Euler circle and Hamilton cycles Given $K_n$ define sequence $S_4$:  writing all vertices in order of traversing it by Euler circle
How many distinct Hamilton cycles are in this sequence $S$?
 A: It can certainly vary.  As an Eulerean path uses all $\frac{n(n-1)}{2}$ edges and a Hamiltonian path uses $n$, you can't have more than $\frac{n-1}{2}$.  Can you have that many?  There aren't any Eulerean paths if $n$ is even and $>2$.  Depending upon your definition, I can define an Eulearean path that has no segment that is Hamiltonian 
A: The question does not seem to be very precise.
In any case,
It is well known that $K_{2k+1}$ can be decomposed into $k$ hamiltonian cycles.
So we can find an Euler tour with exactly $\dfrac{n-1}{2}$ hamiltonian cycles for odd $n$.
For even $n$, we can do an "incomplete tour" with  $\dfrac{n}{2}-1$ (See link above) hamiltonian cycles.
A: There aren't necessarily any Hamiltonian cycles in your sequence.
As Ross already pointed out, provided $n$ is even and $n > 2$, then all vertices have odd degree and no Eulerian circuit exists.  So, we need only worry about when $n$ is odd and $n > 3$.
For $K_5$ arranged as in the picture, an Eulerian circuit with no Hamiltonian cycles is given by
 
$$1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 2 \rightarrow 5 \rightarrow 4 \rightarrow 1 \rightarrow 5 \rightarrow 3 \rightarrow 1.$$
Now suppose $K_{n-2}$ has an Eulerian circuit with no Hamiltonian cycles.  First label the vertices of $K_n$ as $1, \ldots, n$.  Our Eulerian circuit with no Hamiltonian cycles for $K_n$ is given by first travelling $1 \rightarrow 2$, then travelling the Eulerian circuit with no Hamiltonian cycles for $K_{n-2}$ across the vertices $2, \ldots, n-1$, arriving back at $2$.  Then for $k = 2, 4, 6, \ldots, n - 3$, we travel $$k \rightarrow n \rightarrow k + 1 \rightarrow 1 \rightarrow k + 2.$$  Finally, we travel $n-1 \rightarrow n \rightarrow 1.$
It follows from induction that $K_n$ contains an Eulerian circuit with no Hamiltonian cycles for all odd $n$.
