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equidistribution of Moebius function

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  • Kermit Rose
    1. Re: The distribution of Moebius(n) Posted by: Phil Carmody thefatphil@yahoo.co.uk thefatphil Date: Sun Jun 18, 2006 10:13 am (PDT) The thing that is not
    Message 1 of 4 , Jun 19, 2006
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      1. Re: The distribution of Moebius(n)
      Posted by: "Phil Carmody" thefatphil@... thefatphil
      Date: Sun Jun 18, 2006 10:13 am (PDT)


      The thing that is not intuitively clear to me is that those that remain,
      those
      with mu=+1 and -1, should be equidistributed. If anyone has a good heuristic
      argument I'm all ears.

      Phil


      *****

      Equidistribution is related to the equidistribution of odd and even in
      binomial formula

      (1 - 1 )^ m = 0

      Sum of positive coeficients = magnitude of sum of negative coeficients.


      The square free integers are the coeficients in the formal expansion of

      (1 + 2) (1 + 3)(1 +5)(1+7)...(1+p)......

      = 1 + 2 + 3 + 6 + 5 + 10 + 15 + 30 +
      7 + 14 + 21 + 42 + 35 + 70 + 105 + 210 + ...


      The number of terms in the formal expansion of

      m products ( 1 + p1)(1 + p2) (1+p3)....(1 + pm) is 2^m.

      half of these terms correspond to a product of an even number of primes,
      and
      half of these terms correspond to a product of an odd number of primes.

      Kermit < kermit@... >
    • Phil Carmody
      ... You ve re-ordered two subsequences of the squarefree integers, and provided a bijection between those two subsequences. Regarding the re-ordering, note
      Message 2 of 4 , Jun 20, 2006
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        --- Kermit Rose <kermit@...> wrote:
        >> The thing that is not intuitively clear to me is that those that remain,
        >> those
        >> with mu=+1 and -1, should be equidistributed. If anyone has a good heuristic
        >> argument I'm all ears.
        >
        > Equidistribution is related to the equidistribution of odd and even in
        > binomial formula
        >
        > (1 - 1 )^ m = 0
        >
        > Sum of positive coeficients = magnitude of sum of negative coeficients.
        >
        > The square free integers are the coeficients in the formal expansion of
        >
        > (1 + 2) (1 + 3)(1 +5)(1+7)...(1+p)......
        >
        > = 1 + 2 + 3 + 6 + 5 + 10 + 15 + 30 +
        > 7 + 14 + 21 + 42 + 35 + 70 + 105 + 210 + ...
        >
        >
        > The number of terms in the formal expansion of
        >
        > m products ( 1 + p1)(1 + p2) (1+p3)....(1 + pm) is 2^m.
        >
        > half of these terms correspond to a product of an even number of primes,
        > and
        > half of these terms correspond to a product of an odd number of primes.

        You've re-ordered two subsequences of the squarefree integers, and provided a
        bijection between those two subsequences. Regarding the re-ordering, note that
        the largest member of the two subsequences are different by a factor of 2.

        This same type of argument can provide a bijection between the primes and the
        square-free non-primes, despite the fact that the former has zero density in
        the squarefree integers.

        So from my PoV, it's not a good enough heuristic as it stands, sorry. It was
        close, though.

        Phil

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      • Bob Gilson
        I doubt that there is a good heuristic. As important as Moebius(n) itself is, the cumulative value is more relevant, i.e. the number you get if you add
        Message 3 of 4 , Jun 22, 2006
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          I doubt that there is a good heuristic.

          As important as Moebius(n) itself is, the cumulative value is more relevant, i.e. the number you get if you add Moebius(1)+Moebius(2)+Moebius(3)...+Moebius(k) for some number k. This is Mertens Function M(k). Its first 10 values for arguments k=1,2,3... up to 10, are 1,0,-1,-1,-2,-1,-2,-2,-2,-1.

          M(k) is a very irregular function, oscillating back and forth in a "random walk".

          For arguments 1000,2000 ...up to 10,000 it has values: 2,5,-6,-9,2,0,-25,-1,1,-23.

          For arguments 1,000,000, 2,000,000 to 10,000,000, it has values 212,-247,107,192,-709,257,-184,-189,-340,1037. Ignoring the signs, it's evident that the size of M(k) increases, but nothing else is clear.

          Cheers

          Bob

          Phil Carmody <thefatphil@...> wrote:
          --- Kermit Rose <kermit@...> wrote:
          >> The thing that is not intuitively clear to me is that those that remain,
          >> those
          >> with mu=+1 and -1, should be equidistributed. If anyone has a good heuristic
          >> argument I'm all ears.
          >
          > Equidistribution is related to the equidistribution of odd and even in
          > binomial formula
          >
          > (1 - 1 )^ m = 0
          >
          > Sum of positive coeficients = magnitude of sum of negative coeficients.
          >
          > The square free integers are the coeficients in the formal expansion of
          >
          > (1 + 2) (1 + 3)(1 +5)(1+7)...(1+p)......
          >
          > = 1 + 2 + 3 + 6 + 5 + 10 + 15 + 30 +
          > 7 + 14 + 21 + 42 + 35 + 70 + 105 + 210 + ...
          >
          >
          > The number of terms in the formal expansion of
          >
          > m products ( 1 + p1)(1 + p2) (1+p3)....(1 + pm) is 2^m.
          >
          > half of these terms correspond to a product of an even number of primes,
          > and
          > half of these terms correspond to a product of an odd number of primes.

          You've re-ordered two subsequences of the squarefree integers, and provided a
          bijection between those two subsequences. Regarding the re-ordering, note that
          the largest member of the two subsequences are different by a factor of 2.

          This same type of argument can provide a bijection between the primes and the
          square-free non-primes, despite the fact that the former has zero density in
          the squarefree integers.

          So from my PoV, it's not a good enough heuristic as it stands, sorry. It was
          close, though.

          Phil

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          [Non-text portions of this message have been removed]
        • Phil Carmody
          ... Good -- elegant -- you might be right. However, I think that I can formulate a trend to look for that would imply the equidistribution of the +1 and -1
          Message 4 of 4 , Jun 23, 2006
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            --- Bob Gilson <bobgillson@...> wrote:
            > I doubt that there is a good heuristic.

            Good -- elegant -- you might be right. However, I think that I can
            formulate a trend to look for that would imply the equidistribution
            of the +1 and -1 values.

            For squarefree numbers in the range [N-N*eps, N+N*eps), create an
            expression in terms of N for the number or the proportion of them
            that have f \in {1, 2, 3, ... } factors. I believe such expressions
            exist, as they certainly do for f=1, the primes. Then accumulate
            the terms for odd and even f's, et viola - perhaps there is an
            expression which has a simple limit.

            Annoyingly, I suspect each term in the expression would (very slowly)
            tend (as a proportion) to zero as N increases, so one could not simply
            take a finite number of terms, nor even forget any earlier terms. I
            think it would be a very nasty expression indeed.

            The error bounds on this expression would also be fairly hideous, and
            unfortunately that might matter. Euler's gamma shows how much the
            primes can connive with their non-randomness.

            Phil

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