Are there incompressible strings $S$ such that $K(SS) \lt K(S)$? This question was inspired by Sune Jakobsen's question and Tsuyoshi's answer. In Kolmogorov-complexity, a string is incompressible if $K(S) \ge |S|$.

Is there incompressible string $S$ such that $K(SS) \lt K(S)$? Is there an infinite set of such strings? What is the theoretical consequence if such infinite set of strings does exist?
A rough argument that there should be infinitely many such strings is as follows.

Consider a machine $M$ that first left-pads its input with zeroes out to some known length $2f(|T|)$ (we will specify $f$ below), then repeatedly performs some length-preserving and random-looking operation on it, halting only when it has produced (essentially by accident) a string consisting of two copies of the same string.  It is not required to halt for every input.  When it does, the output it produces from input $T$ will have complexity satisfying $K(SS) \approx K(T) + K_M$ for some $K_M$, as long as $f(|T|)$ is large enough, and $S$ will have length $f(|T|)$.  The string $S$, moreover, will have complexity $K(S) \approx K(SS) + K_C$ for some $K_C$.  The key final step is to choose the right $f(|T|)$.  It must be large enough to ensure that $SS$ is best described as "the output of $M$ applied to $T$", but small enough to ensure that the description of $S$ as "half the output of $M$ applied to $T$" is not less than $|S| = f(|T|)$.  Taking $f(|T|)=|T|+K_M+K_C-1$ should work.  We conclude that arbitrarily long incompressible inputs $T$, if they halt, will typically produce outputs $SS$ such that $K(SS) &lt; K(S)$ and such that $S$ is incompressible.  The machine $M$ could do any number of things; for instance, the machine $M$ that squares its input and then discards every other bit of the result would probably work.