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Probability of a number being a mobius number...

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  • Sudarshan Iyengar
    Its been a thrill to me to use the pari to factorize large numbers. A very small observation which i made is what follows:- A mobius function M(n) is defined
    Message 1 of 2 , Jul 8, 2005
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      Its been a thrill to me to use the pari to factorize large numbers. A
      very small observation which i made is what follows:-

      A mobius function M(n) is defined as follows:

      M(n) = 0 if p^2|n for some prime number p.
      M(n)= -1 if n has odd number of prime divisors
      M(n)=+1 if n has even number of prime divisors.

      A number n is said to be a mobius number if M(n)=+1 or -1
      meaning there is no p such that p^2|n

      The question:-
      The probability that a given big number n is a mobius number is very
      high. What is the mathematical reasoning behind this.
    • Sarad AV
      hi Sudarshan, ... You might also like to try out MIRACL s factoring package below. http://indigo.ie/~mscott/#other ... According to mathworld (URL below) the
      Message 2 of 2 , Jul 8, 2005
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        hi Sudarshan,

        > Its been a thrill to me to use the pari to factorize
        > large numbers.

        You might also like to try out MIRACL's factoring
        package below.

        http://indigo.ie/~mscott/#other

        > The question:-
        > The probability that a given big number n is a
        > mobius number is very
        > high. What is the mathematical reasoning behind
        > this.

        According to mathworld (URL below) the asymptotic
        density of square free numbers is 6/(pi^2) which is
        approximately 60%. So you should hit a squareful
        number nearly 40% of the time. With what probability
        did you find that the number under test to be
        squarefree and for how many integers did you test?

        http://mathworld.wolfram.com/Squarefree.html


        Sarad.




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