Sequence : $f(f(n))+f(n+1)=2n$ Does there exist a function $f : \mathbb{Z}^+ \to \mathbb{Z}^+$  such that $f(f(n))+f(n+1)=2n$, $\forall n \in \mathbb{Z}^+$ ?
My attempt :
Substitute $n=1$, $f(f(1))+f(2)=2$ so $f(f(1))=f(2)=1$ 
Substitute $n=2$, $f(f(2))+f(3)=4$ so $f(1)+f(3)=4$ 
Substitute $n=3$, $f(f(3))+f(4)=6$ 
Assume that there exists $k >1$ such that $f(k) \geq k$,
then $f(f(k-1))+f(k)=2k$, we have $f(f(k-1))<k$
 A: Yes there exists such a function.
Let's define $g(x)$ as follows:
$$ g(n) = 
             \left\{ \begin{array}{lcc}
             \ \ \ n \ \ \ \ \ \ \ ,  
             & 
             \mbox{ if } n 
             \mbox{ is odd } , \\
             \\ n-1 \ \ \ ,   
             & 
             \mbox{ if } n 
             \mbox{ is even } . 
             \end{array}
   \right.$$ 
One can check easilly that $g(n)$ satisfies the relation 
$g(g(n))+g(n+1)=2n$, $\forall n \in \mathbb{N}$ . 
A: For a positive integer $n$, let $P(n)$ denote the condition that $f\big(f(n)\big)+f(n+1)=2n$.  The condition $P(1)$ implies that $f(2)=1$.  Therefore, $P(2)$ leads to $f(1)\in\{1,2,3\}$. Hence, there are only three functions satisfying the condition, and they are listed below.
If $f(1)=1$, then $P(2)$ gives $f(3)=3$.  Then, $P(3)$ gives $f(4)=3$.  By induction on the positive integer $k$, we can easily see that
$$f(2k-1)=2k-1\text{ and }f(2k)=2k-1$$
for every $k=1,2,3,\ldots$.  In other words,
$$f(n)=2\,\left\lfloor\frac{n}{2}\right\rfloor+1$$
for all $n\in\mathbb{Z}_{>0}$.
If $f(1)=2$, then $P(2)$ implies $f(3)=2$.  Then, $P(3)$, $P(4)$, and $P(5)$ lead to $f(4)=5$, $f(5)=4$, and $f(6)=5$.  It can be proven again by induction that 
$$f(3k-2)=3k-1\,,\,\,f(3k-1)=3k-2\,,\text{ and }f(3k)=3k-1$$
for all $k=1,2,3,\ldots$.  In other words,
$$f(n)=3\,\left\lceil\frac{n}{3}\right\rceil-1-\left\lfloor\frac{n\mod 3}{2}\right\rfloor$$
for $n=1,2,3,\ldots$.
If $f(1)=3$, then $P(1)$ gives $f(3)=1$.  Then, $P(3)$, $P(4)$, and $P(5)$ yield $f(4)=5$, $f(5)=4$, and $f(6)=5$, respectively.  We claim that, for all positive integers $k$,
$$f(3k+1)=3k+2\,,\,\,f(3k+2)=3k+1\,,\text{ and }f(3k+3)=3k+2\,.$$
However, this claim can be easily verified by induction on $k$.  Thus,
$$f(n)=3\,\left\lceil\frac{n}{3}\right\rceil-1-\left\lfloor\frac{n\mod 3}{2}\right\rfloor$$
for $n=4,5,6,\ldots$, with $f(1)=3$, $f(2)=1$, and $f(3)=1$.
