How to generate specific members of a natural number series I have a series I can generate sequentially if I pair elements with Pell numbers. If I pair them with $1,2,5,12,29,...$ (Pell numbers, $p_n$) and call elements of this series $q_n$, then $q_n=q_{n-1}+p_{n-1}$ where $q_0=1$.
$$1\quad 
2\quad 
4\quad 
9\quad 
21\quad 
50\quad 
120\quad 
289\quad 
697\quad 
1682\quad 
4060\quad 
9801\quad 
23661\quad 
57122$$
I would like to generate any $q_n$ directly. I can get $p_n$ directly but so far my formula depends on knowing $q_{n-1}$ beforehand. Can the $n^{th} q$ be generated directly?
 A: One way to find a formula is to convert the system of difference equations into first order difference equations - the Pell numbers $p_n=2p_{n-1}+p_{n-2}$ is a second order difference equation. To do this, we introduce a dummy sequence
$$a_n=p_{n-1}$$
for $n\ge1$; this lets us write
$$\begin{align*}
a_n&=p_{n-1}\\
p_n&=2p_{n-1}+a_{n-1}\\
q_n&=q_{n-1}+p_{n-1}.
\end{align*}$$
Now the reason we like this first order system is that we can define
$$x_n=\begin{pmatrix}a_n\\p_n\\q_n\end{pmatrix}$$
so that for $n\ge 2$
$$x_n=\begin{pmatrix}p_{n-1}\\2p_{n-1}+a_{n-1}\\q_{n-1}+p_{n-1}\end{pmatrix}=\begin{pmatrix}0&1&0\\1&2&0\\0&1&1\end{pmatrix}x_{n-1}.$$
Letting $A=\begin{pmatrix}0&1&0\\1&2&0\\0&1&1\end{pmatrix}$, we get
$$x_n=Ax_{n-1}=A^2x_{n-2}=\cdots=A^{n-1}x_1.$$
Now all that remains is to compute $A^{n-1}$ by writing as Jordan Normal Form, i.e. finding the eigenvalues and eigenvectors of $A$. 
A: The Pell numbers (OEIS A000129 $ 0, 1, 2, 5, 12, 29, 70, ...)$ are usually given by
$P_0=0, P_1=1, P_n=2P_{n-1}+P_{n-2}=\dfrac{(1+\sqrt2)^n-(1-\sqrt2)^n}{2\sqrt2}$.
The Pell-Lucas numbers (OEIS A002203 $2,2,6,14,34,82,198,...)$ are usually given by
$Q_0=2, Q_1=2, Q_n=2Q_{n-1}+Q_{n-2}=(1+\sqrt2)^n+(1-\sqrt2)^n.$
Your $q_n$ (OEIS A024537 $1, 2, 4, 9, 21, 50,...)$ is $\dfrac{Q_{n+1}+2}4.$
