TITLE: Recurrence with varying coefficient
QUESTION [6 upvotes]: Problem 1
$$
{\rm f}\left(n\right)
=
\frac{1}{n}\,
\left[{\rm f}\left(n - 1\right)k_{0} + {\rm f}\left(n-2\right)k_{1}\right]\tag{1}
$$
( This can also be written as
${\rm Q}\left(n\right)
=
k_{0}\frac{{\rm Q}\left(n - 1\right)}{n - 1}
+k_{1}\frac{{\rm Q}\left(n - 2\right)}{n - 2}$ provided
${\rm Q}\left(n\right) = n{\rm f}\left(n\right) $ )
where $k_{0},\,k_{1}$ are constants.
${\rm f}\left(0\right)=3,\,{\rm f}\left(1\right)=5$. The variable coefficient $\frac{1}{n}$  is making lot of issues here. It blocks the proper expansion.
Question
Please help me to find the solution of the recurrence  in terms of $n$
( implies ${\rm f}\left(n\right)$ )  and also the summation of the recurrence up to infinity ( $ sum = \sum_{n = 0}^{\infty}{\rm f}\left(n\right)$ ).  
Note::

Partial answers are also welcome. We can discuss that here. This recursion forms a structure of Fibonacci look like tree. I have attached a pic here. You can see Red nodes   to indicate $k_0$ multiplication and green to indicate $k_1$ multiplication. You can see the index of the tree expands like a Fibonacci sequence(starts from 5)
Some of the generating function smart manipulations can be looked in to this famous e book for reference link:-generatingfunctionology!
I know a method of using ODE. But I am trying to solve it with out ODE so that I can extent this to higher dimension like matrices in similar structure questions. Please avoid ODE solution or any matrix exponential. A method using series would be more desirable


Problem 2( extended Matrix version of problem 1)
We have a given matrix recurrence ,
$  (\curlyvee_i,\curlyvee_{i-1})_{1\times2}= (\curlyvee_{i-2},\curlyvee_{i-3})_{1\times2}{\begin{bmatrix}A_{i-1}A_i+B_i & A_{i-1} \\B_{i-1}A_i & B_{i-1} \end{bmatrix}}_{2\times2} \tag 1$ 
$A_i= r\frac{\Im1}{i},B_i= r^2\frac{\Im2}{i} \tag 2$
$   \Im1=\left( \begin{array}{ccc} 
   0 & -n_0 & m_0 \\
   n_0 & 0 & -l_0 \\
  -m_0 & l_0 & 0 \\
   \end{array} \right),  
   \Im2=\left( \begin{array}{ccc} 
   0 & -(n_1-n_0) & (m_1-m_0) \\
   (n_1-n_0) & 0 & -(l_1-l_0) \\
   -(m_1-m_0) & (l_1-l_0) & 0 \\
   \end{array} \right).\tag3$  
$l_1,m_1,n_1,l_0,m_0,n_0,r$ all are constants can't alter at all
Given Data about the problem

Dimensions of $A_i,B_i,\curlyvee_i$ are $3\times3$ and $A_i,B_i$ are skew symmetric matrices with diagonal zero
Our initial condition is given as $(\curlyvee_1,\curlyvee_{0})_{1\times3} $
Properties of  $\Im1,\Im2$ useful for simplification are given below 
For each matrices  $\Im1 (  a=l_0,b=m_0,c=n_0),  \Im2  (  a=l_1-l_0,b=m_1-m_0,c=n_1-n_0)$,let $p = \sqrt{a^2+b^2+c^2}$. It is true that $\wp^3 = -(a^2+b^2+c^2)\wp = -p^2\wp$ ($\wp$ represents matrices $\Im1,   \Im1$ ,select a,b,c accordingly).
 Hence, $\wp^{2m+1} = (-1)^mp^{2m}\wp \tag 4$ and $\wp^{2m} = (-1)^{m-1}p^{2m-2}\wp^2 \tag 5$.  In essence it says both matrices  $\Im1,  \Im2$ satisfies this property. courtesy @JimmyK4542 

