Determine is there a function $a: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$i)$ $a(0)=0$
$ii)$ $a(n)=n-a(a(n))$, $\forall n \in$ $ \mathbb{N}$.
If exists prove:
$a)$ $a(k)\geq a(k-1)$
$b)$ Does not exist positive integer $k$ such that $a(k-1)=a(k)=a(k+1)$.
gobathegreat wrote:
Determine is there a function $a: \mathbb{N} \rightarrow \mathbb{N}$ such that:
$i)$ $a(0)=0$
$ii)$ $a(n)=n-a(a(n))$, $\forall n \in$ $ \mathbb{N}$.
If exists prove:
$a)$ $a(k)\geq a(k-1)$
$b)$ Does not exist positive integer $k$ such that $a(k-1)=a(k)=a(k+1)$.
I suppose your definition of $\mathbb N$ is $\mathbb Z_{\ge 0}$. If so :
$a(n)$ is trivially injective and $n\ge a(n)\ge 0$
So $a(1)\in\{0,1\}$ and $a(1)\ne a(0)=0$
So $a(1)=1$ which is unfortunately wrong.
So $\boxed{\text{No such function}}$