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Factor patterns

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  • 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 1 of 9 , Jun 30, 2002
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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 2 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 3 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 4 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 5 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 6 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 7 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 8 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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