Determining the closed form of a particular inhomogeneous recursive relation

Question

In my research I have come to the point where I would like to know the expression for $a_n$ which satisfies \begin{equation} a_n=2a_{n+1}-a_{n+2}+4, \hspace{4mm}a_1=3, \hspace{2mm}a_2=9 \end{equation}

Let me tell you where this come from. I have this sequence of numbers: \begin{equation} 3,9,19,33,51,... \end{equation} You can see that the difference between two numbers is equal to the difference between the previous two numbers, increased by 4, e.g., $9-3=6$, $19-9=10=6+4$, ... Hence I can write \begin{equation} a_{n+2}-a_{n+1}=a_{n+1}-a_{n}+4 \end{equation} from which the above equation follows.

I am aware of the general procedure for solving these recursive equations. I started to look for a solution of the homogeneous equation, by setting $a_n=Ac^n$. This gives the equation $c^2-2c+1=0$, solved by $c=1$. Then I looked for a particular solution, say $a_n=b$. However, this gives $b=2b-b+4$, from which $4=0$. Clearly this method does not work here.

Do you have an idea on how to find $a_n$? Does this equation even have a closed solution?

Answer 1

$c=1$ is a double root of your characteristic equation $c^2-2c+1=0$, so that gives not just one but two linearly independent solutions of the homogeneous equation: $a_n = 1$ and $a_n = n$. Since polynomials of degree $0$ and $1$ are solution of the homogeneous equation, your particular solution should be of the form $a_n = b n^2$.

You should end up with $a_n = 2 n^2 + 1$.

BTW, you could also try looking up your sequence in the OEIS.

Answer 2

Note that $$ \eqalign{ & a_{\,n + 2} - 2a_{\,n + 1} + a_{\,n} = \Delta ^{\,2} a_{\,n} = 4\quad \Rightarrow \cr & \Rightarrow \quad a_{\,n} = {4 \over {2!}}n^{\,\underline {\,2\,} } + c_{\,1} n^{\,\underline {\,1\,} } + c_{\,0} n^{\,\underline {\,0\,} } = \cr & = 2n\left( {n - 1} \right) + c_{\,1} n + c_{\,0} \cr} $$ where $n^{\,\underline {\,m\,} }$ indicates the Falling Factorial.