20161224, 19:07  #34 
Nov 2016
5404_{8} Posts 
The correct text file for extended R17 is here. (the above file only lists the k's < 44, we should lists all k's < 49)

20161224, 19:45  #35 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
10010010101101_{2} Posts 
You can surely post 10 messages an hour. You have proven your capacity to type.
You come across as a graphomaniac. But do you think anybody is reading your inane stream of conscience? Why would they  if you are appear to be deaf to other people's messages? 
20161226, 19:26  #36 
Nov 2016
2^{2}×3×5×47 Posts 
The (probable) primes with n > 1000 for the extended Sierpinski/Riesel problems (with bases b <= 32, except b = 2, 3, 6, 15, 22, 24, 28, 30) are:
S4: 186*4^10458+1 S7: (141*7^1044+1)/2 S10: 804*10^5470+1 S12: 404*12^714558+1 378*12^2388+1 S16: (23*16^1074+1)/3 S17: 10*17^1356+1 S18: 122*18^292318+1 381*18^24108+1 291*18^2415+1 S25: (61*25^3104+1)/2 S26: 32*26^318071+1 217*26^11454+1 95*26^1683+1 178*26^1154+1 R4: (106*4^45531)/3 74*4^12761 296*4^12751 ( = 74*4^12761, the two primes are the same, since 296 = 74 * 4) R7: (367*7^151181)/6 (313*7^59071)/6 (159*7^48961)/2 (429*7^38151)/2 (419*7^10521)/2 R12: (298*12^16761)/11 R17: 44*17^64881 (29*17^49041)/4 (13*17^11231)/4 R25: 86*25^10291 R26: 115*26^5202771 32*26^98121 (121*26^15091)/5 Last fiddled with by sweety439 on 20170207 at 15:02 
20161226, 19:34  #37 
Nov 2016
2^{2}·3·5·47 Posts 
It should not be "n>1", it should be "n>=1", we allow n=1 for (k*b^n+1)/gcd(k+1, b1) and (k*b^n1)/gcd(k1, b1), but we do not allow n=0.
Last fiddled with by sweety439 on 20161226 at 19:35 
20161226, 19:41  #38  
Nov 2016
2^{2}·3·5·47 Posts 
Quote:
Thus, in this problem, we say "k=1617 is remaining for the extended Riesel base 3 problem", but in that problem, we say "k=806 is remaining for the base 3 d=1 problem". 

20161227, 16:48  #39 
Nov 2016
2^{2}·3·5·47 Posts 
These problems are just my extending for the original Sierpinski/Riesel problems.
Last fiddled with by sweety439 on 20161227 at 16:48 
20161227, 16:49  #40 
Nov 2016
B04_{16} Posts 
For (269*10^n+1)/9, tested to n=6000, still no (probable) prime found.

20161228, 16:34  #41 
Nov 2016
2^{2}×3×5×47 Posts 
For the two bases that have only one k remaining:
S10, k=269: (269*10^n+1)/9 seems to be tested to n=10000 (I ran the program of 25 hours and 38 minutes!!!), still no (probable) prime found, base released. R7, k=197: See the link: https://www.rosehulman.edu/~rickert/Compositeseq/, (197*7^n1)/2 is already tested to n=15000 with no (probable) prime found. Last fiddled with by sweety439 on 20161228 at 16:35 
20161228, 16:43  #42 
Nov 2016
2^{2}×3×5×47 Posts 
Also tested extended S15, S22 and S24 to k=500.
For extended S15, there are 4 k's <= 500 remain: 219, 225, 341, 343. For extended S22, there are 5 k's <= 500 (excluding the GFNs: i.e. k=22 and k=484) remain: 173, 346, 383, 461, 464. For extended S24, there are 5 k's <= 500 remain: 319, 346, 381, 461, 486. However, all of them are correspond to CRUS primes, since all these k's satisfy that gcd(k+1,241)=1, and according to the CRUS page, the smallest remain k for the original S24 problem is k=656. Thus, there are in fact no k's <= 500 remain for extended S24. Last fiddled with by sweety439 on 20161228 at 17:26 
20161228, 16:51  #43 
Nov 2016
2^{2}·3·5·47 Posts 
If the k satisfies that gcd(k+1, b1)=1, then the correspond prime of the extended Sierpinski/Riesel problem is the same as the correspond prime of the original Sierpinski/Riesel problem.
The extended Sierpinski/Riesel problem is just my extending of the original Sierpinski/Riesel problem to the k's such that gcd(k+1, b1) is not 1. Since k*b^n+1 is always divisible by gcd(k+1, b1), the division is necessary for there to be any chance of finding primes. Last fiddled with by sweety439 on 20161228 at 16:53 
20161228, 17:20  #44 
Nov 2016
B04_{16} Posts 
Also tested extended R15, R22 and R24 to k=500.
For extended R15, there are 6 k's <= 500 remain: 47, 203, 239, 407, 437, 451. For extended R22, there are 4 k's <= 500 (k=185 prime is given by the CRUS page) remain: 208, 211, 355, 436. For extended R24, there are 7 k's <= 500 remain: 69, 201, 339, 346, 364, 389, 461. However, the k's != 1 mod 23 are correspond to CRUS primes, according to the CRUS page, k=389 is the only k <= 500 remain for the original R24 problem. Last fiddled with by sweety439 on 20161228 at 17:22 
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