20200903, 08:33  #969 
Nov 2016
7·11·31 Posts 
2 (probable) primes found for R70:
(376*70^64841)/3 (496*70^49341)/3 k=811 still remains .... 
20200903, 21:37  #970 
Nov 2016
7×11×31 Posts 
https://docs.google.com/document/d/e...Ns5LlL_FA0/pub
Update newest file for Sierpinski problems to include the top 10 (probable) primes for S78, S96, and S126 
20200904, 22:43  #971 
Nov 2016
7·11·31 Posts 
https://docs.google.com/document/d/e...MpCV9458Wi/pub
Update newest file for Sierpinski conjectures, according to GFN2 and GFN10, the test limit of S512 k=2 is (2^541)/91 = 2001599834386886, and the test limit of S10 k=100 is 2^313 = 2147483645 
20200904, 23:03  #972  
Nov 2016
7·11·31 Posts 
Quote:
k=106 R7: (k=197 and 367 are only probable primes) k=79 k=139 (proven by N1method) (certificate for large prime factor for N1) k=159 k=299 k=313 k=391 k=419 k=429 k=437 k=451 R10: k=121 R12: k=298 R17: k=13 k=29 R26: k=121 R31: k=21 k=39 (proven by N1method) (certificate for large prime factor for N1) k=49 k=113 k=115 k=123 k=124 R33: k=213 R35: k=1 (proven by N1method) R37: k=5 (proven by N1method) R39: k=1 (proven by N1method) R43: k=4 (proven by N1method) R45: k=53 Last fiddled with by sweety439 on 20200905 at 12:03 

20200905, 20:17  #973 
Nov 2016
7×11×31 Posts 
Last fiddled with by sweety439 on 20200905 at 20:18 
20200906, 19:49  #974 
Nov 2016
7×11×31 Posts 
Also reserved R88 and found these (probable) primes:
(49*88^22231)/3 (79*88^76651)/3 (235*88^13301)/3 (346*88^29691)/3 (541*88^11871)/3 (544*88^89041)/3 k = 46, 94, 277, 508 are still remaining. 
20200906, 20:13  #975 
Nov 2016
2387_{10} Posts 
Still no (probable) prime found for R70 k=811
If a (probable) prime for R70 k=811 were found, then will produce a group of 11 consecutive proven Riesel conjectures: R67 to R77 
20200906, 23:53  #976  
Nov 2016
7·11·31 Posts 
Quote:
k=100 k=121 k=142 k=256 k=386 R49: k=79 R51: k=1 R57: k=87 (proven by N1method) (certificate for large prime factor for N1) R58: (k=382, 400, and 421 are only probable primes) k=4 (proven by N1method) k=103 k=109 k=142 k=163 k=217 k=271 k=334 k=361 k=379 k=445 k=457 k=487 R61: (k=13 is only probable prime) k=10 k=41 k=77 Last fiddled with by sweety439 on 20200906 at 23:56 

20200907, 00:27  #977 
Nov 2016
7·11·31 Posts 
Update newest file for Riesel problems to include recent status for R70 and R88

20200913, 12:37  #978 
Nov 2016
7×11×31 Posts 
I conjectured that:
If (k,b,c) is integer triple, k>=1, b>=2, c != 0, gcd(k,c)=1, gcd(b,c)=1, if (k*b^n+c)/gcd(k+c,b1) does not have covering set of primes for the n such that: (the set of the n satisfying these conditions must be nonempty, or (k*b^n+c)/gcd(k+c,b1) proven composite by full algebra factors) * if c != +1, (let r be the largest integer such that (c) is perfect rth power) k*b^n is not perfect rth power * if c = 1, k*b^n is not perfect odd power (of the form m^r with odd r>1), except the case: 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 * if c = 1, k*b^n is not perfect power (of the form m^r with r>1), except the case: 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 * 4*k*b^n*c is not perfect 4th power Then there are infinitely many primes of the form (k*b^n+c)/gcd(k+c,b1), except the case: 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. Last fiddled with by sweety439 on 20200913 at 12:39 
20200917, 19:28  #979  
Nov 2016
2387_{10} Posts 
Quote:
nvalue : factors 2 : 3^3 · 9679 4 : 43 · 79 · 83 · 1483 6 : 881 · 759379493 8 : 3 · 356807111111111 10 : 31 · 67883 · 813864335521 12 : 53 · 51703370062893081761 18 : 163 · 68860007363271983640081799591 22 : 4801 · 23279 · 3561827 · 4036715519 · 17881240410679 28 : 210323 · 6302441 · 88788971627962097615055082730651231 30 : 38270136643 · 4920560231486977484668641122451121981831 and it does not appear to be any covering set of primes, so there must be a prime at some point. R40 also has two special remain k: 520 and 11560, 520 = 13 * base, 11560 = 289 * base, and the further searching for k = 11560 is k = 289 with odd n > 1 (since 289 is square, all even n for k = 289 have algebra factors) Another base is R106, which has many k with algebra factors (these k are all squares): 64 = 2^6 (thus, all n == 0 mod 2 and all n == 0 mod 3 have algebra factors) 81 = 3^4 (thus, all n == 0 mod 2 have algebra factors) 400 = 20^2 (thus, all n == 0 mod 2 have algebra factors) 676 = 26^2 (thus, all n == 0 mod 2 have algebra factors) 841 = 29^2 (thus, all n == 0 mod 2 have algebra factors) 1024 = 2^10 (thus, all n == 0 mod 2 and all n == 0 mod 5 have algebra factors) We should check whether they have covering set for the n which do not have algebra factors, like the case for R30 k=1369 and R88 k=400 Last fiddled with by sweety439 on 20200918 at 19:18 

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