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A258615
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The order of the group Aut(Z/nZ)*, or the number of automorphisms of (Z/nZ)*.
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6
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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
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OFFSET
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1,5
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COMMENTS
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(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
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LINKS
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FORMULA
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See Theorem 4.1 in the Hillar and Rhea link.
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EXAMPLE
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|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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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