Suppose I have the following recursive formula:
$$A(n)=-\sum_{k=0}^{n-1}\binom{n}{k}\frac{1+(-1)^{n-k}+2(n-k)(-1)^{n-k-1}}{2}A(k)$$ Then I can combine the negatives to get $$A(n)=-\sum_{k=0}^{n-1}\binom{n}{k}\frac{1+(-1)^{n-k}-2(n-k)(-1)^{n-k}}{2}A(k)$$ $$A(n)=-\sum_{k=0}^{n-1}\binom{n}{k}\frac{1+[1-2(n-k)](-1)^{n-k}}{2}A(k)$$ Taking care of the negative out front gives me $$A(n)=\sum_{k=0}^{n-1}\binom{n}{k}\frac{-1-[1-2(n-k)](-1)^{n-k}}{2}A(k)$$ $$A(n)=\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{n-k}[2(n-k)-1]-1}{2}A(k)$$ It's better, but is there anything else I can do for simplification?
I would separate the constant and $k$ terms like this:
$\begin{array}\\ A(n) &=\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{n-k}[2(n-k)-1]-1}{2}A(k)\\ &=\sum_{k=0}^{n-1}\binom{n}{k}(-1)^{n-k}(n-k)A(k)-\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{n-k}+1}{2}A(k)\\ &=\sum_{k=1}^{n}\binom{n}{k}(-1)^{k}kA(n-k)-\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k}+1}{2}A(n-k)\\ &=\sum_{k=1}^{n}\binom{n}{k}(-1)^{k}kA(n-k)-\sum_{k=1}^{\lfloor n/2 \rfloor}\binom{n}{2k}A(n-2k)\\ \end{array} $
That last simplification is due to $(-1)^{k}+1 = 2$ for even $k$ and $0$ for odd $k$.
As to whether or not this is a simplification is up to you.