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RE: [PrimeNumbers] Factor patterns

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  • Jon Perry
    A swerve. Can you explain why sigma(n-1)+sigma(n)+sigma(n+1) is nearly always even? (PARI/GP) for
    Message 1 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 2 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 3 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 4 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

            __________________________________________________
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            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 5 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 6 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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