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Prime question

  1. Nov 14, 2011 #1
    Is there a way within reasonable errors to say what part of the positive integers are prime and what part is factored greater than one? Oh course one is a factor of all numbers greater than zero.

    Yeats ago playing around a floating constant became known to me. to the tenth decimal place is was .7052311717917 and so on depending on what point in the system a prime was a total position of in; it is a ratio of the number system if factoring greater than 1 is calculated..
    Tthis of course was what i calculated as the factor part of the positive real integers,
    ring a bell to any one?

    Two part question, sorry.

    Andas usual I most likely lost most people. If so my apologies. .

    For once I will try and explain.
    Simply put some one years ago asked what part of the whole positive integer system was prime. I of course could not answer such. Yet it made me think.
    I came up with : if I can not come up with the total prime part could it be possible to come up with the factor part of the system greater than one, of course leaving the primes within a reasonable error factor. Oh course I am sure I am not the first to come up with the idea, So were can I find this?
    The number I gave above is what I calculated the system to be of the factored part, other than one of course.
  2. jcsd
  3. Nov 14, 2011 #2
    See http://en.wikipedia.org/wiki/Prime-counting_function
  4. Nov 16, 2011 #3
    Thanks for the reply.

    Of course a system that could tell were the next prime is, would be great for what I did. Instead of a table of known primes such could be used to calculate what I did here to some un believable lengths.

    if one is a periodic position on the entire number line that for the positive integers it is an intersection will all the prime integers,
    then 2 is also a periodic intersection of the whole set of positive integers such that its intersection is 1/2 the system plus or minus 1 divided by infinity. ( the last part due to we do not know if the system would end in odd or even.)
    So what part of the system does 3 intersect?
  5. Nov 16, 2011 #4
    Weell strange I dont find anything on this

    2 is 1/2 of the sytem covered.
    3 unto itself would be 1/3 of the sytem covered or 1/3
    yet every other 3 is already covered by 2, so 1/3 dived by 2 = 1/6
    So 1/2 plus 1/6 equals 4/6.

    Asimple pater happens
    2 is 1/2 , 3 is 4/6, 5 is 21/ 30, 7 is 148/210, and so on..

    Note it is easier to leave the fraction ( ratio as whole instead of reducing them
    This is simply due to both the numerator and denominator can be multiplied by the next prime. of course.
    The numerator is then just plus 1.

    It is such the changes in the gain of each prime is added to the situation yet using just a 1 over the denominator ( the actual coverage of the previous prime intersecting already. to the new intersected portions of the system to see how they change rapidly.)

    2 is 1/2 then 3 is 1/6, 5 is 1/30, 7 is 1/210, 11 is 1/ 2310,13 is1 30030, and so on. It gets to be a very small increase very rapidly.

    If some one sees were I am wrong or gust plain crazy let me know.

    It is easy to see that the larger a prime number used for factor, it rapidly unto itself becomes smaller, ( 1/2, 1/3. 1/5, 1/7, 1/11 1/13. With the part that is intersected by previews numbers and no gain by such intersection, to removing part of the system: then the method I have used should be valid.

    The question then becomes does it converge??

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