Deriving the formula of number of geometrical isomers in symmetric polyenes For example, if we take the case of Hexa-2.4-diene we have 3 isomers which will have cis-cis, trans-trans and cis-trans arrangements at the two double bonds present. The aim is to get a formula for the number of geometrical isomers of a symmetric polyene with n double bonds, when n is odd and when it is even.
I observed that cis-trans and trans-cis would mean the same thing for two double bonds at same position from the two sides. When I try to think what's going to happen after this it just gets too overwhelming because I think that one would have to consider a lot of cases.
I also read this post but it does not seem to help.
Edit- The answer is-


*

*$2^{n-1}+2^{\frac{n-1}{2}}$ ; when $n$ is odd.

*$2^{n-1}+2^{\frac{n}{2}-1}$ ; when $n$ is even.
 A: Each double bond can be either cis or trans independently. Two sequences of cis/trans values that are related by inversion (as in your example cis-trans and trans-cis) describe the same isomer. Thus we need to find the number of equivalence classes of binary sequences of length $n$ under inversion.
There are $2^n$ binary sequences of length $n$. For $n$ even, there are $2^\frac n2$ palindromic binary sequences of length $n$ (the slots come in pairs, and we can independently choose a value for each pair), whereas for $n$ odd there are $2^\frac{n+1}2$ (the central slot is a singleton, and the remaining $n-1$ slots come in pairs, so we can choose $\frac{n-1}2+1=\frac{n+1}2$ values independently). We need to subtract half the non-palindromic sequences, since they form equivalence classes in pairs. Thus if there are $p$ palindromic sequences, there are $2^n-\frac12\left(2^n-p\right)=2^{n-1}+\frac p2$ geometric isomers, in agreement with the answer you quote.
In the present case, the count could be performed with an elementary argument; in more complicated cases, one would perform such a count using Burnside’s lemma.
