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¡¡¡¡Now the problem naturally arises when ϕ∗(N) divides N. We can easily find some instances (see also https://oeis.org/A319481):
N1 = 1, N2 = 2, N3 = 2 · 3, N4 = 2¡°2· 3,
N5 = 2¡°3· 3 · 7,
N6 = 2¡°4· 3 · 5,
N7 = 2¡°5· 3 · 5 · 31,


N8 = 2¡°8· 3 · 5 · 17,
N9 = 2¡°11· 3 · 5 · 112· 23 · 89, N10 = 2¡°16· 3 · 5 · 17 · 257,
N11 = 2¡°17· 3 · 5 · 17 · 257 · 131071, N12 = 2¡°32· 3 · 5 · 17 · 257 · 65537.

All of these examples satisfy N = 2ϕ∗(N) except N = 1 = ϕ∗(1) and N = 6 = 3ϕ∗(6).
Such integers were implicitly referred in [6] as possible orders of groups with perfect order subsets, groups G with the number of elements in each order subset dividing |G|. Integers equal to twice of its unitary totient had been introduced in the OEIS in 1999 by Yasutoshi Kohmoto
https://oeis.org/A030163.

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