FindCycle not working as expected for multigraph

Say I have a simple multigraph like so:

edges = {DirectedEdge[a, b], DirectedEdge[b, c], DirectedEdge[c, a]};
g = Graph[Join[edges, edges]]


FindCycle has no trouble discovering a length-3 cycle, however it unexpectedly fails to find the length-6 cycle that takes "two laps" around the vertices.

FindCycle[g, {3}]
FindCycle[g, {6}]

(* output:
{{a \[DirectedEdge] b, b \[DirectedEdge] c, c \[DirectedEdge] a}}
{}
*)


I initially just assumed Mathematica did not have the capability to fully support multigraphs like this, but FindEulerianCycle happily identifies the length-6 cycle as expected:

FindEulerianCycle[g]

(* output:
{{a \[DirectedEdge] b, b \[DirectedEdge] c, c \[DirectedEdge] a,
a \[DirectedEdge] b, b \[DirectedEdge] c, c \[DirectedEdge] a}}
*)


It seems bizarre that FindEulerianCycle[g] can find a length-n cycle yet FindCycle[g, {n}] returns nothing.

So my questions are: Is there a good way (using FindCycle or otherwise) to properly handle "multi-cycles" (which are not Eulerian in general)? Is the observed behavior a bug or is there something convincing in the documentation that indicates it's by design?

-
I presume Mathematica searches for "simple cycles," i.e., no repetitions of vertexes, even though the EulerianCycle is somehow found. An Eulerian cycle in a directed graph can of course pass through the same vertex more than once. – David G. Stork Jan 3 '15 at 1:03
Note too: FindPostmanTour[g] gives the full Eulerian cycle for your case. – David G. Stork Jan 3 '15 at 1:09
in documentation, FindCycle returns simple cycles, while FindHamiltonianCycle, FindEulerianCycle, and FindFundamentalCycles return specific types of cycles. – halmir Jan 4 '15 at 16:18
Ah, indeed, there it is buried at the end of the "Background" section. Nice find, @halmir. If you want to submit that as an answer I'll accept it. – latkin Jan 4 '15 at 22:17