20200711, 03:58  #1 
May 2020
50_{8} Posts 
Numbers Sierpinski to multiple bases
Hello! This question errs on the side of notnumbertheory, so I'll ask it here.
Given that Brier numbers (numbers that are Sierpinski and Riesel) exist and can be made by considering different covering sets, I was curious as to how far it could be reasonably extended. For a start, could there be a number that is Sierpinski to 2 and 3? I imagine it's relatively easy  just find the right sorts of covering sets and zip them together  but how small could we reasonably expect the smallest one to be? On the scale of known Brier numbers? What if we extend it to 2, 3, and 5? How about arbitrarily many primes? Can this sort of process of multibase Sierpinski numbers be able to be extended forever for the set of primes, or is there a hard limit? Is there a way to use Aurifeuilleanbased Sierpinski numbers to guarentee covering sets that work together well for the primes? I'm aware that "the set of powers of 2" is already an infinite set for which a k exists that is Sierpinski to all of them (or, really, any set of powers of some base), but how much can we do with other sets? Thanks, Gelly 
20200714, 07:28  #2 
"Jeppe"
Jan 2016
Denmark
5^{2}×7 Posts 
I agree with you it should be possible to find "simultaneous" Sierpinski numbers to multiple bases, by picking disjoint covering sets (one covering set for the first base, and a nonoverlapping covering set for the second base).
The precise definition of Sierpinski to base b is important, of course. For base b=3, most people exclude the trivial examples where k is odd. Because then a set with just one prime in it covers, namely { 2 }. So to be Sierpinski to base 3 and base 2, you would have to accept an even Sierpinski number to base 2. Brunner, Caldwell, Krywaruczenko and Lownsdale require a Sierpinski k to base b to have the property gcd(k+1, b1) = 1 (to avoid covering sets with one prime) and also the property that k is not a rational power of b. The second property is to avoid generalized Fermat sequences for which only exponents of the form two to something, can produce primes. For example, if b = 1000, then taking k = 10000, leading to 10000*1000^n + 1, would be excluded by the definition by Caldwell et al. Because 10000 = 1000^(4/3). And in essence you would be looking at candidates 10^(2^m) + 1, and we do not think they will yield primes for m > 1 (although we cannot prove it). Enough smalltalk with no answer to your question! Come on, everybody, have you ever located a number that was Sierpinski to two "nonrelated" bases? /JeppeSN 
20200714, 14:45  #3 
May 2020
2^{3}·5 Posts 
Just in case it's missed elsewhere, I had a minor discussion/revelation about the "infinite set" problem that has a pretty trivial solution.
For bases of the form 15n+14, k=4 is Sierpinski to all of those bases. This is easily realized by the fact that 15n+14 is 1 mod 3 and 1 mod 5, where 4 is 1 mod 3 and 1 mod 5. For odd powers of the base, you have (1)*(1)+1 = 0 mod 3, and for even powers of the base you have (1)*(1)+1 = 0 mod 5. Fitting the gcd property below, k+1 is 5 and b1 is 15n+13, so they will always be coprime as well. Therefore, 4 is Sierpinski to all bases of the form 15n+14, which is not only an infinite set, but also a nonzero density one, too. Using different two prime covering sets can also lead to k with other sets, such as k = 20 (covering set {3, 19} and bases 57n+56, n =/= 7m+1). Perhaps you can combine some of these trivially to make much larger sets, but that might be asking too much. I'm sure this is trivial for anyone who spends an iota of time on things like CRUS, but it's a pleasant revelation for me and answers one of my questions. 
20200714, 15:22  #4  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3115_{10} Posts 
Quote:
Last fiddled with by Uncwilly on 20200714 at 20:11 Reason: TRIM YOUR QUOTES! 

20200714, 15:30  #5  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
5·7·89 Posts 
Quote:
Last fiddled with by Uncwilly on 20200714 at 20:12 Reason: TRIM YOUR QUOTES! 

20200714, 15:56  #6 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
3115_{10} Posts 
This is the conjectured smallest Sierpinski/Riesel number in new definition to base b for 2<=b<=2048, "NA" if the smallest such number is > 10^6
Original definition: k's such that k*b^n+1 (+ for Sierpinski,  for Riesel) is not prime for n>=1 and gcd(k+1,b1) (+ for Sierpinski,  for Riesel) = 1 New definition: k's such that (k*b^n+1)/gcd(k+1,b1) (+ for Sierpinski,  for Riesel) is not prime for n>=1 Brunner, Caldwell, Krywaruczenko and Lownsdale exclude the k's which is a rational power of b, I do not exclude these k's but exclude the k's making a full covering set with all or partial algebraic factors (e.g. (1*4^n1)/3, (1*9^n1)/8, 4*9^n1, (4*19^n1)/3, 4*24^n1, (4*34^n1)/3, 9*4^n1, 9*14^n1, 9*24^n1, 6*24^n1, 6*54^n1, (1*8^n1)/7, 27*8^n1, 8*27^n1, 1*8^n+1, (1*27^n+1)/2, 1*32^n+1, (1*32^n1)/31, (27*8^n+1)/7, 8*27^n+1, (4*16^n+1)/5, (4*81^n+1)/5, 4*625^n+1, (324*16^n+1)/5, (64*81^n+1)/5, 2500*16^n+1, 2500*81^n+1, etc.), I also exclude the k's that do not make a full covering set with all or partial algebraic factors but have no possible prime (e.g. 8*128^n+1, 32*128^n+1, (27*2187^n+1)/2, etc.) I include this k for this Sierpinski/Riesel base b if and only if this k may have infinitely many primes. Last fiddled with by sweety439 on 20200714 at 16:06 
20200714, 16:01  #7 
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
5·7·89 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). 
20200714, 17:50  #8  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
C2B_{16} Posts 
Quote:
Last fiddled with by Uncwilly on 20200714 at 20:12 Reason: TRIM YOUR QUOTES! 

