On a recent ((at least, recent at the time of writing; the SF alone knows how far into the future this will run)) episode of Wrong, But Useful, Dave noted that 33 of the first 100 Fibonacci numbers were even, 333 of the first 1000, and so on.

My reaction wasn’t quite as it should have been: I said something like “well, yes, obviously”. While there’s a not-too-difficult reason behind it (which I’ll explain shortly), it’s a pretty nice exercise in algebraic proof, and the kind of thing that GCSE students ought to be playing with.

## So why is it obvious?

The Fibonacci sequence starts 1, 1, 2, 3, 5, 8, …, with each term the sum of the previous two.

As soon as you start looking at the pattern of odds and evens, the reason becomes clear: the parities go odd, odd, even and repeat. That’s natural, especially if you think of how the sequence is generated:

• First term: odd
• Second term: odd
• Third term: odd + odd (so even)
• Fourth term: odd + even (so odd)
• Fifth term: even + odd (so odd)
• etc.

Every third term turns out to be even!

So, of the first 99 Fibonacci numbers, 33 are even (and the 100th is odd). Similarly, of the first 999, 333 are even (and the 1000th is odd).

## My reaction

I don’t like my reaction to this, which is that it’s not an especially surprising observation ((And not just because Dave made it.)) I think it’s the straightforwardness of the pattern that does it - it doesn’t feel much more interesting than noting that 50 of the first hundred integers are even, for example.

For this sort of thing to make me go ‘oo!’ - like, for example, finding $\pi$ in colliding blocks - I think the pattern needs to be a bit more convoluted.

So, while it’s probably a good GCSE exercise, I’m afraid it just doesn’t tickle me.

And in the meantime, you whippersnappers can get off of my lawn.