How to solve the following system? 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?

Following Rotwang, define $b(k)=c(k)/k!$, then note that both the second and fifth can be satisfied as long as $\sum{b(2k)}=0$.  Then the first and third are redundant and define $d(k)=b(2k+1)$.  Now we have $\sum{d(k)}=1, \sum{(-1)^kd(k)}=-1$.  Adding and subtracting, $\sum{d(2k)}=0, \sum{d(2k+1)}=1.$  So the final is that we must have $$\begin{align} \sum\frac{c(2k)}{k!}&=0\\
   \sum\frac{c(4k+1)}{(4k+1)!}&=0\\
   \sum\frac{c(4k+3)}{(4k+3)!}&=1  \end{align}$$
and any $c(k)$ that satisfies this will work.
