OFFSET
1,2
COMMENTS
Let G_n be an n-vertex simple graph, with a(G_n) automorphisms. Then l(G_n) = n!/a(G_n) is the number of labeled copies of G_n. So a(n) is the number of G_n for which n does not divide l(G_n).
For prime p, a(p) is the number of circulants of order p.
The number of circulants of order n is A049287(n).
REFERENCES
John P. McSorley, Smallest labelled class (and largest automorphism group) of a tree T_{s,t} and good labellings of a graph, preprint, (2016).
R. C. Read, R. J. Wilson, An Atlas of Graphs, Oxford Science Publications, Oxford University Press, (1998).
James Turner, Point-symmetric graphs with a prime number of points, Journal of Combinatorial Theory, vol. 3 (1967), 136-145.
EXAMPLE
If n=3 then both G_3 = K_3 and its complement have a(G_3) = 6, so l(G_3) = 3!/6 = 1, and so 3 does not divide l(G_3); no other graphs G_3 satisfy this, so a(3)=2.
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
John P. McSorley, Dec 29 2015
STATUS
approved