Which is more; even or odd positive factors?

Question

Suppose $f(n)=$ $\{$ ( number of $n$'s positive even factors) $-$ (number of $n$'s positive odd factors) $\}$

How can we prove/disprove the below statement?

$f(n)< 0 $ for half or more than numbers from $1$ to $2012^{2013}$

 

There is a positive number for $n$ such that $f(n)=2013$

Please edit the question by giving a good language.

Answer 1

For the first statement: if $n$ is odd, $n$ has no even factors. Therefore, $f(n) = (0 - \text{number of n's positive odd factors}) \lt 0$ for odd $n$. $2012^{2013}$ is even, so exactly half of the possible $n \in [1, 2012^{2013}]$ are odd, implying that the first statement is true.

Answer 2

As to the second question, let $n=2^23^{60}5^{10}7^2$. Then the odd divisors of $n$ are those numbers of the form $3^a\times 5^b\times 7^c$ where $a\in \{0,60\}$, $b\in \{0,10\}$ and $c\in \{0,2\}$. thus there are $61$ choices for $a$, $11$ for $b$ and $3$ for $c$ so the total number of odd divisors of $n$ is $61\times11\times3=2013$. As to the even divisors, they are of the form $2d$ or $4d$ where $d$ is an odd divisor of $n$. Thus there are $2\times 2013$ even divisors. Hence $f(n)=2013$.

Answer 3

For the second statement, let $n=2^{2014}$.