Asymptotic approximation of an integer sequence

We can write $$ - \frac{1}{K(x)} = -\frac{2}{\pi} + \sum_{n=1}^{\infty} \frac{2}{\pi} f_{2n} \, x^n, $$ where $K(x)$ is the complete elliptic integral of the first kind. From the behaviour of $K(x)$ as $x \to 1^-$, we can infer that $$ - \frac{1}{K(x)} = \sum_{k=1}^{\infty} \frac{2^{2k-1} \log^{k-1} 2}{\log^k(1-x)} \left( 1 + \mathcal{O}_k(1-x) \right), $$ as $x \to 1^-$. This implies \begin{align*} \frac{2}{\pi} f_{2n} &= \left[ x^n \right] \sum_{k=1}^{N} \frac{2^{2k-1} \log^{k-1} 2}{\log^k(1-x)} + \mathcal{O}\!\left( \frac{1}{n \log^{N+2} n} \right) + \mathcal{O}\!\left( \frac{1}{n^2 \log^2 n} \right) \\ &= \sum_{k=1}^{N} 2^{2k-1} \log^{k-1} 2 \, \frac{(-1)^n B_{n+k}^{(n+1)}(1)}{(n+k)!} + \mathcal{O}\!\left( \frac{1}{n \log^{N+2} n} \right) + \mathcal{O}\!\left( \frac{1}{n^2 \log^2 n} \right), \end{align*} for any fixed $N \ge 0$ as $n \to +\infty$. Here, $B_n^{(\alpha)}(x)$ denotes the generalised Bernoulli polynomials. By a result of Nörlund, we have \begin{align*} \frac{(-1)^n B_{n+k}^{(n+1)}(1)}{(n+k)!} &= \frac{(-1)^k}{(n+k) \log^k (n+k)} \sum_{m=1}^{M} \binom{m+k-1}{m} \left[ \frac{\mathrm{d}^m}{\mathrm{d}t^m} \frac{1}{\Gamma(t)} \right]_{t=0} \frac{(-1)^m}{\log^m(n+k)} \\ &\quad + \mathcal{O}\!\left( \frac{1}{n \log^{M+k+1} n} \right) \\ &= \frac{(-1)^k}{n \log^k n} \sum_{m=1}^{M} \binom{m+k-1}{m} \left[ \frac{\mathrm{d}^m}{\mathrm{d}t^m} \frac{1}{\Gamma(t)} \right]_{t=0} \frac{(-1)^m}{\log^m n} \\ &\quad + \mathcal{O}\!\left( \frac{1}{n \log^{M+k+1} n} \right) + \mathcal{O}\!\left( \frac{1}{n^2 \log^{k+1} n} \right), \end{align*} for any fixed $k \ge 1$, $M \ge 0$ as $n \to +\infty$. Accordingly, $$\boxed{ f_{2n} = \frac{\pi}{n \log^2 n} \left( \sum_{k=0}^{N-1} \frac{c_k}{\log^k n} + \mathcal{O}\!\left( \frac{1}{\log^N n} \right) \right) + \mathcal{O}\!\left( \frac{1}{n^2 \log^2 n} \right),} $$ for any fixed $N \ge 0$ as $n \to +\infty$, where $$ \boxed{c_k = (-1)^k\sum_{m=0}^{k} 2^{2m} \log^m 2 \, \binom{k+1}{m} \left[ \frac{\mathrm{d}^{k-m+1}}{\mathrm{d}t^{k-m+1}} \frac{1}{\Gamma(t)} \right]_{t=0}.} $$ The first three coefficients are \begin{align*} c_0 &= 1, \\ c_1 &= -2 \gamma - 8 \log 2, \\ c_2 &= 3 \gamma^2 + 24 \gamma \log 2 + 48 \log^2 2 - \frac{\pi^2}{2}, \end{align*} in agreement with those given at $\text{A}054474$.

This may not be a complete answer but perhaps provides some insight.

