A transcendental number from the diophantine equation $x+2y+3z=n$

Question

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.

Answer 1

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.

Answer 2

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.