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I would like to ask the community for feedback regarding the following two conjectures of mine:

$\textbf{Conjecture 1}$ Let $\mathcal{F}_N^- = \{F_n:-N \leq n < 0\}$, i.e. be the set of Fibonacci numbers with negative index up to $-N$. Then we have

$$ \lim_{N \rightarrow \infty} \frac{E\left[\mathcal{F}_N^-\right]}{E\left[\mathcal{F}_{N-1}^-\right]} = \phi, $$

with $E$ being the expectation operator and $\phi$ the golden ratio. $\Box$

$\textbf{Proof Approach}$ Define

$$ \bar{F}_N := E\left[\mathcal{F}_N^-\right] = \frac{1}{N} \sum_{i=-1}^{-N} F_i. $$

Then by the closed form solution of the Fibonacci sequence we have

\begin{eqnarray} \bar{F}_N &=& \frac{1}{\sqrt{5}N}\left[ \sum_{i=-1}^{-N}\phi^i - \sum_{i=-1}^{-N}\Phi^i \right], \end{eqnarray}

with $\Phi = -\frac{1}\phi = 1 - \phi$. The terms inside the brackets (denote them $\phi_N$ and $\Phi_N$ respectively) are then

\begin{eqnarray} \phi_N &=& \phi^{-1} + \phi^{-2} + \ldots + \phi^{-N}\\ &=& \phi - \phi^{1-N} \;\;\; \text{ and}\\ \Phi_{N_{even}} &=& -\phi + \phi^2 - \phi^3 + \ldots - \phi^{N-1} + \phi^N\\ &=& (\phi^2 - \phi) + (\phi^4 - \phi^3) + \ldots + (\phi^N - \phi^{N-1})\\ &=& 1 + \phi^2 + \phi^4 + \ldots + \phi^{N-2} \;\;\; \text{ and}\\ \Phi_{N_{odd}} &=& 1 + \phi^2 + \phi^4 + \ldots + \phi^{N-3} - \phi^N. \end{eqnarray}

We rewrite $\phi^n = F_{n-1} + F_n\phi$, thus

\begin{eqnarray} \Phi_{N_{even}} &=& 1 + (F_1 + F_2\phi) + (F_3 + F_4\phi) + \ldots + (F_{N-3} + F_{N-2}\phi)\\ &=& 1 + (F_1 + F_3 + \ldots + F_{N-3}) + \phi (F_2 + F_4 + \ldots + F_{N-2})\\ &=& 1 + \left(\vphantom{x^2}F_1 + (F_1 + F_2) + (F_3 + F_4) + \ldots + (F_{N-5} + F_{N-4})\right) + \phi\left(\vphantom{x^2}F_1+(F_2+F_3) + \ldots + (F_{N-4} + F_{N-3}) \right)\\ &=& 2 + \sum_{i=1}^{N-4} F_i + \phi \sum_{i=1}^{N-4} F_i + \phi F_{N-3}\\ &=& 2 + \phi^2\sum_{i=1}^{N-4} F_i + \phi F_{N-3} \\ &=& 2 + \phi^2\sum_{i=1}^{N-3} F_i - F_{N-3} \;\;\; \text{ and}\\ \Phi_{N_{odd}} &=& 2 + \phi^2\sum_{i=1}^{N-4} F_i - F_{N-4} - \phi^N \end{eqnarray}

Since $F_1 = F_2 = 1$ and $\phi^n = \phi^{n-1} + \phi^{n-2}$. Now I would assume as I am interested in the limit, that I can calculate $\Phi_N = \frac{1}2 \left(\Phi_{N_{even}} + \Phi_{N_{odd}}\right)$. I define $N_{even} = N$ and $N_{odd} = M$ and I assume that wlog $N + 1 = M$ and

\begin{eqnarray} \Phi_N &=& \frac{1}2 \left(\Phi_{N_{even}} + \Phi_{N_{odd}}\right)\\ &=& 2 + \phi^2\sum_{i=1}^{N-4} F_i + \frac{1}2 \left(\phi^2 F_{N-3} - F_{N-3} - F_{N-4}\right) - \frac{\phi^{N+1}}{2}. \end{eqnarray}

Plugging everything back into $\bar{F}_N$ and taking the ratio we have

\begin{eqnarray} \frac{\bar{F}_N}{\bar{F}_{N-1}} &=& \left(1 - \frac{1}{N} \right) \frac{\phi_N - \Phi_N}{\phi_{N-1} - \Phi_{N-1}}. \end{eqnarray}

Before I begin calculating this monstrum I would like to ask if the steps are correct?


$\textbf{Conjecture 2}$ Let the settings be as above. Then we have

$$ \lim_{N \rightarrow \infty} \frac{Var\left[\mathcal{F}_N^-\right]}{Var\left[\mathcal{F}_{N-1}^-\right]} = \phi^2 \text{ or } e, $$

with $Var$ being the variance and $e$ the Euler number. $\Box$

I strongly suspect it to be $\phi^2$ but numerical simulations show that the convergence is very slow and thus it is hard to tell because $\phi^2$ and $e$ are very close to each other.

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