprime candidate. There are many such k's for R4, R16, R19, R24, R25, and some other Riesel bases. In fact, the R4 problem is the same as the twin Proth/Riesel primes problem, since if k is square and gcd(k-1,4-1) = 1, then k is divisible by 3, and let k = m^2, k*4^n-1 factors to (m*2^n-1) * (m*2^n+1), this m is also divisible by 3, and find an n such that the two numbers are both primes is the same as the twin Proth/Riesel primes problem, see http://mersenneforum.org/showthread.php?t=8479. According to this website, this problem still has 7 k's remain for k's with algebra factors in the original problem and k<39939: 12321, 15129, 23409, 25281, 29241, 33489, 35721. (these k's equal 111^2, 123^2, 153^2, 159^2, 171^2, 183^2, 189^2) Last fiddled with by sweety439 on 2017-04-17 at 18:02
2018-12-14, 21:04   #2
sweety439

Nov 2016

281910 Posts

Quote:
 Originally Posted by sweety439 For some k's without covering set, k*b^n+-1 still cannot be prime because of the algebra factors, thus, we can choose the solve another problem, if k*b^n+-1 has r-almost prime candidate but no (r-1)-almost prime candidate, then we want to find an r-almost prime (this r-almost prime should not have small prime factor) of the form k*b^n+-1. For the k's with full algebra factors, we need to find an n such that all the factors are primes. For the k's with partial algebra factors, we need to find an n with algebra factors but all the factors are primes. These are the status for some k's with full/partial algebra factors for Sierpinski/Riesel bases b<=64: S8, k=1: factors to (1*2^n + 1) * (1*4^n - 1*2^n + 1), and the two numbers are both primes for n = 1, 2, 4, ... S8, k=8: factors to (2*2^n + 1) * (4*4^n - 2*2^n + 1), and the two numbers are both primes for n = 1, 3, ... S16, k=2500: factors to (50*4^n + 10*2^n + 1) * (50*4^n - 10*2^n + 1), and the two numbers are both primes for n = 6, ... S16, k=40000: factors to (200*4^n + 20*2^n + 1) * (200*4^n - 20*2^n + 1), and the two numbers are both primes for n = 5, ... S27, k=8: factors to (2*3^n + 1) * (4*9^n - 2*3^n + 1), and the two numbers are both primes for n = 1, 2, 4, ... S27, k=216: factors to (6*3^n + 1) * (36*9^n - 6*3^n + 1), and the two numbers are both primes for n = 1, 3, ... S27, k=512: factors to (8*3^n + 1) * (64*9^n - 8*3^n + 1), and the two numbers are both primes for n = 2, ... S32, k=1: factors to (1*2^n + 1) * (1*16^n - 1*8^n + 1*4^n - 1*2^n + 1), and the two numbers are both prime for n = 1, 4, 8, ... S63, k=3511808: if n=3q, factors to (152*63^q + 1) * (23104*3969^q - 152*63^q + 1), the two numbers are not both prime for all q<=1000. S63, k=27000000: if n=3q, factors to (300*63^q + 1) * (90000*3969^q - 300*63^q + 1), the two numbers are not both prime for all q<=1000. S64, k=1: factors to (1*4^n + 1) * (1*16^n - 1*4^n + 1), and the two numbers are both primes for n = 1, 2, ... R9, k=4: factors to (2*3^n - 1) * (2*3^n + 1), and the two numbers are both primes for n = 1, 2, ... R9, k=16: factors to (4*3^n - 1) * (4*3^n + 1), and the two numbers are both primes for n = 1, 3, 15, ... R9, k=36: factors to (6*3^n - 1) * (6*3^n + 1), and the two numbers are both primes for n = 1, ... R9, k=64: factors to (8*3^n - 1) * (8*3^n + 1), and the two numbers are both primes for n = 2, 10, ... R12, k=25: if n=2q, factors to (5*12^q - 1) * (5*12^q + 1), and the two numbers are both primes for q = 1, ... R12, k=27: if n=2q+1, factors to (18*12^q - 1) * (18*12^q + 1), and the two numbers are both primes for q = 1, 3, ... R12, k=64: if n=2q, factors to (8*12^q - 1) * (8*12^q + 1), and the two numbers are both primes for q = 2, ... R12, k=300: if n=2q+1, factors to (60*12^q - 1) * (60*12^q + 1), the two numbers are not both prime for all q<=1000. R12, k=324: if n=2q, factors to (18*12^q - 1) * (18*12^q + 1), and the two numbers are both primes for q = 2, ... For all the bases above except S63 and R12, this problem is proven for the k's with algebra factors in the original problems. However, S16, S27 and S32 are still not proven since their original problem are not proven, all other bases above are completely proven. (note that I do not exclude GFNs in the conjectures, thus, in my definition, S32 is not proven, and it has only k=4 remain) That is, there is known semiprime of all the forms above but 3511808*63^n+1 (n%3=0), 27000000*63^n+1 (n%3=0), 300*12^n-1 (n%2=1), all the forms above have no prime candidate, but all have semiprime candidate. There are many such k's for R4, R16, R19, R24, R25, and some other Riesel bases. In fact, the R4 problem is the same as the twin Proth/Riesel primes problem, since if k is square and gcd(k-1,4-1) = 1, then k is divisible by 3, and let k = m^2, k*4^n-1 factors to (m*2^n-1) * (m*2^n+1), this m is also divisible by 3, and find an n such that the two numbers are both primes is the same as the twin Proth/Riesel primes problem, see http://mersenneforum.org/showthread.php?t=8479. According to this website, this problem still has 7 k's remain for k's with algebra factors in the original problem and k<39939: 12321, 15129, 23409, 25281, 29241, 33489, 35721. (these k's equal 111^2, 123^2, 153^2, 159^2, 171^2, 183^2, 189^2)
We use the extension of the thread https://mersenneforum.org/showthread.php?t=23866, e.g. for R4 k=25, we need n such that 5*2^n-1 and (5*2^n+1)/3 (for even n) (5*2^n-1)/3 and 5*2^n+1 (for odd n) are both primes.

