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Re: [PrimeNumbers] Interesting Pattern

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  • 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 1 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 2 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 3 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 4 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 5 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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