I have the Pell sequence $a_n = 2a_{n-1} + a_{n-2}$ for $n \geq 2$ and $a_0 = 1, a_1=2$. I am trying to employ counting logic from Delannoy paths to show that $a_n = \sum\frac{(i+j+k)!}{i!j!k!}$ for non-negative integer triples $(i,j,k)$ where $i+j+2k=n$. I have tried using that for each triple $(i,j,k)$, $\frac{(i+j+k)!}{i!j!k!}$ counts the number of possible three dimensional paths for that triple. It seems that I may be missing out on a useful identity to help with what I am trying to show.
2025-01-12 21:56:04.1736718964
Delannoy Paths and Pell Sequence Relation
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It’s not hard to see that the Pell numbers as defined in the question count the paths with steps $\langle 1,1\rangle$, $\langle 1,-1\rangle$, and $\langle 2,0\rangle$ from $\langle 0,0\rangle$ to the line $x=n$: the $2a_{n-1}$ term in the recurrence accounts for those that end with a step $\langle 1,\pm1\rangle$, and the $a_{n-2}$ accounts for those that end with a step $\langle 2,0\rangle$.
Now suppose that $P$ is a path from the origin to the line $x=n$ that uses only these steps. Say it uses $i_P$ steps $\langle 1,1\rangle$, $j_P$ steps $\langle 1,-1\rangle$, and $k_P$ steps $\langle 2,0\rangle$; clearly $i_P+j_P+2k_P=n$. Conversely, if $i+j+2k=n$, we may arrange $i$ steps $\langle 1,1\rangle$, $j$ steps of $\langle 1,-1\rangle$, and $k$ steps $\langle 2,0\rangle$ in any order to get a path $P$ such that $i_P=i$, $j_P=j$, and $k_P=k$. Thus, there are $\binom{i+j+k}{i,\,j,\,k}$ paths $P$ such that $i_P=i$, $j_P=j$, and $k_P=k$.