login
A258615
The order of the group Aut(Z/nZ)*, or the number of automorphisms of (Z/nZ)*.
6
1, 1, 1, 1, 2, 1, 2, 6, 2, 2, 4, 6, 4, 2, 8, 8, 8, 2, 6, 8, 12, 4, 10, 168, 8, 4, 6, 12, 12, 8, 8, 16, 24, 8, 16, 12, 12, 6, 16, 192, 16, 12, 12, 24, 16, 10, 22, 192, 12, 8, 32, 16, 24, 6, 32, 336, 36, 12, 28, 192, 16, 8, 288, 32, 192, 24, 20, 32, 60, 16, 24, 336, 24, 12, 32, 36, 48, 16, 24, 1536, 18, 16, 40, 336, 256
OFFSET
1,5
COMMENTS
(Z/nZ)* represents the multiplicative group of units mod n and this sequence gives the number of automorphisms of (Z/nZ)*.
A formula for this sequence can be found in the Hillar and Rhea reference.
Or equivalently, a(n) is the order of Aut(Aut(C_n)), where C_n is the cyclic group of order n. - Jianing Song, Apr 06 2019
LINKS
Andrew Howroyd, Table of n, a(n) for n = 1..10000 (terms 1..500 from Dominic Milioto)
C. J. Hillar and D. Rhea, Automorphisms of finite abelian groups
C. J. Hillar and D. Rhea, Automorphisms of finite abelian groups, arXiv:math/0605185 [math.GR], 2006.
Dominic Milioto, The size of an Automorphism Group, Wolfram Demonstrations Project.
FORMULA
See Theorem 4.1 in the Hillar and Rhea link.
a(p^k) = A000010(A000010(p^k)) for p an odd prime and k>0. - Tom Edgar, Jun 05 2015
EXAMPLE
|Aut((Z/1Z)*)|=1. |Aut(Z/28Z)*| = 12.
(Z/5Z)* is isomorphic to Z/4Z, which has two automorphisms, so a(5) = 2. - Tom Edgar, Jun 05 2015
PROG
(PARI)
zp(g)={sum(i=1, #g, my(f=factor(g[i])); sum(j=1, #f~, x^f[j, 1]*y^f[j, 2]))}
aut(p, q)={my(s=0, d=0, m=1); forstep(i=poldegree(q), 1, -1, my(c=polcoeff(q, i)); if(c, s+=i*c*d + (i-1)*c*(d+c); m*=prod(i=1, c, p^i-1); d+=c)); s+=d*(d-1)/2; m*p^s}
a(n)={my(p=zp(znstar(n).cyc)); prod(i=1, poldegree(p), aut(i, polcoeff(p, i)))} \\ Andrew Howroyd, Jun 30 2018
CROSSREFS
Cf. A000010.
Sequence in context: A265870 A341409 A284466 * A152431 A143965 A182073
KEYWORD
nonn
AUTHOR
Dominic Milioto, Jun 05 2015
STATUS
approved