Avoiding the boxed permutation 123

import itertools

def has_box_123(n,p):
    for a in range(n):
        for c in range(a+1,n):
            A, C = p[a], p[c]
            if A > C:
                continue
            found = False
            for b in range(a+1,c):
                if A < p[b] < C:
                    if found:
                        break
                    else:
                        found = True
            else:
                if found:
                    return True
    return False

for n in range(11):
    count = 0
    for p in itertools.permutations(range(n)):
        if not has_box_123(n,p):
            count += 1
    print(n,count)