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

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?

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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

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