Problem

Source: Vietnam TST 2019 Day 2 P6

Tags: combinatorics



In the real axis, there is bug standing at coordinate $x=1$. Each step, from the position $x=a$, the bug can jump to either $x=a+2$ or $x=\frac{a}{2}$. Show that there are precisely $F_{n+4}-(n+4)$ positions (including the initial position) that the bug can jump to by at most $n$ steps. Recall that $F_n$ is the $n^{th}$ element of the Fibonacci sequence, defined by $F_0=F_1=1$, $F_{n+1}=F_n+F_{n-1}$ for all $n\geq 1$.