<s>[INST] How to solve recurrence equation $f(n) = f(n-5) + f(n-10)$? How to solve recurrence equation $f(n) = f(n-5) + f(n-10)$? For something like fibonacci sequence $f(n+1) = f(n) + f(n-1)$ I can solve for the quadratic equation $x^2-x-1=0$ then $f(n) = A x_1 + A^\prime x_2$. But what should I do for this one?
 [/INST] Although a topic in Discrete math, I attempted this question using the tools I learned in linear algebra.
Rewrite it as $f(n+10) = f(n+5) +f(n)$, this can be represented as
\[ \left( \begin{array}{c} 
f_{n+10} \\
f_{n+5} \end{array} \right) = \left( \begin{array}{cc} 
1 &1 \\
1& 0 \end{array} \right) \left( \begin{array} {c} 
f_{n+5} \\
f_n \end{array} \right)
 \]
\[ \left( \begin{array}{c} 
f_{n+10} \\
f_{n+5} \end{array} \right) = \left( \begin{array}{cc} 
1 &1 \\
1& 0 \end{array} \right)^{n+5} \left( \begin{array} {c} 
f_{1} \\
f_{-4} \end{array} \right)
 \] 
Then diagonalize  $\left( \begin{array}{cc} 
1 &1 \\
1& 0 \end{array} \right)$ to find eigenvalues, which is the same as the ones for fibonacci, $1/2 (1-\sqrt5), 1/2 (1+\sqrt5)$
then $f_{n+5} = \alpha (1/2 (1-\sqrt5))^{n+5} +\beta (1/2 (1+\sqrt5))^{n+5} $ 
Is this correct? If not please advise. 
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