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RE: [PrimeNumbers] Factor patterns

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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 1 of 9 , Jul 2, 2002
      --- 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 2 of 9 , Jul 2, 2002
        --- 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 3 of 9 , Jul 2, 2002
          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

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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 4 of 9 , Jul 2, 2002
            > 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 5 of 9 , Jul 3, 2002
              (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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