Closed form of the recursive function $F(1):=1,\;F(n):=\sum_{k=1}^{n-1}-F(k)\sin\left(\pi/2^{n-k+1}\right)$ Suppose that $F$ is defined via the recurrence relation
$$F(1)=1, \qquad F(n)=\sum_{k=1}^{n-1}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr)$$ What is $F(N)$? I don't have any idea how to solve this problem. Only one thing that I've noticed is that: $$ 0=\sum_{k=1}^{n}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr).$$

Edit:
From the comment section:
I was trying to rewrite $\sin\left(\frac{\pi}{2^n}\right)$ by the formula for a double argument and I've ended up with $$\sin\left(\frac{\pi}{2^n}\right)=\frac{\sin\left(\frac{\pi}{2}\right)}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)}=\frac{1}{2^{n-1}\prod_{k=2}^n\cos\left(\frac{\pi}{2^k}\right)},n\gt 1,$$
but it doesn't seem to help.
$F$ is used in another formula. It should be true for most of the functions
$$g(x)=\sum_{n=1}^\infty \left(\sin\left(x2^{n-1}\right)\sum_{k=1}^n\left(F(k)g\left(\frac{\pi}{2^{n-k+1}}\right)\right)\right),\;x\in\left(0,\frac{\pi}{2}\right)$$
First four values of $F$ are:
\begin{align*}F(1)&=1\\F(2)&=-\sin\left(\frac{\pi}{4}\right)\\F(3)&=-\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{4}\right)\\F(4)&=-\sin\left(\frac{\pi}{16}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)-\sin^3\left(\frac{\pi}{4}\right)\end{align*}
These terms look like if they created some pattern, but the fifth term which is
\begin{align*}F(5)=-\sin\left(\frac{\pi}{32}\right)+2\sin\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{16}\right)-3\sin^2\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{8}\right)+\sin^2\left(\frac{\pi}{8}\right)+\sin^4\left(\frac{\pi}{4}\right)\end{align*}
meses the pattern up.

The $F(n+1)$ is the coefficient of $x^n$ in the power series expansion of
$$
\dfrac{1}{{\displaystyle\sum\limits_{n = 0}^\infty  {\sin \left( {\dfrac{\pi }{{2^{n + 1} }}} \right)x^n } }}.
$$
Consequently,
$$
F(n + 1) = \sum\limits_{\substack{k_1  + 2k_2  +  \cdots  + nk_n  = n \\ 
  k_1 ,k_2 , \ldots ,k_n  \in \mathbb{Z}_{ \ge 0} }} { \frac{{(k_1  + k_2  +  \cdots  + k_n )!}}{{k_1 !k_2 ! \cdots k_n !}}\prod\limits_{j = 1}^n {( - 1)^{k_j } \sin ^{k_j } \left( {\frac{\pi }{{2^{j + 1} }}} \right)}} ,
$$
for $n\geq 0$. I do not think there is an explicit formula simpler than this.
