The compound shown is octa-2,4,6-triene drawn without stereochemistry. How many geometrical isomers does it have? Give an integer.

All three double bonds can be $E$ or $Z$, giving $2^3=8$ labelled combinations, but the molecule is symmetric end to end, so the sets $(2E,4Z,6Z)$ and $(2Z,4Z,6E)$ are the same compound, as are $(2E,4E,6Z)$ and $(2Z,4E,6E)$. The distinct isomers are $(E,E,E)$, $(Z,Z,Z)$, $(E,Z,E)$, $(Z,E,Z)$, $(E,E,Z)$ and $(E,Z,Z)$: 6 in total. Formula check for a symmetric chain with $n=3$ double bonds (odd): $2^{n-1}+2^{(n-1)/2}=4+2=6$.