Probability of a number being a mobius number...
- 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 probability that a given big number n is a mobius number is very
high. What is the mathematical reasoning behind this.
- hi Sudarshan,
> Its been a thrill to me to use the pari to factorizeYou might also like to try out MIRACL's factoring
> large numbers.
> The question:-According to mathworld (URL below) the asymptotic
> The probability that a given big number n is a
> mobius number is very
> high. What is the mathematical reasoning behind
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?
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