20200423, 19:13  #34  
Nov 2016
1932_{10} Posts 
Quote:
Generally, if the number pair {k,b} (k>=1, b>=2) such that there exists an n (n can be 0, n also can be negative integer) such that all prime factors of "the numerator of the absolute value of (k*b^n+1)/gcd(k+1,b1) (or (k*b^n1)/gcd(k1,b1))" are also prime factors of k*b, then (k*b^n+1)/gcd(k+1,b1) (or (k*b^n1)/gcd(k1,b1)) cannot have covering set. If (k*b^n+1)/gcd(k+1,b1) (or (k*b^n1)/gcd(k1,b1)) have algebra factors, then there might be no primes of the form (k*b^n+1)/gcd(k+1,b1) (or (k*b^n1)/gcd(k1,b1)) even have it has no covering set, however, if there is an n (n can be 0, n also can be negative integer) such that "k*b^n is not perfect power of rational number" (for (k*b^n1)/gcd(k1,b1)) or "k*b^n is neither perfect odd power of rational number nor of the form 4*m^4 with rational number m" (for (k*b^n+1)/gcd(k+1,b1)) and all prime factors of "the numerator of the absolute value of (k*b^n+1)/gcd(k+1,b1) (or (k*b^n1)/gcd(k1,b1))" are also prime factors of k*b, then we believe it will eventually have a prime (and there is conjectured to be infinitely many primes of this form), but it is an unsolved problem. For example, for (4*b^n1)/gcd(41,b1), (9*b^n1)/gcd(91,b1), and (16*b^n1)/gcd(161,b1) for nonsquare b, if there is an (positive or negative) odd number n such that all prime factors of "the numerator of the absolute value of (k*b^n1)/gcd(k1,b1)" are also prime factors of k*b, then we believe they will eventually have a prime (and there is conjectured to be infinitely many primes of this form) the case (4*b^n+1)/gcd(4+1,b1), we require the n not divisible by 4, but (9*b^n+1)/gcd(9+1,b1) and (16*b^n+1)/gcd(16+1,b1) does not have any requirement of n, besides, the cases (8*b^n+1)/gcd(8+1,b1) and (8*b^n1)/gcd(81,b1) require n not divisible by 3, (64*b^n1)/gcd(641,b1) require n divisible by neither 2 nor 3, (64*b^n+1)/gcd(64+1,b1) require n divisible by neither 3 nor 4, etc. See post https://mersenneforum.org/showpost.p...&postcount=675 and https://mersenneforum.org/showpost.p...&postcount=353 for more information. Last fiddled with by sweety439 on 20200423 at 19:23 

20200423, 19:14  #35  
Nov 2016
2^{2}·3·7·23 Posts 
Quote:
3*2^n1 (the dual form 2^n3, for n=1, the value is 1) 5*2^n1 (the dual form 2^n5, for n=2, the value is 1) 9*2^n1 (the dual form 2^n9, for n=3, the value is 1) 17*2^n1 (the dual form 2^n17, for n=4, the value is 1) 33*2^n1 (the dual form 2^n33, for n=5, the value is 1) etc. 

20200423, 19:28  #36  
Nov 2016
2^{2}×3×7×23 Posts 
Quote:
2*581^n1 (the value of n=0 is 1, and all prime factors of 1 are also prime factors of 2*581) 3*718^n+1 (the value of n=0 is 4, and all prime factors of 4 are also prime factors of 3*718) 294*213^n1 (the value of n=1 is 27/71, and all prime factors of 27 are also prime factors of 294*213) 122*123^n+1 (the value of n=0 is 123, and all prime factors of 123 are also prime factors of 122*123) 267*268^n1 (the value of n=1 is 1/268, and all prime factors of 1 are also prime factors of 267*268) 106*214^n+1 (the value of n=0 is 107, and all prime factors of 107 are also prime factors of 106*214) 859*430^n+1 (the value of n=0 is 860, and all prime factors of 860 are also prime factors of 859*430) 354*352^n1 (the value of n=1 is 1/176, and all prime factors of 1 are also prime factors of 354*352) etc. 

20200506, 15:31  #37  
Nov 2016
3614_{8} Posts 
Quote:


20200506, 16:54  #38 
"Curtis"
Feb 2005
Riverside, CA
47×89 Posts 

20200507, 12:34  #39 
Nov 2016
2^{2}×3×7×23 Posts 

20200507, 15:56  #40 
Nov 2016
2^{2}·3·7·23 Posts 
R2080 found CK=83320516600067570, also using exponent=144
I know that, in th conjecture of R2080, all k where k = m^2 and m = = 102 or 1979 mod 2081, and all k where k = 65*m^2 and m == 316 or 1765 mod 2081, are proven composite by partial algebra factors. 
20200528, 09:54  #41 
Nov 2016
2^{2}·3·7·23 Posts 
Found the CK for bases 20492200, 0 if the CK is > 5M, now I am finding them.

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