Is it possible to find the explicit formula for the recurrence relation $a_n = b_{n}a_{n-1} + a_{n-2}$, where $b_{n}$ are known?

Question

Given $a_n = b_{n}a_{n-1} + a_{n-2}$
with starting conditions: $a_0 = 0,\ a_1 = 1$
I need the explicit form of $a_n$.

$b_n$ are known. For example, let's say $[b_2,b_3,b_4,b_5] = [1,2,3,1]$.
The first few terms of the sequence are: $0,\ 1,\ b_2,\ b_3b_2+1,\ b_4(b_3b_2+1)+b_2,\ ...$

For the simpler case where $a_{n-1}$ is always multiplied by the same $b$,
$a_n = ba_{n-1} + a_{n-2}$,
the solution is known. It can be found by guessing $a_n=r^n$ and solving the resultant characteristic equation.

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Here's how I tried to solve it:
I know the solution to a geometric sequence $a_n = ba_{n-1}$ is
$a_n = a_0b^n$

And I reasoned if we instead have $a_n = b_na_{n-1}$ it becomes
$a_n = a_0\prod^n_1b_n$

So, I thought maybe I should follow the logic used to solve $a_n = ba_{n-1} + a_{n-2}$, but instead of guessing $a_n=r^n$, trying
$a_n=\prod^n_1r_i = r_1\cdot r_2\cdot\ ...\ \cdot r_n$

Plugging this into the recursion relation $a_n = b_{n}a_{n-1} + a_{n-2}$ gives
$\prod^n_1r_i = b_{n}\prod^{n-1}_1r_i + \prod^{n-2}_1r_i $

$\Rightarrow (r_{n}r_{n-1} - b_{n}r_{n-1} - 1)\prod^{n-2}_1r_i = 0$

I think because $a_{n-2}\ne 0$ in general, this implies
${r_{n}r_{n-1} - b_{n}r_{n-1} - 1\ =\ 0}\ $

This is where I get stuck. Instead of a solution for $r_n$, I have a recurrence relation.
$r_n = r_{n-1}^{-1} + b_n$
I'm not sure what the base value $r_1$ would be. Also, I expect to find two possible answers for $r_n$, so that $a_n$ would be a linear combination of those, with coefficients determined by "$a_0 = 0,\ a_1 = 1$".

It occurred to me that if I had another equation relating $r_n$ and $r_{n-1}$, two solutions could be found. For instance, if I had $r_n = r_{n-1}$, I could substitute that in and solve $r_n = r_{n}^{-1} + b_n$, a quadratic with two solutions for $r_n$. However, I don't have that equation, and it can't be true because $b_{n}\ne b_{n-1}$ in general. It's as if I'm missing a vital equation, and I don't know where to find it.

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I also tried the telescoping technique:
The LHS of $a_n - a_{n-2} = b_{n}a_{n-1}$ has a lot of cancelling terms when all added up.

$\sum_2^n(a_i - a_{i-2}) = a_2 - a_0 + a_3 - a_1 + a_4 - a_2 + a_5 - a_3\ +\ ...\ +\ a_{n-1} - a_{n-3} + a_n - a_{n-2}$
$\Rightarrow - a_0 - a_1 + a_{n-1} + a_n = \sum_2^nb_{i}a_{i-1}$

$\Rightarrow a_n = a_0 + (b_2 + 1)a_1 + \sum_3^{n-1}b_{i}a_{i-1} + (b_n - 1)a_{n-1}$

But even if I've done this correctly, it's not a win because now I have a formula in terms of all preceding $a_n$s instead of just the previous two.

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I'm aware that any solution for $a_n$ would have in it some combination of all previous $b_n$s. That's fine, because they are known. Has this problem already been solved? Is it possible to solve, and if not why not?

Answer 1

The recurrence can be written in matrix form as:

$$ A_n = \begin{pmatrix} a_{n} \\ a_{n-1} \end{pmatrix} = \begin{pmatrix} b_n & 1 \\ 1 & 0 \end{pmatrix} \cdot \begin{pmatrix} a_{n-1} \\ a_{n-2} \end{pmatrix} = B_n \cdot A_{n-1} \quad\quad \text{with}\;\; A_1 = \begin{pmatrix} a_{1} \\ a_{0} \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} $$

Then, telescoping:

$$ A_n = B_n \cdot A_{n-1} = B_n \cdot B_{n-1} \cdot A_{n-2} = \dots = B_n \cdot B_{n-1} \cdots B_2 \cdot A_1 = P_n \cdot A_1 $$

The product $\,P_n = \prod_{k=2}^n B_k\,$ can be calculated recursively, and a "more closed" form may exist for particular sequences $\,b_n\,$.

Answer 2

Essentially, $a_n$ is the $n$th partial numerator of the continued fraction $$0+\dfrac{1}{1+\dfrac{1}{b_2+\dfrac{1}{b_3+\cdots}}}\ .$$ There is no known simple formula for $a_n$ in terms of $b_n$. However, Euler's rule may be of interest.

• Write down the string $b_2b_3b_4\cdots b_n$.
• Write down all strings you can obtain by deleting pairs of consecutive letters from this expression. This includes the original string (deleting no letters). If $n$ is odd, it is possible to delete all letters, and this expression will be taken to mean $1$.
• Add all the results from the previous step.

Example: $n=4$. Starting with $b_2b_3b_4$ you get $$b_2b_3b_4+b_4+b_2\ ,$$ which matches your $a_4$.

Another example: $n=7$. Starting with $b_2b_3b_4b_5b_6b_7$, you get the strings (red letters are not really there, they are deleted) $$\displaylines{b_2b_3b_4b_5b_6b_7,\ \color{red}{b_2b_3}b_4b_5b_6b_7,\ b_2\color{red}{b_3b_4}b_5b_6b_7,\ b_2b_3\color{red}{b_4b_5}b_6b_7,\ b_2b_3b_4\color{red}{b_5b_6}b_7,\cr b_2b_3b_4b_5\color{red}{b_6b_7},\ \color{red}{b_2b_3}\color{red}{b_4b_5}b_6b_7,\ \color{red}{b_2b_3}b_4\color{red}{b_5b_6}b_7,\ \color{red}{b_2b_3}b_4b_5\color{red}{b_6b_7},\ b_2\color{red}{b_3b_4}\color{red}{b_5b_6}b_7,\cr b_2\color{red}{b_3b_4}b_5\color{red}{b_6b_7},\ b_2b_3\color{red}{b_4b_5}\color{red}{b_6b_7},\ \color{red}{b_2b_3}\color{red}{b_4b_5}\color{red}{b_6b_7}\cr}$$ and so $$\eqalign{a_7&=b_2b_3b_4b_5b_6b_7+b_4b_5b_6b_7+b_2b_5b_6b_7+b_2b_3b_6b_7+b_2b_3b_4b_7\cr &\qquad+b_2b_3b_4b_5+b_6b_7+b_4b_7+b_4b_5+b_2b_7+b_2b_5+b_2b_3+1\ .\cr}$$

It is not hard to prove by induction that this rule works.