Last fiddled with by sweety439 on 2018-12-14 at 21:08

2018-12-14, 21:10   #3
sweety439

Nov 2016

2,819 Posts

Quote:
 Originally Posted by sweety439 We use the extension of the thread https://mersenneforum.org/showthread.php?t=23866, e.g. for R4 k=25, we need n such that 5*2^n-1 and (5*2^n+1)/3 (for even n) (5*2^n-1)/3 and 5*2^n+1 (for odd n) are both primes.
Thus, for R4: (we should find n such that all of these formulas take prime values)

Code:
k  formula(s)
1  {(1*2^n-1)/3,1*2^n+1} (for even n) or {1*2^n-1,(1*2^n+1)/3} (for odd n)
2  {2*4^n-1}
3  {3*4^n-1}
4  {2*2^n-1,(2*2^n+1)/3} (for even n) or {(2*2^n-1)/3,2*2^n+1} (for odd n)
5  {5*4^n-1}
6  {6*4^n-1}
7  {(7*4^n-1)/3}
8  {8*4^n-1}
9  {3*2^n-1,3*2^n+1}
10 {(10*4^n-1)/3}
11 {11*4^n-1}
12 {12*4^n-1}
13 {(13*4^n-1)/3}
14 {14*4^n-1}
15 {15*4^n-1}
16 {(4*2^n-1)/3,4*2^n+1} (for even n) or {4*2^n-1,(4*2^n+1)/3} (for odd n)
17 {17*4^n-1}
18 {18*4^n-1}
19 {(19*4^n-1)/3}
20 {20*4^n-1}
21 {21*4^n-1}
22 {(22*4^n-1)/3}
23 {23*4^n-1}
24 {24*4^n-1}
25 {5*2^n-1,(5*2^n+1)/3} (for even n) or {(5*2^n-1)/3,5*2^n+1} (for odd n)

