OFFSET
1,2
COMMENTS
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.
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.
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.
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.
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, Note on the map f_k : M_k -> Ab_k
EXAMPLE
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.
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.
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.
PROG
(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.
CROSSREFS
KEYWORD
nonn
AUTHOR
Miles Englezou, Mar 24 2026
STATUS
approved