20170121, 11:01  #1 
Nov 2016
2^{2}×3×5×47 Posts 
The dual Sierpinski/Riesel problem
There is a "Five or bust!" project, to search the primes of the form 2^n+k for all odd numbers k<78557, this is because 2^n+k is the dual of k*2^n+1. For k*2^n1, the dual of it is 2^nk.
The project found the probable primes 2^1518191+75353, 2^2249255+28433, 2^4583176+2131, 2^5146295+41693, and 2^9092392+40291. (see http://mersenneforum.org/showthread.php?t=10761) There is a "mixed Sierpinski theorem", which is that for all odd numbers k<78557, there is a prime of the form either k*2^n+1 or 2^n+k. At past, there was 3 k's < 78557 such that there is neither known prime of the form k*2^n+1 nor known prime of the form 2^n+k, namely 19249, 28433 and 67607 (for 67607, there was only probable prime known at that time, this probable prime is 2^16389+67607, but this number is now proven prime). Several years ago, a probable prime 2^551542+19249 was found, since it is only a probable prime and not a proven prime, we can not actually say that 19249 can be removed from the list. However, people were looking for prime for the only remaining number, 28433, and found the prime 28433*2^7830457+1, since it is a proven prime, we can remove 28433 from the list. In March 2007, the prime 19249*2^13018586+1 was found, and "the mixed Sierpinski problem" become a theorem as soon as this prime was found. (The smallest odd k such that there is no known prime of the form k*2^n+1 is 21181, and the smallest odd k such that there is no known proven prime of the form 2^n+k is 2131, it has only probable prime known: 2^4583176+2131, all k's < 78557 has a known (probable) prime of the form 2^n+k) For the mixed Riesel problem, there are still 5 odd k's < 509203 (these k's are 2293, 342847, 344759, 386801, 444637) with neither known prime of the form k*2^n1 nor known (probable) prime of the form 2^nk, the first such k is 2293. (note that 2293 is also the smallest odd k such that there is no known prime of the form k*2^n1, 2293 is also the smallest odd k such that there is no known (probable) prime of the form 2^nk, and all k's < 2293 has a known proven prime of the form 2^nk) For k = 363343 and 384539, only probable prime of the form 2^nk is known, the probable primes are 2^13957363343 and 2^32672384539. See http://mersenneforum.org/showthread.php?t=10754 for the dual Sierpinski problem and see http://mersenneforum.org/showthread.php?t=6545 for the dual Riesel problem. For generalized Sierpinski/Riesel conjecture to base b, the dual of k*b^n+1 is (b^n+k)/gcd(b^n, k), and the dual of k*b^n1 is (b^nk)/gcd(b^n, k). These are the dual (probable) primes that I found of the Sierpinski/Riesel conjectures that has only one k remaining for bases b<=144: (See http://mersenneforum.org/showpost.ph...42&postcount=1) Code:
form dual form least prime for the "dual form" dual n 2036*9^n+1 9^n+2036 9^4+2036 4 7666*10^n+1 (10^n+7666)/2 (10^67+7666)/2 67 244*17^n+1 17^n+244 17^838+244 838 5128*22^n+1 (22^n+5128)/8 (22^11+5128)/8 11 398*27^n+1 27^n+398 27^7+398 7 166*43^n+1 43^n+166 ? (>1000) 17*68^n+1 (68^n+17)/17 (68^1+17)/17 1 1312*75^n+1 75^n+1312 ? (>1000) 8*86^n+1 (86^n+8)/8 (86^205+8)/8 205 32*87^n+1 87^n+32 ? (>1000) 1696*112^n+1 (112^n+1696)/32 (112^44+1696)/32 44 48*118^n+1 (118^n+48)/16 (118^57+48)/16 57 34*122^n+1 (122^n+34)/2 (122^2+34)/2 2 40*128^n+1 (128^n+40)/8 (128^2+40)/8 2 Code:
form dual form least prime for the "dual form" dual n 1597*6^n1 6^n1597 6^31597 3 4421*10^n1 10^n4421 10^2124421 212 3656*22^n1 (22^n3656)/8 ? (>1000) 404*23^n1 23^n404 23^568404 568 706*27^n1 27^n706 27^2706 2 424*93^n1 93^n424 93^1424 1 29*94^n1 94^n29 94^229 2 924*103^n1 103^n924 103^1924 1 84*109^n1 109^n84 109^684 6 24*123^n1 (123^n24)/3 (123^524)/3 5 926*133^n1 133^n926 133^2926 2 Last fiddled with by sweety439 on 20170121 at 12:18 
20170421, 16:07  #2 
Nov 2016
2^{2}·3·5·47 Posts 
The n's such that the dual forms are (probable) primes:
9^n+2036: 4, 208, ... (10^n+7666)/2: 67, 103, ... 17^n+244: 838, ... (22^n+5128)/8: 11, 63, 519, ... 27^n+398: 7, 375, ... 43^n+166: (no known such n) (68^n+17)/17: 1, 7, ... 75^n+1312: (no known such n) (86^n+8)/8: 205, ... 87^n+32: (no known such n) (112^n+1696)/32: 44, ... (118^n+48)/16: 57, ... (122^n+34)/2: 2, 98, ... (128^n+40)/8: 2, 8, ... 6^n1597: 3, 1731, .,. 10^n4421: 212, 284, ... (22^n3656)/8: (no known such n) 23^n404: 568, ... 27^n706: 2, 10, 786, ... 93^n424: 1, 133, 151, 397, ... 94^n29: 2, ... 103^n924: 1, 97, ... 109^n84: 6, 18, 20, 362, ... (123^n24)/3: 5, 84, ... 133^n926: 2, 111, 155, ... 
20170421, 16:28  #3 
Nov 2016
2^{2}×3×5×47 Posts 
Thus, the "mix Sierpinski/Riesel problem" of these bases are proven except S43, S75, S87 and R22.