Noting that

$$U(x)=\sum_{n=0}^{\infty} u_{2 n}\, x^{2 n}=\sum_{n=0}^{\infty} \frac{\binom{2 n}{n}^2}{4^{2 n}}\, x^{2 n}=\frac{2 K\left(x^2\right)}{\pi}\,,\quad |x|<1\tag{1}$$

where $K(x)$ is the complete elliptic integral of the first kind leads to

$$F(x)=1-\frac{1}{U(x)}=1-\frac{\pi}{2\, K\left(x^2\right)}=\sum_{n=0}^{\infty} f_{2 n}\, x^{2 n}\,,\quad |x|<1\tag{2}.$$

The following table illustrates several values of $f_{2 n}$. I believe $16^n f_{2 n}$ is an integer but I was not able to find any related OEIS entries.

$$\begin{array}{ccc} n & f(2 n) & \log _2(\text{Denominator}[f(2 n)]) \\ 0 & 0 & 0 \\ 1 & \frac{1}{4} & 2 \\ 2 & \frac{5}{64} & 6 \\ 3 & \frac{11}{256} & 8 \\ 4 & \frac{469}{16384} & 14 \\ 5 & \frac{1379}{65536} & 16 \\ 6 & \frac{17223}{1048576} & 20 \\ 7 & \frac{56001}{4194304} & 22 \\ 8 & \frac{11998869}{1073741824} & 30 \\ 9 & \frac{41064827}{4294967296} & 32 \\ 10 & \frac{571915951}{68719476736} & 36 \\ 11 & \frac{2018982161}{274877906944} & 38 \\ 12 & \frac{115338112823}{17592186044416} & 44 \\ 13 & \frac{415720532641}{70368744177664} & 46 \\ 14 & \frac{6041874952949}{1125899906842624} & 50 \\ 15 & \frac{22103950817043}{4503599627370496} & 52 \\ 16 & \frac{20825721430968213}{4611686018427387904} & 62 \\ 17 & \frac{77047750289886219}{18446744073709551616} & 64 \\ 18 & \frac{1145470055108455527}{295147905179352825856} & 68 \\ 19 & \frac{4274935497276922857}{1180591620717411303424} & 70 \\ 20 & \frac{256206642255178772127}{75557863725914323419136} & 76 \\ \end{array}$$

Figure (1) below illustrates a discrete plot of $f_{2 n}$ in blue and $\frac{1}{2 \pi n}$ in orange. I don't believe $f_{2 n}\approx\frac{1}{2 \pi n}$ is as good of an approximation as $u_{2 n}\approx\frac{1}{\pi n}$, so it could probably stand some improvement.

Figure (1): Illustration of $f_{2 n}$ in blue and $\frac{1}{2 \pi n}$ in orange

This other answer gives the approximation $f_{2n}\approx C_5 \frac{1}{n \log^2(n)}$ which is consistent with the dominant term in the more complicated approximation

$$a(n)\approx \frac{\pi\, 16^n}{n} \left(\frac{1}{\log^2(n)}-\frac{2 \gamma +8 \log(2)}{\log^3(n)}+\frac{3 \gamma^2-\frac{\pi ^2}{2}+48 \log^2(2)+24 \gamma \log(2)}{\log^4(n)}\right)\tag{3}$$

given by OEIS Entry A054474 which implies $f_{2 n}=\frac{A054474(n)}{16^n}\approx \frac{a_n}{16^n}$ for $n>0$.

Figure (2) below illustrates $f_{2n}\approx C_6 \frac{1}{n \log(n)}$ appears to be a better approximation than $f_{2n}\approx C_5 \frac{1}{n \log^2(n)}$ or $f_{2 n}\approx \frac{a_n}{16^n}$ for small values of $n$ since $f_{2 n} \cdot \left(n \log^2(n)\right)$ illustrated in blue and $f_{2 n} \cdot \frac{16^n}{a_n}$ illustrated in orange both seem to exhibit growth as $n$ increases whereas $f_{2 n} \cdot (n \log(n))$ illustrated in green seems to be relatively constant.

Figure (2): Illustration of $f_{2 n} \cdot \left(n \log^2(n)\right)$ in blue, $f_{2 n} \cdot \frac{16^n}{a_n}$ in orange, and $f_{2 n} \cdot (n \log(n))$ in green

But perhaps Figure (2) above is misleading since if OEIS Entry A054474 is correct $f_{2n}\approx C_5 \frac{1}{n \log^2(n)}$ and $f_{2 n}\approx \frac{a_n}{16^n}$ must become better estimates than $f_{2n}\approx C_6 \frac{1}{n \log(n)}$ as $n\to\infty$.