Guess The Golden Function

Putting n=1 in the functional equation we get $h(h(1))+h(2)=3$

This shows $h(2)≤2$ and $h(h(1))≤2$

Suppose $h(2)=1$ and $h(h(1))=2$

Put $h(1)=k$ so that $h(k)=2$ .Taking $n=2$ in the functional equation we get $h(h(2))+h(3)=4$

We thus see $h(3)=4-h(1)=4-k$ . Since $h(3)≥1$ , we conclude $k≤3$

If $k=1$ ,then we get $2=h(h(1))=h(k)=h(1)=k=1$ contradiction

By putting $k=2,3$ you will get similar contradictions.

We conclude that $h(2)=1$ and $h(h(1))=2$ is not possible

So, $h(2)=2$ and $h(h(1))=1$ and hence $h(3)=4-h(h(2))=2$ .

Inductively we get $h(4)=5-h(h(3)0=5-h(2)=5-2=3$ ,

$h(5)=6-h(h(4))=6-h(3)=4$

$h(6)=7-h(h(5))=7-h(4)=4$ and so on . See that $h(n)≥2$ for $n≥2$

Suppose $h(1)=k≥2$ .Then $3=h(h(1))+h(2)=h(k)+2≥2+2=4$ which is impossible.

Thus $h(1)=1$ and $h(2)=2$

We claim that $h(n)=[n\phi]-n+1$ where $\phi=\frac{1+\sqrt{5}}{2}$

Lemma 1 . For any natural number $n$ , $[\phi([n\phi]-n+1]= n or n+1$

Proof $[\phi([n\phi]-n+1]$ $<$ $\phi(n\phi-n+1$ $=$ $n(\phi^{2}-\phi)+\phi$ $=$ $n+\phi$ $<$ $n+2$

and $[\phi([n\phi]-n+1]$ $>$ $\phi(n\phi-1-n+1)-1$ $=$ $n(\phi^{2}-\phi)-1$ $=$ $n-1$

Lemma 2 For all natural numbers $n$ , $[(n+1)\phi]$ $=$ $[n\phi]+2$ if $[\phi([n\phi]-n+1]$ $=$ $n$ and $[n\phi]+1$ , otherwise

Proof Suppose $[(n+1)\phi$ $=$ $[n\phi]+1$ .Then we get

$[\phi([n\phi]-n+1]= [\phi([(n+1)\phi]-n)]>\phi(\phi(n+1)-1-n)-1=n$

so that $[\phi([n\phi]-n+1)]=n+1$ by lemma 1.On the other hand it can be easily proved if

$[(n+1)\phi]=[n\phi]+2$ , then $[\phi([n\phi]-n+1)]=n$

Suppose the claim is true for $1≤j≤n$

$h(n+1)= n+2-h(h(n))=n+2-h([n\phi]-n+1)$ $=n+2-[\phi([n\phi]-n+1)]+([n\phi]-n+1)-1$ , since $[n\phi]-n+1<2n-n+1=n+1$

This reduces to $h(n+1)=[n\phi]+2-[\phi([n\phi]-n+1)]$

Suppose $n$ is such that $[\phi([n\phi]-n+1)]=n$ . Then we have $[(n+1)\phi]=[n\phi]+2$ and hence $h(n+1)=[(n+1)\phi]-n$

Considering the other possibility, you will also get $h(n+1)=[(n+1)\phi]-n$

This completes our induction step.

So, $h(2014)=[2014\phi]-2014+1=3258-2014+1=1245$