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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 |
[QUOTE=robert44444uk;358380]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... [/QUOTE] 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?" [QUOTE] I'm trying to think of a use. [/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. [QUOTE] If the p is a base in k.p^n+/-1, [/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. |
[QUOTE=R.D. Silverman;358388]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?" [/QUOTE] 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. [/QUOTE] 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 [/QUOTE] 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. |
[QUOTE=robert44444uk;358456]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. [/QUOTE] 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?? |
+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) |
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 |
[QUOTE=kar_bon;358462]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[/QUOTE] 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?? |
Background is the thread [URL=http://www.mersenneforum.org/showthread.php?t=18407]k=22544089918041953*E(130) generates 215 known primes[/url] 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. |
[QUOTE=R.D. Silverman;358457]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??[/QUOTE] 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. 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 [url]http://www.mersenneforum.org/showthread.php?t=18407[/url] 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. |
[QUOTE=robert44444uk;358466]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. [/QUOTE] 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? Please show a proof. If this can be done it would be an interesting piece of mathematics. |
....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. |
[QUOTE=robert44444uk;358468]....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 [/QUOTE] That question is (or seems to be) purely numerical. AFAIK, this sub-forum is for the discussion of the mathematics [i]behind[/i] all of the computations, rather than the actual computations. 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 off-topic. Perhaps I assume too much? Do people not understand the distinction between the math and the computing? |
[QUOTE=R.D. Silverman;358467]
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? [/QUOTE] No, y is fixed. Payam numbers deal with either - or +. So it is for all n for either - or +. 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. |
[QUOTE=robert44444uk;358470][QUOTE=R.D. Silverman;358467]
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? QUOTE] No, y is fixed. Payam numbers deal with either - or +. So it is for all n for either - or +. [/QUOTE] Even with y fixed it is still an infinite set, so the query about an existence proof still remains. |
[QUOTE=R.D. Silverman;358471][QUOTE=robert44444uk;358470]
Even with y fixed it is still an infinite set, so the query about an existence proof still remains.[/QUOTE] Its the same line of argument around covering sets that works for Sierpinski numbers but the other way around. Sierpinski works off the recurrence of certain factors as n increases and uses CRM to define a k where there is cover at all possible n values in the covering period. 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. |
[QUOTE=robert44444uk;358477][QUOTE=R.D. Silverman;358471]
Its the same line of argument around covering sets that works for Sierpinski numbers but the other way around. Sierpinski works off the recurrence of certain factors as n increases and uses CRM to define a k where there is cover at all possible n values in the covering period. 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. [/QUOTE] The mystery (to me) is how one can generate a covering set where each element has a [i]bounded order[/i]. Reference please. |
[QUOTE=R.D. Silverman;358479][QUOTE=robert44444uk;358477]
The mystery (to me) is how one can generate a covering set where each element has a [i]bounded order[/i]. Reference please.[/QUOTE] Payam numbers do not aim to generate a covering set. Just to exclude certain primes as factors, contrasting the way that Sierpinski ensures their inclusion. 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. |
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 p-1 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. |
[QUOTE=R.D. Silverman;358469]
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 off-topic. [/QUOTE] Thinking a bit more about this, except base 4 and base 9, bases that are squares exhibit the property of having no prime members in M(x), other than 2 and 3. And bases 4 and 9 have one of those two primes in M(x) 2 and 3 are always members given the members of M(x) exhibit the property of being multiplicative order of p-1 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. |
[QUOTE=robert44444uk;358629]Thinking a bit more about this, except base 4 and base 9, bases that are squares exhibit the property of having no prime members in M(x), other than 2 and 3. And bases 4 and 9 have one of those two primes in M(x)
2 and 3 are always members given the members of M(x) exhibit the property of being multiplicative order of p-1 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.[/QUOTE] This is still just numerology. Is there any mathematics that you wish to discuss? 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. |
[QUOTE=R.D. Silverman;358633]This is still just numerology. Is there any mathematics that you wish to discuss?
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.[/QUOTE] Perhaps the moderators should set up a sub-forum for discussion of purely numerical OBSERVATIONS, free from any mathematical reasoning??? This thread would more properly belong there. |
[QUOTE=R.D. Silverman;358634]Perhaps the moderators should set up a sub-forum for discussion of
purely numerical OBSERVATIONS, free from any mathematical reasoning??? This thread would more properly belong there.[/QUOTE] I think you are probably right. This is all about observations, there is not a lot of maths here. 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 p-1 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 (p-1)/2 base b^2 in this second power series and thereby no longer qualify to be part of M(x) for that series as p-1 =/= (p-1)/2. The fact that it is (p-1)/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 (p-1)/2 base b, then it appears to have order (p-1)/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=/=(p-1) 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 p-1 is the highest possible value for the order for a prime p, then p-1 is no longer a valid multiplicative order for base b^2. Think 2 and 3 are special cases as mentioned before. |
[QUOTE=robert44444uk;358380]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[/QUOTE] The sequence (when duplicated term removed) should be [URL="https://oeis.org/A029932"]A029932[/URL], your sequence is wrong. the right sequence is 3, 7, 23, 41, 109, 191, 191, 191, 271, 271, 271, 271, 271, 271, 2791, 2791, 11971, 11971, 11971, 11971, 11971, 11971, 31771, 31771, 31771, 31771, 31771, 190321, 190321 271 and 11971 are really in the sequence, but your sequence is wrong. |
Could you please stop necroposting?
This is your last warning. |
[URL="https://www.mersenneforum.org/showpost.php?p=563216&postcount=3"]Déjà[/URL] [URL="https://www.mersenneforum.org/showthread.php?p=563218"]vu[/URL]?
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Sweety is not a bot, we had "human" PM exchanges.
But his posts should not be in math forum, he should be restricted to his blog, or banned. |
[QUOTE=LaurV;568711]But his posts should not be in math forum, he should be restricted to his blog, or banned.[/QUOTE]
Do what we do with COVID patients: mask (with a one-user usergroup) and quarantine (to his blog). Sorry, couldn't resist. :razz: |
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