Friday, 19 July 2013

Mill's Constant - Awesome Number

The latest Numberphile video reveals a number I had never heard of - Mills' Constant.

It by plugging it into a simple equation, it generates an infinite number prime numbers.

But there's a catch:



After posting the video, a lot of people asked for the proof.

Dr James Grime - who stars in the video - has since sent me a link to the proof.

HERE IT IS.

Want more prime number videos? Here is our prime playlist.
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11 comments:

  1. Unknown19 July 2013 at 10:01

    I would like to ask a few questions regarding the proof that we are given.

    1. Why does this proof depend on the Riemann's Hypothesis?
    2. In (4) and (5), why is P_(n+1)^(3-n-1) greater than P_n^(3-n)? In addition, why is [P_(n+1)+1]^(3-n-1) smaller than [P_n+1]^(3-n)?
    3. Why does it follow, in the end of the proof, that A<v_n?

    I apologize if the answer seems to be trivial. Please answer my questions. Many thanks.

    ReplyDelete
  2. Unknown19 July 2013 at 18:14

    I found a flaw in this proof.

    (4) states that for all non-negative integer n, P_(n+1)^(3-n-1) > P_n^(3-n).

    However, when n=3,
    P_(n+1)^(3-n-1)
    =P_(3+1)^(3-3-1)
    =P_4^(-1)
    <1
    =P_3^0
    =P_3^(3-3)
    =P_n^(3-n)

    Therefore (4) in the proof is wrong, thus the whole proof is wrong.

    ReplyDelete
    Replies
    1. Rig19 July 2013 at 20:31

      This comment has been removed by the author.

      Delete
    2. Unknown19 July 2013 at 22:35

      This comment has been removed by the author.

      Delete
    3. Unknown19 July 2013 at 22:42

      This is only true for n less than or equal to three. But it is true for all n greater than 3, so you can still confirm (by taking the limit as n goes to infinity) that it is a bounded sequence, which was the point of showing that it is a decreasing function.

      Delete
    4. Unknown19 July 2013 at 22:45

      Basically it says some nth prime to some power x is less than the next prime in the sequence raised to one less power than x. Which is true of the sequence in 2). This paper has been published in one of the largest mathematical journals for over 60 years, I might suggest holding back a little bit on thinking you've disproving it next time!

      Delete
  3. Rig19 July 2013 at 20:37

    I believe the the comments referring to a flaw are a due to a typesetting problem with the proof (.pdf) that was linked. The proof shows terms like P_(n+1)^(3-n-1) where it should really be P_(n+1)^3^(-n-1).

    It looks like every place where the exponent 3-n is used, it should read 3^-n, and every place where the exponent 3-n-1 is used it should read 3^-n-1.

    Of course, it's late and I could be wrong

    ReplyDelete
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