I hope this question is appropriate for MO:

Let $n$ be a natural number, $U_n := \{ d | d \text{ divides } n, \gcd(d,n/d)=1\}$ be the set of unitary divisors.

We can make $U_n$ to a boolean ring:

$$a \oplus b := \frac{ab}{\gcd(a,b)^2} = \frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}$$ and $$a \otimes b := \gcd(a,b)$$

Let $\Pi(n) := \{ p | p \text{ is prime}, p| n\}$ be the set of prime divisors of $n$. We can define a topology on this set where the open sets are

$$ \{ \Pi(d) | d \text{ divides } n \}$$

then $$\Pi(\operatorname{rad}(ab)) = \Pi(a) \cup \Pi(b)$$ and $$\Pi(\gcd(a,b)) = \Pi(a) \cap \Pi(b)$$

where $\operatorname{rad}(x) = \prod_{p|x}p$ is the radical of $x$.

To each open set $U$ we define a number

$$\operatorname{rad}(U):= \prod_{p \in U}p$$

The open sets build also a boolean ring with:

$$U \oplus V := U \Delta V$$ where $\Delta$ denotes the symmetric difference, and $$U \otimes V := U \cap V$$

Then $\operatorname{rad}$ is a isomorphism of boolean rings:

$$\operatorname{rad}(U \oplus V) = \operatorname{rad}(U) \oplus \operatorname{rad}(V)$$ $$\operatorname{rad}(U \otimes V) = \operatorname{rad}(U) \otimes \operatorname{rad}(V)$$ Also $\operatorname{rad}(\emptyset) = 1$, where $1$ is the zero in $U_{\operatorname{rad}(n)}$ and $\operatorname{rad}(\Pi(n)) = \operatorname{rad}(n)$, where $\operatorname{rad}(n)$ is the one in $U_{\operatorname{rad}(n)}$.

Furthermore, since $k(a,b) = \frac{\gcd(a,b)^2}{ab} = \frac{1}{a\oplus b}$ is a positive definite function on the natural numbers and a simililarity, we can embedd this boolean ring $U_n$ isometrically in Euclidean space $\mathbb{R}^{2^{\omega(n)}}$ (on the sphere of radius one with center $0$) where $\omega(n)$ counts the distinct prime divisors of $n$ and we can define a distance between two unitary divisors:

$$ d(a,b) = \sqrt{k(a,a)+k(b,b)-2k(a,b)} = \sqrt{2(1-\frac{1}{a\oplus b})}$$

Also for all $a,b,c \in U_n$ we have:

$$k(c\oplus a , c \oplus b ) = k(a,b)$$

My (soft) question is this:

Is this of any use for anything, maybe in number theory? :) Thanks for your help.


1 Answer 1


This is too long for a comment, so I am writing an answer after two years. :-) Finally something useful.

Idea: Integer composition <-> Subsets of a finite set <-> Bolean ring of unitary divisors

Integer compositions: https://en.wikipedia.org/wiki/Composition_(combinatorics)

Sagemath: https://doc.sagemath.org/html/en/reference/combinat/sage/combinat/composition.html

The integer compositions could be useful in algorithmic composition/music for manipulation of durations in a bar:

(For instance: One could measure with nearest neighbors how far / near two bars are from the durations (/integer compositions) perspective using the kernel above.)

Algorithmic composition:

To every bar associate to the durations of the bar the composition of an integer.

For example:

 Durations of bar:                1/4,1/4,1/8,1/8,1/4 

 composition of the integer n=8:   2,   2,  1 , 1 , 2 ( 2+2+1+1+2  = 8)                

 subset of {1,2,..,n-1} :         {2,4,5,6}

 unitary divisors of P_{n-1} = 2*3*...*p_{n-1}:     3*7*11*13 = 3003

In case one needs to compute a direct embedding (without requiring to compute the expensive Cholesky decomposition) of these feature vectors, here is an example how to do it:

Let $e_d$ be the $d$-th standard-basis vector in the Hilbert space $H=l_2(\mathbb{N})$. Let $h(n) = J_2(n)$ be the second Jordan totient function. Define:

$$\phi(n) = \frac{1}{n} \sum_{d|n}\sqrt{h(d)} e_d$$.

Then we have:

$$ \left < \phi(a),\phi(b) \right > = \frac{\gcd(a,b)^2}{ab}=:k(a,b)$$

The vectors $\phi(a_i)$ are linearly independent for each finite set $a_1,\cdots,a_n$ of natural numbers, since

$$\det(G_n) = \prod_{i=1}^n \frac{h(a_i)}{a_i^2} $$ is not zero, where $G_n$ denotes the Gram matrix.

Here is some sagemath code which does the translation between unitary divisors of the primorial numbers and integer compositions:

Sage Cell Server with code


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