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I'm interested in this series: $$\mathcal S_p=\sum_{n=1}^\infty\frac{\left(F_n\right)^p}{2^{np}},\quad p\in\mathbb N,\tag1$$ where $F_n$ are the Fibonacci numbers, defined by the recurrence $$F_1=1,\quad F_2=1,\quad F_n=F_{n-1}+F_{n-2}\tag2$$ or by the explicit formula $$F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5},\quad\phi=\frac{1+\sqrt5}2.\tag3$$ We can find that $$\mathcal S_1=2,\quad\mathcal S_2=\frac{12}{25},\quad\mathcal S_3=\frac{376}{2201},\quad\mathcal S_4=\frac{16048}{221125},\quad\mathcal S_5=\frac{25697312}{765370111}.\tag4$$

  • Can we prove that $\mathcal S_p$ is always a rational?
  • Can we find a general formula, recurrence or generating function for $\mathcal S_p$?

Update: I was able to get the following formula involving finite summation: $$\mathcal S_p=\sum_{m=0}^p\binom{p}{m}\left(\vphantom{\Large|}\left(2\sqrt5\right)^p\,\phi^{p-2 m}+(-1)^{p-m+1}\right)^{-1}.\tag5$$

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An easy answer for the first half: It's known that for any $p$, the sequence $\mathcal{F}_p = \{(F_n)^p\}$ satisfies a linear recurrence relation with rational coefficients. (See, for instance, http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/Fibonomials.html ) This implies that its generating function (i.e., $\mathcal{F}_p(x)=\sum_n (F_n)^px^n$) is a rational function of $x$ (also with rational coefficients). Your sum is then just the value $\mathcal{F}_p(2^{-p})$ and so it's likewise rational. You should even be able to adapt the methods in the link to give you an explicit form for the generating function and thus for your $\mathcal{S}_p$.

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