 mersenneforum.org Conjectured K question
 Register FAQ Search Today's Posts Mark Forums Read  2020-04-23, 19:13   #34
sweety439

Nov 2016

2×13×79 Posts Quote:
 Originally Posted by sweety439 9*1024^n-1 is unlikely to have a covering set, like the simple cases: 3*2^n+-1, 5*2^n+-1, 7*2^n+-1, 9*2^n+-1, 11*2^n+-1, etc. they are unlikely to have a covering set, but also no proof.
In fact, 9*1024^n-1 cannot have a covering set, since for n=0, the value of 9*1024^n-1 is 8, and the only prime factor of 8 is 2, thus if 9*1024^n-1 have a covering set, then the prime 2 must be in the covering set, however, 9*1024^n-1 cannot be divisible by 2 for any n>=1, a contradiction!

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,b-1) (or (k*b^n-1)/gcd(k-1,b-1))" are also prime factors of k*b, then (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1)) cannot have covering set.

If (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1)) have algebra factors, then there might be no primes of the form (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1)) 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^n-1)/gcd(k-1,b-1)) 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,b-1)) and all prime factors of "the numerator of the absolute value of (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1))" 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^n-1)/gcd(4-1,b-1), (9*b^n-1)/gcd(9-1,b-1), and (16*b^n-1)/gcd(16-1,b-1) 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^n-1)/gcd(k-1,b-1)" 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,b-1), we require the n not divisible by 4, but (9*b^n+1)/gcd(9+1,b-1) and (16*b^n+1)/gcd(16+1,b-1) does not have any requirement of n, besides, the cases (8*b^n+1)/gcd(8+1,b-1) and (8*b^n-1)/gcd(8-1,b-1) require n not divisible by 3, (64*b^n-1)/gcd(64-1,b-1) require n divisible by neither 2 nor 3, (64*b^n+1)/gcd(64+1,b-1) require n divisible by neither 3 nor 4, etc.

Last fiddled with by sweety439 on 2020-04-23 at 19:23   2020-04-23, 19:14   #35
sweety439

Nov 2016

2·13·79 Posts Quote:
 Originally Posted by sweety439 Well... since all prime factors of k*2^n+-1 are odd, thus if there exist an n such that k*2^n+-1 or its dual (|2^n+-k|) is power of 2 (including 1), then k*2^n+-1 cannot have a covering set. Thus these k*2^n+-1 cannot have a covering set: 1*2^n+1 (for n=0, the value is 2) 3*2^n+1 (for n=0, the value is 4) 7*2^n+1 (for n=0, the value is 8) 15*2^n+1 (for n=0, the value is 16) 31*2^n+1 (for n=0, the values is 32) 1*2^n-1 (the dual form |2^n-1|, for n=1, the value is 1) 3*2^n-1 (for n=0, the value is 2, also the dual form |2^n-3|, for n=2, the value is 1) 5*2^n-1 (for n=0, the value is 4) 7*2^n-1 (the dual form |2^n-7|, for n=3, the value is 1) 9*2^n-1 (for n=0, the value is 8) 15*2^n-1 (the dual form |2^n-15|, for n=4, the value is 1) 17*2^n-1 (for n=0, the value is 16) 31*2^n-1 (the dual form |2^n-31|, for n=5, the value is 1) 33*2^n-1 (for n=0, the value is 32) etc. However, there is no proof that 5*2^n+1, 9*2^n+1, 11*2^n+-1, 13*2^n+-1, 17*2^n+1, ... have no covering set (i.e. the sequence of the smallest prime factor of k*2^n+-1 is unbounded above).
Also,

3*2^n-1 (the dual form |2^n-3|, for n=1, the value is 1)
5*2^n-1 (the dual form |2^n-5|, for n=2, the value is 1)
9*2^n-1 (the dual form |2^n-9|, for n=3, the value is 1)
17*2^n-1 (the dual form |2^n-17|, for n=4, the value is 1)
33*2^n-1 (the dual form |2^n-33|, for n=5, the value is 1)
etc.   2020-04-23, 19:28   #36
sweety439

