So, we can find the answer using the generating function: $$f(x)=(1+x+x^{-1})^N=x^{-N}\sum_{k=0}^{N}\sum_{m=0}^{k}{N \choose k}{k \choose m}x^kx^m$$ and the number of N-tuples for each integer $s$ is the coefficient of the term $x^s$. So, by taking $s=k+m-N$ and using the fact that $m$ is uniquely determined by the condition $s=k+m-N$, the number of N-tuples for each $s$ is $$\sum_{k=0}^{N}{N \choose k}{k \choose s-k+N}$$
Is my proof acceptable? Since $s$ can become negative, if $s=-N$, then $s-k+N$ also becomes negative except for the case of $k=0$. So, whenever binomial coefficient takes "illegal" value on the top or the bottom entry, it automatically becomes $0$, doesn't it?
Using other method, I got a different answer; I'm sure about my answer
$$E=\text{Coefficient of } \large \alpha^s\text{ in }(\alpha^0+\alpha^1+\alpha^{-1})^N\\ =\text{Coefficient of } \large \alpha^s\text{ in }\alpha^{-N}(1+\alpha^1+\alpha^2)^N$$ Let power of $\alpha$ from bracket be k,then since $s=-N+k$,$k=s+N$ $$ E=\text{Coefficient of } \large \alpha^{s+N}\text{ in }(1-\alpha^3)^N(1-\alpha)^{-N}\\ =\text{Coefficient of } \large \alpha^{s+N}\text{ in }\left(\binom N0-\binom N1\alpha^1\cdots\right)(1-\alpha)^{-N}\\$$ Let powerof $\alpha$ from first bracket is $k_1$ and from second bracket is $k_2$, then $k_1+k_2=s+N$.So, $k_2=s+N-k_1$
As: $$(1-x)^{-n}=\sum_{k=0}^m(-1)^k\binom {N+k-1}{k}x^k$$ Therefore, $$E=\sum_{k=0}^{N}\binom Nk(-1)^{N+(s+N-k)-1}\binom {N+(s+N-k)-1}{s+N-k}$$