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?
 A: In this case it is exactly Fibonacci recurrences, except separate ones for $n$ congruent to $0$ modulo $5$, $n$ congruent to $1$ modulo $5$, and so on up to $n$ congruent to $4$ modulo $5$. Knowing the values of $f(0)$ and $f(5)$ will let you compute $f$ at multiples of $5$, and nowhere else. Similarly, knowing $f(1)$ and $f(6)$ will let you compute $f(11)$, $f(16)$, $f(21)$, and so on, but nowhere else. So it is natural to separate the problem into cases. 
So let $g_0(n)=f(5n)$, We have the familiar recurrence $g_0(n)=g_0(n-1)+g_0(n-2)$.
Let $g_1(n)=f(5n+1)$. We have the familiar recurrence $g_1(n)=g_1(n-1)+g_1(n-2)$.
And so on.  For a concrete solution we will need initial values $f(0)$ up to $f(9)$.
Remark: You could however proceed "as usual," getting the characteristic equation $x^{10}-x^5-1=0$. This is a quadratic in $x^5$. Solve for $x^5$. We get the usual roots. For the values of $x$, take the ordinary fifth roots of the equation $y^2-y-1=0$, and multiply by fifth roots of unity (four of these are non-real.) After a while, one can get a general formula that will involve sines and cosines of $2n\pi/5$.  However, this is very much less pleasant than the division into cases suggested above.
A: 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. 
A: I am not quite sure if this is what you were looking for but as was answered previously if you begin at $n=0,5,10,15,20,25,30,...$ and use the Binet form below you get the Fibonacci Numbers 
$$\frac{1}{\sqrt{5}}\left(\sqrt[5]{\frac{1}{2}+\frac{\sqrt{5}}{2}}\right)^{n+5}-\frac{1}{\sqrt{5}}\left(\sqrt[5]{\frac{1}{2}-\frac{\sqrt{5}}{2}}\right)^{n+5}$$
If you use Excel you can paste the formula below in for A1=0, A2=5, A3=10..... and verify that you get the Fibonacci Numbers
(((1/2+SQRT(5)/2))^(1/5)^(A1+5)-((1/2-SQRT(5)/2))^(1/5)^(A1+5))/SQRT(5)

