#A proof for A189912
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The sequence A189912 is defined as \[a_n \coloneqq \sum_{k=0}^n \frac{n!}{(n-k)! (\lfloor k/2 \rfloor!)^2 (\lfloor k/2 \rfloor +1)}.\] Let us split this sum into even and odd \(k.\) We get \[\begin{aligned} &\sum_{k=0}^n \frac{n!}{(n-2k)! (\lfloor 2k/2 \rfloor!)^2 (\lfloor 2k/2 \rfloor +1)} + \\ &\sum_{k=0}^n \frac{n!}{(n-(2k+1))! (\lfloor (2k+1)/2 \rfloor!)^2 (\lfloor (2k+1)/2 \rfloor +1)}. \end{aligned}\] Simplification and reindexing leads to \[\sum_{k=0}^n \left( \frac{n!}{(n-2k)! (k!)^2 (k+1)} + \frac{n!}{(n-2k-1)! (k!)^2 (k+1)} \right).\] Rewriting gives \[\sum_{k=0}^n \frac{n!}{ (k!) (k+1)!}\left( \frac{1}{(n-2k)!} + \frac{1}{(n-2k-1)!} \right) = \sum_{k=0}^n \frac{n!}{(k!) (k+1)!}\left( \frac{1}{(n-2k)!} + \frac{n-2k}{(n-2k)!} \right),\] so we end up with \[\sum_{k=0}^n (n+1-2k) \frac{n!}{ (k!) (k+1)! (n-2k)!}.\] The expression \(\frac{n!}{(k!) (k+1)! (n-2k)!}\) is exactly A055151, so this verifies the conjecture by W. Schulte, Oct 23 2016.