#A proof for A189912
This page is part of the OEIS proof collection.
The sequence A189912 is defined as \[a_n \coloneqq \sum_{k=0}^n \frac{n!}{(n-k)! (\lfloor k/2 \rfloor!)^2 (\lfloor k/2 \rfloor +1)}.\] Splitting the sum into even and odd indices gives \[\begin{aligned} a_n &=\sum_{j=0}^{\lfloor n/2\rfloor} \frac{n!}{(n-2j)!(j!)^2(j+1)} \\ &\quad+ \sum_{j=0}^{\lfloor(n-1)/2\rfloor} \frac{n!}{(n-2j-1)!(j!)^2(j+1)}. \end{aligned}\] Rewrite the odd summand using \(1/(n-2j-1)!=(n-2j)/(n-2j)!.\) When \(n\) is even, extending that sum to \(j=n/2\) adds a zero term. Hence \[a_n=\sum_{j=0}^{\lfloor n/2\rfloor} (n+1-2j)\frac{n!}{j!(j+1)!(n-2j)!}.\] The factor \(n!/(j!(j+1)!(n-2j)!)\) is the triangular array A055151. This proves W. Schulte’s conjecture from October 23, 2016.