python3 - <<'EOF'
import itertools
def det(A):
 n=len(A)
 A=[list(r) for r in A]
 for c in range(n):
  p=next((r for r in range(c,n) if A[r][c]),None)
  if p is None:return 0
  A[c],A[p]=A[p],A[c]
  for r in range(c+1,n):
   if A[r][c]:A[r]=[x^y for x,y in zip(A[r],A[c])]
 return 1
def count_formula(n):
 s=n-2
 pairs=[(i,j) for i in range(s) for j in range(i+1,s)]
 cnt=0
 for bits in itertools.product([0,1],repeat=len(pairs)+s):
  M=[[0]*s for _ in range(s)]
  for i in range(s):M[i][i]=bits[i]
  for (i,j),b in zip(pairs,bits[s:]):M[i][j]=M[j][i]=b
  v=[M[i][i] for i in range(s)]
  if not any(v):continue
  if not det(M):continue
  p=sum(v)%2
  x=1^p
  one=[sum(M[i][j] for j in range(s))%2 for i in range(s)]
  w=[(x*v[i]^one[i]) for i in range(s)]
  D=[[0]*n for _ in range(n)]
  D[0][1]=D[1][0]=p
  for i in range(s):
   D[0][i+2]=D[i+2][0]=v[i]
   D[1][i+2]=D[i+2][1]=w[i]
   for j in range(s):
    D[i+2][j+2]=(M[i][j]^(v[i]&v[j])^(i==j))
  if det(D):cnt+=1
 return cnt
print('n4',count_formula(4))
print('n6',count_formula(6))
def count_direct(n):
 pairs=[(i,j) for i in range(n) for j in range(i+1,n)]
 cnt=0
 for bits in itertools.product([0,1],repeat=len(pairs)):
  C=[[0]*n for _ in range(n)]
  C[0][0]=C[1][1]=1
  for (i,j),b in zip(pairs,bits):C[i][j]=C[j][i]=b
  if not det(C):continue