As it is well known fibonacci numbers satisfy the recurrence relation $$F_{n}=F_{n-1}+F_{n-2}$$ with initial conditions $F_{0}=0$ and $F_{1}=1$.
While playing around with numbers,I noticed the following recurrence relation for the fibonacci numbers
$$F_{n}=4F_{n+1}+F_{n-4}-(2F_{n+2}+F_{n-2})$$ Valid for $n\geq 4$
Can anyone enjoy the challenge of proving this simple relation?
It can be seen that: $$F_{n-4} = F_{n-2} - F_{n-3} = - F_{n-1} + 2 F_{n-2} = 2 F_{n} - 3 F_{n-1} = - 3 F_{n+1} + 5 F_{n},$$ $$ F_{n-2} = F_{n} - F_{n-1} = - F_{n+1} + 2 F_{n}$$ and leads to: \begin{align} 4F_{n+1}+F_{n-4}-(a F_{n+2}+F_{n-2}) &= 4 F_{n+1} - 3 F_{n+1} + 5 F_{n} - a F_{n+2} + F_{n+1} - 2 F_{n} \\ &= 2 F_{n+1} - a F_{n+2} + 3 F_{n}\\ &= (2 - a) F_{n+2} + F_{n}. \end{align} When $a =2$ this reduces to $$ 4 F_{n+1} + F_{n-4} - 2 F_{n+2} - F_{n-2} = F_{n}.$$ When $a = 1$ this reduces to $$ 4 F_{n+1} + F_{n-4} - F_{n+2} - F_{n-2} = L_{n+1}$$ where $L_{n}$ are the Lucas numbers.