Intuitively speaking, the Kolmogorov complexity of an integer n is the least number of bits one needs to describe n in some suitable language. For instance, one could use the language of Turing machines: an integer n has Kolmogorov complexity at most k if there exists a Turing machine with length at most k which, when run, will halt to produce n as output.
Thus, for instance, any k-bit integer has Kolmogorov complexity at most k+O(1) (the O(1) overhead being for the trivial program that outputs the remaining k bits of the Turing machine). On the other hand, it is obvious that there are at most [math]2^k[/math] integers of Kolmogorov complexity at most k (we'll refer to this as the counting argument). As a consequence, most k-bit integers have Kolmogorov complexity close to k.
On the other hand, a deterministic algorithm which takes k as input and runs in time polynomial in k can only produce integers of Kolmogorov complexity [math]O(\log k)[/math], since any number produced can be described by a Turing machine of length [math]O(\log k) + O(1)[/math] (the O(1) is to describe the algorithm, and the [math]O(\log k)[/math] bits are to describe how long the program is to run and what k is).