20200809, 15:06  #936 
Nov 2016
2522_{10} Posts 
S13:
2 (2) 8 (4) 11 (564) 29 (10574) 281? (>5000) S14: 2 (1) 6 (6) 22 (16) 29 (23) 61 (126) 73 (1182) 208 (>5000) S15: 2 (1) 5 (2) 13 (10) 29 (30) 49 (112) 189 (190) 197 (464) 219 (1129) 341 (>5000) S16: 2 (1) 3 (2) 5 (3) 18 (4) 23 (1074) 89 (>20000) R13: 1 (5) 20 (10) 25 (15) 43 (77) 127 (95) 154 (469) 288 (109217) 337? (>5000) R14: 1 (3) 2 (4) 5 (19698) 617? (>5000) R15: 1 (3) 14 (14) 39 (16) 47 (>5000) R16: 2 (1) 11 (2) 18 (3) 31 (12) 48 (15) 74 (638) 322 (4624) 443 (>1500000) Last fiddled with by sweety439 on 20200814 at 14:15 
20200809, 16:40  #937 
Nov 2016
2522_{10} Posts 
Sierpinski k=2:
3 (1) 12 (3) 17 (47) 38 (2729) 101 (192275) 218 (333925) 365? (>200000) Sierpinski k=3: 2 (1) 5 (2) 18 (3) 28 (7) 43 (171) 79 (875) 83 (>8000) Sierpinski k=4: 3 (1) 5 (2) 17 (6) 23 (342) 53 (>1610000) Sierpinski k=5: 2 (1) 3 (2) 16 (3) 19 (78) 31 (1026) 137 (>2000) Sierpinski k=6: 2 (1) 4 (2) 14 (6) 19 (14) 20 (15) 48 (27) 53 (143) 67 (4532) 108 (16317) 129 (16796) 212 (>400000) Riesel k=1: 2 (2) 3 (3) 7 (5) 11 (17) 19 (19) 35 (313) 39 (349) 51 (4229) 91 (4421) 152 (270217) 185? (>66337) Riesel k=2: 2 (1) 5 (4) 20 (10) 29 (136) 67 (768) 107 (21910) 170 (166428) 581 (>200000) Riesel k=3: 2 (1) 3 (2) 23 (6) 31 (18) 42 (2523) 107 (4900) 295 (5270) 347 (>25000) Riesel k=4: 2 (1) 7 (3) 23 (5) 43 (279) 47 (1555) 72 (1119849) 178? (>5000) Riesel k=5: 2 (2) 8 (4) 14 (19698) 31? (>6000) Riesel k=6: 2 (1) 13 (2) 21 (3) 48 (294) 119 (665) 154 (1989) 234 (>400000) Last fiddled with by sweety439 on 20200810 at 03:40 
20200810, 06:48  #938 
Nov 2016
100111011010_{2} Posts 
Update PARI program files.

20200811, 16:42  #939 
Nov 2016
2·13·97 Posts 
The "ground base" of (k*b^n+1)/gcd(k+1,b1) is the smallest value of b' such that
* b' is a root of b (or b itself) * b'1 is divisible by gcd(k+1,b1) * The relative complement of (k*b^n+1)/gcd(k+1,b1) in (k'*b'^n+1)/gcd(k'+1,b'1) contain no primes e.g. the "ground base" of (k*4^n+1)/gcd(k+1,41) is 2 for k == 1 mod 3 (since for these k, the relative complement of (k*4^n+1)/gcd(k+1,41) in (k*2^n+1)/gcd(k+1,21) is (2*k*4^n+1)/gcd(k+1,41), and all numbers of the form (2*k*4^n+1)/gcd(k+1,41) is divisible by 3, thus not prime), but it is 4 for k == 0 or 2 mod 3, similarly, the "ground base" of (k*4^n1)/gcd(k1,41) is 2 for k == 2 mod 3, but 4 for k == 0 or 1 mod 3 More examples: the "ground base" for (k*36^n1)/gcd(k1,361) is 6 for k == 6 mod 7, but 36 for other k the "ground base" for (k*25^n1)/gcd(k1,251) is 5 for k == 2 mod 3 or k == 5 mod 8, but 25 for other k the "ground base" for 222*625^n+1 is 25 (since all primes of the form 222*25^n+1 are of the form 222*625^n+1, numbers of the form 5550*625^n+1 are divisible by 13, thus cannot be prime, and 25 is the smallest possible base, numbers of the form 1110*25^n+1 can be prime) the "ground base" for 366*625^n1 and 9150^625^n1 are both 625 (since both of them can contain primes) the "ground base" for 29*1024^n1 and 74*1024^n1 are both 32 the "ground base" for 40*25^n+1 and (61*25^n+1)/2 are both 5 the "ground base" for 2036*9^n+1, 302*9^n1, 386*9^n1, and 744*9^n1 are all 9 the "ground base" for 443*16^n1, 2297*16^n1, 13380*16^n1, 13703*16^n1, 2908*16^n+1, 6663*16^n+1, and 10183*16^n+1 are all 16 the "ground base" for 9519*16^n1 and 19464*16^n1 are both 4 the "ground base" for 18344*16^n1, 23669*16^n1, 31859*16^n1, and 21181*16^n+1 are all 2 the "ground base" for 244*529^n+1, 376*529^n+1, and 394*529^n+1 are all 23 the "ground base" for 426*529^n+1 is 529 the "ground base" for 62*576^n+1, 227*576^n+1, and 1077*576^n+1 are all 576 the "ground base" for 656*576^n+1, 1851*576^n+1, and 2351*576^n+1 are all 24 the "ground base" for 94*8^n+1 is 2 (this is a case for a nonsquare power, since 47*8^n+1 has covering set {3, 5, 13} and 188*8^n+1 has trivial factor of 7, only 94*8^n+1 can be prime) the "ground base" for 37*8^n1 is 8 (since both 37*8^n1 and 74*8^n1 can be prime) the "ground base" for 450*100^n1 is 10 the "ground base" for 653*100^n1, 74*100^n1, and 470*100^n1 are all 100 the "ground base" for 5*196^n1 is 14 the "ground base" for 198*196^n1 is 196 Last fiddled with by sweety439 on 20200816 at 21:04 
20200811, 16:57  #940  
Nov 2016
2×13×97 Posts 
Quote:


20200811, 17:07  #941  
Nov 2016
2·13·97 Posts 
Quote:
If this conjecture is true, then k*2^n+1 has no covering set for all k<78557, since all odd k<78557 have at least one known proven prime of the form either k*2^n+1 (n>=1) or of the form 2^n+k (n>=1), and if k*2^n+1 has covering set, then 2^n+k must have the same covering set, since the dual of (k*b^n+c)/gcd(k+c, b1) is (c*b^n+k)/gcd(k+c, b1)/gcd(k, b^n), and if a form has covering set, then the dual form must have the same covering set Last fiddled with by sweety439 on 20200811 at 17:14 

20200812, 16:34  #942 
Nov 2016
2×13×97 Posts 
fixed the program so it can check whether (k*b^n+1)/gcd(k+1,b1) has:
* covering set (for primes <= 50000, check to exponent = 5000) * full algebra factors (k and b are both rth powers for (r>1 Riesel case) (odd r>1 Sierpinski case)) (4*k and b are both 4th powers Sierpinski case) * partial algebra factors for algebra factors of difference of squares for (even/odd) n and one fixed prime factor for (odd/even) n for Riesel case It current cannot check these cases: * Partial algebra factors for algebra factors of difference of squares for (even/odd) n and covering set of >=2 primes for (odd/even) n for Riesel case: ** 1369*30^n1 ({7, 13, 19} for odd n) ** (400*88^n1)/3 ({3, 7, 13} for odd n) ** 324*95^n1 ({7, 13, 229} for odd n) ** 93025*498^n1 ({13, 67, 241} for odd n) ** 61009*540^n1 ({17, 1009} for odd n) ** Partial algebra factors with period = 2 for square Sierpinski base: ** 114244*225^n+1 * Partial algebra factors for period > 2: ** (343*10^n1)/9 (period = 3) ** 3511808*63^n+1 (period = 3) ** 27000000*63^n+1 (period = 3) ** (64*847^n1)/9 (period = 3) ** 64*957^n1 (period = 3) ** 2500*13^n+1 (period = 4) ** 2500*55^n+1 (period = 4) ** 16*200^n+1 (period = 4) ** (324*1101^n+1)/5 (period = 4) ** 324*2070^n+1 (period = 4) ** 64*936^n1 (period = 6) * b = q^m, k = q^r, where q is not of the form t^s with odd s>1, and m and r have no common odd prime factor, and the exponent of highest power of 2 dividing r >= the exponent of highest power of 2 dividing m, and the equation 2^x = r (mod m) has no solution ** 8*128^n+1 ** 32*128^n+1 ** 64*128^n+1 ** (27*2187^n+1)/2 ** (243*2187^n+1)/2 ** (729*2187^n+1)/2 ** 64*16384^n+1 ** 1024*16384^n+1 ** 4096*16384^n+1 ** 128*32768^n+1 ** 2048*32768^n+1 ** 8192*32768^n+1 ** 16384*32768^n+1 ** 8*131072^n+1 ** 32*131072^n+1 ** 64*131072^n+1 ** 128*131072^n+1 ** 1024*131072^n+1 ** 2048*131072^n+1 ** 4096*131072^n+1 ** 16384*131072^n+1 But since these cases are very rarely happen, these programs are >99% sufficient Last fiddled with by sweety439 on 20200812 at 16:39 
