Benoit Cloitre offered two 'mirror sequences', which allow to compute $\pi$ and $e$ in similar ways:
$$u_{n+2}=u_{n+1}+\frac{u_n}{n}$$
$$v_{n+2}=\frac{v_{n+1}}{n}+v_{n}$$
$$u_1=v_1=0$$
$$u_2=v_2=1$$
$$\lim_{n \to \infty} \frac{n}{u_n}=e$$
$$\lim_{n \to \infty} \frac{2n}{v_n^2}=\pi$$
The formulation and the proof can be seen here.
What do you think - is it just a coincidence or is there some deeper meaning in this mirror algorithm about the connection of two constants?
By @EricStucky in the comment, the better question:
Is there any connection between $e$ and $π$ which is essentially different than Euler's formula?
Of course, I expect an answer related to my own question about this 'mirror sequence'
If, on the other hand, someone shows a clear relation between this sequence and Euler's formula, that's fine too
As always, with me, please double check. The sequence is quite obvious.
We will transform both to differential equations by the "Method of Coefficients". This is similar to one of Benoit Cloitre's methods but a little more direct.
Using the two OGF forms: $$V\left(x\right)={\displaystyle \sum_{k=0}^{\infty}} v_{k}x^{k}, U\left(x\right)={\displaystyle \sum_{k=0}^{\infty}}u_{k}x^{k}$$ The alignment/conversion techniques are:
$$\left[x^{k}\right]\frac{V\left(x\right)}{x^{2}}=v_{k+2};\left[x^{k}\right]\frac{V\left(x\right)}{x}=v_{k+1};\left[x^{k}\right]x\cdot\frac{\partial V(x)}{\partial x}=n\cdot v_{k}$$ We line up the $[x^{k}]$ terms for $V_{k}\left(x\right),U_{k}\left(x\right)$ and flatten the recursions: $$n\cdot v_{n+2}-v_{n+1}-n\cdot v_{n}=0$$ $$n\cdot u_{n+2}-n\cdot u_{n+1}-u_{n}=0$$ $x\cdot\frac{\partial\left(\frac{U\left(x\right)}{x^{2}}\right)}{\partial x}-x\cdot\frac{\partial\left(\frac{U(x)}{x}\right)}{\partial x}-U\left(x\right)=0$
$x\cdot\frac{\partial\left(\frac{U\left(x\right)}{x^{2}}\right)}{\partial x}-x\cdot\frac{\partial\left(\frac{U(x)}{x}\right)}{\partial x}-U\left(x\right)=0$
$\left(\frac{1}{x}-1\right)\frac{\partial U\left(x\right)}{\partial x}-\left(\frac{2}{x^{2}}-\frac{1}{x}+1\right)U(x)=0$
$x\cdot\left(1-x\right)\frac{\partial U\left(x\right)}{\partial x}-\left(2-x+x^{2}\right)U(x)=0$
$\frac{1}{U\left(x\right)}\frac{\partial U\left(x\right)}{\partial x}-\left(\frac{\left(x^{2}-x+2\right)}{x\cdot\left(1-x\right)}\right)=0$
$U\left(x\right)=\frac{e^{-x}\cdot x^{2}}{\left(1-x\right)^{2}}$
Where the integration constant is evaluated by the first three terms of the taylor series expansion.
$x\cdot\frac{\partial\left(\frac{V\left(x\right)}{x^{2}}\right)}{\partial x}-\frac{V\left(x\right)}{x}-x\cdot\frac{\partial V\left(x\right)}{\partial x}=0$
$x\cdot\frac{-2}{x^{3}}V\left(x\right)+x\cdot\frac{1}{x^{2}}\frac{\partial V\left(x\right)}{\partial x}-\frac{V\left(x\right)}{x}-x\cdot\frac{\partial V\left(x\right)}{\partial x}=0$
$\left(\frac{1}{x}-x\right)\frac{\partial V\left(x\right)}{\partial x}-\left(\frac{1}{x}+\frac{2}{x^{2}}\right)V\left(x\right)=0$
$\left(1-x^{2}\right)\cdot x\cdot\frac{\partial V\left(x\right)}{\partial x}-\left(x+2\right)\cdot V\left(x\right)=0$
$\frac{1}{V\left(x\right)}\frac{\partial V\left(x\right)}{\partial x}-\left(\frac{2}{x}-\frac{1}{2\cdot\left(x+1\right)}-\frac{3}{2\cdot\left(x-1\right)}\right)=0$
$ln\left(V\left(x\right)\right)=ln\left(x^{2}\right)-ln\left(\left(x+1\right)^{\frac{1}{2}}\right)-ln\left(\left(\left(x-1\right)^{\frac{3}{2}}\right)\right)+C$
$V\left(x\right)=\frac{x^{2}}{\left(x-1\right)^{\frac{1}{2}}\cdot\left(x+1\right)^{\frac{3}{2}}}$