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This is not a proper answer since it involves several heuristics. I will use the generating function $f(x)$ found by @skbmoore. For convenience, let $g(x)=f(2x)$. I would like to use Hayman's formula for the asymptotics of the Maclaurin coefficients of certain entire functions. The first non-trivial assumption is that $g(x)$ is admissible in the sense of Hayman, i.e., we can use his formula. Note that the Maclaurin coefficients of $g(x)$ are $2^n a_n$.

First, we estimate the growth of the logarithmic derivative. Numerics suggests that $$ a(r) := r\frac{{g'(r)}}{{g(r)}} = 2r + \sum\limits_{k = 0}^\infty {\left( {\frac{{r/2^k }}{{e^{r/2^k } - 1}} - 1} \right)} = 2r - \frac{{\log r}}{{\log 2}} - \frac{1}{2} + \mathcal{O}\!\left( {\frac{1}{r}} \right) $$ as $r\to +\infty$. Hayman defines the sequence $r_n$ via $a(r_n ) = n$. Using the above, I found $$ r_n = \frac{n}{2} + \frac{{\log n}}{{2\log 2}} - \frac{1}{4} + o(1) $$ as $n\to +\infty$. We also have $b(r) := ra'(r) = 2r + \mathcal{O}(1)$ for large $r$. Now $$ \log g(r) = \int_1^r {\frac{{a(t)}}{t}dt} + \log g(1) = 2r - \frac{{\log ^2 r}}{{2\log 2}} - \frac{{\log r}}{2} + C + o(1) $$ for large $r$ with some real constant $C$. Therefore, by Hayman's formula, \begin{align*} \log (2^n a_n ) & = - n\log r_n + \log g(r_n ) - \frac{1}{2}\log (2\pi ) - \frac{1}{2}\log b(r_n ) + o(1) \\ & = - n\log \left( {\frac{n}{2}} \right) + n - \frac{{\log ^2 n}}{{2\log 2}} + C' + o(1) \end{align*} as $n\to +\infty$ with some real constant $C'$. Equivalently, $$ \log _2 a_n = - n\log _2 n + \frac{n}{{\log 2}} - \frac{{\log _2^2 n}}{2} + C'' + o(1) $$ as $n\to +\infty$ with some real constant $C''$. Up to the value of the constant and the weaker error term, this is the same result as yours.

This is not a proper answer since it involves several heuristics. I will use the generating function $f(x)$ found by @skbmoore. For convenience, let $g(x)=f(2x)$. I would like to use Hayman's formula for the asymptotics of the Maclaurin coefficients of certain entire functions. The first non-trivial assumption is that $g(x)$ is admissible in the sense of Hayman, i.e., we can use his formula. Note that the Maclaurin coefficients of $g(x)$ are $2^n a_n$.

First, we estimate the growth of the logarithmic derivative. It can be shown that \begin{align*} a(r) :\!&= r\frac{{g'(r)}}{{g(r)}} = 2r + \sum\limits_{k = 0}^\infty {\left( {\frac{{r/2^k }}{{e^{r/2^k } - 1}} - 1} \right)} \\ & = 2r - \frac{{\log r}}{{\log 2}} - \frac{1}{2} - \sum\limits_{k = 1}^\infty {\frac{{2^k r}}{{e^{2^k r} - 1}}} +c(r), \end{align*} where the function $c(r)$ is bounded and staisfies $c(r)=c(2r)$. Numerics suggests that $|c(r)|<2\times 10^{-5}$. Hayman defines the sequence $r_n$ via $a(r_n ) = n$. Using the above, I found $$ r_n = \frac{n}{2} + \frac{{\log n}}{{2\log 2}} - \frac{1}{4} + o(1) $$ as $n\to +\infty$. We also have $b(r) := ra'(r) = 2r + \mathcal{O}(1)$ for large $r$. Now \begin{align*} \log g(r) & = \int_1^r {\frac{{a(t)}}{t}dt} + \log g(1) \\ & = 2r - \frac{{\log ^2 r}}{{2\log 2}} - \frac{{\log r}}{2} + C_0 - \sum\limits_{k = 1}^\infty \log (1 - e^{ - 2^k r} ) + d(r), \end{align*} with some real constant $C_0$ and with a bounded function $d(r)$ which staisfies $d(r)=d(2r)$. Therefore, by Hayman's formula, \begin{align*} \log (2^n a_n ) & = - n\log r_n + \log g(r_n ) - \frac{1}{2}\log (2\pi ) - \frac{1}{2}\log b(r_n ) + o(1) \\ & = - n\log \left( {\frac{n}{2}} \right) + n - \frac{{\log ^2 n}}{{2\log 2}} + C_1 +d(r_n)+ o(1) \end{align*} as $n\to +\infty$ with some real constant $C_1$. Equivalently, $$ \log _2 a_n = - n\log _2 n + \frac{n}{{\log 2}} - \frac{{\log _2^2 n}}{2} + C_2 +d(r_n)+ o(1) $$ as $n\to +\infty$ with some real constant $C_2$.

