Generating function for the number of ordered triples Let an be the number of ordered triples $(i, j, k)$ of integer numbers
such that i ≥ 0, j? ≥ 1, k ≥ 1, and $i+3j +3k = n$. Find the generating
function of the sequence (a0, a1, a2, . . .) and calculate a formula for $a_{n}$.
First I used polynomials to express the possible combinations of $i,j,k$ and got $(\frac{1}{1-x})(\frac{1}{1-x^3}-1)(\frac{1}{1-x^3}-1)$, with $\frac{1}{1-x}$ standing for a sequence 1, 1, 1, 1 and the two other sums for possible combinations of $j$ and $k$. I subtracted 1 from each sum since I can't have $j=0$ or $k=0$. Is my calculation correct? How can I go about calculating the formula for $a_{n}$?

Yes, your generating function is correct. To extract its coefficients, there are many things you could do. The factor of $\frac1{1-x}$ always leads to sums of coefficients of other terms. For the term $\left(\frac{x^3}{1-x^3}\right)^2$, you can first remove the $x^6$ in the numerator and then deal with the denominator. One thing you can always do is just work out the first few terms in the series by hand and then hope to see a pattern. For instance,
\begin{eqnarray*}
\frac1{(1-x^3)^2}&=&(1+x^3+x^6+x^9+\cdots)(1+x^3+x^6+x^9+\cdots)\\
&=&1+2x^3+3x^6+4x^9+\cdots.
\end{eqnarray*}
Perhaps that is enough of a start so that you can finish on your own.
