A transcendental number from the diophantine equation $x+2y+3z=n$ Let $\displaystyle n=1,2,3,\cdots.$ We denote by $D_n$ the number of non-negative integer solutions of the diophantine equation 
$$x+2y+3z=n$$

Prove that 
  $$
\sum_{n=0}^{\infty} \frac{1}{D_{2n+1}}
$$
  is a transcendental number.


If I have not mistaken something,
$$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)}{3}, $$
when $n\equiv 0,2\pmod{3}$, and
$$ [x^{2n+1}]\frac{1}{(1-x)(1-x^2)(1-x^3)} = \frac{(n+1)(n+3)+1}{3} $$
when $n\equiv 1\pmod{3}$, hence:
$$\begin{eqnarray*}\sum_{n=0}^{+\infty}\frac{1}{D_{2n+1}}&=&\sum_{n=0}^{+\infty}\frac{3}{(n+1)(n+3)}-\!\!\!\!\sum_{n\equiv 1\!\!\pmod{\!\!3}}\frac{3}{(n+1)(n+3)(n+2)^2}\\&=&\frac{9}{4}-\frac{1}{3}\left(\frac{9}{2}-\frac{\sqrt{3}\,\pi}{2}-\frac{\pi^2}{6}\right)\\&=&\frac{1}{36}\left(27+2\pi\left(\pi+3\sqrt{3}\right)\right),\end{eqnarray*}$$
and we only need to show that $u=\pi(\pi+3\sqrt{3})$ is a trascendental number.
But if $u$ were algebraic, then
$$\pi = \frac{-3\sqrt{3}+\sqrt{27+4u}}{2}$$
would be algebraic too, contradiction.
