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The Fibonacci numbers denoted by $F_i$ for $i\ge1$ are $$1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,\cdots$$ where they satisfy the property $F_{i+2}=F_{i+1}+F_i$.

I have listed the first $15$ numbers of the sequence as they will be useful for reference later.

Now define $\theta_i$ as the concatenation of $1.F_i$ so that $1<\theta_i<2$.

So for example, $$\theta_1=1.1,\quad\theta_6=1.13,\quad\theta_{15}=1.987.$$

Next, consider the following $3\times3$ system of equations: $$\begin{bmatrix}\theta_i&\theta_{i+1}&\theta_{i+2}\\\theta_{i+4}&\theta_{i+5}&\theta_{i+6}\\\theta_{i+8}&\theta_{i+9}&\theta_{i+10}\end{bmatrix}\begin{bmatrix}X\\Y\\Z\end{bmatrix}=\begin{bmatrix}\theta_{i+3}\\\theta_{i+7}\\\theta_{i+11}\end{bmatrix}$$ from which we can solve for $[X\quad Y\quad Z]^T$ but of course this would be very tedious.

When $i=1,2$, both $X$ and $Y$ are negative but $Z$ is positive. When $i=3$, the opposite occurs.

Question: For $i\le n$, what is the probability that exactly one of $X,Y,Z$ will be negative?

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  • $\begingroup$ Isn't the sequence $1,1,2,\cdots $ ? $\endgroup$ – tarit goswami Sep 13 '18 at 18:37
  • $\begingroup$ Dang @TheSimpliFire doesn't know something? I didn't know such a thing was possible. $\endgroup$ – Don Thousand Sep 13 '18 at 18:38
  • $\begingroup$ @taritgoswami Conventionally, but in this question I have omitted the first term to avoid duplication. $\endgroup$ – TheSimpliFire Sep 13 '18 at 18:38
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    $\begingroup$ @taritgoswami It doesn't change the fundamentals of the problem, just an off by 1 shift. Focus on the problem instead $\endgroup$ – Don Thousand Sep 13 '18 at 18:41
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    $\begingroup$ One big problem is that it is difficult to give an explicit formula for $\theta_i$ since the number of digits in $F_i$ depends on the integer part of $i\log_{10}((1+\sqrt{5})/2)$. $\endgroup$ – marty cohen Sep 13 '18 at 19:54

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