20131105, 14:28  #12  
"Bob Silverman"
Nov 2003
North of Boston
7635_{10} Posts 
Quote:
If there is some theoretical reason to believe that a base exists with your desired properties, then the theory would certainly merit discussion here. But a discussion of a purely numerical search is offtopic. Perhaps I assume too much? Do people not understand the distinction between the math and the computing? 

20131105, 14:30  #13  
Jun 2003
Suva, Fiji
11111111010_{2} Posts 
Quote:
I did look into "AntiBriers" which are the complement of Brier numbers. Here there are y which meet the requirement for both  and + at the same time, but this is still for fixed y. Last fiddled with by robert44444uk on 20131105 at 14:32 

20131105, 14:32  #14  
"Bob Silverman"
Nov 2003
North of Boston
3×5×509 Posts 
[QUOTE=robert44444uk;358470]
Quote:
proof still remains. 

20131105, 15:43  #15  
Jun 2003
Suva, Fiji
2·1,021 Posts 
[QUOTE=R.D. Silverman;358471]
Quote:
If we have a look at the opposite..the Payam situation; 7 can only be a factor of a member of the power series k*2^n+1 if k is 3,5 or 6 mod 7. If k=11 for example, i.e k=4mod7, then 7 is never a factor, albeit n goes to infinity. All that we are doing in defining a payam number is picking a k using CRM so that all primes with a multiplicative order base 2 of less than a given value are never factors. If they are never factors, then the smallest factor that any member of the series can have is greater than the given value. This changes the chance of the member being prime, and overall, the series should produce on average more primes than a given k of a similar size. And so they do. 

20131105, 15:54  #16  
"Bob Silverman"
Nov 2003
North of Boston
3·5·509 Posts 
[QUOTE=robert44444uk;358477]
Quote:
element has a bounded order. Reference please. 

20131105, 16:36  #17  
Jun 2003
Suva, Fiji
2·1,021 Posts 
[QUOTE=R.D. Silverman;358479]
Quote:
Payam factors are excluded for all n, Sierpinski factors are included and are factors of different members of a covering set. Sierpinski will guarantee zero primes, Payam certainly does not guarantee infinite primes or even a guaranteed higher density. For example, I think it is unlikely but there could be a covering set of non Payam factors (all greater than the E level) that generate a Sierpinski covering set and guarantee that a Payam power series has zero primes. But, maybe I haven't answered the question you are asking. 

20131106, 11:03  #18 
Jun 2003
Suva, Fiji
7FA_{16} Posts 
In fact what I found is a rather useless property, the two primes in question are the first instance primes p that are not multiplicative order p1 for all prime bases less than 43 and 73 respectively.
In terms of the Payam number formula, that means that the two primes are not part of the M(x) formula, for those prime bases mentioned above In terms of what I was trying to achieve, i.e. M(x)=1, I might have a look at that when I get a little more time. 
20131107, 10:43  #19  
Jun 2003
Suva, Fiji
3772_{8} Posts 
Quote:
2 and 3 are always members given the members of M(x) exhibit the property of being multiplicative order of p1 the base, or 1 the base. 2 and 3 always show these properties. Putting aside the question of non square composite bases, the question remains whether there is a prime base which has no other members in M(x) other than 2 and 3. 

20131107, 12:23  #20  
"Bob Silverman"
Nov 2003
North of Boston
3·5·509 Posts 
Quote:
In particular, the assertion in the second paragraph "2 and 3 always show these properties" needs PROOF. Even a heuristic/probabilistic argument suggesting why the assertion should be true would be something. The question in paragraph 3 is also somewhat interesting. Or at least it would be if it were backed by some REASONING. 

20131107, 12:26  #21  
"Bob Silverman"
Nov 2003
North of Boston
3·5·509 Posts 
Quote:
purely numerical OBSERVATIONS, free from any mathematical reasoning??? This thread would more properly belong there. 

20131107, 15:01  #22  
Jun 2003
Suva, Fiji
2×1,021 Posts 
Quote:
I can't write proofs to the standards expected by mathematicians. I lack the training to do this. The reasoning behind me saying what I did about square bases was as follows: If a constituent prime number member p of M(x) in the Payam notation for the power series (first series) y*M(x)*b^n+/1, [y fixed, M(x) fixed, n from 1 to infinity], is defined in terms of having multiplicative order p1 base b then if you square b and use this as a base, thereby defining a second power series, then the constituent members of M(x) in the first series will have a multiplicative order of (p1)/2 base b^2 in this second power series and thereby no longer qualify to be part of M(x) for that series as p1 =/= (p1)/2. The fact that it is (p1)/2 can no doubt be proven  this is my observation. If a prime is not in M(x) in the first power series, and has order (p1)/2 base b, then it appears to have order (p1)/2 base b^2. (observation) If a prime is not in M(x) in the first power series, and has order q base b, with q=/=(p1) then it appears to have order q/2 base b^2 or q base b^2. Both observations at this stage. This covers all possibilities of values of multiplicative order  all either decrease the value of the multiplicative order or leave it the same, and therefore as p1 is the highest possible value for the order for a prime p, then p1 is no longer a valid multiplicative order for base b^2. Think 2 and 3 are special cases as mentioned before. Last fiddled with by robert44444uk on 20131107 at 15:03 

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