This is not a proper answer since it uses several heuristics. I will use the generating function $f(x)$ found by @skbmoore. For convenience, let $g(x)=f(2x)$. I would like to use Hayman's formula for the asymptotics of the Maclaurin coefficients of certain entire functions. The first non-trivial assumption is that $g(x)$ is admissible in the sense of Hayman, i.e., we can use his formula. Note that the Maclaurin coefficients of $g(x)$ are $2^n a_n$.

First, we estimate the growth of the logarithmic derivative. Numerics suggests that $$ a(r) := r\frac{{g'(r)}}{{g(r)}} = 2r + \sum\limits_{k = 0}^\infty {\left( {\frac{{r/2^k }}{{e^{r/2^k } - 1}} - 1} \right)} = 2r - \frac{{\log r}}{{\log 2}} - \frac{1}{2} + \mathcal{O}\!\left( {\frac{1}{r}} \right) $$ as $r\to +\infty$. Hayman defines the sequence $r_n$ via $a(r_n ) = n$. Using the above, I found $$ r_n = \frac{n}{2} + \frac{{\log n}}{{2\log 2}} - \frac{1}{4} + o(1) $$ as $n\to +\infty$. We also have $b(r) := ra'(r) = 2r + \mathcal{O}(1)$ for large $r$. Now $$ \log g(r) = \int_1^r {\frac{{a(t)}}{t}dt} + \log g(1) = 2r - \frac{{\log ^2 r}}{{2\log 2}} - \frac{{\log r}}{2} + C + o(1) $$ for large $r$ with some real constant $C$. Therefore, by Hayman's formula, \begin{align*} \log (2^n a_n ) & = - n\log r_n + \log g(r_n ) - \frac{1}{2}\log (2\pi ) - \frac{1}{2}\log b(r_n ) + o(1) \\ & = - n\log \left( {\frac{n}{2}} \right) + n - \frac{{\log ^2 n}}{{2\log 2}} + C' + o(1) \end{align*} as $n\to +\infty$ with some real constant $C'$. Equivalently, $$ \log _2 a_n = - n\log _2 n + \frac{n}{{\log 2}} - \frac{{\log _2^2 n}}{2} + C'' + o(1) $$ as $n\to +\infty$ with some real constant $C''$. Up to the value of the constant and the weaker error term, this is the same result as yours.

This is not a proper answer since it involves several heuristics. I will use the generating function $f(x)$ found by @skbmoore. For convenience, let $g(x)=f(2x)$. I would like to use Hayman's formula for the asymptotics of the Maclaurin coefficients of certain entire functions. The first non-trivial assumption is that $g(x)$ is admissible in the sense of Hayman, i.e., we can use his formula. Note that the Maclaurin coefficients of $g(x)$ are $2^n a_n$.

First, we estimate the growth of the logarithmic derivative. Numerics suggests that $$ a(r) := r\frac{{g'(r)}}{{g(r)}} = 2r + \sum\limits_{k = 0}^\infty {\left( {\frac{{r/2^k }}{{e^{r/2^k } - 1}} - 1} \right)} = 2r - \frac{{\log r}}{{\log 2}} - \frac{1}{2} + \mathcal{O}\!\left( {\frac{1}{r}} \right) $$ as $r\to +\infty$. Hayman defines the sequence $r_n$ via $a(r_n ) = n$. Using the above, I found $$ r_n = \frac{n}{2} + \frac{{\log n}}{{2\log 2}} - \frac{1}{4} + o(1) $$ as $n\to +\infty$. We also have $b(r) := ra'(r) = 2r + \mathcal{O}(1)$ for large $r$. Now $$ \log g(r) = \int_1^r {\frac{{a(t)}}{t}dt} + \log g(1) = 2r - \frac{{\log ^2 r}}{{2\log 2}} - \frac{{\log r}}{2} + C + o(1) $$ for large $r$ with some real constant $C$. Therefore, by Hayman's formula, \begin{align*} \log (2^n a_n ) & = - n\log r_n + \log g(r_n ) - \frac{1}{2}\log (2\pi ) - \frac{1}{2}\log b(r_n ) + o(1) \\ & = - n\log \left( {\frac{n}{2}} \right) + n - \frac{{\log ^2 n}}{{2\log 2}} + C' + o(1) \end{align*} as $n\to +\infty$ with some real constant $C'$. Equivalently, $$ \log _2 a_n = - n\log _2 n + \frac{n}{{\log 2}} - \frac{{\log _2^2 n}}{2} + C'' + o(1) $$ as $n\to +\infty$ with some real constant $C''$. Up to the value of the constant and the weaker error term, this is the same result as yours.

