mersenneforum.org > Math k*b^n+/-1, Bases 271 and 11971
 Register FAQ Search Today's Posts Mark Forums Read

 2013-11-04, 17:26 #1 robert44444uk     Jun 2003 Oxford, UK 193210 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(z-1) have a multiplicative order p-1 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(z-1) primorial, that provides a relatively prime series with integer n increasing, as all members of the series cannot have factors smaller than q(z-1). 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 p-1 order2, but in this new instance M(x)=1 given that all the primes to q(z-1) 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 2013-11-04 at 17:32
2013-11-04, 17:47   #2
R.D. Silverman

Nov 2003

22·5·373 Posts

Quote:
 Originally Posted by robert44444uk 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(z-1) have a multiplicative order p-1 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...
You need to learn the definition of "set". The list above is not a set.

I do not know you personally. But when someone attempting to write
mathematics shows he/she does not know basic, elementary definitions,

"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:
 I'm trying to think of a use.
So am I. It would be helpful to know what motivated this. Please explain where it came from and why you think it might be useful.

Quote:
 If the p is a base in k.p^n+/-1,
??? This phrase is poorly worded at best and total gibberish at worst.
What does it mean to be "a base in k.p^n +/- 1"??????

2013-11-05, 11:33   #3
robert44444uk

Jun 2003
Oxford, UK

22·3·7·23 Posts

Quote:
 Originally Posted by R.D. Silverman You need to learn the definition of "set". The list above is not a set. 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?"
Bob, I am not a mathematician as you well know from former run-ins.

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:
 So am I. It would be helpful to know what motivated this. Please explain where it came from and why you think it might be useful.
Reading further down my text, it came from thinking about the M(x) part of the formula for Payam numbers. M(x) is the multiple of all the primes p less than or equal to x that are p-1 order 2. This is a large number usually and a major contributor to the large size of k that are a feature of Payam numbers.

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:
 ??? This phrase is poorly worded at best and total gibberish at worst. What does it mean to be "a base in k.p^n +/- 1"?????? Please use standard mathematical terminology
the p is the base in the power series k.p^n+/-1. k and p are fixed and n is variable in the series. I can't think this isn't mathematical terminology. Usually we see this as b but I had already referred to it as p throughout.

2013-11-05, 12:03   #4
R.D. Silverman

Nov 2003

11101001001002 Posts

Quote:
 Originally Posted by robert44444uk Bob, I am not a mathematician as you well know from former run-ins. 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. Reading further down my text, it came from thinking about the M(x) part of the formula for Payam numbers. M(x) is the multiple of all the primes p less than or equal to x that are p-1 order 2. This is a large number usually and a major contributor to the large size of k that are a feature of Payam numbers.
You need to define Payam numbers for the audience. You should say
why they might be interesting. You have not stated a purpose for
the question: Why should we care??

Last fiddled with by R.D. Silverman on 2013-11-05 at 12:03 Reason: typo

 2013-11-05, 12:22 #5 LaurV Romulan Interpreter     Jun 2011 Thailand 249416 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 2013-11-05 at 12:24
 2013-11-05, 13:29 #6 kar_bon     Mar 2006 Germany 43·67 Posts Here's a small overview for k*271^n-1: 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
2013-11-05, 13:33   #7
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by kar_bon Here's a small overview for k*271^n-1: 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
I see no mathematics or discussion. All I see is numerology without
explanation. And we still lack an explanation as to why anyone should
be interested. What is the RELEVENCE??

 2013-11-05, 13:43 #8 kar_bon     Mar 2006 Germany 1011010000012 Posts Background is the thread k=22544089918041953*E(130) generates 215 known primes where the Expression k*2^n-1 for k= 22544089918041953*E(130) (which is equivalent to k=1480472640274704456611717878515654164205) with a Nash-weight of 8818 produces 214 primes for n=1 to ~800000. This is the record-holder for numbers of the form k*2^n-1 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 2013-11-05 at 13:47 Reason: 214 corrected
2013-11-05, 13:45   #9
robert44444uk

Jun 2003
Oxford, UK

22×3×7×23 Posts

Quote:
 Originally Posted by R.D. Silverman You need to define Payam numbers for the audience. You should say 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??

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 p-1 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(x-1) 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.

2013-11-05, 14:01   #10
R.D. Silverman

Nov 2003

22×5×373 Posts

Quote:
 Originally Posted by robert44444uk Thank you Bob. I shall try to answer your challenges. 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.
Thanks for the background material.

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^n-1 never has a factor q whose order to the base
2 is less than (a pre-specified) z for ALL n and ALL y?

If this can be done it would be an interesting piece of mathematics.

 2013-11-05, 14:10 #11 robert44444uk     Jun 2003 Oxford, UK 22·3·7·23 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 2013-11-05 at 14:11

 Similar Threads Thread Thread Starter Forum Replies Last Post gd_barnes Conjectures 'R Us 132 2021-01-09 05:58 Stargate38 Lounge 44 2020-10-24 11:33 wblipp GPU Computing 50 2012-10-11 13:23 henryzz Conjectures 'R Us 15 2010-04-18 18:07 roger Information & Answers 1 2007-04-25 14:35

All times are UTC. The time now is 13:07.

Sun Apr 11 13:07:48 UTC 2021 up 3 days, 7:48, 1 user, load averages: 1.56, 1.45, 1.35