I'm curious how one might account for rounding in simple recurrence relations.
$\textbf{Explanation}$
For a specific problem, suppose we have a sequence of positive integers $a_1, a_2, a_3,...$ with where each $a_i$ obeys the following rule $$a_n=a_{n-1}+\text{floor}\Big[\frac{a_{n-4}}{2}\Big]$$ Where "floor" here just means round down. For example if $a_1=5$ and $a_4=7$ then $$a_5=7+\text{floor}\Big[\frac{5}{2}\Big]=9$$
And if we start with $a_1=a_2=a_3=a_4=6$ and continue the sequence we get $$6\rightarrow6\rightarrow6\rightarrow6\rightarrow9\rightarrow12\rightarrow15\rightarrow18\rightarrow22\rightarrow28\rightarrow35\rightarrow44\rightarrow55\rightarrow...$$ What is $a_{100}$? And more importantly, can $a_n$ in general be exactly calculated without calculating all the previous terms in the sequence?
Here's what's been tried so far
$\textbf{Linear Algebra Approximation}$
The update rule for the previous sequence can be approximated with a linear transformation $$ A = \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \frac{1}{2} & 0 & 0 & 0 \\ \end{bmatrix} \quad \quad \text{with} \quad \quad A \cdot \begin{bmatrix} a_n \\ b_n \\ c_n \\ d_n \\ \end{bmatrix} = \begin{bmatrix} a_{n+1} \\ b_{n+1} \\ c_{n+1} \\ d_{n+1} \\ \end{bmatrix} $$ where the $a_n$ in each vector corresponds to the same $a_n$ of the previous sequence example.
This matrix has only one all-positive eigenvector with an eigenvalue $\lambda=1.254...$ which is the growth rate approached by the sequence. One could predict the value of $a_n$ by calculating $$A^n \cdot \begin{bmatrix} a_1 \\ a_1 \\ a_1 \\ a_1 \\ \end{bmatrix}$$
The problem with using a linear transformation is that rounding isn't taken into account. So the predictions one gets inevitably overshoot. And some starting sequences don't even increase when rounding is taken into account. Consider $a_1=a_2=a_3=a_4=1$ $$1\rightarrow1\rightarrow1\rightarrow1\rightarrow1\rightarrow1\rightarrow1\rightarrow1\rightarrow...$$
$\textbf{Error Estimation}$
For initial values of the sequence that highly divisible by $2$, the rounding down won't affect the sequence for a longer time.
More specifically if we start with $$a_1=a_2=a_3=a_4=k2^m \text{ where } k \text{ is odd}$$ Then $a_{4m+5}$ will be the first term affected by the rounding. One could introduce a rounding error function. Something of the form $$f(k, m) = \text{% error approached by the initial value } k2^m$$
It was suggested to use linear dynamical discrete systems to derive an exact formula for $a_n$. How exactly that is done is still unknown. Are there any good example problems similar to this one?
For your specific question about calculating $a_n$ exactly without calculating all the previous terms, the answer is no, in general, unless you are really lucky. In your specific case, if the modulo $2$ behavior of the sequence were periodic then you might have a hope, but it is not.