This is not a proper answer since it uses several heuristics. I will use the generating function $f(x)$ found by @skbmoore. For convenience, let $g(x)=f(2x)$. I would like to use Hayman's formula for the asymptotics of the Maclaurin coefficients of certain entire functions. The first non-trivial assumption is that $g(x)$ is admissible in the sense of Hayman, i.e., we can use his formula. Note that the Maclaurin coefficients of $g(x)$ are $2^n a_n$.

First, we estimate the growth of the logarithmic derivative. Numerics suggests that $$ a(r) := r\frac{{g'(r)}}{{g(r)}} = 2r + \sum\limits_{k = 0}^\infty {\left( {\frac{{r/2^k }}{{e^{r/2^k } - 1}} - 1} \right)} = 2r - \frac{{\log r}}{{\log 2}} - \frac{1}{2} + \mathcal{O}\!\left( {\frac{1}{r}} \right) $$ as $r\to +\infty$. Hayman defines the sequence $r_n$ via $a(r_n ) = n$. Using the above, I found $$ r_n = \frac{n}{2} + \frac{{\log n}}{{2\log 2}} - \frac{1}{4} + o(1) $$ as $n\to +\infty$. We also have $b(r) := ra'(r) = 2r + o(1)$ for large $r$. Now $$ \log g(r) = \int_1^r {\frac{{a(t)}}{t}dt} + \log g(1) = 2r - \frac{{\log ^2 r}}{{2\log 2}} - \frac{{\log r}}{2} + C + o(1) $$ for large $r$ with some real constant $C$. Therefore, by Hayman's formula, \begin{align*} \log (2^n a_n ) & = - n\log r_n + \log g(r_n ) - \frac{1}{2}\log (2\pi ) - \frac{1}{2}\log b(r_n ) + o(1) \\ & = - n\log \left( {\frac{n}{2}} \right) + n - \frac{{\log ^2 n}}{{2\log 2}} + C' + o(1) \end{align*} as $n\to +\infty$ with some real constant $C'$. Equivalently, $$ \log _2 a_n = - n\log _2 n + \frac{n}{{\log 2}} - \frac{{\log _2^2 n}}{2} + C'' + o(1) $$ as $n\to +\infty$ with some real constant $C''$. Up to the value of the constant and the weaker error term, this is the same result as yours.

This is not a proper answer since it uses several heuristics. I will use the generating function $f(x)$ found by @skbmoore. For convenience, let $g(x)=f(2x)$. I would like to use Hayman's formula for the asymptotics of the Maclaurin coefficients of certain entire functions. The first non-trivial assumption is that $g(x)$ is admissible in the sense of Hayman, i.e., we can use his formula. Note that the Maclaurin coefficients of $g(x)$ are $2^n a_n$.

First, we estimate the growth of the logarithmic derivative. Numerics suggests that $$ a(r) := r\frac{{g'(r)}}{{g(r)}} = 2r + \sum\limits_{k = 0}^\infty {\left( {\frac{{r/2^k }}{{e^{r/2^k } - 1}} - 1} \right)} = 2r - \frac{{\log r}}{{\log 2}} - \frac{1}{2} + \mathcal{O}\!\left( {\frac{1}{r}} \right) $$ as $r\to +\infty$. Hayman defines the sequence $r_n$ via $a(r_n ) = n$. Using the above, I found $$ r_n = \frac{n}{2} + \frac{{\log n}}{{2\log 2}} - \frac{1}{4} + o(1) $$ as $n\to +\infty$. We also have $b(r) := ra'(r) = 2r + \mathcal{O}(1)$ for large $r$. Now $$ \log g(r) = \int_1^r {\frac{{a(t)}}{t}dt} + \log g(1) = 2r - \frac{{\log ^2 r}}{{2\log 2}} - \frac{{\log r}}{2} + C + o(1) $$ for large $r$ with some real constant $C$. Therefore, by Hayman's formula, \begin{align*} \log (2^n a_n ) & = - n\log r_n + \log g(r_n ) - \frac{1}{2}\log (2\pi ) - \frac{1}{2}\log b(r_n ) + o(1) \\ & = - n\log \left( {\frac{n}{2}} \right) + n - \frac{{\log ^2 n}}{{2\log 2}} + C' + o(1) \end{align*} as $n\to +\infty$ with some real constant $C'$. Equivalently, $$ \log _2 a_n = - n\log _2 n + \frac{n}{{\log 2}} - \frac{{\log _2^2 n}}{2} + C'' + o(1) $$ as $n\to +\infty$ with some real constant $C''$. Up to the value of the constant and the weaker error term, this is the same result as yours.