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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