Beaton–Brown–Cameron conjecture on independence equivalence of odd cycles

From papers

Let n4n\geq 4 be odd, let CnC_n be the cyclic graph on nn vertices, and let DnD_n be the graph shown in Figure~. Two graphs are independence equivalent when they have the same independence polynomial.

Beaton–Brown–Cameron conjecture. If 3∤n3\not\vert n, then a graph GG is independence equivalent to CnC_n if and only if

G{Cn,Dn}.G\in\{C_n,D_n\}.

This conjecture asks for a complete description of the independence equivalence class of CnC_n for odd n4n\geq 4 not divisible by 33. The cited results establish the analogous classification for even n4n\geq 4, except n=6n=6, and for prime powers n=pkn=p^k with p5p\geq 5 prime; the stated odd-cycle case remains open here.

Progress summary

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Sources & referencesView supporting material

Primary source

Boon Leong Ng, “Independence equivalence classes of cycles”, arXiv:2104.10080 (2021).

Solutions 0

No solutions have been posted yet.