Maximum number of knights on a $2 \times n$ ($n\ge 2$) board

Question

So, the puzzle is what is the maximum number of knights one can place on a $2 \times n$ ($n\ge 2$) board such that no two knights can attack each other.

Apparently there is a formula for this $2\cdot\left(2\cdot(n/4) + \min(n\,\%\,4, 2)\right)$

How can one explain this formula or how does one derive this?

You can get within one or two of this with the simple checkerboard pattern. You can express it as $\begin {cases} n&n%4=0\n+1&n%4=1,3\n+2&n%4=2\end {cases}$ — Ross Millikan, Dec 23 '13 at 18:41

score 3 · Accepted Answer · answered Dec 21 '13 at 13:59

Fill the strip with $2\times 2$ with knights, followed by $2\times 2$ empty squares, rinse and repeat, and you fulfill the formula.

As to why this is the maximum, counting how many fields each knight voids (guards, covers, or whatever term you prefer) should get you at least quite close.

Maximum number of knights on a $2 \times n$ ($n\ge 2$) board

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