# Leading digits of Fibonacci sequence

There is a standard problem to show that the distribution of leading digits of $2^n$ is that the digit $k$ occurs with the frequency $\log_{10}(k+1)-\log_{10}(k)$. (This easily generalises to other bases --- though base 2 is rather pointless!)

Since Fibonacci is also exponential except for an error term''. Is this true for that as well --- or does the error term make it fail?

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The error term is $o(1)$, so whatever you have for $2^n$ should also work for the Fibonacci sequence. –  GH from MO Feb 6 '12 at 17:48
How could the error term possibly affect the distribution of the leading digit? –  Qiaochu Yuan Feb 6 '12 at 17:54
The phrasing with the emphasize on the error term is perhaps confusing, but then I do not (fully) understand the remark of GH as the 'base' is not 2 for Fibinacci. Okay, in the end 'the reason' is 'the same'; I will expand my answer a bit. –  quid Feb 6 '12 at 18:07

The key part of the proof is to show that $n \log \phi$ is uniformly distributed modulo $1$ where $\phi$ is the Golden Ratio (the base of the logarithm being the one for which the law should be established). This is achieved using Weyl's equidistribution theorem.