Negative Linear recurrence Given any (homogeneus) linear recurrence equation of some order (defined for positive integers), and a solution of that sequence. How are related the solution for the extension to negative integers of that equation with the positive solution? Is there any general result?
Trivial example: Fibonaccis Numbers.
Fibonaccis recurrence is defined by the formula:
\begin{equation}
f_n= f_{n-1}+f_{n-2}
\tag{1}
\end{equation}
The solution to this equation (given the usual initial conditions) is:
\begin{equation}
f_n= \frac{\varphi^n+(1-\varphi)^n}{\sqrt5}
\end{equation}
The eqn. (1) can be mapped to:
\begin{equation}
f_{-n}= f_{-n+1}-f_{-n+2}
\end{equation}
for negative integers, and the solution (with the same initial conditions) is related with the "positive solution" as:
\begin{equation}
f_{-n}= (-)^n f_n
\tag{2}
\end{equation}
I would love to know (even if you do not know the solution for positive integers): Is there any formula that link the two solutions? (like in the eqn (2))
 A: We approach your problem with some generalizations of the Fibonacci sequence, e.g., $f_n = af_{n-1} + bf_{n - 2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas, and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a, b \in \mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” Applied Probability Trust; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)
We have extended Maynard's analysis to include arbitrary $f_0, f_1 \in \mathbb{R}$. It is relatively straightforward to show that
$$f_n = \left(f_1 - \frac{af_0}{2}\right) \frac{\alpha^n - \beta^n}{\alpha -
 \beta} + \frac{f_0}{2} (\alpha^n + \beta^n) $$
where $$\alpha, \beta = \frac{a \pm\sqrt{a^2 + 4b}}{2}.$$
The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1 \ \text{and} \ f_0=0, f_1=1$.
We can now look at the negative recurrence, namely
$$f_{-n} = \left(f_1 - \frac{af_0}{2}\right) \frac{\alpha^{-n} - \beta^{-n}}{\alpha -
 \beta} + \frac{f_0}{2} (\alpha^{-n} + \beta^{-n}) $$
Let's simplify this a bit by assuming that we have either Fibonacci-type sequences, $f_0=0$, or Lucas-type sequences, $\left(f_1 - \frac{af_0}{2}\right)=0$. Then we have $f_{-n}=(-1)^nf_n$ provided that
$$
\alpha^{-n}=(-1)^n\beta^n\\\
\beta^{-n}=(-1)^n\alpha^n
$$
You can show that this is true for the Fibonacci and Lucas sequences, since $\alpha=\varphi \text{ and } \beta=-1/\varphi$. It is also true for the Pell and Pell-Lucas sequences for which $\alpha, \beta=1\pm\sqrt{2}$. But it would not be true for Jacobsthal and Jacobsthal-Lucas sequences for which $\alpha, \beta=2, -1$.
This should give you sufficient material to work on negative sequences for the generalized Fibonacci sequence, which you can see depends on the sequence coefficients ($a,b$) as well as the initial conditions ($f_0,f_1$).
