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20170417, 17:28  #1 
Nov 2016
2^{2}·3·5·47 Posts 
Semiprime and nalmost 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 ralmost prime candidate but no (r1)almost prime candidate, then we want to find an ralmost prime (this ralmost 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^n1 (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(k1,41) = 1, then k is divisible by 3, and let k = m^2, k*4^n1 factors to (m*2^n1) * (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 20170417 at 18:02 
20181214, 21:04  #2  
Nov 2016
2^{2}×3×5×47 Posts 
Quote:
Last fiddled with by sweety439 on 20181214 at 21:08 

20181214, 21:10  #3  
Nov 2016
5404_{8} Posts 
Quote:
Code:
k formula(s) 1 {(1*2^n1)/3,1*2^n+1} (for even n) or {1*2^n1,(1*2^n+1)/3} (for odd n) 2 {2*4^n1} 3 {3*4^n1} 4 {2*2^n1,(2*2^n+1)/3} (for even n) or {(2*2^n1)/3,2*2^n+1} (for odd n) 5 {5*4^n1} 6 {6*4^n1} 7 {(7*4^n1)/3} 8 {8*4^n1} 9 {3*2^n1,3*2^n+1} 10 {(10*4^n1)/3} 11 {11*4^n1} 12 {12*4^n1} 13 {(13*4^n1)/3} 14 {14*4^n1} 15 {15*4^n1} 16 {(4*2^n1)/3,4*2^n+1} (for even n) or {4*2^n1,(4*2^n+1)/3} (for odd n) 17 {17*4^n1} 18 {18*4^n1} 19 {(19*4^n1)/3} 20 {20*4^n1} 21 {21*4^n1} 22 {(22*4^n1)/3} 23 {23*4^n1} 24 {24*4^n1} 25 {5*2^n1,(5*2^n+1)/3} (for even n) or {(5*2^n1)/3,5*2^n+1} (for odd n) Last fiddled with by sweety439 on 20181214 at 21:20 

20181214, 21:17  #4  
Nov 2016
2^{2}·3·5·47 Posts 
Quote:
e.g. for k=7, 7*4^n1 has trivial factor of 3, thus we should take out this factor, and this formula is (7*4^n1)/3, we need n such that this formula takes prime value. (7*4^n1 has no algebra factors) e.g. for k=9, 9*4^n1 = (3*2^n1) * (3*2^n+1), thus we need n such that both these two formulas take prime values. (9*4^n1 has no trivial factors) e.g. for k=25, 25*4^n1 has trivial factor of 3, thus we should take out this factor, and this formula is (25*4^n1)/3, besides, (25*4^n1)/3 = (5*2^n1) * (5*2^n+1)/3 (for even n) or (5*2^n1)/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 20181214 at 21:18 

20181214, 21:20  #5  
Nov 2016
2^{2}·3·5·47 Posts 
Quote:


20181214, 21:27  #6 
Nov 2016
2^{2}·3·5·47 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^n1 is always divisible by 5, and for even n, 4*24^n1 = (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^(3q1)*3^q  1 and m*2^(3q1)*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. 
20181214, 21:33  #7 
Nov 2016
2^{2}·3·5·47 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.

20181214, 21:35  #8  
Nov 2016
2^{2}×3×5×47 Posts 
Quote:


20181214, 21:42  #9  
Nov 2016
2^{2}·3·5·47 Posts 
Quote:
Last fiddled with by sweety439 on 20181214 at 21:42 

20181214, 21:48  #10  
Nov 2016
2^{2}·3·5·47 Posts 
Quote:
For R10, the n for k = 343 is 1, since (7*10^11)/3 (= 23) and (49*100^1+7*10^1+1)/3 (= 1657) are both primes. Last fiddled with by sweety439 on 20181214 at 22:01 

20181214, 21:59  #11 
Nov 2016
2820_{10} 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^n1 and (k^2*4^n+k*2^n+1)/7 or (k*2^n1)/7 and k^2*4^n+k*2^n+1 For R9, these formulas are either (k*3^n1)/2 and (k*3^n+1)/4 or (k*3^n1)/4 and (k*3^n+1)/2 For R12, these formulas are k*12^n1 and k*12^n+1 (since for all these addition k (25, 27, 64, 300, 324), gcd(k1,121) = 1) 
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