Let $w(n,l)$ denote the number of closed walks of length $2l$ from a given vertex of the $n$-cube. Then, it is well-known that

$$cosh^n(x)=\sum_{l=0}^{\infty}\frac{w(n,l)}{(2l)!}x^{2l}.$$

Differentiating both sides, we get $$n \cdot cosh^{n-1}(x)\cdot sinh(x) = \displaystyle\sum_{l=1}^{\infty}\frac{w(n,l)}{(2l-1)!}x^{2l-1}.$$ By the Cauchy product of the Maclaurin series of $n*cosh^{n-1}(x)$ and $sinh(x)$ and comparing coefficients of the LHS and RHS, we get the recursion

$$w(n,l)=n\sum_{k=1}^{l}\binom{2l-1}{2k-1}w(n-1,l-k).$$

The above recursion has the following simple combinatorial interpretation. Let us count the total number of closed walks of length $2l$ on the $n$-cube. W.L.O.G, let the initial step be along the 1st dimension. Then, out of the remaining $2l-1$ steps, choose $2k-1$ more places to step back and forth the "1st" dimension. Note that there is exactly one way for this once the $2k-1$ places are chosen. For the remaining $2l-2k$ steps, we take steps in every dimension except the 1st, resulting in $w(n-1,l-k)$ ways. As $k$ is the number of times we walk back and forth the 1st dimension, we sum $k$ from 1 to $l$ ($k>0$ as the initial step is along the 1st dimension). Finally, as the initial step can be taken in $n$ dimensions, we multiply by $n$ and get the above recursion.

My question is the following. To obtain the above recursion, we considered the Cauchy product of the Maclaurin series of $n\cdot cosh^{n-1}(x)$ and $sinh(x)$. This, however, is equivalent to the Cauchy product of the Maclaurin series of $n \cdot cosh^{n}(x)$ and $tanh(x),$ which by the same method gives

$$w(n,l)=n\sum_{k=1}^{l}(-1)^{k+1}\binom{2l-1}{2k-1}A(2k-1)w(n,l-k),$$

in which the "tangent numbers" $A(2k-1)=T_k$ count the number of alternating permutations of $2k-1$ elements (note how the dimension of $w$ is unchanged). I was wondering if a combinatorial interpretation of the above was possible, in a similar fashion to the first recursion. The $(-1)^{k+1}$ term hints inclusion-exclusion, but I'm unable to come up with a satisfactory explanation.

The following post on $w(n,l)$ focuses on a closed-form expression, without mention of recursive formulae. Number of closed walks on an $n$-cube

Let $w(n,l)$ denote the number of closed walks of length $2l$ from a given vertex of the $n$-cube. Then, it is well-known that

$$\cosh^n(x)=\sum_{l=0}^{\infty}\frac{w(n,l)}{(2l)!}x^{2l}.$$

Differentiating both sides, we get $$n \cdot \cosh^{n-1}(x)\cdot \sinh(x) = \displaystyle\sum_{l=1}^{\infty}\frac{w(n,l)}{(2l-1)!}x^{2l-1}.$$ By the Cauchy product of the Maclaurin series of $n\cosh^{n-1}(x)$ and $\sinh(x)$ and comparing coefficients of the LHS and RHS, we get the recursion

$$w(n,l)=n\sum_{k=1}^{l}\binom{2l-1}{2k-1}w(n-1,l-k).$$

The above recursion has the following simple combinatorial interpretation. Let us count the total number of closed walks of length $2l$ on the $n$-cube. W.L.O.G, let the initial step be along the 1st dimension. Then, out of the remaining $2l-1$ steps, choose $2k-1$ more places to step back and forth the "1st" dimension. Note that there is exactly one way for this once the $2k-1$ places are chosen. For the remaining $2l-2k$ steps, we take steps in every dimension except the 1st, resulting in $w(n-1,l-k)$ ways. As $k$ is the number of times we walk back and forth the 1st dimension, we sum $k$ from 1 to $l$ ($k>0$ as the initial step is along the 1st dimension). Finally, as the initial step can be taken in $n$ dimensions, we multiply by $n$ and get the above recursion.

My question is the following. To obtain the above recursion, we considered the Cauchy product of the Maclaurin series of $n\cdot \cosh^{n-1}(x)$ and $\sinh(x)$. This, however, is equivalent to the Cauchy product of the Maclaurin series of $n \cdot \cosh^n(x)$ and $\tanh(x),$ which by the same method gives

$$w(n,l)=n\sum_{k=1}^{l}(-1)^{k+1}\binom{2l-1}{2k-1}A(2k-1)w(n,l-k),$$

in which the "tangent numbers" $A(2k-1)=T_k$ count the number of alternating permutations of $2k-1$ elements (note how the dimension of $w$ is unchanged). I was wondering if a combinatorial interpretation of the above was possible, in a similar fashion to the first recursion. The $(-1)^{k+1}$ term hints inclusion-exclusion, but I'm unable to come up with a satisfactory explanation.

The following post on $w(n,l)$ focuses on a closed-form expression, without mention of recursive formulae. Number of closed walks on an $n$-cube