Estimating how many of the first $10,000$ Fibonacci numbers start with the digit $9$

Question

Consider the problem of estimating how many of the first $10,000$ Fibonacci numbers begin with the digit $9$.

The only ideas I have so far:

• Obviously, if we assume that the every first digit is equally likely, the answer is around $1000$ (Note: Dietrich Burde points out that is wrong. $0$ can't be the first digit, so I should divide by $9$, not $10$).
• Listing out the Fibonacci numbers: $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, ...$, we can see that their first digits, $1, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 6, 9, 1, ...$ seem to follow a pattern somewhat similar to $1, 2, 3, 5, 8$, with $9$'s introduced less often. So perhaps the answer is less than $1000$.

I'll put the answer here:

456 of the first $10,000$ Fibonacci numbers start with the digit $9$.

Any ideas/hints of how to estimate or compute this analytically?

Answer 1

The Fibonacci numbers are roughly $\varphi^n \cdot \frac{1}{\sqrt{5}}$. Asking whether these numbers start with a 9 is the same as asking whether their logs mod 1 fall between log 9 and log 10. But those logs are $n \log(\varphi) - .5 \log(5)$. Since $\log(\varphi)$ is irrational, we expect multiples of it (mod 1) to be evenly distributed around the mod 1 circle.

Answer 2

According to Benford's law, at least from a statistical point of view, you should get around $$10000\left(\log10-\log9\right)=457.5749\dots$$ which is quite close to your calculated value. That being said, it's not an exact calculation.