

A137315


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


1



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

Table of n, a(n) for n=1..48.
John N. Bray and Robert A. Wilson, On the orders of automorphism groups of finite groups, Bull. London Math. Soc. 37 (2005) 381385.
W. Ledermann, B. H. Neumann, On the order of the automorphism group of a finite group. I, Proc. Roy. Soc. Lon., 233A(1195) (1956), 494506
D. MacHale and R. Sheehy, Finite groups with few automorphisms, Math. Proc. Roy. Irish Acad., 104A(2) (2004), 231238.
Benjamin Sambale, On a theorem of Ledermann and Neumann, arXiv:1909.13220 [math.GR], 2019.


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).
Sequence in context: A024612 A227025 A073881 * A139795 A290649 A118259
Adjacent sequences: A137312 A137313 A137314 * A137316 A137317 A137318


KEYWORD

nonn,hard,more


AUTHOR

Benoit Jubin, Apr 06 2008, May 26 2008


STATUS

approved



