## Probability of a number being a mobius number...

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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
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.
• 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