Sorting "k-tonic" sequences I hope somebody knows a ref to this, so I do not have to read the literature...

Consider a sequence of numbers $x_1, \ldots, x_n$. Think about the sequence as $n-1$ intervals $[x_1, x_2], [x_2, x_3], \ldots, [x_{n-1},x_n]$. Clearly, the original sequence is bitonic if any point on the real line stabs at most 2 intervals. We will refer to a sequence where a point stabs at most $k$ intervals as being $k$-idiotic. Visually, if you draw the graph of the sequence (i.e., connect the points $p_i =(i,x_i)$ in order), then the above corresponds to the condition that no horizontal line intersects the graph more than $k$ times.

It is not too hard (but not too easy, either) to see that $k$-idiotic sequences can be sorted in $O( n \log k )$ time, which is clearly optimal.

Question: This result should be known. Do you know any appropriate ref?
Here's a Levcopoulos-Petersson sorting algorithm reference, but a different one somewhat older than the one in Andreas' answer:

Levcopoulos, Christos; Petersson, Ola (1989), "Heapsort - Adapted for Presorted Files", WADS '89: Proceedings of the Workshop on Algorithms and Data Structures, Lecture Notes in Computer Science, 382, London, UK: Springer-Verlag, pp. 499–509, doi:10.1007/3-540-51542-9_41.

There's a description of the algorithm in http://en.wikipedia.org/wiki/Cartesian_tree#Application_in_sorting from which the O(n log k) bound is easy to see. More precisely the time for the algorithm is $O(\sum\log k_i)$ where $k_i$ is the number of intervals containing input item $x_i$. In a $k$-idiotic sequence, each $k_i$ is uniformly bounded by $k$ so the total time is just $O(n\log k)$.