1

Define $f(n)=\dfrac{n^2}{48}+\dfrac{n}{8}(1+x_2(n))+X(n),$ where $$X(n)=-\dfrac{7}{48}+\dfrac{9}{16}x_2(n)+\dfrac{1}{4}x_2(\lfloor \dfrac{n+1}{2} \rfloor)+\dfrac{1}{3}x_3(n),$$

and $\begin{eqnarray} x_2(n)= \begin{cases} 1 & 2\mid n \\ 0 & 2\nmid n \end{cases} \end{eqnarray},$ $\begin{eqnarray} x_3(n)= \begin{cases} 1 & 3\mid n \\ 0 & 3\nmid n \end{cases} \end{eqnarray}.$

  • Prove that the equation $2x+3y+4z=n,(x,y,z\in \mathbb N)$ has exactly $f(n)$ solutions.

  • The equation $x+2y+3z=n,(x,y,z\in \mathbb N)$ has exactly $\lfloor \dfrac{n^2}{12}+\dfrac{n}{2}+1 \rfloor$ solutions. Can you simplify $f(n)$ in this form?

Kunnysan
  • 2,050
lsr314
  • 15,806

1 Answers1

2

Being motivated by Torsten Hĕrculĕ Cärlemän's comment I can tell you that this could be a way to have the solution.

Notice that, $$F(q)=\displaystyle\sum_{n=1}^\infty f(n)q^n=\left(\displaystyle\sum_{n=1}^\infty q^{2n}\right)\left(\displaystyle\sum_{n=1}^\infty q^{3n}\right)\left(\displaystyle\sum_{n=1}^\infty q^{4n}\right)=\frac{q^9}{(1-q^2)(1-q^3)(1-q^4)}$$ where $q=\large e^{2\pi iz}$.

So, $$f(n)=\frac{1}{2\pi i}\displaystyle\int_0^1F(q)\large e^{-2in\pi z} dz=\displaystyle\int_\gamma\frac{q^{8-n}dq}{(1-q^2)(1-q^3)(1-q^4)}$$

where $\gamma$ is unit circle centered at $0$.

So, $$f(n)=-\displaystyle\int_\gamma\frac{q^{8-n}dq}{(q-1)^3(q+1)^2(q-i)(q+i)(q-\omega)(q-\omega^2)}$$

Hence, $$f(n)=-2\pi i\left(Res(z=1)+Res(z=-1)+Res(z=i)+Res(z=-i)+Res(z=\omega)+Res(z=\omega^2)\right)$$.

$Res(z=\pm i)=-\frac{(\mp i)^n(1 \pm i)}{16}$.................. (Here your $x_2(n)$ should come and $x_3(n)$ comes from residues at $\omega$ and $\omega^2$ terms).

You may proceed like this way and I believe that result would be obtained in this way, but I am leaving the rest of the solution to you as I am not relying on accuracy of my calculation.

Kunnysan
  • 2,050