Mrthfx

Suppose that you have a large supply of red, white, green, and blue poker chips. You want to
make a vertical stack of n chips in such a way that the stack does not contain any consecutive
blue chips.

I've already found the recurrence relation for $a_{n}$ (where an denotes the number of ways you can make
such a stack of n poker chips):
$a_{n}=3a_{n-1}-3a_{n-2}$

But I am unsure of how to go about solving it. After solving it, I'm supposed to consider the specific case where you want to count only the stacks that use exactly 10 chips, and count $a_{10}$ directly.

Any help in the right direction is greatly appreciated!

For situations involving linear, homogenous recurrence relations, the characteristic polynomial method works best.

Let us have the recurrence relation, for constants $c_i$ and $k>0$,

$$a_n = c_1 a_{n-1} + c_2 a_{n-2} + ... + c_{n} a_{n-k}$$

Much like with ordinary differential equations, we develop a characteristic polynomial for $a_n$. Let $a_n = alpha^n$. Then we get

$$alpha^n = c_1 alpha^{n-1} + c_2 alpha^{n-2} + ... + c_{k} alpha^{n-k}$$

Divide through by $alpha^{n-k}$ next:

$$alpha^{k} = c_1 alpha^{k-1} + c_2 alpha^{k-2} + ... + c_{k}$$

Bring everything to the same side:

$$alpha^{k} - c_1 alpha^{k-1} - c_2 alpha^{k-2} - ... - c_{k} = 0$$

This is a polynomial, and we seek the roots to it. Let them be $alpha_1,...,alpha_k$. (If $alpha_i$ is a duplicate root, replace the first duplicate with $nalpha_i$, the second with $n^2 alpha_k$, and so on.)

Then, for these roots, up to constants $A_1,...,A_k$ depending on your initial conditions, we have

$$a_n = A_1 alpha_1^n + ... + A_k alpha_k^n$$

Footnotes & Caveats:

As you might imagine, it is hypothetically possible for the recurrence to have complex roots. I do not know how to handle those situations since, as I noted in a few past answers, I'm taking a combinatorics class this semester and this stuff is relatively new to me, so I'm guessing they're keeping us to the "basic" stuff. They might do the same for you, I don't know. It probably depends on the class/text whether the examples are "nice enough" in that respect.

Also, a nice tidbit: it's a good paranoia check to double-check your solution. Once you have the explicit form for $a_n$, check that your initial solutions are valid, and perhaps a few other values you obtain from the recurrence relation. In examples like this where you have to derive the recurrence relation yourself instead of simply being given it, you should be able to get some values by brute force for $n=1,2,3,$ and so on, for ever-how-many initial conditions you need to use. (You need as many initial conditions as there are previous values that determine $a_n$.)

Also bear in mind that this method only works for linear, homogenous recurrence relations. For nonhomogenous ones, I've spoken to you on solving them. For nonlinear ones, we need something more elaborate (such as generating functions) but such discussion is well beyond the scope of this post.

A simple example to motivate this method:

Example: Let us find the solution to the Fibonacci recurrence
$$a_n = a_{n-1} + a_{n-2}$$
where $a_0 = 0,$ and $a_1 = 1$.

(Bear in mind that while here each $a_{text{something}}$ has coefficient $1$, they need not be, and as in the previous explanation the coefficients "carry over" to the characteristic polynomial. The Fibonacci relation is simply a common first example.)

Here, the characteristic polynomial is given by

$$alpha^n = alpha^{n-1} + alpha^{n-2}$$

Divide through by $alpha^{n-2}$:

$$alpha^2 = alpha + 1 implies alpha^2 - alpha - 1 = 0$$

This quadratic can be solved by the quadratic formula. It's a well known result that the two roots to this are the golden ratio $varphi$ and its conjugate $overline varphi$:

$$varphi = frac{1 + sqrt 5}{2} ;;;;; overline varphi = frac{1 - sqrt 5}{2}$$

Thus, up to constants $A_1,A_2$, we can claim

$$a_n = A_1 varphi ^n + A_2 overline{varphi}^n = A_1 left( frac{1 + sqrt 5}{2} right)^n + A_2 left( frac{1 - sqrt 5}{2} right)^n$$

What remains is to determine the constants $A_1, A_2$. To do this, substitute your initial conditions. Thus, you get a system of equations as below. In $a_0$, you let $n=0$ in your solution for $a_n$ above; similarly, $n=1$ in the $a_1$ case.

$$left{begin{matrix}
a_0 = 0\
a_1 = 1
end{matrix}right. implies left{begin{matrix}
A_1 + A_2 = 0\
A_1 varphi + A_2 overline{varphi} = 1
end{matrix}right.$$

To solve this is a fairly typical exercise in solving systems of equations, or linear algebra if you're faced with the awful situation of many initial conditions. I'll skip the boring bits, leaving the algebra to you, simply saying you should get $A_1 = 1/sqrt 5, A_2 = -1/sqrt5$.

And thus we get a general formula for the Fibonacci relation!

$$a_n = frac{ varphi ^n}{sqrt 5} - frac{overline{varphi}^n}{sqrt 5} = frac{ 1}{sqrt 5}left(frac{1 + sqrt 5}{2}right)^n - frac{1}{sqrt 5}left( frac{1 + sqrt 5}{2} right)^n$$

By induction, your recurrence relation can be written as
$$begin{pmatrix}
a_n\
a_{n-1}
end{pmatrix}
=
begin{pmatrix}
3&-3\
0&1
end{pmatrix}
begin{pmatrix}
a_{n-1}\
a_{n-2}
end{pmatrix}
=
begin{pmatrix}
3&-3\
0&1
end{pmatrix}^{n-1}
begin{pmatrix}
a_1\
a_0
end{pmatrix}
.$$
The Jordan decomposition of this matrix allows for simple closed forms for the coefficients of the powers of this matrix.

Whenever you have a recurrence relation of the form $u_{n+2}=alpha u_{n+1}+beta u_n$, you want to find a basis of the set of solutions. One good idea is to look for geometric sequences. If $r$ is the rate, then $r$ verifies
$$r^2=alpha r+beta$$
If $r_1$ and $r_2$ are the (complex) solutions, then every sequence is of the form
$$u_n=Ar_1^n+Br_2^n$$
and you find $A$ and $B$ by looking at the initial values.

Thanks for an interesting question.

The recurrence relation given in the question is not correct, albeit with only a sign error.

It should be;
$$a_n=3a_{n-1}+3a_{n-2}$$
$$a_0=1 : a_1=4$$
This gives rise to the generating series,
$$1+4x+15x^2+57x^3+216x^4+819x^5+dots+169209x^9+641520x^{10}+2432187x^{11}+dots$$
So, assuming the initial term is $a_0$, $a_{10}=641520$, which you asked for.

This generating series has generating function,
$$frac{1+x}{1-3x-3x^2}$$
Applying partial fractions to this gives, after some algebra,
$$frac{1+x}{1-3x-3x^2}=frac{21-5 sqrt{21}}{42 big( 1-frac{3-sqrt{21}}{2}x big)}+frac{21+5 sqrt{21}}{42 big( 1-frac{3+sqrt{21}}{2}x big)}$$
which are recognisable as standard bits directly translating into a formula for the $n^{th}$ term,
$$T_n=left( frac{1}{2}-frac{5 sqrt {21}}{42} right)left( frac{3 - sqrt {21}}{2} right)^n + left( frac{1}{2}+frac{5 sqrt {21}}{42} right)left( frac{3 + sqrt {21}}{2} right)^n$$

Happy to elaborate on any of the detail if necessary.