Nov 2016

2×13×79 Posts Quote:
 Originally Posted by sweety439 In fact, 9*1024^n-1 cannot have a covering set, since for n=0, the value of 9*1024^n-1 is 8, and the only prime factor of 8 is 2, thus if 9*1024^n-1 have a covering set, then the prime 2 must be in the covering set, however, 9*1024^n-1 cannot be divisible by 2 for any n>=1, a contradiction! 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,b-1) (or (k*b^n-1)/gcd(k-1,b-1))" are also prime factors of k*b^n, then (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1)) cannot have covering set. If (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1)) have algebra factors, then there might be no primes of the form (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1)) 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^n-1)/gcd(k-1,b-1)) 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,b-1)) and all prime factors of "the numerator of the absolute value of (k*b^n+1)/gcd(k+1,b-1) (or (k*b^n-1)/gcd(k-1,b-1))" are also prime factors of k*b^n, 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^n-1)/gcd(4-1,b-1), (9*b^n-1)/gcd(9-1,b-1), and (16*b^n-1)/gcd(16-1,b-1) 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^n-1)/gcd(k-1,b-1)" are also prime factors of k*b^n, 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,b-1), we require the n not divisible by 4, but (9*b^n+1)/gcd(9+1,b-1) and (16*b^n+1)/gcd(16+1,b-1) does not have any requirement of n, besides, the cases (8*b^n+1)/gcd(8+1,b-1) and (8*b^n-1)/gcd(8-1,b-1) require n not divisible by 3, (64*b^n-1)/gcd(64-1,b-1) require n divisible by neither 2 nor 3, (64*b^n+1)/gcd(64+1,b-1) 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.
Some remain k's in CRUS are proven to have no covering set: (and for these cases, there is no n such that k*b^n is perfect power, thus these k*b^n+-1 cannot have any algebra factors)

2*581^n-1 (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^n-1 (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^n-1 (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^n-1 (the value of n=-1 is 1/176, and all prime factors of 1 are also prime factors of 354*352)

etc.   2020-05-06, 15:31   #37
sweety439

Nov 2016

2×13×79 Posts Quote:
 Originally Posted by NHoodMath Just as a note so if dannyridel is interested, use covering.exe for small CK (say k<2^32) and bigcovering.exe for larger CK than 2^32. There is a bug in the covering.exe program for k's greater than some upper bound, I don't know what that bound was, I just remember it exists. (Side note, I know people back in the early days of CK searching were interested to see if there were any CK > base 280's for bases >2048. Didn't take much searching, b=2080 has a much higher CK than 280, about 10x greater. That's a base that'll be a lot of fun for the annual base searchers to look at in 60 years LOL!)
Well, what are the CK for R2080 and S2080? I am curious about it.   2020-05-06, 16:54   #38
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

11·383 Posts Quote:
 Originally Posted by sweety439 Well, what are the CK for R2080 and S2080? I am curious about it.
Then run the programs and find them!   2020-05-07, 12:34   #39
sweety439

Nov 2016

80616 Posts Quote:
 Originally Posted by VBCurtis Then run the programs and find them!
Is (the CK for S2080) 13836346724406967? I use exponent 144, however there may be a bug, like SR280 using exponent 144

Of course I ran bigcover.exe, not cover.exe   2020-05-07, 15:56 #40 sweety439   Nov 2016 2·13·79 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.   2020-05-28, 09:54   #41
sweety439

Nov 2016

1000000001102 Posts Found the CK for bases 2049-2200, 0 if the CK is > 5M, now I am finding them.
Attached Files Sierpinski CK bases 2049-2200.txt (1.5 KB, 10 views) Riesel CK bases 2049-2200.txt (1.5 KB, 10 views)   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post T.Rex Miscellaneous Math 27 2015-10-16 19:02 primus Miscellaneous Math 1 2014-10-12 09:25 primus Computer Science & Computational Number Theory 16 2014-08-15 01:15 T.Rex Math 75 2007-09-04 07:53

All times are UTC. The time now is 23:44.

Thu Jul 9 23:44:46 UTC 2020 up 106 days, 21:17, 0 users, load averages: 1.86, 1.72, 1.62