Does there exist a function whose range is the complement of the range of this function? Does this function have a complement on the positive integers in the sense that the range of the complement should be the complement of the range of the function?    
$$\frac{1}{8} \left(1-(-1)^n+2 n (2+n)\right)$$
Here is the inverse/complement which I got from Generic Human:
$$\left\lfloor \sqrt{4 n-1}\right\rfloor +n+1$$    
Edit per @robjohn:  
Domain is $\mathbb{N}$. I'm trying to use the Lambek-Moser theorem to generate complimentary sets.  
The formula generates the pronic numbers interleaved with the squares.
$\{1,2,4,6,9,12,16,20,25,30\}$
I want to generate the complementary set of everything else.
$\{3,5,7,8,10,11,13,14,15,17\}$
The Lambek-Moser theorem uses the inverse to generate the complement.
My question is: am I interpreting this scenario properly?
 A: $f(-2)=f(-1)=f(0)=0$ so if the domain is the integers it does not have an inverse. In fact, $f(n)=f(-2-n)$
However, this function is monotonic increasing on the non-negative integers, so it does have an inverse if the domain is the non-negative integers.

Suppose
$$
m=\frac18\left(1-(-1)^n+2n(2+n)\right)
$$
If $n$ is even, then $(n+1)^2=4m+1$
If $n$ is odd, then $(n+1)^2=4m$
Therefore, here is the inverse:
If $4m$ is a perfect square, then $n=\sqrt{4m}-1$
If $4m+1$ is a perfect square, then $n=\sqrt{4m+1}-1$
If neither $4m$ nor $4m+1$ is a perfect square, then $m$ is not in the range of the function.
A: write $a_n = \frac{1}{8} \left(1-(-1)^n+2 n (2+n)\right)$ then 
$a_{n+1}-a_n=\frac{1}{8} \left(1-(-1)^n+2(n+1)(3+n)-(1-(-1)^n+2 n (2+n)\right)
=$ $\frac{1}{8} ( (-1)^{n+2}+{-1}^n+10n+6)) \geq \frac{1}{8} (-2+6+10n) >0$ since $a_n$ is increasing it is $1-1$ and it can be inverted using as domain its range.
A: Note: If all you care about is existence of such a function, then the Lambek-Moser theorem guarantees the existence of such a function simply because $F$ is increasing. But if you want a nice closed-form expression, then Lambek-Moser doesn't tell you that it exists, it can just help you find it.
In the context of the Lambek-Moser theorem, you have the increasing function from $\mathbb N^*\to\mathbb N^*$:
$$F(2k)=k(k+1)\\
F(2k+1)=(k+1)^2$$
Define $f(n)=F(n)-n$, which happens to coincide with $F(n-2)$:
$$f(2k)=k(k-1)\\
f(2k+1)=(k-1)^2$$
Since $f(2k)=(k-1/2)^2-1/4$, we can write $$f(n)=\left\lfloor \frac{(n-1)^2}{4} \right\rfloor$$
Given $n\ge 1$, we're looking for $p\ge 0$ such that
$$f(p)<n\le f(p+1)$$
$$\frac{(p-1)^2}{4}<n\le \frac{p^2}{4}$$
$$(p-1)^2<4n\le p^2$$
$$(p-1)^2\le 4n-1<p^2$$
$$p-1=\lfloor \sqrt{4n-1} \rfloor$$
giving us finally:
$$f^*(n)=1+\lfloor \sqrt{4n-1} \rfloor$$
$$\boxed{G(n)=1+n+\left\lfloor \sqrt{4n-1} \right\rfloor}$$
so that $\{F(n):n\ge 1\}$ and $\{G(n):n\ge 1\}$ form a partition of $\mathbb N^*$.
