4
$\begingroup$

So, on another math forum, someone posted a question about the leftmost digit of a randomly chosen Fibonacci number. It likely follows Benford's Law for the distribution of the leftmost digit.

I know the rightmost digit is cyclic with a cycle of 60, so I wanted to check if there was any kind of repetition for the leftmost digit. It seems that $F(n+67)$ and $F(n)$ have the same first digit a surprisingly high percentage of the time (my calculations put it at around 97% for the first 1000 Fibonacci numbers). I tried to see if this was understood at all, or just another mystery of the Fibonacci numbers, but after a few hours of searching, I did not come across anything.

You can see it in action with this graph:

http://www.wolframalpha.com/input/?i=FLOOR%5BFibonacci%5Bn%2B67%5D%2F10%5E(FLOOR(Log%5BFibonacci%5Bn%2B67%5D%5D%2FLog%5B10%5D))%5D-FLOOR%5BFibonacci%5Bn%5D%2F10%5E(FLOOR(Log%5BFibonacci%5Bn%5D%5D%2FLog%5B10%5D))%5D

Anyway, back to my question. Does anyone know of any research related to this?

$\endgroup$
  • 8
    $\begingroup$ Your observation stems from the fact that $\phi^{67}$ is close to a power of 10. $\endgroup$ – arkeet Jun 6 '18 at 21:00
  • 3
    $\begingroup$ For large $n$ you can approximate $F_n\approx \frac 1{\sqrt 5}\times \left(\frac {1+\sqrt 5}2 \right)^n$ and you can read off the first digit by taking $\log_{10}$ of the right hand. $\endgroup$ – lulu Jun 6 '18 at 21:04
  • $\begingroup$ That makes a lot of sense. Thank you both! $\endgroup$ – InterstellarProbe Jun 6 '18 at 21:09
0
$\begingroup$

Community wiki answer so the question can be marked as answered:

The question has been answered in the comments. The observed regularity is due to the proximity of $\phi^{67}$ to a power of $10$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.