OFFSET
1,9
COMMENTS
Every group of units is abelian.
If a(k) = 0, then k has a prime factorization k = s*2^r, where s is odd squarefree and r >= 1. See the Englezou link for a proof.
It also follows from the proof that A000688(k) = A380578(k) = A000041(r) for those k = s*2^r, since the prime signature of k is (r, 1, ..., 1); the equation follows by the partition product formula for the number of abelian groups of a given order. Thus a(k) = 0 only for group orders whose number of abelian groups is a partition number.
LINKS
Miles Englezou, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems (invphi.gp).
Miles Englezou, Proof
EXAMPLE
a(2) = 0 since C_2 is the only group of order 2 and C_2 appears as the group of units of the rings Z/3Z, Z/4Z, and Z/6Z, which are all the rings Z/kZ such that phi(k) = 2.
a(3) = 1 since C_3 is the only group of order 3 and C_3 does not appear as the group of units for any ring Z/kZ because phi(k) = 3 has no solution.
a(8) = 1 since of the 3 abelian groups of order 8, only C_8 does not appear as the group of units of the ring Z/kZ for any k such that phi(k) = 8.
a(16) = 2 since of the 5 abelian groups of order 16, only C_2 x C_2 x C_2 x C_2 and C_4 x C_4 do not appear as the group of units of the ring Z/kZ for any k such that phi(k) = 16.
PROG
(PARI) a(n) = my(f = factor(n)[, 2]); prod(i = 1, #f, numbpart(f[i])) - #Set(apply(x -> znstar(x)[2], invphi(n))) \\ [using Max Alekseyev's invphi.gp; see link]
CROSSREFS
KEYWORD
nonn
AUTHOR
Miles Englezou, Mar 19 2026
STATUS
approved