20131104, 17:26  #1 
Jun 2003
Oxford, UK
2^{3}×241 Posts 
k*b^n+/1, Bases 271 and 11971
I have given a little (I mean very little) thought to a set that is defined in terms of the following:
Set member z is the prime p which is the first instance of primes such that none of the primes q up to q(z1) have a multiplicative order p1 base q. The set (I think) goes as follows  the first 21 members: 7,11,11,59,131,131,181,181,271,271,271,271,271,1531,2791,11971,11971,11971,11971,11971,11971... corresponding to the primes 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73... An example: 271 is 135order2, 30order3, 27order5, 135order7, 135 order11, 18order13, 135 order17, 30 order19, 18 order23, 6 order29, 45 order 31, 135 order37, 45 order41, but 270 order 43. The list is not on OEIS so happy for someone to put it up if they like. I'm trying to think of a use. If the p is a base in k.p^n+/1, then it should be possible to define k much smaller than q(z1) primorial, that provides a relatively prime series with integer n increasing, as all members of the series cannot have factors smaller than q(z1). The efficient k values are found by applying CRM. This is a bit akin to the Payam series, y.M(x)*2^n+/1 with M(x) the multiple of primes that are p1 order2, but in this new instance M(x)=1 given that all the primes to q(z1) are considered as part of the CRM. It might be interesting to find a few efficient k and find some primes with the bases 271 and 11971 Comments/ observations/ continuation/ efficient k/ subsequent primes welcome Last fiddled with by robert44444uk on 20131104 at 17:32 
20131104, 17:47  #2  
Nov 2003
2^{2}·5·373 Posts 
Quote:
I do not know you personally. But when someone attempting to write mathematics shows he/she does not know basic, elementary definitions, it discourages others from further reading. One must ask (rhetorically) "If an author is totally mixed up regarding elementary concepts, why should anyone else believe that the author knows what he/she is doing?" Quote:
Quote:
What does it mean to be "a base in k.p^n +/ 1"?????? Please use standard mathematical terminology. 

20131105, 11:33  #3  
Jun 2003
Oxford, UK
2^{3}×241 Posts 
Quote:
Ah, it is a list. I appreciate from reading this morning about sets that it is not a set, as the member are not distinct. Each member is correlated to a prime and in order of the primes, and the primes are a set, but I do see this now. Quote:
The two primes in the title of the text provide M(x) equal to 1 when x=42 in the case of the base 271 and x=72 in the case of base 11971. Quote:


20131105, 12:03  #4  
Nov 2003
2^{2}·5·373 Posts 
Quote:
why they might be interesting. You have not stated a purpose for your discussion. Absent this information, your discussion becomes nothing more than numerical curiosities. Motivate your audience. Answer the question: Why should we care?? Last fiddled with by R.D. Silverman on 20131105 at 12:03 Reason: typo 

20131105, 12:22  #5 
Romulan Interpreter
Jun 2011
Thailand
243B_{16} Posts 
+1 for the last post of RDS (different for his usual brute style, and very well formulated, in a positive way; additionally, I have no idea what those numbers are, just to talk in my own name...)
(edit: don't point me to lmgtfy, I did that already :P) Last fiddled with by LaurV on 20131105 at 12:24 
20131105, 13:29  #6 
Mar 2006
Germany
2×1,439 Posts 
Here's a small overview for k*271^n1:
1 <= k <= 60 1 <= n <= 60 primes: k  prime for n 2  1,22,40 8  2 12  1,3 14  1,7,19 18  1,5,7,11 20  1 24  2,28 30  3,17 32  2,7 38  3,5,17 42  38 44  1,5,12 48  1,2,19 50  16,22,46 54  1,21,34, 60  2 
20131105, 13:33  #7  
Nov 2003
7460_{10} Posts 
Quote:
explanation. And we still lack an explanation as to why anyone should be interested. What is the RELEVENCE?? 

20131105, 13:43  #8 
Mar 2006
Germany
2×1,439 Posts 
Background is the thread k=22544089918041953*E(130) generates 215 known primes where the Expression k*2^n1 for k= 22544089918041953*E(130) (which is equivalent to k=1480472640274704456611717878515654164205) with a Nashweight of 8818 produces 214 primes for n=1 to ~800000.
This is the recordholder for numbers of the form k*2^n1 for the same k. The table above only shows there're much less primes than for base 2 in that ranges. Last fiddled with by kar_bon on 20131105 at 13:47 Reason: 214 corrected 
20131105, 13:45  #9  
Jun 2003
Oxford, UK
3610_{8} Posts 
Quote:
I will define a Payam number as meeting the following (I do hope I have got this right!!): If factoring all of the members of the power series generated by y*M(x)*2^n+/1 (where y is an integer and M(x) the product of primes defined below, n variable), never produces a factor that meets the condition of having a multiplicative order of less than z order 2, then the integer y*M(x) is called a Payam number at the z level. The y value can be found through applying CRM using information on the primes with multiplicative order modulo 2 <= z that are not in the M(x) product. By necessity certain primes p are required to be prime factors of the Payam number where they are <= z+1 and are p1 order 2 and these are in the M(x) product. By convention, the z level is normally depicted as E(z), and the certain primes p are collected together as a product in the M(x) function. I have seen the M(x) function also shown as E(x1) on various sites  the reason for the E and M naming convention is to distinguish between a multiplier and the level at which the series will have no factors. The definitions of Payam Numbers (for example on Mathworld refer only to the smallest Payam number of a given level). Why are they interesting? Payam numbers generate very prime series. For example http://www.mersenneforum.org/showthread.php?t=18407 shows a Payam number that has generated 215 primes in the 1 power series to date. This is the complement to Sierpinski numbers that generate no primes in the power series. There is no use for these numbers that I can see other than to provide recreation for prime hunters, which abound on this site. The purpose of the discussion is to provide ideas for those prime hunters seeking new/ different challenges. Payam numbers have rarely been in fashion with prime hunters because they are too large for LLR software to handle superefficiently. Checking one candidate at a given n takes 3 times longer than for a small k in k*2^n+/1. Only one large Payam number appears in the top 5000 currently and this belongs to the 215 record holder. 

20131105, 14:01  #10  
Nov 2003
2^{2}·5·373 Posts 
Quote:
What you write above is unclear. What does "never produces a factor" mean? Do you mean that a factor q does not exist such that q divides yM(x)2^n +/1 and the order of 2 mod q is less than z??? Please clarify. If so, one must then ask: All of the members? This is an infinite set. How does one show that yM(x)2^n+1 or yM(x)2^n1 never has a factor q whose order to the base 2 is less than (a prespecified) z for ALL n and ALL y? Please show a proof. If this can be done it would be an interesting piece of mathematics. 

20131105, 14:10  #11 
Jun 2003
Oxford, UK
3610_{8} Posts 
....Without having seen Bob's reply.....
Sorry folks, I think I am going totally down the wrong line here with this. It is not going to go down the lines I was hoping for, which was to determine a base for which M(x)=1 Apologies for time spent by all. I will have a look at Bob's reply to see if I can answer his specific points. Last fiddled with by robert44444uk on 20131105 at 14:11 
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