What is the generating function for $\{ a_k \}$ where $a_k$ is the number of solutions of $x_1 + x_2+x_3=k$ when $x_1,x_2,x_3$ are integers with $x_1 \geq2$ , $0\leq x_2 \leq 3$ , and $2 \leq x_3 \leq 5$ ?

My answer comes to be $\frac { x^4 -x^{12} - 2x^8 }{(1-x)^3}$

which evaluates to $(x^4 -x^{12} - 2x^8) \sum_{r=0}^{\infty} \binom{r+2}{r} x^r$ but rosen's key says it's $\frac{x^4(1+x+x^2+x^3)^2}{(1-x)}$

which one is correct? And when in the next part they have asked the value of $a_6$ whatever I derived, using that I got correct value for $a_6$ .So I am in doubt whether my anser is correct. Please help.


Encode the condition $x_{1}\geq 2$ using $$ x^{2}+x^{3}+\dotsb=\frac{x^2}{1-x}. $$ Similarly, encode the condition $0\leq x_2 \leq 3$ using $$ 1+x^1+x^2+x^3=\frac{1-x^4}{1-x} $$ and encode the condition $2\leq x_{3} \leq 5$ using $$ x^2+x^3+x^4+x^5=x^2\frac{1-x^{4}}{1-x} $$ Thus $$ \sum_{k\geq 0 }a_{k}x^k=\frac{x^2}{1-x}\times \frac{1-x^4}{1-x} \times x^2\frac{1-x^{4}}{1-x} =\frac{x^4}{1-x}(1+x+x^2+x^3)^2. $$

  • 3
    $\begingroup$ +1. @OP: From this representation, you can see that you only had a sign error in your result; if you flip the sign of your $x^{12}$ term in the numerator, the result is the representation in the centre of the last equation here, with the numerator multiplied out. $\endgroup$ – joriki Jun 25 '18 at 16:25
  • $\begingroup$ @Foobaz,joriki-Thanks for the clarification.! :) $\endgroup$ – user3767495 Jun 26 '18 at 2:57

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