Closed form of the recursive function $F(1):=1,\;F(n):=\sum_{k=1}^{n-1}-F(k)\sin\left(\pi/2^{n-k+1}\right)$

Question

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: $$ 0=\sum_{k=1}^{n}-F(k)\sin\biggl(\frac{\pi}{2^{n-k+1}}\biggr).$$

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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.

Answer 1

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.

Answer 2

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 $G(n))$.