Software engineers use the Fibonacci numbers in two equally important contexts: to demonstrate their knowledge of recursion during interviews, and to plan their work using story points, which are time estimates but with plausible deniability. It was in the latter context that I noticed something about the Fibonacci sequence that I hadn’t before: each number is a little less than twice the preceding number, and in fact the difference is one of the earlier numbers in the sequence. If $F_n$ is the $n$th Fibonacci number, then

$F_n = 2 F_{n-1} - F_{n-3}$

I always find these kinds of things easier to understand with examples—and I was in a JavaScript-y mood this morning—so here’s an interactive visualization of this fact.

This property is easy to prove. Let’s number the Fibonacci numbers $F_n$ such that $F_1 = 1$ and $F_2 = 1$, and let’s only consider the case where $n \geq 4$, since our result is nonsensical if we go any earlier in the sequence than that. The $n$th Fibonacci number is defined as

$F_n = F_{n-1} + F_{n-2} . \tag{1}$

Likewise, the previous number is given by

$F_{n-1} = F_{n-2} + F_{n-3} .$

Rearranging these terms, $F_{n-2} = F_{n-1} - F_{n-3} .$

If we plug this expression for $F_{n-2}$ back into (1), we get

\begin{align*} F_n &= F_{n-1} + \left( F_{n-1} - F_{n-3} \right) \\ &= 2 F_{n-1} - F_{n-3} , \end{align*}

exactly as we wanted to prove. Nothing earth-shattering here, but hey—you try to make it through a backlog grooming meeting without doing any recreational mathematics.