For each integer $n$, let $a(n)$ be the number of nonnegative integer triples $(x,y,z)$ such that
$$x+2y+3z=n$$
From the data, the following recursion appears to hold
$$
a(n)=
\begin{cases}
\text{if}\;n<0,\;\text{then}\\[3.5pt]
\qquad 0\\[2.5pt]
\text{else if}\;n=0,\;\text{then}\\[3.5pt]
\qquad 1\\[.6pt]
\text{else}\\[.4pt]
\qquad a(n-1)+a(n-2)-a(n-4)-a(n-5)+a(n-6)\\
\end{cases}
$$
In particular, for $0\le n\le 15$, we get
$$\begin{array}
{
c|c|
c|c|c|c|c|
c|c|c|c|c|
c|c|c|c|c|
}
\hline
n
& 0
& 1
& 2
& 3
& 4
& 5
& 6
& 7
& 8
& 9
& 10
& 11
& 12
& 13
& 14
& 15
\\
\hline
a(n)
& 1
& 1
& 2
& 3
& 4
& 5
& 7
& 8
& 10
& 12
& 14
& 16
& 19
& 21
& 24
& 27
\\
\hline
\end{array}$$
Note:
I initially thought the recursion could be justified via a straightforward application of the principle of inclusion-exclusion, but the argument eludes me now.
Update:
Using the OP's generating function approach, together with a key idea from the solution by @Christian Blatter, the claimed recursion can be justified as follows . . .
Clearly, $a(0)=1$, and $a(n)=0$ when $n < 0$.
Working formally, we get
\begin{align*}
&\sum_{n\in\mathbb{Z}}\;a(n)x^n
=
1+\sum_{n=1}^\infty\;a(n)x^n
\\[6pt]
&
\phantom{\sum_{n\in\mathbb{Z}}\;a(n)x^n}
\,=
\left(\prod_{i=0}^\infty x^i\right)
\left(\prod_{i=0}^\infty x^{2i}\right)
\left(\prod_{i=0}^\infty x^{3i}\right)
\\[6pt]
&
\phantom{\sum_{n\in\mathbb{Z}}\;a(n)x^n}
\,=
\left(\frac{1}{1-x}\right)
\left(\frac{1}{1-x^2}\right)
\left(\frac{1}{1-x^3}\right)
\\[6pt]
\implies\;&\left(\sum_{n\in\mathbb{Z}}\;a(n)x^n\right)\bigl((1-x)(1-x^2)(1-x^3)\bigr)=1
\\[6pt]
\implies\;&\left(\sum_{n\in\mathbb{Z}}\;a(n)x^n\right)(1-x-x^2+x^4+x^5-x^6)=1
\\[6pt]
\implies\;&a(n)-a(n-1)-a(n-2)+a(n-4)+a(n-5)-a(n-6)=0,\;\text{for all}\;n \ge 1
\\[6pt]
\end{align*}
which confirms the claimed recursion.
New Update:
Here's another way to justify the claimed recursion . . .
As previously noted, it's clear that for $n < 0$, we have $a(n)=0$.
By direct evaluation, we get the values
$$\begin{array}
{
c|c|c|c|c|c|c|
}
\hline
n
& 0
& 1
& 2
& 3
& 4
& 5
\\
\hline
a(n)
& 1
& 1
& 2
& 3
& 4
& 5
\\
\hline
\end{array}$$
and it's then easily verified that the claimed recursion holds for $n\le 5$.
Thus, in what follows, assume $n\ge 6$.
Let $b(n)$ be the number of nonnegative integer ordered pairs $(x,y)$ such that $x+2y=n$.
Then for $a(n)$, we have the recursion
$$a(n)=a(n-3)+b(n)\tag{eq1}$$
and for $b(n)$ we have the recursion
$$b(n)=b(n-2)+1\tag{eq2}$$
Then from $(\text{eq}1)$, we get
$$b(n)=a(n)-a(n-3)\tag{eq3}$$
hence
$$b(n-2)=a(n-2)-a(n-5)\tag{eq4}$$
Using $(\text{eq}3)$ and $(\text{eq}4)$ to make replacements for $b(n)$ and $b(n-2)$ in $(\text{eq}2)$ and then solving for $a(n)$, we get
$$a(n)=a(n-2)+a(n-3)-a(n-5)+1\tag{eq5}$$
hence
$$a(n-1)=a(n-3)+a(n-4)-a(n-6)+1\tag{eq6}$$
Subtracting $(\text{eq}6)$ from $(\text{eq}5)$ and then solving for $a(n)$, we get
$$a(n)=a(n-1)+a(n-2)-a(n-4)-a(n-5)+a(n-6)$$
which completes the proof of the claimed recursion.