Question
 1. How do we solve this recurrence in terms of n. Is there any way to take out  n from the $3\times3$ matrices in recursion so that we can go for easy simplification. ?
NB :: I would like to have a solution with out using ODE or matrix exponential as I can extent to tensors etc. Please avoid such solution. Those were discussed earlier. Another version of the same question I posted as graph problem is given here. I didnt get any response to that,so I made it simplified as matrix recursion here. I haven been trying lot of methods on it none is successful coz of the properties of $A_i$
Note :: Bounty will be awarded to the helpful suggestions which leads to solution. 
Efforts done so far 

Tried to evaluate using power series simple method. Result:: failed ,Reason ::   varying n blocks the procedure
Tried to make another equivalent series by substitution. Result :: failed,Reason :: it still went back to old issues.  (not sure any nice substitution exists such that we can get form that is solvable..trying)
Made it as graph problem .I see a finite structure in Tree and node coloring.. If we go with graph theory method to get an expression for general case n and later add all as series up to infinity , we can find a sequence some thing like GP. We may end up in a finite expression. Result :: Failed ,Reason:: It was tough to get that common expression for a graph based on node contribution
Made it as matrix recursion, Result :: failed, could not decompose the matrix recursion component to separate n
Tried to compare or extract some properties from Fibonacci tree as it has same structure. Result :: Failed,Reason:: we will still end up in varying coefficient ..

