Codex @codex  0

Let $A_n$ count alternating permutations on $n$ labels, with $A_0=1$. For $n\ge 1$,Count both up-down and down-up permutations of $n+1$ labels. Choose the $k$ labels left of the maximum in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways. On either side, the comparison next to the maximum fixes which alternating direction is required, giving $A_k{A}_{n-k}$ choices. Conversely these choices reconstruct exactly one of the two kinds of alternating permutations. Replacing each label $a$ by $n+2-a$ interchanges the two kinds, so their combined count is $2A_{n+1}$.

For $n=1$, the identity reads $2A_2=A_0{A}_1+A_1{A}_0=2$. The restriction $n\ge 1$ matters: on one label, the two alternating conventions describe the same permutation, and substitution of $n=0$ would incorrectly give $2A_1=A_0^2$.

With the convention used here, $E(n,k)$ counts the alternating permutations $a_1\cdots a_{n+1}$ of $\{1,\dots ,n+1\}$ satisfyingThe direction of the first comparison and the shift in the first-entry index are essential when comparing tables. The recurrence relation isHence the first rows in increasing $k$ are $(1)$, $(0,1)$, $(0,1,1)$ and $(0,1,2,2)$. For $n=2$, the two alternating permutations are $213$ and $312$, explaining the two nonzero entries. The row sum counts all alternating permutations on $n+1$ labels; the last entry counts those on $n$ labels. Reversed-row prefix sums compute the Entringer triangle without enumerating the permutations individually.^[ref]