Let $U$ be a universal Turing Machine. Suppose I have a Kolmogorov incompressible string $s$ of length $n$. Let $A:\{1,...,n\} \to \{0,1\}$ be an algorithm such that $A(i) = s_i$.
I believe that the time it takes for the universal Turing machine to evaluate $A$ on input $i$ should be $\Theta(n)$. I'm not familiar with Kolmogorov complexity, and I wanted to ask if the following intuition is correct
An upper bound on the running time of $A$ is $O(n)$, since the description of $A$ can just list the whole string $s$ and return $s_i$ on input $i$. The universal Turing machine on inputs $(A,i)$ simply reads the description of $A$ (Which is $O(n)$ in length) and outputs the $i^{th}$ coordinate (which takes time $\log n$)
A lower bound on the running time of $A$ can be proved as follows.
a. The description length of $A$ must be at least $n - log n - c$ for a constant $c > 0$. This is because the program $B$: "Run $A$ on every input from $1$ to $n$" has description length $DL(B) = DL(A) + log(n) + c$ and generates the string $s$. Thus $DL(B) \geq n$ and $DL(A) \geq n- \log(n) - c$.
b. The universal Turing machine $U$ on input $(A,i)$ needs to read both $A$ and $i$, which requires at least $DL(A) + \log n \geq n-c$ operations.
Is this intuition correct? Or am I missing something big about Kolmogorov complexity?