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RE: [PrimeNumbers] Re: Disection of n by roots

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  • Jon Perry
    ... I ve just tried this: ? rf(371293) %7 = [0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] ? rf(371294) %8 = [1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0,
    Message 1 of 7 , Oct 26, 2002
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      > 371293 = 13^5
      > 371294 = 2*7*11*2411
      >
      >Each of these two numbers has a [1/5,1/6) factor.

      I've just tried this:

      ? rf(371293)
      %7 = [0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
      ? rf(371294)
      %8 = [1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0]

      However:

      ? for (n=371290,400000,if (rf(n)[5]>0 && rf(n+1)[5]>0,print(n)))
      371305
      371436
      371448
      371579
      371591

      gives some results.

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


      -----Original Message-----
      From: Jon Perry [mailto:perry@...]
      Sent: 24 October 2002 22:19
      To: primenumbers@yahoogroups.com
      Subject: RE: [PrimeNumbers] Re: Disection of n by roots


      >Say that N & N+1 each have a [1/5,1/6) factor. (I don't think the
      >word "root" is what you want...)

      Yeah, you're right.

      > 371293 = 13^5
      > 371294 = 2*7*11*2411
      >
      >Each of these two numbers has a [1/5,1/6) factor.

      Good work. Could you possibly find an N such that it has two common
      [1/n,1/(n+1)] factors with N+1?

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


      -----Original Message-----
      From: jbrennen [mailto:jack@...]
      Sent: 24 October 2002 20:07
      To: primenumbers@yahoogroups.com
      Subject: [PrimeNumbers] Re: Disection of n by roots


      --- In primenumbers@y..., "Jon Perry" <perry@g...> wrote:
      > I personally am having trouble finding a pair of consecutive
      > integers which both have a [1/5,1/6) root:
      >
      > ... This doesn't (tested to ~200000)

      Say that N & N+1 each have a [1/5,1/6) factor. (I don't think the
      word "root" is what you want...)

      Then we must have two distinct primes p & q, with p<q, such that:

      p^6 >= N+1

      and

      q^5 <= N+1

      Which of course implies that p^6 >= q^5.

      It is easy to see that the smallest such (p,q) pair is (11,13).

      Plug in q=13 and get q^5 == 371293 <= N+1.

      So N >= 371292.

      Quick examination shows the first such pair:

      371293 = 13^5
      371294 = 2*7*11*2411

      Each of these two numbers has a [1/5,1/6) factor.




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