Consider the completely additive function $\eta(n) := \sum_{p\mid n} v_p(n)p$ defined on natural numbers, with values in natural numbers. For literature, on this function, see [the corresponding OEIS sequence][1]. Consider the Gram matrix:
$$G(n) := (k(a,b))_{1 \le a,b \le n}$$
where
$$k(a,b) = \eta(\gcd(a,b))$$
is a positive semi definite kernel on the natural numbers, because with $\phi(n) := \sum_{p\mid n} \sqrt{v_p(n)p} e_p$ we can embedd $n$ in the Hilbert space of series over $\mathbb{R}$ or $\mathbb{C}$ and we have $k(a,b) = \langle \phi(a),\phi(b) \rangle$
The game for the natural numbers is like this:
Given two natural numbers $a,b$, what is the minimal cost of a series transformation from $a$ to $b$. A transformation is multiplication / division with a prime number $p$ and its cost is $p$. For a series of transformations, the cost is the sum of transformations in the series. (Spoiler: The answer is independent from the chosen series and is
>!$\eta(\frac{ab}{\gcd(a,b)^2})$
which can be seen by introducing a graph $G = (\mathbb{N},E)$ on the natural numbers, where the edges $a \approx b$ are defined through
$$(a,b) \in E \iff \frac{a}{b} \text{ or } \frac{b}{a} \text{ is prime}$$
with weights on the edges equal to $w((a,b)) := \max\{a/b,b/a\} = $ a prime number.
A path from $a$ to $b$ corresponds to a series of transformations in the game above, and a shortest path, corresponds to a series of transformations of minimal cost.
The connection to the graph and the Gram matrix is given by the shortest path distance:
>! $$d_s(a,b) = \eta( \frac{ab}{\gcd(a,b)^2} ) = \eta(a)+\eta(b)-2\eta(\gcd(a,b)) = |\phi(a) - \phi(b)|_H^2 = d_H(a,b)^2$$
where the subscript $H$ denotes the Hilbert space of series.
It is fun and somewhat surprising to look at the spectrum $\sigma(G(n))$ of these Gram matrices:
[![Here we see the spectrum of $G(169)$][2]][2]. One can recognize three parts:
$$\sigma(G(n)) = \sigma_p \cup \sigma_c \cup \sigma_0$$
where $\sigma_p := \{ \lambda \in \sigma(G(n)) \mid \lambda \in \mathbb{N} , n/2 < \lambda \le n \}$ and
$\sigma_c := \{ \lambda \in \sigma(G(n)) \mid \lambda \notin \mathbb{N} , n/2 < \lambda \le n \}$ and $\sigma_0 := \{ \lambda \in \sigma(G(n)) \mid 0 \le \lambda \le n/2 \}$
Furthermore, looking at the numbers of $\sigma_p$ we recognize that these are exactly the prime numbers $p$ with $n/2 < p \le n$. That these prime numbers are a subset of $\sigma_p$ can be proven by observing that the $p$-th row and column in the Gram matrix consists of $0$s everywhere except at the diagonal where it has entry $p$. So by permuting this $p$-th column and then $p$-th row. We see that the matrix becomes a block-diagonal and can see that $n/2 < p \le n$ are eigenvalues. The off diagonal zeros can be explained because $k(a,p) = \eta(\gcd(a,p)) = \eta(1) = 0$.
Using the embedding $\phi(n)$ we can define a linear operator $R(e_n) := \phi(e_n)$ on the standard basis $(e_n)_{n \in \mathbb{N}}$ of $H$ and with this, we can define an unbounded, self-adjoint, linear and positive semidefinite operator on $H$:
$$O = R^* R $$
Searching for theorems and usages of unbounded, self-adjoint operators , I found Theorem 5.7 on Schmüdgen's "Unbounded self-adjoint operators on Hilber spaces":
Let $A$ be a self-adjoint operator on a Hilbert space $H$. Then there exists a unique spectral measure $E = E_A$ on the Borel Sigma-algebra $B(\mathbb{R})$, such that:
$$A = \int_{\mathbb{R}} \lambda \, d E_A(\lambda)$$
The idea is to transform $t$ to $t(1+t\cdot t)^{-1/2}$ to get from unboundend to bounded and backwards. However I am not sure, **if this theorem can be applied for $A=O$ as above, and if it can applied, if it is possible to compute the integral above directly? (Q1)**
In the German Wikipedia article of "spectra measure" there is an interesting observation from Quantum Mechanics of observables, which can be roughly divided in three parts:
1) the point spectrum ( one uses sums) [This would correspond I think to the prime numbers defined above]
2) the continuous spectrum (one uses integrals) [This would correspond to the $\sigma_c$ I think, $c$ for continuous.
3) very seldom there is a singular-continuous spetral part, which I think would correspond to $\sigma_0$.
**I am curious, if any of these observations can be proven? (Q2)**
The properties defining the kernel $k(a,b)=\eta(\gcd(a,b))$ are:
Suppose $k$ is a p.s.d kernel function (symmetric) on the natural numbers, such that:
1) $\gcd(a,b) = 1 \iff k(a,b) = 0$
2) $k(p^a,p^a) = ap$ for all primes $p$ and $a\ge 0$
3) $k\left(\frac{ac}{\gcd(a,c)^2},b\right) = k\left(\frac{a}{\gcd(a,c)},b\right) + k\left(\frac{c}{\gcd(a,c)},b\right)$
It is a little bit fun and crazy to think about what conjectures this function $\eta$ might also satisfy:
$$\forall a,b : \frac{a+b}{\gcd(a,b)} < \eta\left(\frac{ab(a+b)}{\gcd(a,b)^3}\right)^2$$
which is obviously inspired by the abc-conjecture, and as a consequence, would maybe have an application for FLT:
$$z^n < \eta((xyz)^n)^2 = n^2 \eta(xyz)^2 = n^2( \eta(x) + \eta(y) + \eta(z))^2 $$
$$\le^{\text{ because } \eta(a)\le a} n^2 ( x+y+z)^2 \le n^2(3z)^2 = 9 n^2 z^2$$
and since $n>2$ we would get:
$$z^{n-2} \le 9n^2$$
and again since $n-2>0$ we would derive:
$$z < \exp\left(\frac{\log(5n^2)}{n-2}\right)$$
But the right hand side of this inequality is $<2$ for $n\ge 13$. So this would mean that
$$x^n+y^n = z^n, 1 \le x \le y \le z, \gcd(x,y)=1, n>2$$
would have no solution for $n\ge 13$. This is of course wild speculation, because the abc-inequality for $\eta$ has yet to be tested if it is true or not, but let me get a little bit crazier and get back to my favourite topic, p.s.d kernels defined over the natural numbers, and point to the symmetric function, which also seems to be a p.s.d kernel:
$$K(a,b):=\frac{a+b}{\gcd(a,b) \eta( \frac{ab(a+b)}{\gcd(a,b)})^2}$$
The abc-conjecture for $\eta$ might than be restated as:
$$\forall a,b \in \mathbb{N}: K(a,b) < 1$$
**Q3) It would be nice to check this inequality for many numbers and report with or without code a counterexample or that it has succeded for these examples?**
**Edit**: The p.s.d conjecture for $K$ is wrong for $N=110$ and the matrix $K_N := (K(a,b))_{ 1 \le a,b \le N}$ is not positive semidefinite.
[1]: https://oeis.org/A001414
[2]: https://i.sstatic.net/YAmXF.png