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A Does this imply infinite twin primes?

  1. Apr 28, 2016 #1
    I can prove the twin prime counting function has this form:

    [tex]\pi_2(n)=f(n)+\pi(n)+\pi(n+2)-n-1,[/tex]

    where [itex]\pi_2(n)[/itex] is the twin prime counting function, [itex]f(n)[/itex] is the number of twin composites less than or equal to [itex]n[/itex] and [itex]\pi(n)[/itex] is the prime counting function.

    At [itex]n=p_n,[/itex] this becomes

    [tex]\pi_2(p_n) = f(p_n) + \pi(p_n) + \pi(p_n + 2) - p_n - 1.[/tex]

    With this form, can I make the following argument?: Assume the twin prime counting function becomes a constant [itex]c[/itex], then I can change the twin prime counting function to [itex]c[/itex] in the equation. The prime counting function [itex]\pi(n)[/itex] at the prime sequence [itex]p_n[/itex] is just [itex]n[/itex], so I can change that to [itex]n[/itex]. Because I'm assuming no more twin primes, [itex]p_n+2[/itex] is not a prime so [itex]\pi(p_n+2)[/itex] will also become [itex]n[/itex], the equation directly above this paragraph can therefore be simplified to:

    [tex]c = f(p_n) + 2n - p_n - 1.[/tex]

    Adding [itex]1[/itex] to both sides of this and rearranging it gives,

    [tex]p_n - f(p_n) = 2n - b[/tex], where [itex]b=c+1.[/itex]

    The right side of [itex]p_n - f(p_n) = 2n - b[/itex]

    has only one possible parity, either odd or even because it is an even number [itex]2n[/itex] minus a constant [itex]b.[/itex]

    But, the left side can be both odd and even many times over because [itex]f(p_n)[/itex] can be odd or even and is subtracted from [itex]p_n[/itex] which is odd for [itex]p>2.[/itex]
    So, the left side will change parity for different values of [itex]n,[/itex] while the right side of the equation will remain one parity. Therefore, the two sides cannot be equal for all [itex]n.[/itex]

    This seems to show the twin prime counting function cannot become constant and therefore, there are infinite twin primes. Now assuming I can prove the form given at the beginning of this question, does that argument hold water?
     
  2. jcsd
  3. Apr 29, 2016 #2

    andrewkirk

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    It is not clear that the LHS of the final equation continues to change sign, once the largest twin-prime T is surpassed. Can you prove that there exist ##m,n\geq T## such that ##p_n-f(p_n)## and ##p_m-f(p_m)## have different parity? If not, the argument from parity collapses.
    Also, it is unclear exactly what the twin-prime and twin composite counting functions do. Does f(10) count the pair (8,10) as one or as two? Does f(9) include both, one or neither of that pair? The same questions need to be answered for ##\pi_2##.
     
  4. Apr 29, 2016 #3
    Only counting once. Twin prime counting function is counting "smaller" twin primes and twin composite version is counting "smaller" twin composites. As an example for the composite version: given 22,23,24,25,26,27 it would count every even number and include 25, but it would not count 27 because 29 is prime. So the odd composites have the real effect on the count, since all even [itex]n>2[/itex] are counted. As for proving ##p_n-f(p_n)## and ##p_n-f(p_n)## have different parity, it would come down to proving [itex]f(p_n)[/itex] changes parity infinitely many times. No proof for that at present. Not sure if I can, I will try. It would seem impossible otherwise though. The latter would imply that the number of smaller twin composites [itex]<=n[/itex] becomes always even or always odd. It would seem that what I have done is showed that the twin prime conjecture is equivalent to proving [itex]f(p_n)[/itex] changes parity infinitely many times. Thank you for responding by the way. :)
     
    Last edited: Apr 29, 2016
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