Is the Kolmogorov complexity of a number always its logarithm?

Question

if I have a natural number $a(n,m)$ that depends on some $n$ and $m$, where $m$ is fixed,

isn't then the Kolmogorov complexity of it simply its logarithm?

Answer 1

The Kolmogorov complexity of a number is the length of the shortest program that prints out that number. If your number has $n$ digits, then there is a program of length approximately $n$ which prints out that number by printing all of its digits, but sometimes there are much shorter programs.

For example, a for loop going from $1$ to $k$ can print out the number $2^k$, so the Kolmogorov complexity of $n = 2^k$ is at most approximately $\log k$, or $\log \log n$. (I'm ignoring constants throughout this discussion.)