I'm inspired in the so-called Erdős squarefree conjecture, this section from Wikipedia, to state in this post a question, involving a different arithmetic function, that due its difficulty I feel as an exercise (in the Appendix I show another example), and in my next post I am going to add a different variation that I think that due its difficulty should be a conjecture.

In this post we denote the Euler's totient function as $\varphi(n)$, and it is well-known that satisfies $$\varphi(n)=n\prod_{p\mid n}\left(1-\frac{1}{p}\right)$$ for integers $n>1$.

Also is well-known the multiplicative formula for factorials, this section of Wikipedia.

Computational fact. In the segment of positive integers $1\leq n\leq 20000$, the integer $n=160$ is the last integer for which $$\binom{2n}{\varphi(n)}$$ is a square-free integer (has no repeated prime factors).

The sequence of binomial coefficients $$\binom{2n}{\varphi(n)}$$ having some repeated prime factors (when $n\geq 1$ runs over integers) starts as $$4,28,1820,18564\ldots$$

Question. I belive that the following claim is true:

The binomial coefficient $\binom{2n}{\varphi(n)}$ is never squarefree for $n>160$.

Can you prove it or provide me hints? Many thanks.

Now I am going to add in next appendix a different example that I think more easy than previous, with the purpose to provide it to everyone that is interested in this kind of questions.

Appendix: We (exclude the case $n=2$) consider the arithmetic function $\binom{n\varphi(n)}{\sigma(n)}$, where $\sigma(n)=\sum_{d\mid n}d$ denotes the sum of divisors function. Then I believe that the following claim is true (you can to study it in your home, but feel free to add a counterexample as a comment if you find it).

Claim. The binomial coefficient $$\binom{n\varphi(n)}{\sigma(n)}$$ is never squarefree for $n>22$.

  • $\begingroup$ Concerning the second claim : I did not find such a number yet not divisible by one of the numbers $4,9,25,49,121,361$. My search is currently at $n=11\ 000$ $\endgroup$ – Peter Apr 4 '18 at 21:53
  • $\begingroup$ Can you explain your comment in different words @Peter ? I am saying it since English isn't my native tongue and I don't understand it (you use the expression with two negations I did not find such a number yet not divisible by one). Many thanks. $\endgroup$ – user243301 Apr 4 '18 at 21:55
  • 1
    $\begingroup$ That means that for $23\le n\le 11\ 000$, every number of the form in the second claim is divisible by one of the mentioned numbers, hence not squarefree. $\endgroup$ – Peter Apr 4 '18 at 21:57
  • 1
    $\begingroup$ This is of course no proof, but it seems that a much stronger claim holds. I am now at $18\ 000$ $\endgroup$ – Peter Apr 4 '18 at 22:00
  • 1
    $\begingroup$ I was disconnected many thanks @Peter $\endgroup$ – user243301 Apr 4 '18 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy