Is there a theory about functions whose results are "negative" to the results of other functions? Let's say we have a function $x \cdot 2$, which gives results:
$2,4,6,8,10,\ldots$. We can construct "negative" function $x \cdot 2-1$, which results will be: $1,3,5,7,9,\ldots$. So the all natural numbers will be a sum of both results. But could be such a "negative" function be created for $x \cdot 3$? Is there a theory which analyze it?
Added: One more try in explaining what do I mean:
Every function have a set of results. Those sets can be combined like any sets. Now the question is: has been invented any methods and/or theory behind them, that would allow for transforming one function into another one, based on the relation of their result sets. So in my example: I know the function $x*2$ and its result set, and I would like to find another function which set would "complete" the natural numbers. I know that would be $x*2-1$, but it's just intuition. How could I find that kind of function for $x*3$? I'm seeking for theory/methods that could guide me to the solution to those kind of problems.
 A: Let me try to formalize your question.
Call $A \subseteq \mathbb{N}$ cofinal iff for all natural numbers $n$, there exists $a \in A$ such that $a \geq n$. Call $A$ doubly cofinal iff both $A$ and $A^c$ are cofinal. Whenever $A \subseteq \mathbb{N}$ is cofinal, lets write $\underline{A} : \mathbb{N} \rightarrow \mathbb{N}$ for corresponding the "enumerating function."
For instance:
$$\underline{2\mathbb{N}}(n) = 2n, \qquad \underline{(2\mathbb{N})^c}(n) = 2n-1$$
Your question is essentially:

Question. Is there any theory surrounding the following problem?
Given a set $A \subseteq \mathbb{N}$ that is doubly cofinal, such that a simple and straightforward expression is known for the function $\underline{A},$ find a simple and straightforward expression for $\underline{A^c}.$

For instance, its clear that:
$$\underline{3\mathbb{N}}(n) = 3n$$
But you'd like to be able to write: $$\underline{(3\mathbb{N})^c}(n) = \;???$$ where the question marks are replaced by something simple and useful.
I don't know the answer to this question, but it seems like a natural and worthwhile thing to think about. +1
Edit. It seems possible that 
$$\underline{(3\mathbb{N})^c}(n) = \mathrm{floor}\frac{n+2}{3}+\mathrm{floor}\frac{n+1}{3}$$
A: Write $\mathbb N^+$ to denote the set of integers that are greater than zero
(the positive integers).
Suppose $f: \mathbb N^+ \to \mathbb N^+$ is an increasing function, so that it enumerates a subset of $\mathbb N^+$ in increasing order.
Write $f(\mathbb N^+)$ to denote the image of $f$. You want a function
$f^\complement: \mathbb N^+ \to \mathbb N^+$ whose image is the complement of
the image of $f$ in $\mathbb N^+$, that is,
$f^\complement(\mathbb N^+)\cup f(\mathbb N^+) = \mathbb N^+$
and $f^\complement(\mathbb N^+)\cap f(\mathbb N^+) = \emptyset$.
The set of the $f(n)$ smallest elements of $\mathbb N^+$ is the union
of the set of the $n$ smallest elements of $f(\mathbb N^+)$
(the $n$ smallest numbers you can produce by taking $f(x)$)
and the $f(n) - n$ smallest elements of $f^\complement(\mathbb N^+)$.
That is, if $f(n) > n$,
the integers $\{1,\ldots,f(n)\}$ include the integers
$\{f^\complement(1),\ldots,f^\complement(f(n)-n)\}$
but not $f^\complement(f(n)-n+1)$.
That is, $f^\complement(x) > f(n)$ if and only if $x > f(n) - n$,
that is, $x + n > f(n)$
So one way we can be sure to find out the value of $f^\complement(x)$
is to find the minimum $n$ such that $x + n \leq f(n)$.
This ensures that for this value of $n$,
$f^\complement(x) < f(n)$ 
(the definition of $f^\complement$ rules out equality)
and if $n > 1$, then $f(n - 1) < f^\complement(x)$.
For $n > 1$, then, there are exactly $n - 1$ members of
$f(\mathbb N^+)$ that are less than $f^\complement(x)$, that is, the function
$f^\complement$ "skips" $n-1$ numbers less than $f^\complement(x)$.
It follows that $f^\complement(x) = x + n - 1.$
And if $n = 1$, then $f^\complement(x) < f(1)$, no numbers are "skipped",
and $f^\complement(x) = x = x + n - 1$.
The conclusion is: if $n$ is the least integer such that $x + n \leq f(n)$,
then $f^\complement(x) = x + n - 1.$
The sentence "$n$ is the least integer such that $x + n \leq f(n)$"
can be written
$$
n = \min \{t\in\mathbb N^+ \mid x + t \leq f(t) \}
$$
and so
$$
f^\complement(x) = x + \min \{t\in\mathbb N^+ \mid x + t \leq f(t) \} - 1.
$$
This might not be the way you were hoping to write the sought-for function,
but it is actually a relatively simple formula.
A formula in the particular case $f: x \mapsto 3x$ that you might like is
$$
f^\complement(x) = \left\lfloor \frac{3x - 1}{2} \right\rfloor,
$$
where $\lfloor y\rfloor$ means the greatest integer that is not
greater than $y$.
The way I came upon this formula was to consider that in the first $3n$
positive integers, $n$ of them are in the image of $f$ and $2n$ of them
are in the image of $f^\complement$; on average, $f^\complement(x)$
must increase by $3$ for each time we add $2$ to $x$.
This gives the approximation $f^\complement(x) \approx \frac32 x$.
To make it exact, we apply $\lfloor\cdot\rfloor$
(which makes all the results be integers) and "adjust" the results by 
adding or subtracting a constant term inside the brackets
so the answer comes out right. 
In particular, $\left\lfloor\frac32\cdot2\right\rfloor = 3$,
but $f^\complement(2) = 2$, indicating that we should subtract something
from $\frac32\cdot2$; but $f^\complement(1) = 1$, so the most we can
subtract is $\frac12$ (because $\frac32\cdot1 - 1 = \frac12$).
So we try $\left\lfloor\frac32 x - \frac12\right\rfloor$, simplify it to
$\left\lfloor\frac{3x - 1}{2}\right\rfloor$, 
and check that it works, which it does.
