A364129 Order of Aut^3(C_n) = Aut(Aut(Aut(C_n))), where C_n is the cyclic group of order n.

1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 2, 6, 6, 1, 8, 8, 8, 1, 2, 8, 12, 2, 4, 336, 8, 6, 2, 12, 12, 8, 8, 64, 24, 8, 64, 12, 12, 2, 64, 1152, 192, 12, 12, 24, 64, 4, 10, 1152, 12, 8, 768, 64, 16, 2, 128, 336, 24, 12, 12, 1152, 192, 8, 576, 768, 768, 24, 24, 768, 48, 64, 16, 336, 336, 12, 128, 24, 192, 64, 16, 6144

Offset: 1

Author

Jianing Song, Aug 13 2023

Keywords

• nonn
• hard

Examples

			a(24) = 336 since Aut(C_24) = C_2 X C_2 X C_2, Aut^2(C_24) = PSL(2,7) and Aut(Aut(Aut(C_24))) = PGL(2,7).
a(32) = 64 since Aut(C_32) = C_2 X C_8, Aut^2(C_32) = C_2 X D_8 and Aut^3(C_32) = SmallGroup(64,138).
a(40) = 1152 since Aut(C_40) = C_2 X C_2 X C_4, Aut^2(C_40) = SmallGroup(192,1493) and Aut^3(C_40) = C_2 X SmallGroup(576,8654).
		

Links

• Jianing Song, Table of n, a(n) for n = 1..200
• Jianing Song, Structure and SmallGroupId of Aut^3(C_n) for n <= 100

Crossrefs

Cf. A000010 (order of Aut(C_n)), A258615 (order of Aut^2(C_n)), A364944 (order of Aut^4(C_n)), A364917 (order of Aut^k(C_n) for all sufficiently large k).

Programs

GAP

A364129 := function(n)
local G, i, L;
G := CyclicGroup(n);
for i in [1..3] do
G := AutomorphismGroup(G);
if i = 3 then return Size(G); fi;
L := DirectFactorsOfGroup(G);
if List(L, x->IdGroupsAvailable(Size(x))) = List(L, x->true) then
L := List(L, x->IdGroup(x));
G := DirectProduct(List(L, x->SmallGroup(x))); # It's more efficient to operate on abstract groups when the abstract structure is available
fi; od; end;