Exceptional zeros of a convolutional inverse

Let $$\kappa:\lbrace 1,2,3,\ldots\rbrace\longrightarrow \mathbb Z$$ be the convolutional inverse in the Dirichlet ring of $$n\longmapsto {n+1\choose 2}$$. It is defined by $$\kappa(1)=1$$ and by the functional equation $$\sum_{d\vert n}{d+1\choose 2}\kappa(n/d)=0$$ for $$n\geq 2$$.

It is not hard to show that $$\kappa(3\cdot 5\cdot p^2)=0$$ for every prime number $$p\geq 7$$.

There seem to be very few other values $$n$$ such that $$\kappa(n)=0$$. My computer found only two : $$2^4\cdot 13\cdot 23$$ and $$5^3\cdot 7\cdot 37$$.

Is the set of such exceptional zeros of $$\kappa$$ finite or not?

There are many more examples. An infinite series is for $$n=5^8\cdot7\cdot p^2$$ for primes $$p$$ different from $$5$$ and $$7$$.
Another infinite series is $$n=5\cdot11\cdot17^3\cdot p^2$$ for primes $$p$$ different from $$5,11,17$$. And there are many more infinite series.
• Nice! Thank you very much. There is however an intriguing pattern : series of the form $A p^2$ with primes $p$ coprime to $A$. My two examples seem not to fall into this category. Commented Jun 6, 2023 at 11:56
• @RolandBacher Indeed, all the infinite series I found have this pattern $n=Ap^2$ with $p\nmid A$. However, if $p,q,r$ are distinct primes, then $\kappa(pqr^3)=0$ once $(p + q - 1)r = pq - p - q$. Probably there are infinitely many such triples. Commented Jun 6, 2023 at 13:31