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Factor patterns

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  • 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 1 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 2 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 3 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 4 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 5 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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              Unsubscribe by an email to: primenumbers-unsubscribe@egroups.com
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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 6 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 7 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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