Expert: Paul Klarreich - 7/31/2009

Question
Hi , I have question in number theory but all experts are on vacation , I thought you may help me in this question :
let ω(n) denote the number of distinct prime divisor of n>1, with ω(1)=0,for instance ω(360)=ω(2^3.3^2.5)=3
a)show that 2^ω(n)is a multipicative function
b)for a positive integer n, establish the formula
τ(n^2)=∑ 2^ω(d)...the summation start from d/n.Thank you sooooo much

Answer
Questioner: Mike
Country: United States
Category: Advanced Math
Private: No
Subject: Math
Question: Hi , I have question in number theory but all experts are on vacation , I thought you may help me in this question :
let ω(n) denote the number of distinct prime divisors of n>1, with ω(1)=0,for instance ω(360)=ω(2^3.3^2.5)=3
a)show that 2^ω(n)is a multipicative function
b)for a positive integer n, establish the formula
τ(n^2)=∑ 2^ω(d)...the summation start from d/n.Thank you sooooo much
..................................
Let's see if I understand this:

w(6) = 2,  because   6 = 2 3
w(14) = 2, because  14 = 2 7
w(84) = 3, because  84 = 2^2 3 7

Now your function f(n) = 2^w(n), and so:

f(6) = 2^2 = 4
f(14) = 2^2 = 4

f(84) = 2^3 = 8 and
f(6) f(14) = 4*4 = 8, right?

Perhaps something went wrong. Perhaps I am not understanding what you wrote.