I need to find the function c(k), knowing that
$$\sum_{k=0}^{\infty} \frac{c(k)}{k!}=1$$
$$\sum_{k=0}^{\infty} \frac{c(2k)}{(2k)!}=0$$
$$\sum_{k=0}^{\infty} \frac{c(2k+1)}{(2k+1)!}=1$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k+1)}{(2k+1)!}=-1$$
$$\sum_{k=0}^{\infty} \frac{(-1)^k c(2k)}{(2k)!}=0$$
Is it possible?