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Interesting Pattern

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  • Milton Brown
    I think that the following is an interesting pattern. (It may have been studied before, I would appreciate a reference.) 10^18+1 is divisble by 10^6+1 =
    Message 1 of 6 , Aug 2, 2001
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      I think that the following is an interesting pattern.
      (It may have been studied before, I would appreciate
      a reference.)

      10^18+1 is divisble by 10^6+1 = 101*9901
      10^20+1 " " 10^4+1 = 73*137
      10^24+1 " " 10^8+1 = 17*5882353
      10^35+1 " " 10^5+1 = 11* 9091

      and these patterns seem to repeat often.

      Milton L. Brown
      miltbrown@...



      [Non-text portions of this message have been removed]
    • Jack Brennen
      ... All of these are algebraic factorizations. Specifically, a^(2n+1)+1 == (a+1) * (a^2n - a^(2n-1) + ... + a^2 - a + 1) For instance, set a=10^5 and n=3:
      Message 2 of 6 , Aug 2, 2001
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        Milton Brown wrote:
        >
        > I think that the following is an interesting pattern.
        > (It may have been studied before, I would appreciate
        > a reference.)
        >
        > 10^18+1 is divisble by 10^6+1 = 101*9901
        > 10^20+1 " " 10^4+1 = 73*137
        > 10^24+1 " " 10^8+1 = 17*5882353
        > 10^35+1 " " 10^5+1 = 11* 9091

        All of these are algebraic factorizations. Specifically,

        a^(2n+1)+1 == (a+1) * (a^2n - a^(2n-1) + ... + a^2 - a + 1)

        For instance, set a=10^5 and n=3:

        10^35+1 == (10^5+1) * (10^30-10^25+10^20-10^15+10^10-10^5+1)

        Note that a^n+1 (with a >= 2, n >= 2) can only be prime if n is a power
        of two, as a direct consequence of the above algebraic factorization;
        if n has an odd non-trivial factor k, then a^n+1 has the non-trivial
        factor a^(n/k)+1.
      • Jud McCranie
        ... These are algebraic factors. +---------------------------------------------------+ ... +---------------------------------------------------+
        Message 3 of 6 , Aug 2, 2001
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          At 08:14 PM 8/2/2001 -0700, Milton Brown wrote:
          >I think that the following is an interesting pattern.
          >(It may have been studied before, I would appreciate
          >a reference.)
          >
          >10^18+1 is divisble by 10^6+1 = 101*9901
          >10^20+1 " " 10^4+1 = 73*137
          >10^24+1 " " 10^8+1 = 17*5882353
          >10^35+1 " " 10^5+1 = 11* 9091

          These are algebraic factors.
          +---------------------------------------------------+
          | "In order to think outside the box one first must |
          | know what is inside the box." - Michael Shermer |
          | |
          | Jud McCranie |
          +---------------------------------------------------+
        • d.broadhurst@open.ac.uk
          Jack Brennen wrote ... and it has further algebraic factorizations because x^n + 1 = product_{d|n,d is odd} Phi(2*n/d,x) which with n=35 gives (x + 1)* (x^4 -
          Message 4 of 6 , Aug 2, 2001
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            Jack Brennen wrote

            > 10^35+1 == (10^5+1) * (10^30-10^25+10^20-10^15+10^10-10^5+1)

            and it has further algebraic factorizations because

            x^n + 1 = product_{d|n,d is odd} Phi(2*n/d,x)

            which with n=35 gives

            (x + 1)*
            (x^4 - x^3 + x^2 - x + 1)
            (x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)*
            (x^24 + x^23 - x^19 - x^18 - x^17 - x^16 + x^14 + x^13 + x^12 + x^11
            + x^10 - x^8 - x^7 - x^6 - x^5 + x + 1)

            With x=10, the last algebraic factor is composite.
            The Cunningham project thus records:

            > 35 (1,5,7) 4147571.265212793249617641
          • Milton Brown
            Yes, yours is a nice analysis. I thought that an interesting part of the patterns, was also 9901 (9909901,...) and 9091 (90901,...) which also quite often are
            Message 5 of 6 , Aug 2, 2001
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              Yes, yours is a nice analysis.

              I thought that an interesting part of the patterns,
              was also 9901 (9909901,...) and 9091 (90901,...)
              which also quite often are prime. I think David
              indicated where these have been studied before.

              Milton L. Brown
              miltbrown@...


              ----- Original Message -----
              From: "Jack Brennen" <jack@...>
              To: <primenumbers@yahoogroups.com>
              Sent: Thursday, August 02, 2001 8:26 PM
              Subject: Re: [PrimeNumbers] Interesting Pattern


              > Milton Brown wrote:
              > >
              > > I think that the following is an interesting pattern.
              > > (It may have been studied before, I would appreciate
              > > a reference.)
              > >
              > > 10^18+1 is divisble by 10^6+1 = 101*9901
              > > 10^20+1 " " 10^4+1 = 73*137
              > > 10^24+1 " " 10^8+1 = 17*5882353
              > > 10^35+1 " " 10^5+1 = 11* 9091
              >
              > All of these are algebraic factorizations. Specifically,
              >
              > a^(2n+1)+1 == (a+1) * (a^2n - a^(2n-1) + ... + a^2 - a + 1)
              >
              > For instance, set a=10^5 and n=3:
              >
              > 10^35+1 == (10^5+1) * (10^30-10^25+10^20-10^15+10^10-10^5+1)
              >
              > Note that a^n+1 (with a >= 2, n >= 2) can only be prime if n is a power
              > of two, as a direct consequence of the above algebraic factorization;
              > if n has an odd non-trivial factor k, then a^n+1 has the non-trivial
              > factor a^(n/k)+1.
              >
              >
              > 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/
              >
              >
            • d.broadhurst@open.ac.uk
              Milton Brown wrote ... The latter is ID Number: A001562 (Formerly M3767 and N1537) Sequence: 5,7,19,31,53,67,293,641,2137,3011 Name: (10^p + 1)/11 is
              Message 6 of 6 , Aug 3, 2001
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                Milton Brown wrote

                > 9901 (9909901,...) and 9091 (90901,...)
                > which also quite often are prime. I think David
                > indicated where these have been studied before.

                The latter is

                ID Number: A001562 (Formerly M3767 and N1537)
                Sequence: 5,7,19,31,53,67,293,641,2137,3011
                Name: (10^p + 1)/11 is prime.
                References J. Brillhart et al., Factorizations of b^n +- 1.

                in the EIS. But the former does not figure in the
                Cunningham project tables for cyclotomic
                factors of 10^n +/- 1.

                Perhaps you meant

                f(n)=(10^(6*n)+1)/(10^(2*n)+1)

                f(1)= 9901 is prime
                f(2)= 99990001 is prime
                f(3)= 999999000001 is prime
                f(4)= 9999999900000001 is prime

                Puzzle: Find the next prime of this form :-)

                David
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