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Factor patterns

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  • Markus Frind
    Given 12n-1 is it possible to prove that all factors of 12n-1 add up to 12*k ? Note this breaks down when you have more then 2 factors, but if you if you
    Message 1 of 9 , Jun 30 6:35 PM
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      Given 12n-1 is it possible to prove that all factors of 12n-1 add up to
      12*k ? Note this breaks down when you have more then 2 factors, but if
      you if you limit yourself to 2 factors (prime or composite) it works. The
      same also holds true for 6n-1 does this have to do with the fact that all
      primes can be expressed as either 6k-1 or 6k+1 ?

      12*45 -1 = 7 * 7 *11
      49 + 11 = 60
      7 + 77 = 84

      12*8-1 = 95
      19+5 = 24
      etc
    • djbroadhurst
      Markus observed: If a*b=-1 mod 12 then a+b=0 mod 12. Proof is easy: The only residue pairs for {a,b} mod 12 are {1,11} or {5,7}. It s true for both of these
      Message 2 of 9 , Jun 30 8:53 PM
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        Markus observed:
        If a*b=-1 mod 12 then a+b=0 mod 12.
        Proof is easy: The only residue pairs
        for {a,b} mod 12 are {1,11} or {5,7}.
        It's true for both of these pairs,
        so we are done.
        David
      • djbroadhurst
        Likewise, if a*b=-1 mod 24 then a+b=0 mod 24 with residue pairs {1,23},{5,19},{7,17},{11,13} since n^2=1 mod 24 for all n with gcd(n,24)=1 It obviously fails
        Message 3 of 9 , Jul 1, 2002
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          Likewise, if a*b=-1 mod 24 then a+b=0 mod 24
          with residue pairs {1,23},{5,19},{7,17},{11,13}
          since n^2=1 mod 24 for all n with gcd(n,24)=1

          It obviously fails mod 48:
          5*19 = -1 mod 48
          5+19 != 0 mod 48
          and mod 120:
          7*137 = -1 mod 120
          7+137 != 0 mod 120
          (yes, I know I could have used {7,17},
          but 137 took my fancy:-)

          David
        • Jon Perry
          A swerve. Can you explain why sigma(n-1)+sigma(n)+sigma(n+1) is nearly always even? (PARI/GP) for
          Message 4 of 9 , Jul 1, 2002
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            A swerve.

            Can you explain why sigma(n-1)+sigma(n)+sigma(n+1) is nearly always even?

            (PARI/GP)

            for (n=2,1000,write("sigmax.txt",(sigma(n-1)+sigma(n)+sigma(n+1))%2))

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


            -----Original Message-----
            From: Markus Frind [mailto:flames@...]
            Sent: 01 July 2002 02:35
            To: primenumbers@yahoogroups.com
            Subject: [PrimeNumbers] Factor patterns


            Given 12n-1 is it possible to prove that all factors of 12n-1 add up to
            12*k ? Note this breaks down when you have more then 2 factors, but if
            you if you limit yourself to 2 factors (prime or composite) it works. The
            same also holds true for 6n-1 does this have to do with the fact that all
            primes can be expressed as either 6k-1 or 6k+1 ?

            12*45 -1 = 7 * 7 *11
            49 + 11 = 60
            7 + 77 = 84

            12*8-1 = 95
            19+5 = 24
            etc



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          • Phil Carmody
            ... Look at sigma(n)%2. Mostly 0. Each of the many remotely isolated 1s will cause 3 consecutive 1s in your expression. 101 causes a 11011 in yours, and 11
            Message 5 of 9 , Jul 2, 2002
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              --- Jon Perry <perry@...> wrote:
              > A swerve.
              >
              > Can you explain why sigma(n-1)+sigma(n)+sigma(n+1) is nearly always
              > even?
              >
              > (PARI/GP)
              >
              > for
              > (n=2,1000,write("sigmax.txt",(sigma(n-1)+sigma(n)+sigma(n+1))%2))


              Look at sigma(n)%2.

              Mostly 0.

              Each of the many remotely isolated 1s will cause 3 consecutive 1s in
              your expression. 101 causes a 11011 in yours, and 11 causes 1001 in
              yours.

              No magic.

              Phil

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            • cashogor
              ... Are there any more numbers sharing this property? N.
              Message 6 of 9 , Jul 2, 2002
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                --- In primenumbers@y..., "djbroadhurst" <d.broadhurst@o...> wrote:
                > Likewise, if a*b=-1 mod 24 then a+b=0 mod 24
                > with residue pairs {1,23},{5,19},{7,17},{11,13}
                > since n^2=1 mod 24 for all n with gcd(n,24)=1


                Are there any more numbers sharing this property?

                N.
              • Jon Perry
                Jack wrote: Because sigma(x) is odd if and only if x is a square or twice a square. The density of x such that sigma(x) is odd == 0. [jp :
                Message 7 of 9 , Jul 2, 2002
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                  Jack wrote:

                  Because sigma(x) is odd if and only if x is a square or
                  twice a square.

