Let $\gamma$ be Euler constant and $W$ Lambert W function.


One can show:


$$-2/3\,{\frac {\gamma+\ln \left( \pi \right) }{W \left( 
-1/3\,{\frac { \left( \gamma+\ln \left( \pi \right) \right) {
{\rm e}^{-1/3\,\gamma}}}{\sqrt [3]{\pi }}} \right) }} = 2 \qquad (1) $$


This means at least one of $ \left( 
-1/3\,{\frac { \left( \gamma+\ln \left( \pi \right) \right) {
{\rm e}^{-1/3\,\gamma}}}{\sqrt [3]{\pi }}} \right)$ and
$\gamma+\ln \left( \pi \right)$ is transcendental.



> 
> Q1 Is this known?
> 
> 
> Q2 Can one solve (1) for $\gamma$ or $\pi$? (sage, maple and
>  Wolfram Alpha couldn't).
> 
> 
> 


In machine readable form:



```
 def eulerpi():
      import mpmath
      from mpmath import lambertw,euler,pi,log,exp
      mpmath.mp.dps=10**4
      a= -2*(euler + log(pi))/lambertw(-1/3*( (euler+log(pi))*exp(-euler/3))/pi**(1/3))/3-2
      print mpmath.chop(a)

```


Identity between Euler gamma and pi
By the definition of the Lambert function 
$W\left(-\frac13\alpha e^{-\alpha/3}\right)=-\frac\alpha3\,$ for any $\alpha$, so 
$$
-2/3\frac\alpha{W\left(-\frac13\alpha e^{-\alpha/3}\right)}=2.
$$
Putting here $\alpha=\gamma+\ln\pi$ gives your formula. 


