

A137315


a(n) is the least number m such that any finite group of order at least m has at least n automorphisms.


3



1, 3, 7, 7, 13, 13, 19, 19, 31, 31, 31, 31, 43, 43, 43, 43, 61, 61, 61, 61, 67, 67, 67, 67, 91, 91, 91, 91, 91, 91, 91, 91, 121, 121, 121, 121, 127, 127, 127, 127, 151, 151, 151, 151, 151, 151, 151, 151
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OFFSET

1,2


COMMENTS

a(n) <= (n1)^(n + (n2)[log_2(n1)]) for n > 4 [Ledermann, Neumann, Thm. 6.6].
a(n) is odd [MacHale, Sheehy, Thm. 15].
a(2n1) = a(2n) for 1 < n < 204 [ibid.].
The case of cyclic groups shows that a(n)>=A139795(n). This inequality can be strict: if M denotes the Mathieu group M_{22} of order 2^7.3^2.5.7.11, then Aut(12.M) = M.2, so that a(2^8.3^2.5.7.11 + 1) > 2^9.3^3.5.7.11, but A139795(2^8.3^2.5.7.11 + 1) = 2.3.5.7^2.11.13.23 + 1 < 2^9.3^3.5.7.11.


LINKS



EXAMPLE

a(3) = a(4) = 7 because every finite group with at least 7 elements has at least 4 automorphisms while the cyclic group of order 6 has only phi(6)=2 automorphisms.


CROSSREFS

Different from A139795 (see Comments).


KEYWORD

nonn,hard,more


AUTHOR



STATUS

approved