                  The density of x such that sigma(x) is odd == 0.

                  [jp : http://mathworld.wolfram.com/DivisorFunction.html (eqn28)


                  Try finding n such that n divides sigma(n-1)+sigma(n)+sigma(n+1).
                  Heuristics would indicate an infinite number of such n...

                  2, 5, 7, 33, 18336, 19262, 38184, 54722, ...

                  I've just submitted this to the EIS.

                  ---

                  From the link, the equations can be morphed into:

                  (n-1).sigma(n-1) = 2 + i.phi(n-1)
                  n.sigma(n) = 2 + j.phi(n)
                  (n+1).sigma(n+1) = 2 + k.phi(n+1)

                  n(n^2-1).sigma(n(n^2-1)) = [2 + i.phi(n-1)].[2 + j.phi(n)].[2 + k.phi(n+1)]

                  so the answers occur whenever the RHS contains n^2 as a factor.

                  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 July 2002 13:35
                  To: primenumbers
                  Subject: RE: [PrimeNumbers] Factor patterns


                  --- Jon Perry <perry@...> wrote:
                  > A swerve.
                  >
                  > Can you explain why sigma(n-1)+sigma(n)+sigma(n+1) is nearly always
                  > even?
                  >
                  > (PARI/GP)
                  >
                  > for
                  > (n=2,1000,write("sigmax.txt",(sigma(n-1)+sigma(n)+sigma(n+1))%2))


                  Look at sigma(n)%2.

                  Mostly 0.

                  Each of the many remotely isolated 1s will cause 3 consecutive 1s in
                  your expression. 101 causes a 11011 in yours, and 11 causes 1001 in
                  yours.

                  No magic.

                  Phil

                  =====
                  --
                  "One cannot delete the Web browser from KDE without
                  losing the ability to manage files on the user's own
                  hard disk." - Prof. Stuart E Madnick, MIT.
                  So called "expert" witness for Microsoft. 2002/05/02

                  __________________________________________________
                  Do You Yahoo!?
                  Sign up for SBC Yahoo! Dial - First Month Free
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                • djbroadhurst
                  ... I believe not; you can do it mod 6, mod 12, mod 24 and then there are exceptions. Easy to see why: we require a modulus M such that n^2=1 mod M for all n
                  Message 8 of 9 , Jul 2, 2002
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                    > Are there any more numbers sharing this property?
                    I believe not; you can do it mod 6, mod 12, mod 24
                    and then there are exceptions.
                    Easy to see why: we require a modulus M such that
                    n^2=1 mod M for all n with gcd(n,M)=1
                    M=24 is easy.
                    We cannot have any larger M with gcd(M,5)=1,since
                    5^2-1=24.
                    So try M=12*5.
                    Now we have problems with
                    7^2-1=48
                    unless M=12*5*7.
                    Now we have problems with
                    11^2-1=120
                    It's clear enough that we are are losing out,
                    though I do not have a proof..
                    David
                  • Jon Perry
                    (sorry about the complete tripe in previous post - the n.sigma(n) eqv. 2 mod phi(n) is only true for n=p, and n=4,6,22). -- Continuing Jack s work: for
                    Message 9 of 9 , Jul 3, 2002
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                      (sorry about the complete tripe in previous post - the n.sigma(n) eqv. 2 mod
                      phi(n) is only true for n=p, and n=4,6,22).

                      --

                      Continuing Jack's work:

                      for (n=2,100000,if ((sigma(n-1)+sigma(n)+sigma(n+1))%(n-1)==0,
                      write("sigmaconn.txt",n,":",sigma(n-1)%(n-1),":",+sigma(n)%(n-1),":",sigma(n
                      +1)%(n-1),":",(sigma(n-1)+sigma(n)+sigma(n+1)))))

                      2,3,24,64,227,291,784,1883,7731,18547,25723,30397,94358

                      for (n=2,100000,if ((sigma(n-1)+sigma(n)+sigma(n+1))%(n+1)==0,
                      write("sigmaconp.txt",n,":",sigma(n-1)%(n+1),":",+sigma(n)%(n+1),":",sigma(n
                      +1)%(n+1),":",(sigma(n-1)+sigma(n)+sigma(n+1)))))

                      8,21,22,23,57,157,505,1053,2147,2273,3311,4679,5931,7898,22682

                      --

                      A related question - when does sigma(n,2)%sigma(n,1)==0?

                      If n is a square, and also:

                      20,50,117,180,200,242,325,450,468,500,578,605,650,800,968,980

                      for (n=2,10000, if (sigma(n,2)%sigma(n,1)==0 &&
                      !issquare(n),write("sigmasigmasq.txt",n)))

                      Anyone spot the missing link?

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