20200710, 17:43  #881  
Nov 2016
4462_{8} Posts 
Quote:
These bases have (odd n has factor of 19 and n == 2 mod 4 has factor of 17): 132, 208, 455, 531, 778, 854, 1101, 1177, 1424, 1500, 1747, 1823, 2070, 2146, 2393, 2469, 2716, 2792, 3039, 3115, 3362, 3438, 3685, 3761, 4008, 4084, 4331, 4407, 4654, 4730, 4977, 5053, 5300, 5376, 5623, 5699, 5946, 6022, 6269, 6345, 6592, 6668, 6915, 6991, 7238, 7314, 7561, 7637, 7884, 7960, 8207, 8283, 8530, 8606, 8853, 8929, 9176, 9252, 9499, 9575, 9822, 9898, ... 

20200710, 18:34  #882 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3^{3}×17×19 Posts 
Stop quoting a previous post in full for no reason.

20200710, 18:47  #883  
Nov 2016
2·11·107 Posts 
Quote:
R58 k=400: since 400 is square, all even n have algebra factors, and we only want to know whether it has a covering set of primes for all odd n, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. nvalue : factors 1 : 11 · 37 3 : 7^2 · 27943 5 : 3 · 23 · 66753803 7 : 61 · 254010257987 9 : 7 · 184913 · 40269069377 11 : 3^2 · 11 · 1627649 · 1088170916957 13 : 947 · 894431 · 696389525163251 15 : 7 · 71 · 7193 · 555058285756249213567 25 : 1171 · 10639 · 450563 · 211297508330411 · 720737916824065571 43 : 9901 · (a 73digit prime) 79 : 257 · 4990187 · 365846287 · (a 123digit prime) and it does not appear to be any covering set of primes, so there must be a prime at some point. R93 k=125: since 125 is cube, all n divisible by 3 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 3, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. nvalue : factors 1 : 2 · 1453 2 : 11 · 24571 4 : 271 · 8626061 5 : 2 · 4201 · 4861 · 5323 7 : 2^4 · 11 · 10683609208231 8 : 109 · 52981 · 30280773239 10 : 1259 · 38851 · 30920860779409 14 : 131381 · 748966512379 · 1149784204819 16 : 431 · 3881 · 834208399 · 701264295413691479 20 : 79 · 394653205936499 · 2347827938794175966096911 34 : 469429 · (a 63digit prime) 40 : 271 · (a 78digit prime) 64 : 751 · 766651 · (a 119digit composite with no known prime factor) 70 : 191 · (a 138digit prime) 74 : 179 · (a 145digit prime) 86 : (a 171digit composite with no known prime factor) and it does not appear to be any covering set of primes, so there must be a prime at some point. R96 k=1681: since 1681 is square, all even n have algebra factors, and we only want to know whether it has a covering set of primes for all odd n, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. nvalue : factors 1 : 5^2 · 1291 3 : 43 · 53 · 130517 5 : 23 · 311 · 383235427 7 : 29 · 23159 · 37616513449 9 : 11117 · 20943593541080351 13 : 67273 · 309220057 · 950638256285203 19 : 164429 · 94139680772968423679537510579981183 25 : 34584287 · 4026877213339 · 87002417657719496636646465818327 33 : 28051 · 77261 · (a 59digit prime) 37 : (a 28digit prime) · (a 49digit prime) 55 : 53 · 191 · (a 108digit prime) 59 : 313 · 11953 · (a 113digit prime) and it does not appear to be any covering set of primes, so there must be a prime at some point. Last fiddled with by sweety439 on 20200710 at 19:48 

20200711, 17:30  #884 
Nov 2016
2·11·107 Posts 

20200711, 19:45  #885  
Nov 2016
100100110010_{2} Posts 
Quote:
k is square and k == 1 mod p and b == 1 mod p for some odd prime p (this situation only exists for p == 1 mod 4) or k is square and k == 2^(r1)+1 mod 2^r and b == 2^(r1)+1 mod 2^r for some r >= 2 (this situation only exists for r >= 4) Then this k proven composite by partial algebraic factors (has algebraic factors (difference of two squares) for even n and divisible by a prime (p or 2, respectively) for odd n) 

