OFFSET
1,3
COMMENTS
Let G be a group of order n, let N = {1, 2, ..., n}, and let f: G -> N be a bijection whereby f(G) = I is an index set of G. An automorphism phi of G is a permutation of N via the map f*phi*f^-1: N -> N. It is tempting to ask the question 'how many permutations of N are group automorphisms?'. However this question is not well-defined; to be answerable there would need to be a natural single choice of bijection for every group of order n, which in general does not exist. Any enumeration of permutations of N which are automorphisms for every isomorphism class G will therefore depend on the choice of bijection for every G. a(n) is an upper bound for any such enumeration of permutations of size n since a(n) is either: the maximum enumeration when the choice of bijections ensures that all permutations are distinct; or a(n) is the enumeration including all multiplicities when the choice of bijections leads to permutations which are not distinct.
FORMULA
a(n) is the sum of the k-th row in A137316.
EXAMPLE
a(3) = 2 since for the one group of order 3, C_3, |Aut(C_3)| = 2.
a(4) = 8 since for the two groups of order 4, C_4 and C_2 x C_2, |Aut(C_4)| + |Aut(C_2 x C_2)| = 2 + 6 = 8.
a(8) = 212 since for the five groups of order 8, the sum of the orders of their automorphism groups is 4 + 8 + 8 + 24 + 168 = 212.
PROG
(GAP)
a := function(n)
local T, k;
T := [];
for k in [1..NrSmallGroups(n)] do
T := Concatenation(T, [Order(AutomorphismGroup(SmallGroup(n, k)))]);
od;
return Sum(T);
end;
CROSSREFS
KEYWORD
nonn
AUTHOR
Miles Englezou, Jun 30 2025
EXTENSIONS
Name and comment clarified by Miles Englezou, Jan 08 2026
STATUS
approved