20200715, 16:58  #9  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
110000101011_{2} Posts 
Quote:


20200715, 17:09  #10  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
5×7×89 Posts 
Quote:
25*12^n1 27*12^n1 64*12^n1 (81*17^n1)/16 144*28^n1 (289*28^n1)/9 (175*28^n1)/3 16*33^n1 (169*33^n1)/8 (225*33^n1)/32 (289*33^n1)/32 81*40^n1 (1024*40^n1)/3 16*50^n1 18*50^n1 (529*52^n1)/3 900*52^n1 (637*52^n1)/3 (9*57^n1)/8 (25*57^n1)/8 (121*57^n1)/8 121*60^n1 1369*30^n1 (400*88^n1)/3 324*95^n1 (343*10^n1)/9 64*957^n1 3511808*63^n+1 27000000*63^n+1 324*2070^n+1 2500*13^n+1 2500*55^n+1 114244*225^n+1 16*200^n+1 (64*391^n1)/3 64*936^n1 

20200715, 20:51  #11 
May 2020
2^{3}·5 Posts 
I think I found a solution for b = 2 and b = 3, by doing some hacky work to combine covering sets. I constructed the covering sets such that I could share the most primes between them, and to fill the holes with whatever else I could get away with.
Covering set for 2: {3, 5, 13, 17, 97, 241, 257} modulus 48 Covering set for 3: {5, 7, 13, 17, 41, 73, 193, 577, 769, 6481} modulus 48 In order to reconcile the three shared primes between the covering sets, I actually handpicked the moduli so that I could get the most gas out of using them. Here’s the solutions for the covered primes given the k I calculated, k = 310905992968447463275644916 For base 2: k2^(2n+1)+1 is divisible by 3 (k = 1 mod 3) k2^(4n+2)+1 is divisible by 5 (k = 1 mod 5) k2^(12n)+1 is divisible by 13 (k = 12 mod 13) k2^(8n+4)+1 is divisible by 17 (k = 1 mod 17) k2^(48n+16)+1 is divisible by 97 (k = 62 mod 97) k2^(24n+8)+1 is divisible by 241 (k = 16 mod 241) k2^(16n+8)+1 is divisible by 257 (k = 1 mod 257) This covers the entire space for modulo 48*, and so this number is Sierpinski base 2. For base 3: k is not a trivial case (k = 0 mod 2) k3^(4n+2)+1 is divisible by 5 (k = 1 mod 5) k3^(6n+1)+1 is divisible by 7 (k = 2 mod 7) k3^(3n)+1 is divisible by 13 (k = 12 mod 13) k3^(16n+8)+1 is divisible by 17 (k = 1 mod 17) k3^(8n+4)+1 is divisible by 41 (k = 1 mod 41) k3^(12n+11)+1 is divisible by 73 (k = 70 mod 73) k3^(16n)+1 is divisible by 193 (k = 192 mod 193) k3^(48n+29)+1 is divisible by 577 (k = 19 mod 577) k3^(48n+5)+1 is divisible by 769 (k = 250 mod 769) k3^(24n+17)+1 is divisible by 6481 (k = 4294 mod 6481) This covers the entire space for modulo 48*, and so this number is Sierpinski base 3. I’ve tried to make sure that this is all up to snuff, double checked against all entries for both base 2 and 3 containing the factors I’d expect them to, even triple checked this post, and it all looks good to me. I imagine someone else has done this first, but I’m not sure if smaller. This definitely can be optimized to be smaller  I handpicked every modulus, even the ones that weren’t necessary, so it’s definitely likely that one can do a lot better with a small computer search. You would have to make sure that for the shared primes in the sets, you get maximal coverage with modulus selections, but otherwise you can pick the rest as you would with any other Sierpinski search. I would hope that the smallest one would be smaller than the smallest Brier number, since you should be able to do better with sharing more primes, but I’m not certain. EDIT: JeppeSN helped me realize the Sierpinski modulus for each covering set was smaller than I thought. Thanks! Last fiddled with by Gelly on 20200715 at 21:23 
Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Numbers in Other Bases are Belong to Us  Stargate38  Lounge  44  20201024 11:33 
prime power Sierpinski numbers  snorton  Prime Sierpinski Project  1  20150325 03:07 
A multiple k/c sieve for Sierpinski/Riesel problems  geoff  Sierpinski/Riesel Base 5  522  20091028 05:46 
How did they prove that Riesel/Sierpinski numbers.  Unregistered  Homework Help  17  20091008 18:08 
Sierpinski/ Riesel bases 6 to 18  robert44444uk  Conjectures 'R Us  139  20071217 05:17 