# Leftmost digit of Fibonacci sequence

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?

• Your observation stems from the fact that $\phi^{67}$ is close to a power of 10. – arkeet Jun 6 '18 at 21:00
• 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. – lulu Jun 6 '18 at 21:04
• That makes a lot of sense. Thank you both! – InterstellarProbe Jun 6 '18 at 21:09

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