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We have $f(x) = \sum_{i=0}^\infty a_ix^i$. We know that $f(x)(1+2x+2x^2+x^3)=\frac{1}{(1-x)^3}$. We need to find $a_0, a_1, a_2$. Also we need to find numbers $r, s, t$ from $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3. I already get that $f(x) =\sum_{i=0}^\infty x^i \times \sum_{k=0}^\infty x^{2k} \times \sum_{j=0}^\infty x^{3j}$. I am sure that this problem is connected to this. Also there is a problem connected with two first ones: To write generating function for calculating number of integer solutions of the equation: $x_1 + 2x_2 + 3x_3 = n$. Thank you and merry Christmas!

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For a start.

Consider that $$f(x)(1+2x+2x^2+x^3)(1-x)^3=1$$ must hold for any $x$ that is say $$(1-x-x^2+x^4+x^5-x^6)f(x)=1$$ Replace $f(x)$ by a few terms (since you just want $a_0,a_1,a_2$) and identify the coefficients.

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Your finding is correct. Indeed: $$f(x)(1+2x+2x^2+x^3)=\frac{1}{(1-x)^3} \Rightarrow \\\\ f(x)(1+x)(1+x+x^2)=\frac{1}{(1-x)^3} \Rightarrow \\ f(x)=\frac{1}{(1-x)(1-x^2)(1-x^3)}=\sum_{i=0}^{\infty} x^i\sum_{j=0}^{\infty} x^{2j}\sum_{k=0}^{\infty} x^{3k} \Rightarrow\\ i+2j+3k=n, \ \ \ i,j,k,n\ge 0.$$ So, if you need to find a few terms: $$\begin{align}n&=0\Rightarrow (0,0,0) \Rightarrow a_0=1;\\ n&=1 \Rightarrow (1,0,0) \Rightarrow a_1=1;\\ n&=2 \Rightarrow (2,0,0), (0,1,0) \Rightarrow a_2=2;\\ n&=3 \Rightarrow (3,0,0), (1,1,0), (0,0,1) \Rightarrow a_3=3.\end{align}$$ Wolfram answer.

For more, you can see here.

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  • $\begingroup$ Thank you very much for your comment. Could I have one more minute of your attention for explaining how can i get to $a_n=D(3,n)-ra_{n-1}-sa_{n-2}-ta_{n-3}$ for n≥3? $\endgroup$ – O.Joen Dec 24 '18 at 10:03

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