Inspired by this question I tried to find a recurrence relation for the number of rectangles in a $d$-dimensional hypercube $C$. Let call this number $r_d$. It is known that $r_d$ has a closed form, but I'm curious if my approach leads somewhere.
TL DR:
There are $2^{d-2} d$ rectangles that are not contained in any $d-1$ face. As there are $2d$ $(d-1)$-faces of the cube I'm considering $2^{d-2} d + 2d r_{d-1}$ and I am subtracting all the rectangles that i have counted more than once (i.e. the rectangles that are in the intersection of two or more $d-1$ faces). I'm struggling to find the number of rectangle that i have counted $3$ or more times.
I think the recurrence relation is given by
$$ r_d= 2^{d-2} d+ 2d r_{d-1}- d(d-1) r_{d-2}- \sum_{k=3}^{d-2} f_{d-k} r_{d-k} $$ where $f_{d-k}$ are positive coefficients that depends only on the dimension of the hypercube and $r_d=0$ for $d \le 1$, I'm struggling to find the expressions for such coefficients.
I will denote with $F_{k}$ a generic $k$-dimensional face of $C$ and I will call $k$-diagonals the main diagonals of $F_k$
My approach is based on three results. I will not give the proof of them here to avoid that the question becames too long, but they are proved in my (flawed) answer on the linked question
lemma 1
if $\ell$ is a line segment of $C$ such that $\ell \not \subset F_{d-1}$ for all the $d-1$-dimensional faces then there isn't any line segment parallel and distinct to $\ell$
lemma 2
if $\ell$ is a $d-1$ diagonal than there exists only 1 line segment $\ell'$ that is parallel and distinct from $\ell$. We have $\ell' \subset F'_{d-1}$ where $F'_{d-1}$ is the $d-1$ dimensional face parallel to $F_{d-1}$
lemma 3
Let $R$ a rectangle whose vertex are vertex of $C$ if $R \not \subset F_{d-1}$ for any $d-1$ face than two of the sides of $R$ are parallel $d-1$ diagonals. Conversely the enpoints of a pair of $d-1$ diagonal forms a rectangle that is not contained in any $d-1$ face
As there are $2^{d-2} d $ pairs of parallel $(d-1)$ diagonals, by using lemma 3 we obtain that there are $2^{d-2}d$ rectangles that are not contained in any $(d-1)$ face.
In any $d-1$ dimensional face there are $r_{d-1}$ rectangles, by definition.
Any pair of parallel $d-1$ faces of the hypercube intesect with $2(d-1)$ distinct faces. So there are $$ 2\sum_{k=1}^{d-1} (d-i)=d(d-1) $$ So there are $d(d-1)r_{d-2}$ rectangles that are contained in the intersection of two $n-1$ faces
The idea is that $f_{d-k}$ is the number of intersection of $k$ planes and that the number of rectangles that are contained in at least one $n-1$ dimensional face is $$ 2d r_{d-1}- d(d-1) r_{d-2} - \sum_{k=3}^{d-1} f_{d-k} r_{d-k} $$ By checking with the closed form for $d=5$ it looks like $f_{d-3}= 2^{d-3} \binom{d}{3}=160$, but this formula does not generalize for higher dimension.
Is there any way to find the expression for such coefficients or is my approach flawed?