Is it possible to find an expression of the form
$$\lfloor 2^n \sqrt{2}\rfloor = \sum_{k=1}^r (\alpha_k + \beta_ki)\cdot\Big(a_k+b_k i\Big)^n, $$
where $i^2 = -1$ and $\alpha_k, \beta_k, a_k, b_k$ real numbers? Maybe with $r< 5$ if possible. I am trying to compute the mean value (average) of $\{ 2^n \sqrt{2}\}= 2^n \sqrt{2} – \lfloor 2^n \sqrt{2}\rfloor$ over $n=0, 1,2, \cdots$. Let us denote as $p$ this mean value: $p$ is the proportion of binary digits of $\sqrt{2}$ equal to one, and $p$ is widely believed to be equal to $\frac{1}{2}$.
Note that
$$p=\lim_{n\rightarrow\infty} p_n, \mbox{ with } p_n = \frac{1}{n}\sum_{k=1}^n \Big( 2^k \sqrt{2} – \lfloor 2^k \sqrt{2}\rfloor\Big).$$
Assuming we have a simple closed form for $\lfloor 2^k \sqrt{2}\rfloor$ as suggested in the first formula, then we can have a closed form for $p_n$, involving a finite number of terms. I don't expect this to be true if you replace $\sqrt{2}$ by (say) $\pi$ or $\log 2$, but I would expect this to work with any quadratic irrational.
Of course you can always use the Fourrier series to represent the fractional part function, but I don't think it would be easy to handle: it involves an infinite number of terms ($r=\infty$.)
Best Answer
No it's not possible.
A sequence of complex numbers $A_n$ having a closed form of the type $ A_n = \sum_{i=1}^k a_i b_i^n$ with the $a_i$'s and $b_i$'s complex numbers implies it satisfies a recursion $A_n = c_1A_{n-1}+\dots +c_kA_{n-k}$ for some fixed complex numbers $c_1, c_2, \dots, c_k$.
Moreover, if that sequence $A_n$ is actually a sequence of integers then the constants $c_1, c_2, \dots, c_k$ in the recursion are actually rational numbers. Clearing the denominators gives a recursion $c_0'A_n = c_1'A_{n-1}+\dots +c_k'A_{n-k}$ where now the $c_i'$s are integers.
Now note that the parity of $A_n$ only depends on the parity of $A_{n-1}, A_{n-2},\dots$ and $A_{n-k}$. In particular, this sequence must become periodic mod 2 as once the same length $k$ string of parities occurs everything after it repeats as well, and there are finitely many $(2^k)$ length $k$ strings.
This is a problem though, as the parity of $\lfloor 2^n \sqrt{2} \rfloor$ determines the $n$th binary digit of $\sqrt{2}$. This sequence being periodic implies that $\sqrt{2}$ is rational, which we know is false.
This same argument shows that $\lfloor a^n b \rfloor$ with $a \in \mathbb{N}$ can be written in this form iff $b$ is rational.