This can't be a coincidence

\[ \begin{eqnarray} 0. && 00000 \quad 00001 \quad 00001 \quad 00002 \quad 00003 \quad 00005 \quad 00008 \\ && 00013 \quad 00021 \quad 00034 \quad 00055 \quad 00089\quad 00144\quad \ldots \\ \end{eqnarray} \]

The above shows the first few digits (actually 65) of the decimal representation of the fraction \( \large \frac1{9,999,899,999}. \) If we split the digits into partitions of 5, we can see that the numbers form a Fibonacci sequence: \(0,1,1,2,3,5,8,13,\ldots \). How many positive Fibonacci numbers can we find before the pattern breaks off?

Details and Assumptions:

  • For example, suppose that the fraction equals \(0.00000 \quad 00001 \quad 00001 \quad 00002 \quad 00003 \quad 00009 \ldots \) instead of the one given at the top. Then you could only find the first five Fibonacci numbers, namely \(0,1,1,2,3\). So your answer would then be that there are 4 positive Fibonacci numbers before the pattern breaks off.

Bonus: Generalize this.

Try Daniel Liu's problem that was inspired by this problem.


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