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

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 $$1000$$.

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

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

• But the number cannot start with digit $0$, so dividing by $10$ is not right for the first point. See also quora for your question, using Benford's law. Aug 27, 2023 at 18:22
• Look at Benford's law : en.wikipedia.org/wiki/Benford%27s_law ; Numbers starting wigh 9 should be 4.6%, and you find 4.56%, totally conform to the law. Aug 27, 2023 at 18:27
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.
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.