Does anyone recognize this function? I am looking for a function $f(n)$ that satisfies the following two conditions at the same time $$ \frac{f(n-1)}{f(n)}=(-1)^n\quad ,\quad \frac{f(n+1)}{f(n)}=(+1)^n\equiv 1,\quad \forall  n\in\mathbb{N}\ . $$
As I am not even sure if such a function exists, I'd appreciate any help or comment. Thank you very much in advance!
 A: No such function exists. For the requirements to be well-defined, we need $f(n)\neq 0$, so in particular $f(1)=k\neq 0$. Then $f(2)=f(1+1)=(+1)^1f(1)=k$, but this gives 
$$k=f(1)=f(2-1)=(-1)^nf(2)=-k$$
hence $k=0$, a contradiction.
A: Another way of recognizing the impossibility without doing any calculation (and one that shows why a lot of similar problems are inherently impossible) is by juggling indices.  Your second condition, $\displaystyle\frac{f(n+1)}{f(n)}=1$, can also be written (since the right side is always invertible) as $\displaystyle\frac{f(n)}{f(n+1)}=1$, and then by setting $m=n+1$, as $\displaystyle\frac{f(m-1)}{f(m)}=1$ - and this is obviously incompatible with the first condition.  In general, the two expressions you've written are dependent (as sequences) in that one is just a shifted inverse of the other, so for exactly the same reason you couldn't have e.g. $\displaystyle\frac{f(n-1)}{f(n)}=2$ and $\displaystyle\frac{f(n+1)}{f(n)}=\frac{1}{3}$ for all $n$ in some suitable domain.
