# Capacity of Cycle Graphs

Shannon capacity $\Theta(G)$ of pentagon is achieved at $2$-fold strong product of the pentagon.

It is also known that the Lov\'asz theta $\vartheta(G)^m\neq\alpha(G^{\boxtimes m})$ for any finite positive integer $m$ if $G$ is an odd cycle of length $>5$.

Is it known that $\Theta(G)^m\neq\alpha(G^{\boxtimes m})$ for any finite positive integer $m$ if $G$ is an odd cycle of length $>5$? That is $\limsup_{m\rightarrow\infty}\alpha(G^{\boxtimes m})^{\frac{1}{m}}$ is not attained at a finite positive integer $m$.

Note that the the above statement is true would still not be strong enough to decide $\vartheta(G)=\Theta(G)$.

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As far as I know, nothing is known here. I.e., it's possible that the Shannon capacity of an odd cycle is always attained by some finite power, it's possible that it never is beyond the pentagon, and it's possible that it sometimes is and sometimes isn't. – Henry Cohn Oct 4 '13 at 13:42
The last phrase "....it's possible that it sometimes is and sometimes isn't." seems to echo the first phrase "..is always attained by some finite power" since if it is attained at a finite power, then it is attained at all multiples of the finite power. Isn't this the case? – Turbo Oct 4 '13 at 15:35
I see what you mean is for some graphs it is and for some graphs it is not; and not what I interpreted (for some powers it is and for some powers of the same graph it is not). – Turbo Oct 4 '13 at 16:08