Question

Is there a way to indicate the # of primes in the prime factorization of a number?

Answer 1

ie n=(p1 ^a1)(p2^a2)...(pk^ak)

is there a function that returns k?

Answer 2

wht grade is tht

Answer 3

Number theory

Answer 4

College

Answer 5

wooow, i m just in 9th grade.. lol

Answer 6

any way i'll try to answer ur question

Answer 7

Haha it's fine. I don't think it's anything too complicated for someone's who taken the course, or even someone who's particularly good at google. I'm just drawing a blank

Answer 8

http://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic
is this the stuff

Answer 9

can u check

Answer 10

Right, I need a function that returns the # of primes in one of those lists though

Answer 11

is it similar to ur stuffff

Answer 12

Like, say 9 has prime factorization 3x3, I would need f(9) = 2
or say 27 has prime factorisation 3x3x3, I would need f(27) =3

Answer 13

It's in the general vicinity yeah

Answer 14

i know wht does f(x) mean

Answer 15

I need f(x) = number of prime factorizations

Answer 16

Whatever that function would be

Answer 17

I need that as a step in my homework, but I can't figure out how to get that notation. I suspect it would be some manipulation of the Euler-Phi function

Answer 18

@iGreen

Answer 19

@ribhu

Answer 20

i'm tryna invite people so they can help u

Answer 21

Yeah I see thanks. I guess if they knew they'd pop in and help anyways haha

Answer 22

wait a while

Answer 23

some one came

Answer 24

@iGreen I'm studying Mobius inversion, Euler phi functions, Mersenne primes etc.

I can't figure out some manipulation to give a function that returns exactly the number of prime divisors of a function.

Answer 25

Okay I gotta commute to class. If anyone sees this while I'm gone please feel free to answer, I'll look at it before class and give medal c:

Answer 26

dude can u just send me a message for a sec

Answer 27

@Hitaro9

Answer 28

what you're looking for doesn't exist, but i think what you need is the function \(\omega\)
see http://mathworld.wolfram.com/DistinctPrimeFactors.html

Answer 29

Just a quick thing to add on cause I kept forgetting but this helped me remember these functions better.

Capital Omega represents the total number of prime factors and lowercase omega returns the number of distinct primes. So we can set up this nice inequality that is cute cause it's suggestive based on just the size of the letters haha. \[\Large \Omega(n) \ge \omega(n)\] So for a quick example: \[\Large \omega(45)=\omega(3^25^1)=2 \\ \Large \Omega(45)=\Omega(3^25^1)=3\]
Too bad there's not really any way to evaluate these except by factoring a number. :/