Injectivity and Surjectivity of a Function We have the set $S$ that is the set of odd positive integers. A function $F:\mathbb{N} \rightarrow S$ is defined by $F(n)=k$ for each $n\in \mathbb{N}$, where $k$ is an odd positive integer for which $3n+1=2^mk$ for some non-negative integer $m$. Prove or disprove the following:
(1) $F$ is injective. (2) $F$ is surjective.
I have thought of ways I could start to answer this question, but my attempts have been fruitless. Help would very much be appreciated for this question. 
Million thanks in advance.

For any integer $n$, calculate $3n+1$. Once you have $3n+1$ write that as a product by taking out as many factors of $2$ as possible. The remaining odd factor is the value of $F(n)$.
If $n=1$, then $3n+1 = 4$, and $4=2^2 \times 1$, so $m=2$ and $k=1$. Hence $F(1)=1$
If $n=2$, then $3n+1 = 7$, and $7 = 2^0 \times 7$, so $m=0$ and $k=7$. Hence $F(2)=7$
If $n=3$, then $3n+1 = 10$, and $10 = 2^1 \times 5$, so $m=1$ and $k=5$. Hence $F(3)=5$
If $n=4$, then $3n+1 = 13$, and $13 = 2^0 \times 13$, so $m=0$ and $k=13$. Hence $F(4) = 13$
If $n=5$, then $3n+1 = 16$, and $16 = 2^4 \times 1$, so $m=4$ and $k=1$. Hence $F(5) = 1$
