Does $E^2 \; ( E \approx 1.2640847\ldots)$ equal $D \approx 1.5979102\ldots$?

Question

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

Answer 1

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 $D=E^2$.

Answer 2

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.