A394532 - OEIS

%I #31 Apr 02 2026 20:59:07

%S 1,2,4,6,10,12,20,22,30,40,42,46,58,60,66,70,78,82,88,102,106,110,130,

%T 138,156,166,178,190,210,222,226,232,238,240,262,276,280,282,310,312,

%U 330,346,348,358,366,382,408,418,420,430,438,442,460,462,466,478,498,502,506,520,546

%N Numbers k such that every abelian group of order k appears as a group of units of the ring Z/mZ for at least one m.

%C Equivalently, numbers k such that A394390(k) = 0. By the proof in A394390, for k > 4, every k = s*2^r, where s is odd squarefree and r >= 1.

%C Let Ab_k be the set of the nonisomorphic abelian groups of order k, and let M_k be the set of those m such that phi(m) = k. Let f_k : M_k -> Ab_k be the map such that f_k(m) = G is the group of units of Z/mZ. Then k is a term if f_k surjective.

%C It's also true that f_k is noninjective for every k in this sequence. This is because phi(m) = k has a solution m = p^e, otherwise the cyclic group C_k would not appear as a group of units. Consequently 2*p^e is also a solution and the cardinality of the preimage f^-1(C_k) is at least 2. For further information about the preimage cardinalities, see the Englezou link.

%C For general totient numbers t, the map f_t is injective for t = 2^s, s >= 33, assuming five Fermat primes. See the Englezou link for a proof. This raises the interesting question whether f_r is injective for r <> 2^n. A negative claim would constitute a generalization of Carmichael's conjecture.

%H Miles Englezou, <a href="/A394532/b394532.txt">Table of n, a(n) for n = 1..10000</a>

%H Max Alekseyev, <a href="https://oeis.org/wiki/User:Max_Alekseyev/gpscripts">PARI/GP Scripts for Miscellaneous Math Problems</a> (invphi.gp).

%H Miles Englezou, <a href="/A394532/a394532_1.txt">Note on the map f_k : M_k -> Ab_k</a>

%e 2 is a term since C_2 is the only abelian group of order 2 and C_2 is the group of units of Z/mZ for m = 3, 4, 6. These are all the m such that phi(m) = 2.

%e 12 is a term since there are two abelian groups of order 12, and both groups appear at least once as a group of units of Z/mZ for all m such that phi(m) = 12.

%e 40 is a term since there are three abelian groups of order 40, and every group appears at least once as a group of units of Z/mZ for all m such that phi(m) = 40.

%o (PARI) isok(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.

%Y Cf. A000688, A002202, A390394.

%K nonn

%O 1,2

%A _Miles Englezou_, Mar 24 2026