$f(f(n))+f(n+1)=n+2$, where $n \in \mathbb N^*_+$

Question

Does there exist a function $ f:\mathbb{N}_+^* \to \mathbb{N}_+^*$ such that$ f(f(n))+f(n+1)=n+2$, $\forall n\in\mathbb{N}_+^*$ ?

What I found is that $f(1)=1$ because
We have $f(n)\leq n$ Suppose $f(1)=a>1$ For $n=1 f(a)+f(2)=3$ So

1. $f(a)=1 $
   For $n=a-1$ We have $f(f(a-1))=a+1$ which is a contradiction Or

2. $f(a)=2$. For $n=a-1$. We have $$f(f(a-1))=a$$ which is a contradiction

Answer 1

Yes. As you say, $f(1)=1$ and $f(n)\leqslant n$. Write the functional equation as: $$f(f(n-1))+ f(n)= n+1$$ and rearrange: $$f(n) = n+1-f(f(n-1))$$ This gives us a way to iterate through $f(n)$ based on previous values. Note that $f(n-1) \leqslant n-1$ and so is known and so $f(f(n-1)) \leqslant n-1$ and so $f(n)$ will always be $>0$. The pattern seems slightly chaotic: