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.
 A: By putting
$$
G(n) = F(n + 1)
$$
we can rewrite the recurrence as
$$ \bbox[lightyellow] {  
\sum\limits_{k = 0}^n {G(k)\sin \left( {{{\pi /2} \over {2^{\,n - k} }}} \right)}
  = \sum\limits_{k = 0}^n {G(n - k)\sin \left( {{{\pi /2} \over {2^{\,k} }}} \right)}  = \left[ {0 = n} \right]
 } \tag{1}$$
where $[P]$ denotes the Iverson bracket,
here equivalent to the Kronecker delta.
The recurrence can be unfolded to give
$$
\left\{ \matrix{
  G(0) = 1 \hfill \cr 
  G(n) =  - {1 \over {\sin \left( {\pi /2} \right)}}
 \sum\limits_{k = 1}^n {G(n - k)\sin \left( {{{\pi /2} \over {2^{\,k} }}} \right)}  \hfill \cr}  \right.
$$
Recursion (1) clearly indicates that we are dealing with a convolution, in particular with a multiplicative inversion (aka. reversion) in terms
of power series.
So if we put
$$
H(x,y) = \sum\limits_{0\, \le \,n} {\sin \left( {{x \over {2^{\,n} }}} \right)y^{\,n} } 
$$
then we readily have the ogf for $G(n)$ as
$$
\eqalign{
  & \sum\limits_{0\, \le \,n} {G\left( n \right)y^{\,n} }  = {1 \over {H(\pi /2,y)}}
 = {1 \over {\sum\limits_{0\, \le \,n} {\sin \left( {{\pi  \over {2^{\,n + 1} }}} \right)y^{\,n} } }} =   \cr 
  &  = 1 - {{\sqrt 2 } \over 2}y + \left( {{{1 - \sqrt {2 - \sqrt 2 } } \over 2}} \right)y^{\,2}  +   \cr 
  &  - \left( {\sin \left( {{\pi  \over {16}}} \right)
 + {{\sqrt 2 \left( {1 - 2\sqrt {2 - \sqrt 2 } } \right)} \over 4}} \right)y^{\,3}
  + O\left( {y^{\,4} } \right) \cr} 
$$
To try and find a closed form for $G$ let's try one of the various reversion approaches, which hing
on viewing identity (1) as a system of linear equations with $G(n)$ as unknowns.
$$ \bbox[lightyellow] {  
\left( {\matrix{
   {\sin \left( {x/2^{\,0} } \right)} & 0 & 0 & 0  \cr 
   {\sin \left( {x/2^{\,1} } \right)} & {\sin \left( {x/2^{\,0} } \right)} & 0 & 0  \cr 
   {\sin \left( {x/2^{\,2} } \right)} & {\sin \left( {x/2^{\,1} } \right)} & {\sin \left( {x/2^{\,0} } \right)} & 0  \cr 
   \vdots  &  \ddots  &  \ddots  &  \ddots   \cr 
 } } \right)\left( {\matrix{
   {G(0)}  \cr 
   {G(1)}  \cr 
   {G(2)}  \cr 
    \vdots   \cr 
 } } \right) = \left( {\matrix{
   1  \cr 
   0  \cr 
   0  \cr 
    \vdots   \cr 
 } } \right)
 } \tag{2}$$
where the matrix is a lower triangular Toeplitz matrix.   
However it doesn't look that these methods might lead to a closed expression for the $G(n))$.
A: 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.
