Hey, Tommy!
So I had noticed that you implicitly assumed \( V(s+1) = V(s) \) in your original analysis. I didn't notice it right away, but I noticed it afterwards when I saw it would imply \( 2 \pi i \) periodicity. I looked at it some more, and was pretty sure you were on to something--but never been too good with the Lambert function. This makes much much more sense quite frankly. Especially if we think of the branch cuts appearing at \( \Im(s) = 2\pi k \) for \( k \in \mathbb{Z} \). This is where our \( \phi \) function will recycle, and a cluster of singularities will force non-analycity of \( \tau \). I'm not sure how helpful I'd be at proving this using the Lambert function; but it's always helpful to have alternative representations.
EDIT:
Oh and definitely along the lines \( \Im(s) = 2\pi i k \) there is a neighborhood in which \( |\phi(s)| = 1 \). And these neighborhoods definitely appear closer and closer to \( \mathbb{R} + 2\pi i k \). Just think Picard's theorem for this one. You might appreciate this.
\( e \uparrow \uparrow s \)
Must get arbitrarily large in the right half plane and must get arbitrarily close to every large enough complex value (Picard). Because of this, we get arbitrarily close to \( \phi(s_0) = -s_0 + i\ell \) which makes \( \phi(s+1) = e^{i\ell} \). The trouble is making sure they cluster and don't just appear all willy nilly out of the blue. I implicitly assumed this before without even realizing it. So I was at best hal-right. I think I'm closer now.
So I had noticed that you implicitly assumed \( V(s+1) = V(s) \) in your original analysis. I didn't notice it right away, but I noticed it afterwards when I saw it would imply \( 2 \pi i \) periodicity. I looked at it some more, and was pretty sure you were on to something--but never been too good with the Lambert function. This makes much much more sense quite frankly. Especially if we think of the branch cuts appearing at \( \Im(s) = 2\pi k \) for \( k \in \mathbb{Z} \). This is where our \( \phi \) function will recycle, and a cluster of singularities will force non-analycity of \( \tau \). I'm not sure how helpful I'd be at proving this using the Lambert function; but it's always helpful to have alternative representations.
EDIT:
Oh and definitely along the lines \( \Im(s) = 2\pi i k \) there is a neighborhood in which \( |\phi(s)| = 1 \). And these neighborhoods definitely appear closer and closer to \( \mathbb{R} + 2\pi i k \). Just think Picard's theorem for this one. You might appreciate this.
\( e \uparrow \uparrow s \)
Must get arbitrarily large in the right half plane and must get arbitrarily close to every large enough complex value (Picard). Because of this, we get arbitrarily close to \( \phi(s_0) = -s_0 + i\ell \) which makes \( \phi(s+1) = e^{i\ell} \). The trouble is making sure they cluster and don't just appear all willy nilly out of the blue. I implicitly assumed this before without even realizing it. So I was at best hal-right. I think I'm closer now.