Last fiddled with by sweety439 on 2018-12-14 at 21:20

2018-12-14, 21:17   #4
sweety439

Nov 2016

2,819 Posts

Quote:
 Originally Posted by sweety439 Thus, for R4: (we should find n such that all of these formulas take prime values)
e.g. for k=5, 5*4^n-1 has neither trivial factors nor algebraic factors, thus we need n such that 5*4^n-1 takes prime value.

e.g. for k=7, 7*4^n-1 has trivial factor of 3, thus we should take out this factor, and this formula is (7*4^n-1)/3, we need n such that this formula takes prime value. (7*4^n-1 has no algebra factors)

e.g. for k=9, 9*4^n-1 = (3*2^n-1) * (3*2^n+1), thus we need n such that both these two formulas take prime values. (9*4^n-1 has no trivial factors)

e.g. for k=25, 25*4^n-1 has trivial factor of 3, thus we should take out this factor, and this formula is (25*4^n-1)/3, besides, (25*4^n-1)/3 = (5*2^n-1) * (5*2^n+1)/3 (for even n) or (5*2^n-1)/3 * (5*2^n+1) (for odd n), thus we need n such that both these two formulas take prime values.

Last fiddled with by sweety439 on 2018-12-14 at 21:18

2018-12-14, 21:20   #5
sweety439

Nov 2016

2,819 Posts

Quote:
 Originally Posted by sweety439 Thus, for R4: (we should find n such that all of these formulas take prime values) Code: k formula(s) 1 {(1*2^n-1)/3,1*2^n+1} (for even n) or {1*2^n-1,(1*2^n+1)/3} (for odd n) 2 {2*4^n-1} 3 {3*4^n-1} 4 {2*2^n-1,(2*2^n+1)/3} (for even n) or {(2*2^n-1)/3,2*2^n+1} (for odd n) 5 {5*4^n-1} 6 {6*4^n-1} 7 {(7*4^n-1)/3} 8 {8*4^n-1} 9 {3*2^n-1,3*2^n+1} 10 {(10*4^n-1)/3} 11 {11*4^n-1} 12 {12*4^n-1} 13 {(13*4^n-1)/3} 14 {14*4^n-1} 15 {15*4^n-1} 16 {(4*2^n-1)/3,4*2^n+1} (for even n) or {4*2^n-1,(4*2^n+1)/3} (for odd n) 17 {17*4^n-1} 18 {18*4^n-1} 19 {(19*4^n-1)/3} 20 {20*4^n-1} 21 {21*4^n-1} 22 {(22*4^n-1)/3} 23 {23*4^n-1} 24 {24*4^n-1} 25 {5*2^n-1,(5*2^n+1)/3} (for even n) or {(5*2^n-1)/3,5*2^n+1} (for odd n)
Note that all n must be >= 1 (we allow n = 1, but not allow n = 0 or n < 0)

 2018-12-14, 21:27 #6 sweety439   Nov 2016 2,819 Posts For k's make a full covering set with partial algebraic factors, we only take the n for algebraic factors (i.e. not take the n for covering set for fixed prime factors) e.g. for R24 k=4, we need an even n such that 2*24^(n/2)-1 and 2*24^(n/2)+1 are both primes. (since for odd n, 4*24^n-1 is always divisible by 5, and for even n, 4*24^n-1 = (2*24^(n/2)-1) * (2*24^(n/2)+1), thus we need n such that both these two formulas take prime values) e.g. for R24 k=6, we need an odd n such that 2^(3q-1)*3^q - 1 and m*2^(3q-1)*3^q + 1 are both primes (where q = (n+1)/2). e.g. for R19 k=4, we need an even n such that 2*19^(n/2)-1 and (2*19^(n/2)+1)/3 are both primes.
 2018-12-14, 21:33 #7 sweety439   Nov 2016 B0316 Posts See post https://mersenneforum.org/showpost.p...&postcount=281, the conjectured k's for R4 are 361, 919, 1114, ..., and the conjectured k's for R10 are 334, 1585, 1882, ... and we can prove them.
2018-12-14, 21:35   #8
sweety439