20170520, 15:41  #4 
Nov 2016
2^{2}×3×5×47 Posts 
For generalized Sierpinski/Riesel conjecture to base b, the dual of k*b^n+1 is (b^n+k)/d, where d is the largest number that divides b^n+k for all enough large n, and the dual of k*b^n1 is (b^nk)/d, where d is the largest number that divides b^nk for all enough large n. If this number is not a integer, then we choose its numerator (this number is always a rational number :) )
Last fiddled with by sweety439 on 20170520 at 15:42 
20170601, 14:05  #5  
Nov 2016
5404_{8} Posts 
Quote:
9^n+2036 is already standard (10^n+7666)/2 > 5*10^n+3833 (this n turns to be n1) 17^n+244 is already standard (22^n+5128)/8 > 1331*22^n+641 (this n turns to be n3) 27^n+398 is already standard 43^n+166 is already standard (68^n+17)/17 > 4*68^n+1 (this n turns to be n1) 75^n+1312 is already standard (86^n+8)/8 > 79507*86^n+1 (this n turns to be n3) 87^n+32 is already standard (112^n+1696)/32 > 392*112^n+53 (this n turns to be n2) (118^n+48)/16 > 12117361*118^n+3 (this n turns to be n4) (122^n+34)/2 > 61*122^n+17 (this n turns to be n1) (128^n+40)/8 > 16*128^n+5 (this n turns to be n1) 6^n1597 is already standard 10^n4421 is already standard (22^n3656)/8 > 1331*22^n457 (this n turns to be n3) 23^n404 is already standard 27^n706 is already standard 93^n424 is already standard 94^n29 is already standard 103^n924 is already standard 109^n84 is already standard (123^n24)/3 > 41*123^n8 (this n turns to be n1) 133^n926 is already standard (now, we allow negative primes, such as 5 and 47, thus we can omit the absolute value sign) Last fiddled with by sweety439 on 20170601 at 14:06 

20171011, 17:35  #6 
Nov 2016
2^{2}·3·5·47 Posts 
This file is for the dual Riesel problem for k=2 with bases 2<=b<=400.
The formula is (b^n2)/gcd(b,2) There are 7 bases remain: 278, 296, 305, 338, 353, 386, 397. The original primes for these bases are: Code:
2*278^439081 2*296^361 2*305^21 2*338^121 2*353^21 2*386^21 2*397^181 The dual primes for the remain bases 2<=b<=1024 are: Code:
581^182 (992^902)/2 1019^42 Last fiddled with by sweety439 on 20171018 at 03:12 
20171013, 13:18  #7 
Nov 2016
2^{2}·3·5·47 Posts 
Also update the dual Sierpinski k=2 file, also for bases 2<=b<=400. (in fact, this is "dual extended Sierpinski")
The formula is (b^n+2)/gcd(b,2)/gcd(b1,3) Note: b=128 has no possible prime. Also see the OEIS sequence A138066 Last fiddled with by sweety439 on 20171018 at 03:03 
20171015, 14:06  #9 
Nov 2016
2^{2}·3·5·47 Posts 
These are files for the dual Sierpinski/Riesel problem base 3. (for k<=1024, k even, k not divisible by 3)
i.e. least n>=1 such that 3^n+k or 3^nk is prime. Last fiddled with by sweety439 on 20171018 at 02:59 
20171101, 18:48  #10 
Nov 2016
B04_{16} Posts 
305^n2 is composite for all n<=10000.
File attached. 
20171125, 20:35  #11 
Nov 2016
2^{2}×3×5×47 Posts 

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