20200703, 04:33  #848 
Nov 2016
8C8_{16} Posts 

20200703, 04:49  #849  
Nov 2016
2^{3}·281 Posts 
Quote:
S16: see post #463, already at n=15K S38: reserved by Prime Grid's GFN primes search, already at n=2^241 S50: reserved by Prime Grid's GFN primes search, already at n=2^241 R10: reserved by http://www.worldofnumbers.com/em197.htm (case d=3, k=817) and https://www.rosehulman.edu/~rickert/Compositeseq/ (case b=10, d=3, k=817), already at n=554789 R12: see post #664, already at n=21760 R32: reserved by CRUS (case R1024, k=29), already at n=500K R49: reserved by https://github.com/RaymondDevillers/primes (see the "left49" file) (case b=49 family R{G}), already at n=10K 

20200703, 04:53  #850 
Nov 2016
2^{3}×281 Posts 
Conjecture 1 (the strong Sierpinski conjecture): For b>=2, k>=1, if there is an n such that:
(1) k*b^n is neither a perfect odd power (i.e. k*b^n is not of the form m^r with odd r>1) nor of the form 4*m^4. (2) gcd((k*b^n+1)/gcd(k+1,b1),(b^(945*2^s)1)/(b1)) = 1 for all s, i.e. for all s, every prime factor of (k*b^n+1)/gcd(k+1,b1) does not divide (b^(945*2^s)1)/(b1). (i.e. ord_p(b) is not of the form 2^r (r>=0 if p = 2 or p = 3, r>=1 if p>=5) or m*2^r (m divides 945, r>=0) for every prime factor p of (k*b^n+1)/gcd(k+1,b1)). (3) this (k,b) pair is not 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. (the first 6 Sierpinski bases with k's which are this case are 128, 2187, 16384, 32768, 78125 and 131072) Then there are infinitely many primes of the form (k*b^n+1)/gcd(k+1,b1). Conjecture 2 (the strong Riesel conjecture): For b>=2, k>=1, if there is an n such that: (1) k*b^n is not a perfect power (i.e. k*b^n is not of the form m^r with r>1). (2) gcd((k*b^n1)/gcd(k1,b1),(b^(945*2^s)1)/(b1)) = 1 for all s, i.e. for all s, every prime factor of (k*b^n1)/gcd(k1,b1) does not divide (b^(945*2^s)1)/(b1). (i.e. ord_p(b) is not of the form 2^r (r>=0 if p = 2 or p = 3, r>=1 if p>=5) or m*2^r (m divides 945, r>=0) for every prime factor p of (k*b^n1)/gcd(k1,b1)). Then there are infinitely many primes of the form (k*b^n1)/gcd(k1,b1). 
20200703, 09:44  #851 
Nov 2016
2248_{10} Posts 
(33*27^7876+1)/2 is (probable) prime
the first 4 conjectures of S27 are all proven!!! reserve R57 Last fiddled with by sweety439 on 20200703 at 10:07 
20200703, 12:53  #852 
Nov 2016
2248_{10} Posts 
(281*57^56101)/56 is (probable) prime
the first 4 conjectures of R57 are all proven!!! reserve R49 (the corresponding page only searched it to 10K, I double check it and reserve it for n>10K) Last fiddled with by sweety439 on 20200703 at 13:48 
20200704, 04:53  #853 
Nov 2016
2^{3}×281 Posts 
k is Sierpinski number base b if....
Code:
k b 1 (none) 2 (no such b < 201446503145165177) 3 (no such b < 158503) 4 == 14 mod 15 5 == 11 mod 12 6 == 34 mod 35 7 == 5 mod 24 or == 11 mod 12 8 == 20 mod 21 or == 47, 83 mod 195 or == 467 mod 73815 or == 722 mod 1551615 9 == 19 mod 20 10 == 32 mod 33 11 == 5 mod 24 or == 14 mod 15 or == 19 mod 20 12 == 142 mod 143 or == 296, 901 mod 19019 or 562, 828, 900, 1166 mod 1729 or == 563 mod 250705 or == 597, 1143 mod 1885 13 == 20 mod 21 or == 27 mod 28 or == 132, 293 mod 595 14 == 38 mod 39 or == 64 mod 65 15 == 13 mod 14 but not == 1 mod 16 16 == 38, 47, 98, 242 mod 255 or == 50 mod 51 or == 84 mod 85 17 == 11 mod 12 or == 278, 302 mod 435 or == 283, 355, 367, 607, 907 mod 1638 or == 373, 445, 646, 718 mod 819 18 == 322 mod 323 or == 398, 512 mod 1235 19 == 11 mod 12 or == 14 mod 15 or == 29 mod 40 20 == 56 mod 57 or == 132 mod 133 21 == 43 mod 44 or == 54 mod 55 22 == 68 mod 69 or == 160 mod 161 23 (== 21 mod 22 but not == 1 mod 8) or == 32 mod 33 or == 41 mod 48 or == 83 mod 530 or == 182 mod 795 24 == 114 mod 115 25 == 38 mod 39 or == 51 mod 52 Last fiddled with by sweety439 on 20200704 at 15:56 
20200704, 12:03  #854 
Nov 2016
2^{3}·281 Posts 

20200704, 13:53  #855  
Nov 2016
4310_{8} Posts 
Quote:
Last fiddled with by sweety439 on 20200704 at 13:54 

20200704, 15:17  #856 
Nov 2016
2^{3}×281 Posts 
Special cases of (k*b^n+1)/gcd(k+1,b1) (+ for Sierpinski,  for Riesel):
* gcd(k+1,b1) (+ for Sierpinski,  for Riesel) = 1: the same as the original Sierpinski/Riesel problem in CRUS * Riesel case k=1: the smallest generalized repunit prime base b (see A084740 and http://www.fermatquotient.com/PrimSerien/GenRepu.txt) * Riesel case k=b1: the smallest Williams prime base (b1) * Sierpinski case k=1 and b even: the smallest generalized Fermat prime base b (see http://www.noprimeleftbehind.net/crus/GFNprimes.htm and http://jeppesn.dk/generalizedfermat.html) * Sierpinski case k=1 and b odd: the smallest generalized half Fermat prime base b (see http://www.fermatquotient.com/PrimSerien/GenFermOdd.txt) Last fiddled with by sweety439 on 20200704 at 15:18 
20200704, 15:36  #857 
Nov 2016
2^{3}·281 Posts 
we allow n=1 or n=2 or n=3 or n=4 or ..., but not allow n=0 or n=1 or n=2 or n=3 or ... for (k*b^n+1)/gcd(k+1,b1)

20200704, 16:09  #858 
Nov 2016
2^{3}·281 Posts 
k is Riesel number base b if....
Code:
k b 1 (none) 2 (none) 3 (none) 4 == 14 mod 15 5 == 11 mod 12 6 == 34 mod 35 7 == 11 mod 12 8 == 20 mod 21 or == 83, 307 mod 455 9 == 19 mod 20 or == 29 mod 40 10 == 32 mod 33 11 == 14 mod 15 or == 19 mod 20 12 == 142 mod 143 or == 307 mod 1595 or == 901 mod 19019 13 == 5 mod 24 or == 20 mod 21 or == 27 mod 28 or == 38, 47 mod 255 14 == 8, 47, 83, 122 mod 195 or == 38 mod 39 or == 64 mod 65 15 == 27 mod 28 16 == 50 mod 51 or == 84 mod 85 Last fiddled with by sweety439 on 20200706 at 02:34 
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