3947.sagews

These functions give $S(n,k)$:

def genS(n):
    if(n==0): 
        return 1
    else:
        return (1+x^n)*genS(n-1)            
var('x')
@cached_function 
def S(n,k):
    s = genS(n)
    if(n==0 and k == 0): 
        return 1
    elif(n==0):
        return 0
    else: 
        return s.expand().coeff(x,k)

$H(n,k)$ is the number of $T(n,n,n)$ tilings for odd $n$, with monomers in the top corners that have $k$ horizontal dimers.

\begin{align*}
H(n,k) = 2&\sum_{i=\left\lceil \frac{n-5}{4} \right\rceil}^{\left \lfloor \frac{n-2}{2}\right\rfloor - 1 } \left( \sum_{k_1+k_2 = k-(2i+2)} S(2i+1,k_1)S(n-2i-4,k_2) \right)\\
+ 2&\sum_{i=0}^{\left\lceil \frac{n-3}{4} \right\rceil}\left( \sum_{k_1+k_2 = k - (n-2i-2)} S(2i,k_1)S(n-2i-3,k_2) \right)\\
+&\sum_{k_1+k_2=k} S\left(\frac{n-3}{2},k_1\right)S\left(\frac{n-3}{2},k_2\right) \right).
\end{align*}

var('i,b')
def H(n,k):
    s = 2*sum(
        [sum(
                [S(2*i+1,b)*S(n-2*i-4,k-(2*i+2)-b) 
                 for b in range(k-(2*i+2)+1)]) 
         for i in range(ceil((n-5)/4),floor((n-2)/2))])
    s = s + 2*sum(
        [sum(
                     [S(2*i,b)*S(n-2*i-3,k-(n-2*i-2)-b) 
                      for b in range(k-(n-2*i-2)+1)]) 
         for i in range(floor((n-3)/4)+1)])
    s = s + sum(
             [S((n-3)/2,b)*S((n-3)/2,k-b) for b in range(k+1)])
    return s

$V(n,k)$ is the number of $T(n,n,n)$ tilings for even $n$, with monomers in the top corners that have $k$ vertical dimers.


  \begin{align*}
    V(n,k) = &\,\, 2\sum_{i=0}^{\frac{n-2}{2}-1} \left( \sum_{k_1+k_2 = k - \max\left(2i+1,n-2i-2 \right)} S(2i,k_1)S(n-2i-3,k_2) \right)\\
 + & \sum_{k_1+k_2 = k}
    S\left(\frac{n-2}{2},k_1\right) S\left(\frac{n-2}{2},k_2\right).
  \end{align*}
  
  This is from our paper, and it looks a little nicer because there is left right symmetry that we don't between the top and bottom in the odd case.  The terms are very similar however, and a similar story should unfold when we try to divide out some cyclotomics from $H(n,k)$.

def V(n,k):
    s = 2*sum(
        [sum(
                [S(2*i,b)*S(n-2*i-3,k-max(2*i+1,n-2*i-2)-b)
                 for b in range(k-max(2*i+1,n-2*i-2)+1)])
         for i in range((n-2)/2)])
    s = s + sum(
            [S((n-2)/2,b)*S((n-2)/2,k-b) for b in range(k+1)])
    return s

This is a sanity check for $H(n,k)$.  It shows that all the tilings are accounted for when $H(n,k)$ is summed over all $k$ and a fixed $n$.

H13(x) = sum([ H(13,i)*x^i for i in range((13+1)^2/2)])
print 4*H13(1)

53248