20200711, 19:51  #886  
Nov 2016
2×11×107 Posts 
Quote:
if k*b is square and k*b == 1 mod p and b == 1 mod p for some odd prime p (this situation only exists for p == 1 mod 4) or k*b is square and k*b == 2^(r1)+1 mod 2^r and b == 2^(r1)+1 mod 2^r for some r >= 2 (this situation only exists for r >= 4) Then this k proven composite by partial algebraic factors (divisible by a prime (p or 2, respectively) for even n and has algebraic factors (difference of two squares) for odd n) All such examples for bases <= 64 with k below the CK: (square bases are not counted, since they already have full algebra factors for these square k) R12, k = 3*m^2 and m = = 3 or 10 mod 13 (k = 27 and 300 below the CK) R28, k = 7*m^2 and m = = 5 or 24 mod 29 (k = 175 below the CK) R33, k = 33*m^2 and m = = 4 or 13 mod 17 (k = 528 below the CK) R40, k = 10*m^2 and m = = 18 or 23 mod 41 (k = 3240 and 5290 below the CK) R52, k = 13*m^2 and m = = 7 or 46 mod 53 (k = 637 below the CK) R54, k = 6*m^2 and m = = 1 or 4 mod 5 (k = 6 below the CK) R60, k = 15*m^2 and m = = 22 or 39 mod 61 (k = 7260 below the CK) The smallest Riesel base such that the second situation (k*b is square and k*b == 2^(r1)+1 mod 2^r and b == 2^(r1)+1 mod 2^r for some r >= 2 (this situation only exists for r >= 4)) exists for k below the CK is 153 (square bases are not counted, since they already have full algebra factors for these square k), such k is 17 Last fiddled with by sweety439 on 20200712 at 19:13 

20200712, 15:14  #887  
Nov 2016
4462_{8} Posts 
Quote:
For Sierpinski side ((k*b^n+1)/gcd(k+1,b1)), if k is rth power with odd prime r, then all n divisible by r have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by r, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture, and there must be a prime at some point. For Sierpinski side ((k*b^n+1)/gcd(k+1,b1)), if k is of the form 4*m^4, then all n divisible by 4 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 4, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture, and there must be a prime at some point. Last fiddled with by sweety439 on 20200712 at 15:16 

20200712, 15:15  #888  
Nov 2016
2·11·107 Posts 
Quote:
For Riesel side ((k*b^n1)/gcd(k1,b1)), if k*b^s is rth power with prime r, then all n == s mod r have algebra factors, and we only want to know whether it has a covering set of primes for all n not == s mod r, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture, and there must be a prime at some point. For Sierpinski side ((k*b^n+1)/gcd(k+1,b1)), if k*b^s is rth power with odd prime r, then all n == s mod r have algebra factors, and we only want to know whether it has a covering set of primes for all n not == s mod r, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture, and there must be a prime at some point. For Sierpinski side ((k*b^n+1)/gcd(k+1,b1)), if k*b^s is of the form 4*m^4, then all n == s mod 4 have algebra factors, and we only want to know whether it has a covering set of primes for all n not == s mod 4, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture, and there must be a prime at some point. Last fiddled with by sweety439 on 20200712 at 15:17 

20200712, 15:18  #889  
Nov 2016
2·11·107 Posts 
Quote:


20200712, 15:23  #890 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
3^{3}·17·19 Posts 
Why do you insist on quoting whole posts all the time?

20200712, 16:06  #891  
Nov 2016
2·11·107 Posts 
Quote:
S15 k=343: since 343 is cube, all n divisible by 3 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 3, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. nvalue : factors 1 : 31 · 83 2 : 2^2 · 11 · 877 4 : 2^2 · 809 · 2683 5 : 811 · 160583 7 : 11^2 · 242168453 11 : 31 · 101 · 25357 · 18684739 13 : 397 · 1281101 · 656261029 17 : 11 · 27479311 · 55900668804553 29 : 53 · 197741 · 209188613429183386499227445981 35 : 1337724923 · 18667724069720862256321575167267431 43 : 20943991 · 3055827403675875709696160949928034201885723243 61 : 23539 · (a 61digit prime) and it does not appear to be any covering set of primes, so there must be a prime at some point. S61 k=324: since 324 is of the form 4*m^4, all n divisible by 4 have algebra factors, and we only want to know whether it has a covering set of primes for all n not divisible by 4, if so, then this k makes a full covering set with algebraic factors and be excluded from the conjecture; if not, then this k does not make a full covering set with algebraic factors and be included from the conjecture. nvalue : factors 1 : 59 · 67 2 : 41 · 5881 3 : 13 · 1131413 5 : 5 · 7 · 1563709723 6 : 13 · 256809250661 7 : 23 · 1255679 · 7051433 13 : 191 · 7860337 · 27268229 · 256289843 14 : 1540873 · 1698953 · 244480646906833 31 : 1888149043321 · 441337391577139 · 1721840403480692512106884569347 34 : 10601 · 174221 · (a 54digit prime) and it does not appear to be any covering set of primes, so there must be a prime at some point. 

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