An analog of perfect numbers involving the unitary totient function (Q2288312)

Let \(\varphi\) denote the Euler totient function. It is well known that \(\varphi(N)\) divides \(N\) if and only if \(N\) is of the form \(2^m 3^n\). Such an integer \(N\) may be considered a totient version of a multiperfect number. The present author investigates a unitary analog of this problem. A divisor \(d\) of \(N\) is a unitary divisor if \((d, N/d)=1\). The unitary analog \(\varphi^{\ast}(N)\) of \(\varphi(N)\) is defined as the number of integers \(k\) modulo \(N\) such that \((k, N)^{\ast}=1\), where \((k, N)^{\ast}\) is the greatest divisor of \(k\) which is a unitary divisor of \(N\). A natural question is when \(\varphi^{\ast}(N)\) divides \(N\). In OEIS, twelve such \(N\) are presented. \textit{K. Ford} et al. [Mosc. J. Comb. Number Theory 2, No. 4, 3--18 (2012; Zbl 1295.11100)] showed that such \(N\) must be divisible by \(3\) and \(N/\varphi^{\ast}(N)\le 85\). The present author proves the following three theorems. (1) Such \(N\) has at least eight distinct prime factors (except for those twelve \(N\) presented in OEIS). (2) Such \(N\) must be divisible by an odd prime factor at least \(10^5\) and an odd prime power at least \(10^8\) (except for those twelve \(N\) presented in OEIS). (3) Only finitely many of them are products of consecutive primes. The first two are proved by elementary means and require a large amount of computation, while the third is proved using sieve methods combined with the analytic theory of the distribution of primes in arithmetic progressions.