Nov 2016

2,819 Posts

Quote:
 Originally Posted by sweety439 See post https://mersenneforum.org/showpost.p...&postcount=281, the conjectured k's for R4 are 361, 919, 1114, ..., and the conjectured k's for R10 are 334, 1585, 1882, ... and we can prove them.
Note that for R10 k=343, we should find an n such that (7*10^q-1)/3 and (49*100^q+7*10^q+1)/3 are both primes (the status of R10 k=343 is like the status for S63 k=3511808 and S63 k=27000000).

2018-12-14, 21:42   #9
sweety439

Nov 2016

2,819 Posts

Quote:
 Originally Posted by sweety439 See post https://mersenneforum.org/showpost.p...&postcount=281, the conjectured k's for R4 are 361, 919, 1114, ..., and the conjectured k's for R10 are 334, 1585, 1882, ... and we can prove them.
The 1st, 2nd and 3rd conjectures for R4 and R10 are all proven (the conjectures cover the original conjectures (https://mersenneforum.org/showthread.php?t=21839), the additional k's for R4 are only the square k's except 361 (we need n such that (m*2^q-1)/gcd(m*2^q-1,3) and (m*2^q+1)/gcd(m*2^q+1,3) are both primes (where q=n/2, and m=sqrt(k)) for all q <= 33 except q = 19 (since 19^2 = 361, and (361*4^n-1)/gcd(361-1,4-1) has covering set {3, 5, 7, 13})), and the additional k's for R10 are only k = 343 (we need n such that (7*10^q-1)/3 and (49*100^q+7*10^q+1)/3 are both primes), and we found the n for all of these k's.

Last fiddled with by sweety439 on 2018-12-14 at 21:42

2018-12-14, 21:48   #10
sweety439

Nov 2016

2,819 Posts

Quote:
 Originally Posted by sweety439 The 1st, 2nd and 3rd conjectures for R4 and R10 are all proven (the conjectures cover the original conjectures (https://mersenneforum.org/showthread.php?t=21839), the additional k's for R4 are only the square k's except 361 (we need n such that (m*2^q-1)/gcd(m*2^q-1,3) and (m*2^q+1)/gcd(m*2^q+1,3) are both primes (where q=n/2, and m=sqrt(k)) for all q <= 33 except q = 19 (since 19^2 = 361, and (361*4^n-1)/gcd(361-1,4-1) has covering set {3, 5, 7, 13})), and the additional k's for R10 are only k = 343 (we need n such that (7*10^q-1)/3 and (49*100^q+7*10^q+1)/3 are both primes), and we found the n for all of these k's.
Upload the text file for additional k's for R4, if you need it.

For R10, the n for k = 343 is 1, since (7*10^1-1)/3 (= 23) and (49*100^1+7*10^1+1)/3 (= 1657) are both primes.
Attached Files
 R4.txt (241 Bytes, 164 views)

Last fiddled with by sweety439 on 2018-12-14 at 22:01

2018-12-14, 21:59   #11
sweety439

Nov 2016

2,819 Posts

Update files for R8, R9 and R12, if you need them.

Only R12 k=300 remain.

Note:

For R8, these formulas are either

k*2^n-1 and (k^2*4^n+k*2^n+1)/7

or

(k*2^n-1)/7 and k^2*4^n+k*2^n+1

For R9, these formulas are either

(k*3^n-1)/2 and (k*3^n+1)/4

or

(k*3^n-1)/4 and (k*3^n+1)/2

For R12, these formulas are k*12^n-1 and k*12^n+1 (since for all these addition k (25, 27, 64, 300, 324), gcd(k-1,12-1) = 1)
Attached Files
 R8.txt (22 Bytes, 150 views) R9.txt (66 Bytes, 153 views) R12.txt (38 Bytes, 162 views)

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