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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?

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    $\begingroup$ 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. $\endgroup$ Aug 27, 2023 at 18:22
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    $\begingroup$ 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. $\endgroup$
    – Lourrran
    Aug 27, 2023 at 18:27
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    $\begingroup$ See also A008963 at OEIS. $\endgroup$ Aug 27, 2023 at 19:20

2 Answers 2

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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.

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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.

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