20200814, 14:09  #943 
Nov 2016
2×13×97 Posts 
A PRP number (k*b^n+1)/gcd(k+1,b1) with large n can be proven be prime if and only if:
either * gcd(k+1,b1) = 1 or * gcd(k+1,b1) = k'+1 (k' is the reduced k, i.e. k' = k/(b^r) which r is largest number such that this number is integer (i.e. r = valuation(k,b) in PARI), if k is not multiple of b (MOB), then k' = k) and the number b^n1 (= prod_(dn)Phi_d(b), where Phi is the cyclotomic polynomial) has >= 33.3333% factored Last fiddled with by sweety439 on 20200814 at 14:10 
20200814, 14:22  #944 
Nov 2016
2×13×97 Posts 
Reupdate the files for all Sierpinski/Riesel conjectures for bases 2<=b<=128 and bases b = 256, 512, 1024
Also see https://github.com/xayahrainie4793/E...elconjectures for these files online. 
20200815, 17:48  #945 
Nov 2016
100111011010_{2} Posts 
Some algebra factors for the k's which is not (perfect power for Riesel case) (perfect odd power or of the form 4*m^4 for Sierpinski case):
* S48 k=36 (n = 1 mod 3 and n = 2 mod 4 are algebraic) * S200 k=5 (n = 1 mod 3 are algebraic) * S200 k=16 (n = 2 mod 4 are algebraic, and make this k have full covering set with partial algebraic factors) * S200 k=40 (n = 1 mod 3 are algebraic) * S529 k=184 (n = 1 mod 3 are algebraic) * S968 k=11 (n = 1 mod 3 are algebraic) * R12 k=27 (n = 1 mod 2 and n = 0 mod 3 are algebraic, and make this k have full covering set with partial algebraic factors) * R12 k=300 (n = 1 mod 2 are algebraic, and make this k have full covering set with partial algebraic factors) * R18 k=50 (n = 1 mod 2 are algebraic) * R40 k=490 (n = 1 mod 2 are algebraic) * R80 k=10 (n = 2 mod 3 are algebraic) * R88 k=3773 (n = 2 mod 3 are algebraic) * R392 k=7 (n = 1 mod 3 are algebraic) * R392 k=56 (n = 1 mod 3 are algebraic) * R432 k=3 (n = 1 mod 2 are algebraic) * R578 k=2 (n = 1 mod 2 are algebraic) * R588 k=3 (n = 1 mod 2 are algebraic) * R750 k=6 (n = 2 mod 3 are algebraic) * R800 k=5 (n = 2 mod 5 are algebraic) * R800 k=8 (n = 1 mod 2 and n = 0 mod 3 are algebraic) * R800 k=25 (n = 0 mod 2 and n = 4 mod 5 are algebraic) * R972 k=3 (n = 1 mod 2 are algebraic) * R972 k=8 (n = 0 mod 3 and n = 1 mod 5 are algebraic) * R1152 k=2 (n = 1 mod 2 are algebraic) Last fiddled with by sweety439 on 20200815 at 17:55 
20200815, 17:55  #946  
Nov 2016
2522_{10} Posts 
Quote:
A kvalue which does not have covering set is proven composite by partial algebraic factors if and only if there is covering set for all nvalues which is not algebraic (e.g. R12 k=25, R12 k=27, R19 k=4, R28 k=175, R30 k=1369, S55 k=2500) Both cases of kvalues are excluded from the conjectures. 

Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Semiprime and nalmost prime candidate for the k's with algebra for the Sierpinski/Riesel problem  sweety439  sweety439  11  20200923 01:42 
The reverse Sierpinski/Riesel problem  sweety439  sweety439  20  20200703 17:22 
The dual Sierpinski/Riesel problem  sweety439  sweety439  12  20171201 21:56 
Sierpinski/ Riesel bases 6 to 18  robert44444uk  Conjectures 'R Us  139  20071217 05:17 
Sierpinski/Riesel Base 10  rogue  Conjectures 'R Us  11  20071217 05:08 