We can be sure that number of Hamiltonian paths on the rook graph for any single cell on $n\times2$ chessboard equals $$ H(n+1) = \sum_{k=0}^{n} \binom{n}{k} \binom{k}{\lfloor{\frac{k}{2}\rfloor}} \left(n-\lfloor{\frac{k}{2}\rfloor}\right)! \left(n-\lfloor{\frac{k+1}{2}\rfloor}\right)!$$ Is there a combinatorial proof for it?


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