Fibonacci convergence

We know from Fibonacci and the golden ratio that the n-th term of Fibonacci can be expressed as :

$$F_{n} = \frac{\varphi^{n}}{\sqrt{5}} = \frac{\left(\frac{1+\sqrt{5}}{2} \right)^n}{\sqrt{5}}$$

Searching for a number with at least 1000 digit is the same as searching a for number that is greater than or equal to $10^{999}$

$$\begin{align} \frac{\varphi^{n}}{\sqrt{5}} & >= 10^{999}\\ n * \log(\varphi) - \frac{\log(5)}{2} & >= 999 * log(10)\\ n & >= \frac{\frac{\log(5)}{2} + 999}{\log(\varphi)}\\ n & = \left\lceil\frac{\frac{\log(5)}{2} + 999}{\log(\varphi)} \right\rceil\\ \end{align}$$

Since n must be an integer, it is sufficient to take the ceiling from the previous equation.

From solution2.py:

def thousandth_digit_fibonacci_number(n=1000):
    return ceil((n - 1 + log10(sqrt(5)) / 2) / log10((1 + sqrt(5)) / 2))