Does $E^2 \; ( E \approx 1.2640847\ldots)$ equal $D \approx 1.5979102\ldots$? Does $E^2=D$?
Where $E$ is a constant used in the closed form of the Sylvester Sequence (see: Closed form formula and asymptotics) and $D$ is a constant for the closed formula of the sequence A007018 (see: FORMULA)
$$D = \ln(3/2)+\sum_{n \ge 0} \ln(1+(2a_n+1)^{-2})$$
$$a_n = a_{n-1}^2 + a_{n-1} = \left\lfloor D^{2^n} - \frac12 \right\rfloor$$
$$a_0 = 1$$
(see: A007018, COMMENTS)
$$E=\frac12\sqrt{6}\exp{\sum_{j=1}^\infty2^{-j-1}\ln\left[1+(2e_j-1)^{-2}\right]}$$
$$ e_n = e_{n-1}^2 - e_{n-1} + 1 = \left\lfloor E^{2^{n+1}} + \frac12 \right\rfloor$$
$$ e_0 = 2 $$
(see: this MathWorld Article about the Sylvester Sequence)
Posted by a eager highschool student ;D
 A: For both the Sylvester sequence and the OEIS one, the more difficult part actually lies in showing that the $n$-th term can be expressed in the form of a double exponential. If this is taken for granted, the reasoning is quite simple:
$$D=\lim_{n\rightarrow\infty} \exp{\frac{\ln a_n}{2^n}}$$
$$E^2=\lim_{n\rightarrow\infty}\left(\exp{\frac{\ln e_n}{2^{n+1}}}\right)^2=\lim_{n\rightarrow\infty}\exp{\frac{\ln e_n}{2^n}}$$
Since it's easy to prove by induction that $e_n = a_n+1$, we can also conclude that $D=E^2$.
A: First show by induction that $e_n = a_n + 1$. 
Then:
$$e_n^2 = \Big\lfloor E^{2^{n+1}}+\frac{1}{2}\Big\rfloor^2=E^{2^{n+2}}+ E^{2^{n+1}}+C_1$$ 
Where $C_1$ is some number near zero, say $C_1 \in [-2,2]$. But:
$$e_n^2 = a_n^2 + 2a_n + 1 = \Big\lfloor D^{2^n}-\frac{1}{2}\Big\rfloor^2+ 2\Big\lfloor D^{2^n}-\frac{1}{2}\Big\rfloor+1$$
$$=D^{2^{n+1}}+D^{2^n}+C_2$$
Where again, let's lazily bind $C_2\in[-2,2].$
Now it's clear that as $n\to\infty$, 
$$\lim_{n\to\infty} \left( E^{2^{n+2}}+ E^{2^{n+1}} - D^{2^{n+1}}- D^{2^n}\right)= 0$$ 
Since this cannot hold true for any $D\neq E^2$, we must have $D=E^2$ as hypothesised.
