Coefficient of $x^m$ in the expansion of $(1+x+x^2+x^3)^m$ I want to count the coefficient of $x^m$ in the expansion of $(1+x+x^2+x^3)^m$

we can think of this problem as count possible $m$ steps which consist of $0,1,2,3$ where the order matters and all step sum up to $m$.
First, I have to check how many ways to decompose $m$ into as a sum of $0,1,2,3$.
Second, with the given composition of first step, make them arranged in order. 

How could one do the counting in the first process most easily and efficiently? 
 A: Factorise $(1+x+x^2+x^3)=(1+x)(1+x^2)$ and binomially expand
\begin{eqnarray*}
[x^m]: \left( \sum_{i=0}^{m} \binom{m}{i} x^i \right) \left(\sum_{j=0}^{m} \binom{m}{j} x^{2j} \right) 
\end{eqnarray*}
So we require $i+2j=m$ giving 
\begin{eqnarray*}
 \sum_{j=0}^{ \left\lfloor \frac{m}{2} \right\rfloor } \binom{m}{j}  \binom{m}{m-2j}  
\end{eqnarray*}
It remains to be seen if this can be simplified further.
The first few terms are $1,3,10,31,101,\cdots$ and more details about this sequence can be found here https://oeis.org/search?q=1%2C3%2C10%2C31%2C101&language=english&go=Search
A: This is isomorphic to the problem of how many ways can you roll a total of $m$ on $m$ 4-sided dice (which are numbered from 0 to 3).
Which is, in turn, a "stars and bars" problem.  How many ways can we arrange $m$ stars into $m$ bins such that no bin holds 4 or more stars.
${2m-1\choose m-1} - {m\choose 1}{2m-5\choose m-1} + {m\choose2}{2m-9\choose m-1}\cdots$ stopping before $2m-4k-1$ is less than $m-1$ 
A: $$
\begin{align}
\left[x^m\right]\left(\frac{\color{#C00}{1-x^4}}{\color{#090}{1-x}}\right)^m
&=\left[x^m\right]\left(\color{#C00}{\sum_{k=0}^m\binom{m}{k}(-1)^kx^{4k}}\color{#090}{\sum_{j=0}^\infty\binom{-m}{j}(-1)^jx^j}\right)\tag1\\
&=\sum_{k=0}^{\lfloor m/4\rfloor}(-1)^k\binom{m}{k}(-1)^{m-4k}\binom{-m}{m-4k}\tag2\\
&=\sum_{k=0}^{\lfloor m/4\rfloor}(-1)^k\binom{m}{k}\binom{2m-4k-1}{m-4k}\tag3
\end{align}
$$
Explanation:
$(1)$: Use the Binomial Theorem on the numerator and denominator
$(2)$: we want the coefficients of the terms with exponent $4k+j=m$
$(3)$: $\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$
Applying $(3)$ to $m=\{0,1,2,3,4,5,6,\dots\}$ gives $\{1,1,3,10,31,101,336\dots\}$

Note that in $(3)$, we could use $\binom{2m-4k-1}{m-4k}=\binom{2m-4k-1}{m-1}$, which might simplify things a bit, but that doesn't work for $m=0$.
A: hint: multinomial theorem might be a help. 
A: An alternative approach would be to calculate the $m^{th}$ derivative at $x=0$. By the taylor series you will obtain the coefficient if you devide the  $m^{th}$ derivative at $x=0$ by $m!$.
