Loading ...
Sorry, an error occurred while loading the content.
 

Number crunch

Expand Messages
  • Jon Perry
    We hear a lot about n^2+1 containing a infinite number of primes. A similar function may well be: floor(n^(3/2))+/-k (and obviously 3/2 is a variable too) Does
    Message 1 of 6 , May 1, 2002
      We hear a lot about n^2+1 containing a infinite number of primes.

      A similar function may well be:

      floor(n^(3/2))+/-k

      (and obviously 3/2 is a variable too)

      Does this function, or any variants, contain an infinite nuber of primes?

      Jon Perry
      perry@...
      http://www.users.globalnet.co.uk/~perry/maths
      BrainBench MVP for HTML and JavaScript
      http://www.brainbench.com
    • Phil Carmody
      ... Good question. The forms grow slowly enough for them to statistically be expected to contain an infinite number of primes (anything sub-exponential has
      Message 2 of 6 , May 1, 2002
        --- Jon Perry <perry@...> wrote:
        > We hear a lot about n^2+1 containing a infinite number of primes.
        >
        > A similar function may well be:
        >
        > floor(n^(3/2))+/-k
        >
        > (and obviously 3/2 is a variable too)
        >
        > Does this function, or any variants, contain an infinite nuber of
        > primes?

        Good question. The forms grow slowly enough for them to statistically
        be expected to contain an infinite number of primes (anything
        sub-exponential has this property).

        Thay also probably have no divisor properties which would cause them
        to deviate from a simple statistically predicted density.
        (which isn't true about a lot of forms).

        Phil



        __________________________________________________
        Do You Yahoo!?
        Yahoo! Health - your guide to health and wellness
        http://health.yahoo.com
      • Jon Perry
        I came up with this: integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the primes, and further more: integral(x=1,infinity of sigma(n=1,2 of
        Message 3 of 6 , May 2, 2002
          I came up with this:

          integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the primes, and
          further more:

          integral(x=1,infinity of sigma(n=1,2 of x^n))==sigma(n=1,2 of
          integral(x=1,infinity of x^n)

          Jon Perry
          perry@...
          http://www.users.globalnet.co.uk/~perry/maths
          BrainBench MVP for HTML and JavaScript
          http://www.brainbench.com


          -----Original Message-----
          From: Phil Carmody [mailto:thefatphil@...]
          Sent: 01 May 2002 23:08
          To: primenumbers
          Subject: Re: [PrimeNumbers] Number crunch


          --- Jon Perry <perry@...> wrote:
          > We hear a lot about n^2+1 containing a infinite number of primes.
          >
          > A similar function may well be:
          >
          > floor(n^(3/2))+/-k
          >
          > (and obviously 3/2 is a variable too)
          >
          > Does this function, or any variants, contain an infinite nuber of
          > primes?

          Good question. The forms grow slowly enough for them to statistically
          be expected to contain an infinite number of primes (anything
          sub-exponential has this property).

          Thay also probably have no divisor properties which would cause them
          to deviate from a simple statistically predicted density.
          (which isn't true about a lot of forms).

          Phil



          __________________________________________________
          Do You Yahoo!?
          Yahoo! Health - your guide to health and wellness
          http://health.yahoo.com


          Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
          The Prime Pages : http://www.primepages.org



          Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
        • Phil Carmody
          ... I m fighting your notation somewhat. I assume sigma is summation? Is sigma(n=1,2 of x^n) just x+x^2 = x(x+1) ? The integral from 1..+oo of x(x+1) diverges.
          Message 4 of 6 , May 2, 2002
            --- Jon Perry <perry@...> wrote:
            > I came up with this:
            >
            > integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the
            > primes, and
            > further more:
            >
            > integral(x=1,infinity of sigma(n=1,2 of x^n))==sigma(n=1,2 of
            > integral(x=1,infinity of x^n)

            I'm fighting your notation somewhat. I assume sigma is summation?
            Is sigma(n=1,2 of x^n) just x+x^2 = x(x+1) ?
            The integral from 1..+oo of x(x+1) diverges.
            What is "contains" supposed to mean in the context "+oo contains all
            the primes"?
            So I'm fighting your terminology as well as your notation.

            Phil


            __________________________________________________
            Do You Yahoo!?
            Yahoo! Health - your guide to health and wellness
            http://health.yahoo.com
          • Jon Perry
            First, a couple of typos: Should be intergral(n=1+e,...) as the integral from 1 includes Z, and is trivial. Also the function should be floor(x^n), not x^n,
            Message 5 of 6 , May 3, 2002
              First, a couple of typos:

              Should be intergral(n=1+e,...)

              as the integral from 1 includes Z, and is trivial.

              Also the function should be floor(x^n), not x^n, although it could equally
              be any variety.

              Second, the integral can be refined:

              As n^2>=2n for n>=2, (hence by Bertrand there is a prime), we can use the
              identity:

              n^x=2n

              Taking logs

              xlogn=log2+logn

              x=log2/logn + 1

              Hence we can use the equation:

              sigma(n=1,infinity, of integral(x=1+e,log2/logn + 1, of floor(n^x))

              And this contains all the primes.

              What can the bounds be reduced to?

              Jon Perry
              perry@...
              http://www.users.globalnet.co.uk/~perry/maths
              BrainBench MVP for HTML and JavaScript
              http://www.brainbench.com
            • Jon Perry
              I got a bit twisted. integral(x=1+e,2, of sigma(n=1,infinity, of floor(n^x))) i.e. the numbers: 1^(1+e) through to 1^2 + 2^(1+e) through to 2^2 + ... The point
              Message 6 of 6 , May 3, 2002
                I got a bit twisted.

                integral(x=1+e,2, of sigma(n=1,infinity, of floor(n^x)))

                i.e. the numbers:

                1^(1+e) through to 1^2
                +
                2^(1+e) through to 2^2
                +
                ...

                The point being that by Bertands, every prime is represented by the
                integral, (and the sum is infinite, but I am not currently concerned with
                this).

                The second statement is the identity that the sigma and integral are
                interchangeable:

                sigma(n=1,infinity, of integral(x=1+e,2, of floor(n^x)))

                Jon Perry
                perry@...
                http://www.users.globalnet.co.uk/~perry/maths
                BrainBench MVP for HTML and JavaScript
                http://www.brainbench.com


                -----Original Message-----
                From: Phil Carmody [mailto:thefatphil@...]
                Sent: 02 May 2002 20:05
                To: primenumbers
                Subject: RE: [PrimeNumbers] Number crunch


                --- Jon Perry <perry@...> wrote:
                > I came up with this:
                >
                > integral(x=1,infinity of sigma(n=1,2 of x^n)) contains all the
                > primes, and
                > further more:
                >
                > integral(x=1,infinity of sigma(n=1,2 of x^n))==sigma(n=1,2 of
                > integral(x=1,infinity of x^n)

                I'm fighting your notation somewhat. I assume sigma is summation?
                Is sigma(n=1,2 of x^n) just x+x^2 = x(x+1) ?
                The integral from 1..+oo of x(x+1) diverges.
                What is "contains" supposed to mean in the context "+oo contains all
                the primes"?
                So I'm fighting your terminology as well as your notation.

                Phil


                __________________________________________________
                Do You Yahoo!?
                Yahoo! Health - your guide to health and wellness
                http://health.yahoo.com


                Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
                The Prime Pages : http://www.primepages.org



                Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
              Your message has been successfully submitted and would be delivered to recipients shortly.