Can Anyone solve this number of cases problem? There are $n$ different chairs around the round table, $C_1, C_2, ....C_n$, and one person going to give a number to each chair, $1$, $-1$, $i$ and $-i$. But $1$ can't be placed next to $-1$, and $i$ can't placed next to $-i$. And how can I get the number of the possible cases? 
 A: As an approximation, start with the number of ways to make a line of $n$.  You have $4$ choices for $C_1$ and $3$ choices for all the rest, so $4\cdot 3^{n-1}$. When you try to bend these into a circle, you will fail $\frac 14$ of the time, so so for a circle it is $3^n$  
This is clearly not quite right as the end number is not truly independent of the first, though it will be close very soon.  For $n=1$ the correct answer is $4$ instead of $3$ and for $n=2$ it is $12$, not $9$.  For $n=3$ we have $28$ choices, very close to $3^3=27$  
To be right, we can assume $C_1=1$ and multiply by $4$ at the end for the other starting choices.  Ignoring the circle, let $A(n)$ be the number of strings ending in $1$, $B(n)$ be the number ending in $-1$, and $C(n)$ be the number ending in $i$ or $-1$.  We have $$A(1)=1, B(1)=0, C(1)=0\\
A(n)=A(n-1)+C(n-1)\\B(n)=B(n-1)+C(n-1)\\C(n)=2A(n-1)+2B(n-1)+C(n-1)$$ and our answer is $4(A(n)+C(n))$ because the ends will not conflict.  It turns out the answer is $3^n+2+(-1)^n$
