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Is this question acceptable?: [Closed] Vicinity and range of ascending primes for the ratio 5^{2^{n-1}}/{ p_1∙p_2∙…p_n} to be minimum

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    I am writing this because I want one of my questions
    http://mathoverflow.net/questions/81038/vicinity-and-range-of-ascending-primes-for-the-ratio-52n-1-p-1p-2p-n
    to be reopened. First I have to say ascending primes means when the primes of the form p=a^2+b^2 are in the following sequence 5,13,17,29,37 and so on. Secondly the first prime cannot be smaller than the lower boundary. After we set the first prime the rest of them have to be in ascending order as defined above. After the first prime is set the ratio should be closest to one. Is this sufficient clarification? Lastly I want to bring your attention to the fact that this question is of fundamental importance concerning the distribution of primes.
    • CommentAuthorYemon Choi
    • CommentTimeNov 16th 2011
     

    the fact that this question is of fundamental importance concerning the distribution of primes

    Is this in the same sense that it's a fact that there are nine million bicycles in Beijing?

    • CommentAuthortheojf
    • CommentTimeNov 16th 2011
     

    Hi Vassilis,

    I have left some comments on your question. But the short answer is: I don't understand at all what you're asking, and in order for the question to be reopened it's vitally important that at least the question be clearly stated in standard mathematical language.

  4.  
    I *think* the question is,

    Given n, minimize f(n)=5^{2^{n-1}}/(p_1 x p_2 x ... x p_n), where the p_i are consecutive primes of the form 4 k + 1, and where f(n) is greater than 1.

    If this is really the question you are trying to ask, then please

    1. edit your question so that this is what it says (but in TeX, unlike what I have done),
    2. leave out all the irrelevant stuff about triangles, and sums of squares, and
    3. say something about what this could possibly have to do with the distribution of primes.

    If you do all that, I'll vote to reopen.
    • CommentAuthorHenry Cohn
    • CommentTimeNov 17th 2011
     
    I doubt the question will ever reach a state in which I think it should be re-opened (of course it's hard to be sure, but I don't see a lot of potential here for a question of mainstream research interest). However, here are some brief comments about related mathematics:

    As best I can tell, you want to understand the product of the primes that are 1 mod 4 and lie in a certain interval, and you want to know how well you can control this product by choosing the interval. You can get some weak bounds for this from the prime number theorem (well, the prime number theorem for primes in arithmetic progressions). Specifically, if you take the logarithm you want the sum of log p over primes p congruent to 1 mod 4 and lying in the interval [x,y]. If y and y-x are both large, then this will be (y-x)/2 plus a smaller error term. (Explicit error terms are known, but they are much worse than the roughly square root sized bound one would get from the Riemann hypothesis, which would be the best possible bound.)

    If you exponentiate this, you get a multiplicative factor from exponentiating the error term, so you lose quite a bit of control. That's unavoidable: you'll never get an approximation that is within a 1+o(1) factor of the truth, because you won't have precise enough control over the distribution of primes to know whether a single prime has moved over the endpoint of the interval (and one prime is itself enough to change the product nontrivially). So this all comes down to how you are interpreting "approximately equal to" in your question. For a strong interpretation, it seems hopeless. For a weak interpretation, the prime number theorem may suffice, depending on how weak and how this matches up with the known error bounds.
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    Thank you for your comments. I will respond in ten hours.
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    I will present the question in a way that pleases you not me. Given the number 5^{2^{n-1}} where n any non zero natural number. Find n consecutive primes of the form 4x+1 with the first prime not smaller than the vicinity The vicinity is obtained by breaking the number 5^{2^{n-1}} into n equal factors. The vicinity is equal to one of these factors. To dispute my conjecture it is necessary to prove that the following inequality does not hold for all n.
    {(vicinity)(e^{1/n})}/2ln((vicinity)e{1/n})-{(vicinity)/2ln(vicinity)}>n
    Because the numbers involved are humongous, you will never be able to give a specific numerical example. Even for n=16 we get huge exponentials for vicinity as shown below.
    (5^2048)e^{1/16}/2ln(5^{2048}e^{1/16})-{5^{2048}/2ln(5^{2048})}>16
    You should have in mind that the prime number theorem is not a reliable measuring stick in the above cases and neither is Riemann's method. After these explanations I hope someone will repost the question in a way that pleases you. I will not have any objections to that. In addition, I want to tell you that with the above restrictions the ratio is close to 1. You can challenge yourselves by making the ratio close to 2 or 3, 4, 5, etc. Apart from that try again with base 13.
    Let me know if anything needs clarification.
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    The way you present the question does not please me; it mystifies me. The "vicinity", let's call it V, is, it seems, the n-th root of 5^{2^{n-1}}. Now you start by talking about primes not smaller than V. But then you don't mention these primes any more, so it is not at all clear what role they play. You write of your "conjecture", but you haven't made any conjecture. Or perhaps your conjecture is that the inequality you write, the one involving V (but not involving any primes) holds for all n. Or perhaps your conjecture is that that inequality holds for at least one value of n - it is not clear how your quantifiers go.

