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.


We have that
$$
\displaystyle D_{n} = \left \lfloor \frac {(n+3)^2}{12} \right \rceil
$$
where $\lfloor x \rceil$ is the nearest integer to $x$. You can find a detailed proof here (first pdf link, p.50).
It gives
$$
D_{2n+1} = \left \lfloor \frac {(n+2)^2}{3} \right \rceil.
$$ 
As Jack D'Aurizio explained, we deduce that

$$
\sum_{n=0}^{\infty} \frac{1}{D_{2n+1}}=\frac{3}{4} + \frac{\pi \sqrt{3}}{6} + \frac{\pi^2}{18} 
$$

Since $\sqrt{3}$ is algebraic, this sum is transcendental as $\pi$ is.
Likewise, we obtain

$$
\sum_{n=0}^{\infty} \frac{1}{D_{n}}= \frac{135}{36} - \frac{\pi \sqrt{3}}{6} + \frac{\pi^2}{18} +\frac{\pi \sqrt{3}}{3}\tan \left( \frac{\pi \sqrt{3}}{6} \right),
$$
  $$
\sum_{n=0}^{\infty} \frac{(-1)^n}{D_{n}}=\frac{9}{4} - \frac{\pi \sqrt{3}}{2} + \frac{\pi^2}{18} +\frac{\pi \sqrt{3}}{3}\tan \left( \frac{\pi \sqrt{3}}{6} \right).
$$

For these two last sums, it seems rather difficult to prove they are transcendental, as we may suspect they are.
