If $f(2n)=\frac1{f(n)+1}$ and $f(2n+1)=f(n)+1$ for all $n\in\Bbb N$, then find $n$ such that $f(n)=14/5$.

Question

The set $\mathbb{N}$ is the set of nonnegative integers. Let $ f : \Bbb{N} \rightarrow \Bbb{Q}$ be defined such that

1.) $f(2n) = \dfrac{1}{f(n)+1}$ for all integers $n>0$, and

2.) $f(2n + 1 ) = f(n)+1$ for all $n\in\Bbb{N}$.

If $f(0)=0$, then determine the value of $n\in\mathbb{N}$ such that $f(n)=\dfrac{14}{5}$.

Answer 1

Since $f(n)>0$ for $n>0$, $f(n)>1$ for odd $n$ and $f(n)<1$ for even $n$, you can see that if $f(n)=\frac{14}{5}$, $n$ must be odd, that is $n = 2n_1+1$, where $f(n_1) = \frac 95$. Proceeding in this manner, you get $$ n_1 = 2n_2+1, n_2 = 2n_3, n_3 = 2n_4, n_4=2n_5+1, n_5=2n_6+1, n_6 = 2n_7+1, n_7=0 $$

going back, the answer is $n=115$.

Answer 2

One readily sees that $f(n)>0$ for all $n>0$. It follows that $f(n)>1$ only if $n$ is even and $0<f(n)<1$ only if $n$ is odd. From this, $f(n_1)=\frac{14}5$ implies $n_1=2n_2$ with $f(n_2)=\frac95$. Then $n_2=2n_3$ with $f(n_3)=\frac45$. Then $n_3=2n_4+1$ with $f(n_4)=\frac14$. Then ... proceed until you hit $1$ (and then $0$); this will happen because the sum of numerator and denominator is decreasing.