Question

a) What is the generating function for$\left\{a_k\right\}$, where $a_k$is the number of solutions of $x_1+x_2+x_3=k$when$x_1,x_2$, and $x_3$are integers with$x_1\ge 2,0\le x_2\le 3$, and $2\le x_3\le 5$?

b) Use your answer to part (a) to find$a_6$.

Short answer

(a)The total generating function is$\frac{x^4{\left(1+x+x^2+x^3\right)}^2}{1-x}$.

(b) Therefore,$a_6=6$.

Step-by-step solution

Use Generating Function:

Generating function for the sequence$a_0,a_1,\dots ,a_k,\dots$of real numbers is the infinite series

$G\left(x\right)=a_0+a_{1}x+a_2{x}^2+\dots +a_k{x}^k+\dots =\sum _{k=0}^{+\infty }{a}_k{x}^k$

Extended binomial theorem;

localid="1668603329670" ${(1+x)}^u=\sum _{k=0}^{+\infty }\left(\begin{array}{c}u\\ k\end{array}\right)x^k$

Since$x_1\ge 2$the series representing $x_1$should then contain only terms with a power of at least 2:

$x^2+x^3+x^4+\dots$

Since$0\le x_2\le 3$the series representing $x_2$should then contain only terms with a power of at least 0 and at most 3:

$1+x+x^2+x^3$

Since$2\le x_3\le 5$the series representing $x_3$should then contain only terms with a power of at least 2 and at most 5:

$x^2+x^3+x^4+x^5$

The generating function is the product of the series for each variable:

$$\begin{gathered} \left(1+x+x^2+x^3\right)\left(x^2+x^3+x^4+x^5\right)\left(x^2+x^3+x^4+\dots \right) \\ =x^2{\left(1+x+x^2+x^3\right)}^2\left(x^2\right)\left(1+x+x^2+\dots \right) \\ =x^4{\left(1+x+x^2+x^3\right)}^2\left(1+x+x^2+\dots \right) \\ =x^4{\left(1+x+x^2+x^3\right)}^2\left(\sum _{k=0}^{+\infty }{x}^k\right) \\ =x^4{\left(1+x+x^2+x^3\right)}^2·\frac{1}{1-x} \\ =\frac{x^4{\left(1+x+x^2+x^3\right)}^2}{1-x} \end{gathered}$$

b).

Use Extended binomial theorem:

$$\begin{gathered} \frac{x^4{\left(1+x+x^2+x^3\right)}^2}{1-x}=x^4·{\left(\sum _{k=0}^3{x}^k\right)}^2·\frac{1}{1-x} \\ =x^4·{\left(\frac{1-x^4}{1-x}\right)}^2·\frac{1}{1-x} \\ =x^4·\frac{{\left(1-x^4\right)}^2}{{(1-x)}^3} \\ =x^4·{\left(1+\left(-x^4\right)\right)}^2·(1+(-x){)}^{-3} \end{gathered}$$

By further simplification:

$$\begin{gathered} \frac{x^4{\left(1+x+x^2+x^3\right)}^2}{1-x}=x^4.\sum _{m=0}^{+\infty }\left(\begin{array}{c}2\\ m\end{array}\right){\left(-x^4\right)}^m.\sum _{k=0}^{+\infty }\left(\begin{array}{c}-3\\ k\end{array}\right){\left(-x\right)}^k \\ =x^4.\sum _{m=0}^{+\infty }\left(\begin{array}{c}2\\ m\end{array}\right){\left(-1^4\right)}^m.\sum _{k=0}^{+\infty }\left(\begin{array}{c}-3\\ k\end{array}\right){\left(-1\right)}^k \\ =x^4·\sum _{m=0}^{+\infty }{b}_m{x}^{4m}·\sum _{k=0}^{+\infty }{c}_k{x}^k \end{gathered}$$

Let$b_m=\left(\begin{array}{c}2\\ m\\ \end{array}\right){(-1)}^m$and$c_k=\left(\begin{array}{c}-3\\ k\end{array}\right)(-1{)}^k$

The coefficient of x6a6  if  m=0and k=2:

The coefficient of $x^6$is then the sum of the coefficients for each possible combination of $m$and $k$:

localid="1668605572590" $$\begin{gathered} a_6=b_0{c}_2 \\ =\left(\begin{array}{c}2\\ 0\end{array}\right){\left(-1\right)}^0\left(\begin{array}{c}-3\\ 2\end{array}\right){\left(-1\right)}^2 \\ =6 \end{gathered}$$

Hence$a_6=6$