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

Re: Pythagorean Triplets

Expand Messages
  • Mark Rodenkirch
    ... http://listserv.nodak.edu/scripts/wa.exe?A2=ind0012&L=nmbrthry&P=R529&D=0&H=0&O=T&T=1 I looked at your link and it is quite interesting. I have a couple
    Message 1 of 19 , Feb 26, 2004
    • 0 Attachment
      >
      http://listserv.nodak.edu/scripts/wa.exe?A2=ind0012&L=nmbrthry&P=R529&D=0&H=0&O=T&T=1

      I looked at your link and it is quite interesting. I have a couple of
      comments though. You mention that G0(n) = n*(1+i)^n + 1 and is
      related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
      go on to show primes of that form. Am I missing something? n*(1+i)^n
      + 1 =/= n*2^(n/2) + 1. The same could be said of G2(n) and Woodalls.

      You also have Ne(n) as n^2*2^n + 1, which pre-dates the Hyper-Cullen
      search of Steven Harvey. He has noted your finds as he searches up to
      200000.

      --Mark
    • mikeoakes2@aol.com
      In a message dated 26/02/04 14:20:52 GMT Standard Time, mgrogue@wi.rr.com ... It is if n = 0 mod 8, which was (one of) the values I was talking about.
      Message 2 of 19 , Feb 26, 2004
      • 0 Attachment
        In a message dated 26/02/04 14:20:52 GMT Standard Time, mgrogue@...
        writes:


        > You mention that G0(n) = n*(1+i)^n + 1 and is
        > related to Cullens but later state that G0(n) = n*2^(n/2) + 1 and then
        > go on to show primes of that form. Am I missing something? n*(1+i)^n
        > + 1 =/= n*2^(n/2) + 1.
        >
        It is if n = 0 mod 8, which was (one of) the values I was talking about.
        Remember: (1+i)^2 = 2*i.

        -Mike Oakes


        [Non-text portions of this message have been removed]
      Your message has been successfully submitted and would be delivered to recipients shortly.