Let $f(x)=\sum_{i=0}^{\infty} a_ix^i$. Also $f(x)\cdot(1+2x+2x^2+x^3)=\frac{1}{1-x^3}$. Find numbers $b,c,d$ such that: $$ a_n={3+n-1 \choose n}-ba_{n-1}-ca_{n-2}-da_{n-3} $$ for $n\ge 3$.
The expression can be modified to the following using generating functions: $$ f(x)=\frac{1}{1-x^3} : (1+2x+2x^2+x^3)=\frac{1}{1-x^3}:(1+x)(1+x+x^2)=\frac{(1-x)^2}{(1-x^3)^2(1-x^2)} $$ Then: $$ f(n)=\sum_{n=0}^2 {2\choose n}(-1)^nx^n\cdot \sum_{n=0}^{\infty}{2+n-1\choose n}x^{3n}\cdot \sum_{n=0}^{\infty}x^{2n} $$ For example: $$ a_0={2\choose 0}{1\choose 0}\cdot 1=1\\ a_1=-{2\choose 1}{2\choose 1}\cdot 1=-4\\ a_2={2\choose 2}{3\choose 2}\cdot 1=1 $$ but what happens with $\sum_{n=0}^2 {2\choose n}(-1)^nx^n$ for $a_n$ where $n\ge 3$? Its range for $n$ is only between $0$ and $2$ so does it become $1$ for all $n\ge 3$? If it does then: $$ a_3 = {4\choose 3}\cdot 1=4 $$ Then: $$ a_3={5\choose 3}-3b+4c-d=10-3b+4c+d=4\implies 3b-4c-d=6 \qquad \ast $$ But there's an infinite amount of solutions for the above equation so I don't think I'm in the right direction.
Note: as noticed by @ancientmathematician there was a mistake in the conditions to the problem it should be $f(x)\cdot(1+2x+2x^2+x^3)=\frac{1}{(1-x)^3}$
I believe Claude is correct about the statement of the problem, but I will attempt to provide some useful commentary as is.
So if we substitute $\sum_{i=0}^{\infty} a_{i}x^i$ in for $f(x)$, we obtain
$$\sum_{i=0}^{\infty} a_{i}x^i + 2x\sum_{i=0}^{\infty} a_{i}x^i + 2x^2\sum_{i=0}^{\infty} a_{i}x^i + x^3\sum_{i=0}^{\infty} a_{i}x^i = \sum_{i=0}^{\infty}x^{3i}$$
If you insert the $x$'s into the sums, and peel off all terms with degree $2$ or less, we obtain
$$a_0 + (2a_0+a_1)x + (2a_0 + 2a_1 + a_2)x^2 + \sum_{i=3}^{\infty} (a_{i}+ 2a_{i-1} + 2a_{i-2} + a_{i-3})x^i = \sum_{i=0}^{\infty} x^{3i}$$
Equating coefficients, we can say
$$a_0 = 1$$ $$2a_0+a_1 = 0$$ $$2a_0 + 2a_1 + a_2 = 0$$ $$a_{i}+ 2a_{i-1} + 2a_{i-2} + a_{i-3} = 1, i\geq3$$
Where the last equation holds if $i$ divides $3$, and is $0$ otherwise.
The first three equations give initial values $a_0, a_1, a_2$, and the third statement leads me to believe Claude's comment, since that would yield the desired binomial coefficient in place of the $1$ in the recurrence relation. Hope this helps!
Edit: Notice
$$\frac{1}{(1-x)^3} = \sum_{n\geq 0} \dbinom{-3}{n}(-1)^nx^n = \sum_{n\geq 0} \frac{(-3)(-3-1)...(-3-n+1)}{n!}(-1)^nx^n$$ $$=\sum_{n\geq 0} \frac{(3)(4)...(3+n-1)}{n!}x^n =\sum_{n\geq 0}\frac{(2)(3)(4)...(3+n-1)}{(3-1)n!}x^n = \sum_{n\geq 0}\dbinom{3+n-1}{n}x^n$$
So if you were to have equated coefficients as above, you would get the desired recurrence relation.