A258615 The order of the group Aut(Z/nZ)*, or the number of automorphisms of (Z/nZ)*.

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

Author

Dominic Milioto, Jun 05 2015

Keywords

• nonn

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

Examples

			|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
		

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.
• Jianing Song, Structure and SmallGroupId of Aut((Z/nZ)*) for n <= 200
• Wikipedia, Multiplicative group of integers modulo n

Crossrefs

Cf. A000010.

Programs

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

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