The d(abc)-theorem, the abc-conjecture and positive definite kernels over the natural numbers?

Question

I noticed in the work of Hector Pasten, Th.1.11 the $d(abc)$ theorem and have a question, after doing some experiments with sagemath. Let $s_k(n) = \sum_{d|n}{ d^k }$ be the sum of divisiors $d$ of $n$ to the power $k=0,1,2,3,\cdots$. Then $s_0(n)$ counts the number of diviosrs. $s_1(n) = \sigma(n)$ is the sum of divisors etc. Is it true that for each $k=0,1,2,\cdots$ the following function is a positive definite kernel over the natural numbers? $$K(a,b) = \frac{1} {s_k \left( \frac{ab(a+b)}{\gcd(a,b)^3 } \right )}$$ ? I wrote @Pasten (the author of the theorem) and hopefully I will get an answer, but I thought that meanwhile I could ask the question here, and I hope it is ok to share this question here.

Possible attack on this problem through a related problem: Define $X_a = \{ a/k | 1 \le k \le a \}$. Then $X_a \cap X_b = X_{\gcd(a,b)}$ and $|X_a| = a$. Thus the following functions are proven to be similarities (Encyclopedia of Distances), some of which also are positive definite kernels over the natural numbers:

Example of such similarities are:

$s_{Si}(a,b) = \frac{\gcd(a,b)}{\min(a,b)}$ = Simpson similarity
$s_{BB}(a,b) = \frac{\gcd(a,b)}{\max(a,b)}$ = Braun,Blanquet similarity
$s_{J}(a,b) = \frac{\gcd(a,b)}{a+b-\gcd(a,b)}$ = Jaccard similarity
$s_{S}(a,b) = \frac{2\gcd(a,b)}{a+b}$ = Sorensen similarity
$s_{Cos}(a,b) = \frac{\gcd(a,b)^2}{ab}$ = Squared Cosine similarity

My idea is to define a set $X_a$ for each natural number $a$ and then maybe to make use of known similarities? (This is just an idea and might not lead anywhere).

Thanks for your help!

Related question: The abc-conjecture as an inequality for inner-products?

What does it mean for a function to be a kernel over the natural numbers? — Gerry Myerson, Jan 26 '21 at 22:22
@GerryMyerson: a symmetric, positive definite function over the natural numbers. — mathoverflowUser, Jan 27 '21 at 02:38
Thanks. $K(a,b)$ is obviously symmetric, and obviously positive, so what's left to prove? — Gerry Myerson, Jan 27 '21 at 04:27
OK, I know what it means for a quadratic form to be positive definite – what does it mean for a function to be positive definite, please? — Gerry Myerson, Jan 27 '21 at 11:52
@GerryMyerson: Sorry for not being clear: I mean the matrix $(K(a_i,a_j))_{i,j}$ is positive definite for each choice of natural numbers $a_1,\cdots,a_r$. — mathoverflowUser, Jan 27 '21 at 12:05
@GerryMyerson: https://en.wikipedia.org/wiki/Positive-definite_kernel — mathoverflowUser, Jan 27 '21 at 12:06

score 1 · Accepted Answer · answered Mar 11 '24 at 18:19

1

This is wrong for the divisor counting function $\sigma_0$ and $n=73$ as the Gram matrix has determinant $<0$.

answered Mar 11 '24 at 18:19

mathoverflowUser

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The d(abc)-theorem, the abc-conjecture and positive definite kernels over the natural numbers?

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