Unitary divisor

The number of unitary divisors of a number n is 2^k, where k is the number of distinct prime factors of n.

This is because each integer N > 1 is the product of positive powers p^r_p of distinct prime numbers p. Thus every unitary divisor of N is the product, over a given subset S of the prime divisors {p} of N, of the prime powers p^r_p for p ∈ S. If there are k prime divisors, then there are exactly 2^k subsets S, and the statement follows.

The sum of the unitary divisors of n is odd if n is a power of 2 (including 1), and even otherwise.

Both the count and the sum of the unitary divisors of n are multiplicative functions of n that are not completely multiplicative. The Dirichlet generating function is

${\displaystyle {\frac {\zeta (s)\zeta (s-k)}{\zeta (2s-k)}}=\sum _{n\geq 1}{\frac {\sigma _{k}^{*}(n)}{n^{s}}}.}$

Every divisor of n is unitary if and only if n is square-free.

The sum of the k-th powers of the odd unitary divisors is

${\displaystyle \sigma _{k}^{(o)*}(n)=\sum _{{d\mid n \atop d\equiv 1{\pmod {2}}} \atop \gcd(d,n/d)=1}\!\!d^{k}.}$

It is also multiplicative, with Dirichlet generating function

${\displaystyle {\frac {\zeta (s)\zeta (s-k)(1-2^{k-s})}{\zeta (2s-k)(1-2^{k-2s})}}=\sum _{n\geq 1}{\frac {\sigma _{k}^{(o)*}(n)}{n^{s}}}.}$

A divisor d of n is a bi-unitary divisor if the greatest common unitary divisor of d and n/d is 1. The number of bi-unitary divisors of n is a multiplicative function of n with average order ${\displaystyle A\log x}$ where[1]

${\displaystyle A=\prod _{p}\left({1-{\frac {p-1}{p^{2}(p+1)}}}\right)\ .}$

A bi-unitary perfect number is one equal to the sum of its bi-unitary aliquot divisors. The only such numbers are 6, 60 and 90.[2]

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