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The answer is $a_n=8(n-2)!$ for $n\ge 4$.
Subdivide into cases depending on the values at corners and count. (6 cases give 1, 1 cases give 2, 1 case gives 02 times $(n-2)!$)
The answer is $a_n=8(n-2)!$ for $n\ge 4$.
Subdivide into cases depending on the values at corners and count. (6 cases give 1, 1 cases give 2, 1 case gives 0 times $(n-2)!$)
The answer is $a_n=8(n-2)!$ for $n\ge 4$.
Subdivide into cases depending on the values at corners and count. (6 cases give 1, 1 case gives 2 times $(n-2)!$)
The answer is $a_n=8(n-2)!$ for $n\ge 4$.
Subdivide into cases depending on the values at corners and count. (6 cases give 1, 1 cases give 2, 1 case gives 0 times $(n-2)!$)