    Then you go back to talking about primes, even though you've said nothing to link your inequality to primes. And you toss in "Riemann's method" as if everyone is supposed to know what you mean by that. Then you talk about "the ratio" being close to 1, but since you haven't previously mentioned any ratio it's not clear what you are talking about.

    I'm sorry, but your "explanations" have explained nothing, and everything needs clarification.
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    I am going to give you the following example for clarification.
    If n=5 then V= (5^16)^{1/5}.
    The primes should be taken within the following boundaries.
    (5^16)^{1/5}/e^{1/5}≈141 lower boundary
    (5^16)^{1/5}∙e^{1/5}≈210 upper boundary
    The first primes within these boundaries are p_1=149, p_2=157, p_3=173, p_4=181, p_5=193.
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    @VP - latex does not work here in meta so please don't use $ $. It makes things hard to read. You can edit your posts and take them out if you want to help people read what you have written.

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    OK. The challenge is to prove that within the two boundaries we will always have n primes of the form 4x+1.
    • CommentAuthorgrp
    • CommentTimeNov 18th 2011 edited
     

    Finally, a statement clear enough that has a chance of being understood. I rephrase:

    " I have come across the following pattern. For small n, I am finding (n primes or n consecutive primes, you make clear which) q_i all of the form 4k+1, such that Lower(n) < q_i < Upper(n) . (I'll let you define Lower and Upper in terms of 5^(2^(n-1)) .) Does this hold for all sufficiently large positive integers n? "

    As written, this is a question which is barely suitable for MathOverflow, primarily because the motivation for finding such a constellation is unclear. However, it has the benefit of (after adding the defining details) being clear and precise and understandable by others. Your previous posts did not meet the standards of clarity I would insist upon for a MathOverflow question.

    Now that I understand the challenge, I will repeat the advice others have given. This question is more appropriate for math.stackexchange. The process of asking for and getting help on forming the question is better suited for meta.math.stackexchange. I have not done the calculations, but if you do them, I think you will find that the result is an easy consequence (even for consecutive primes) of the prime k-tuples conjecture, and that you will have more like O(2^(polylog(n)) sequence of primes in the desired interval.

    I will say this one last time: Ask for help (in writing the question down) on meta.math.stackexchange or in some other forum; Ask the question itself on math.stackexchange. If you get no response, ask (on meta.math.stackexchange) for help posting to MathOverflow. Do not post such questions on MathOverflow yourself. Do not ask for help on meta.matheoverflow yourself. ONLY IF YOU GET HELP FROM OUTSIDE MATHOVERFLOW by someone who can help you form a good question, only then should you post on MathOverflow, and say who helped you.

    I don't want to stop your research or your pursuit of mathematics. I encourage you to ask good questions on the right forums. The past few attempts have convinced me that you need more practice on places outside of MathOverflow: you are not ready to post on MathOverflow.

    I wish you well.

    Gerhard "Ask Me About System Design" Paseman, 2011.11.18

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    The question as I posted it in the first place was complete but not understood because of its complexity. If my conjecture is true then with the ratios we will be able to count primes if a quantity is given, in this case the upper boundary as defined above. This is the reason I insist the ratio has to be very close to 1. Lastly I want to tell you there are more chances for a blue sun to pop over North America than for any mathematician on earth to answer the question and my advice to all of you is that you bring me in contact with the best number theorist so I can cooperate with him on this matter.
    @ grp: you never understood this question let alone answered it.
    • CommentAuthorgrp
    • CommentTimeNov 18th 2011
     

    Indeed, it is quite likely that I never understood your question. The evidence I have seen shows no one else (possibly excepting yourself) understood it either. If we bring the best number theorist in, and she or he does not understand it either, where does that get you?

    You have the right to believe the cause is complexity of the question. I have the right to believe that it is matters of exposition and lack of clarity that are responsible for my and for others not understanding your question. I will bet a small amount of money that the people you are interested in having understand the question will face difficulties for the reasons I believe more than for the reasons you believe.

    I stand by my earlier points regarding communication and question asking. Likewise for the last two paragraphs of my earlier post.

    Gerhard Paseman, 2011.11.18

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    I don't think this thread is going anywhere good, and I think the moderators should close it.
  16.  

    Indeed.

    @Vassilis, I think we have established that your question is not suitable to MathOverflow.

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