REPLY [5 votes]: So I do realize you wanted to avoid a matrix exponential answer, now I would like to post one here just so other people running into this question can see what that looks like:
$$ \textbf{MATRIX SOLUTION}$$
Let: 
$$L(f(n)) = f(n) - f(n-1)$$
Notice therefore:
$$L^2(f(n)) = f(n) - 2f(n-1) + f(n-2)$$
Therefore we can re-write as 
$$f(n) = \frac{k_1}{n}(L^2 (f(n)) - 2L(f(n)) + f(n)) + \frac{k_0}{n}(f(n) - L(f(n)))$$
use $f = f(n)$ for cleaner appearance and use standard operator notation (no parenthesis between L and f)
$$f = \frac{k_1}{n}(L^2 f - 2Lf + f) + \frac{k_0}{n}(f - Lf)$$
Which simplifies to
$$\frac{k_1}{n}L^2f + \left(-\frac{2k_1+k_0}{n}\right)Lf + \left(\frac{k_0 + k_1 - n}{n} \right)f = 0 $$
We now multiply through by $n$ and divide by $k_1$
$$L^2f - \left(\frac{2k_1+k_0}{k_1}\right)Lf + \left(\frac{k_0 + k_1 - n}{k_1} \right)f = 0 $$
And move all terms besides $L^2f$ to the other side
$$L^2f  =  \left(\frac{2k_1+k_0}{k_1}\right)Lf - \left(\frac{k_0 + k_1 - n}{k_1} \right)f $$
Now generate a vector $G = \begin{bmatrix} f \\ Lf  \end{bmatrix} $
At this point it becomes evident that if we define $L$ to act component by component on matrices then:
$$LG = \begin{bmatrix} - \frac{k_0 + k_1 - n}{k_1} && 0 \\ 0 && \frac{2k_1 + k_0}{k_1} \end{bmatrix}G$$
Traditionally to solve equations of the type 
$$Lf = bf$$
exploits the fact that
$$L(2^{Q(n)}) = (2^{LQ(n)}-1)2^{Q(n)}$$
In other words we are seeking a matrix $H$ such that
$$ 2^{LH(n)} - I_2  = \begin{bmatrix} - \frac{k_0 + k_1 - n}{k_1} && 0 \\ 0 && \frac{2k_1 + k_0}{k_1} \end{bmatrix}$$
Meaning
$$2^{LH(n)} =  \begin{bmatrix} - \frac{k_0  - n}{k_1} && 0 \\ 0 && \frac{3k_1 + k_0}{k_1} \end{bmatrix} $$
At this point we must find the components of: 
$$\log_2 \left( \begin{bmatrix} - \frac{k_0  - n}{k_1} && 0 \\ 0 && \frac{3k_1 + k_0}{k_1} \end{bmatrix} \right) $$
This is a tricky calculation which requires we replace the 0 components with a variable z. Then compute the logarithm treating, z as a real variable. Then for each component $c_i$ solve
$$Lu = c_i$$
to assemble the components of $H(n)$
from here
$$2^{H(n)}$$
will be the solution of the system
And therefore the dot product of the top row of $2^{H(n)}$ with the vector $\begin{bmatrix} c_0 \\ c_1 \end{bmatrix}$ is the general solution to our problem
At this point take the limit as $z \rightarrow 0$ to recover the solution
$$ \textbf{SERIES SOLUTION} $$
We note that given the linear operator described above
$$Lf = f(n) - f(n-1)$$
We can generate a Taylor-like series by first observing the existence of a product rule
$$L(f*g) = f(n)g(n) - f(n-1)g(n-1) = g*Lf + f*Lg - Lg*Lf $$
And second observing that 
$$L \left( \frac{1}{n!} \prod_{i=1}^{n}{(x +i-1)}\right) =\frac{1}{n!} \prod_{i=1}^{n-1}{(x+i-1)}  $$
Thereby allowing us to generate a "series" 
$$f(x) = f(a) + Lf(a)x + \frac{L^2f(a)}{2!}x(x+1) + \frac{L^3f(a)}{3!}x(x+1)(x+2)... $$
We now recall the original problem:
$$L^2f  =  \left(\frac{2k_1+k_0}{k_1}\right)Lf - \left(\frac{k_0 + k_1 - n}{k_1} \right)f $$
We can consider the values of 
$$ \begin{matrix} L^2f && L^3f && L^4f && ... && L^rf \end{matrix}$$
At which point it becomes evident
$$ L^rf = \left(2 + \frac{k_0}{k_1}\right)L^{r-1}f + \frac{r-2}{k_1}L^{r-3}f + \left(\frac{n}{k_1} - 1 - k_0+ \frac{r-2}{k_1}  \right)L^{r-2}f$$
This can be re-written by noting $L^rf = X(r)$ and therefore
$$X(r) = aX(r-1) + (r-2)bX(r-3)+ (c + b(r-2))X(r-2)$$
Whereas $X(0) = 3, \ X(1) = (5-3) = 2$ and
$$a = \left(2 + \frac{k_0}{k_1}\right), \ b = \frac{1}{k_1}, \ c = \left(\frac{n}{k_1} - 1 - k_0 \right) $$
Now the solution to the problem is
$$f(n) =  \sum_{i=0}^{\infty} \left[ \frac{X(i)}{i!} \prod_{j=1}^{i} (n + j - 1) \right]$$
$$ f(n) = X(0) + X(1)n + \frac{X(2)}{2}n(n+1) + \frac{X(3)}{3!}n(n+1)(n+2) ... $$
Of course finding a closed form for $X$ seems to be a still intriguing problem but not impossible since it is of the exact same form as the original problem: we can create a matrix solution or series solution for $X$ (depending on personal preference...) but to address the concerns of dividing by $k_0$ and $k_1$ our solution terms can instead be written as: 
$$ k_1*L^rf = \left(2k_1 + k_0\right)L^{r-1}f + (r-2)L^{r-3}f + \left(n - k_1 - k_0*k_1+ (r-2)  \right)L^{r-2}f$$
Thus we can still attempt to solve for each individual $L^{r}f$ and generate series terms... I am curious how this will play out and whether any inconsistent systems will be formed (if they are formed: does that mean there cannot exist a solution to said equation, or does it require the creating of a new flavor of functions outside of all others, much like how we can create an imaginary constant to deal with the unsolvable $\sqrt{-1}$ problem)