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?


1 Answer 1


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.

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    $\begingroup$ 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. $\endgroup$ Jun 6 at 11:56
  • $\begingroup$ @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. $\endgroup$ Jun 6 at 13:31

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