20200708, 17:08  #870 
Nov 2016
3·5·137 Posts 
All n must be >= 1.
All MOB kvalues need an n>=1 prime (unless they have a covering set of primes, or make a full covering set with all or partial algebraic factors) MOB kvalues such that (k+1)/gcd(k+1,b1) (+ for Sierpinski,  for Riesel) is not prime are included in the conjectures but excluded from testing. Such kvalues will have the same prime as k / b. There are some MOB kvalues such that (k+1)/gcd(k+1,b1) (+ for Sierpinski,  for Riesel) is prime which do not have an easy prime for n>=1: GFN's and half GFN's: (all with no known primes) S2 k=65536 S3 k=3433683820292512484657849089281 S4 k=65536 S5 k=625 S6 k=1296 S7 k=2401 S8 k=256 and k=65536 S9 k=3433683820292512484657849089281 S10 k=100 S11 k=14641 S12 k=12 Other k's: S2 k=55816: first prime at n=14536 S2 k=90646: no prime with n<=6.6M S2 k=101746: no prime with n<=6.6M S3 k=621: first prime at n=20820 S4 k=176: first prime at n=228 S5 k=40: first prime at n=1036 S6 k=90546 S7 k=21: first prime at n=124 S9 k=1746: first prime at n=1320 S9 k=2007: first prime at n=3942 S10 k=640: first prime at n=120 S24 k=17496 S155 k=310 S333 k=1998 R2 k=74: first prime at n=2552 R2 k=674: first prime at n=11676 R2 k=1094: first prime at n=652 R4 k=19464: no prime with n<=3.3M R6 k=103536: first prime at n=6474 R6 k=106056: first prime at n=3038 R10 k=450: first prime at n=11958 R10 k=16750: no prime with n<=200K R11 k=308: first prime at n=444 R18 k=324: first prime at n=25665 R21 k=84: first prime at n=88 R23 k=230: first prime at n=6228 R27 k=594: first prime at n=36624 R40 k=520: no prime with n<=1K R42 k=1764: first prime at n=1317 R48 k=384: no prime with n<=200K R66 k=1056: no prime with n<=1K R78 k=7800: no prime with n<=1K R88 k=3168: first prime at n=205764 R96 k=9216: first prime at n=3341 R210 k=44100: first prime at n=19817 R306 k=93636: first prime at n=26405 R396 k=156816: no prime with n<=50K R591 k=1182: first prime at n=1190 R954 k=1908: first prime at n=1476 R976 k=1952: first prime at n=1924 R1102 k=2204: first prime at n=52176 R1297 k=2594: first prime at n=19839 R1360 k=2720: first prime at n=74688 Last fiddled with by sweety439 on 20200708 at 17:21 
20200708, 17:58  #871  
Nov 2016
807_{16} Posts 
Quote:
* R30 k=1369: for even n let n=2*q; factors to: (37*30^q  1) * (37*30^q + 1) odd n: covering set 7, 13, 19 * R88 k=400: for even n let n=2*q; factors to: (20*88^q  1) * (20*88^q + 1) odd n: covering set 3, 7, 13 * R95 k=324: for even n let n=2*q; factors to: (18*95^q  1) * (18*95^q + 1) odd n: covering set 7, 13, 229 * S55 k=2500: odd n: factor of 7 n = = 2 mod 4: factor of 17 n = = 0 mod 4: let n=4q and let m=5*55^q; factors to: (2*m^2 + 2m + 1) * (2*m^2  2m + 1) * S200 k=16: odd n: factor of 3 n = = 0 mod 4: factor of 17 n = = 2 mod 4: let n = 4*q  2 and let m = 20^q*10^(q1); factors to: (2*m^2 + 2m + 1) * (2*m^2  2m + 1) * S225 k=114244: for even n let k=4*q^4 and let m=q*15^(n/2); factors to: (2*m^2 + 2m + 1) * (2*m^2  2m + 1) odd n: factor of 113 * R10 k=343: n = = 1 mod 3: factor of 3 n = = 2 mod 3: factor of 37 n = = 0 mod 3: let n=3q and let m=7*10^q; factors to: (m  1) * (m^2 + m + 1) * R957 k=64: n = = 1 mod 3: factor of 73 n = = 2 mod 3: factor of 19 n = = 0 mod 3: let n=3q and let m=4*957^q; factors to: (m  1) * (m^2 + m + 1) * S63 k=3511808: n = = 1 mod 3: factor of 37 n = = 2 mod 3: factor of 109 n = = 0 mod 3: let n=3q and let m=152*63^q; factors to: (m + 1) * (m^2  m + 1) * S63 k=27000000: n = = 1 mod 3: factor of 37 n = = 2 mod 3: factor of 109 n = = 0 mod 3: let n=3q and let m=300*63^q; factors to: (m + 1) * (m^2  m + 1) * R936 k=64: n = = 0 mod 2: let n = 2q; factors to: (8*936^q  1) * (8*936^q + 1) n = = 0 mod 3: let n=3q; factors to: (4*936^q  1) * [16*936^(2q) + 4*936^q + 1] n = = 1 mod 6: factor of 37 n = = 5 mod 6: factor of 109 Last fiddled with by sweety439 on 20200708 at 18:01 

20200708, 18:05  #872  
Nov 2016
3×5×137 Posts 
Quote:
* In the Riesel case if k and b are both rth powers for an r>1 * In the Sierpinski case if k and b are both rth powers for an odd r>1 * In the Sierpinski case if k is of the form 4*m^4, and b is 4th power Then this k proven composite by full algebraic factors 

20200708, 18:10  #873 
Nov 2016
3·5·137 Posts 
The k's that make a full covering set with all or partial algebraic factors are excluded from the conjectures, unless they also have a covering set of primes (e.g. R4 k=361, R8 k=343, R9 k=49, R16 k=100, R49 k=81, R121 k=100, R125 k=8, S125 k=8, R256 k=100, R1024 k=81), thus cannot be the CK

20200708, 18:16  #874  
Nov 2016
3·5·137 Posts 
Quote:
Examples: b = q^7, k = q^r, where r = 3, 5, 6 (mod 7). b = q^14, k = q^r, where r = 6, 10, 12 (mod 14). b = q^15, k = q^r, where r = 7, 11, 13, 14 (mod 15). b = q^17, k = q^r, where r = 3, 5, 6, 7, 10, 11, 12, 14 (mod 17). b = q^21, k = q^r, where r = 5, 10, 13, 17, 19, 20 (mod 21) b = q^23, k = q^r, where r = 5, 7, 10, 11, 14, 15, 17, 19, 20, 21, 22 (mod 23) b = q^28, k = q^r, where r = 12, 20, 24 (mod 28) b = q^30, k = q^r, where r = 14, 22, 26, 28 (mod 30) b = q^31, k = q^r, where r = 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30 (mod 31) b = q^33, k = q^r, where r = 5, 7, 10, 13, 14, 19, 20, 23, 26, 28 (mod 33) etc. (these are all examples for m<=33) Last fiddled with by sweety439 on 20200710 at 06:18 

20200708, 18:22  #875 
6809 > 6502
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Aug 2003
101×103 Posts
43×191 Posts 
There is no need to quote an entire previous post. Trim your quotes.

20200710, 06:31  #876 
Nov 2016
3·5·137 Posts 

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