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Let $[n]=1+q+\dots+q^{n-1}$ and $u(n)=\prod_{j=1}^n \gcd([j],[n])$.

Define $$r(n)=\sum_{d|n,d>1}{(-1)^d \frac{u(n)}{du(\frac{n}{d})^d}r\Big(\frac{n}{d}\Big)^d}+\frac{(1-q)^{n-1}u(n)}{n[n]}$$ with $r(1)=1,$ where $d$ runs over all divisors of $n$ which are greater than $1$.

For $q=1$ the numbers $r(n)$ are integers because in this case $\frac{u(n)}{du(\frac{n}{d})^d}$ is an integer and the right-hand side vanishes for $n>1.$

Computations suggest that in the general case $r(n)$ is a polynomial in $q$ with integer coefficients.

The first elements are $1, 1, -q, 1+q^2+q^3, -q(1-q+q^2), q^2(1+3q+2q^2+2q^3+2q^4+2q^5+q^6), -q(1-q+q^2)^2, \dots.$

For a prime number $p$ we get $r(p)=\sum_{j=1}^{\frac{p-1}{2}}\frac{1}{p}\binom{p}{j}(-q)^j [p-2j]\in \mathbb{Z[q]}$.
Any idea how to prove that $r(n)$ is a polynomial in $q$ with integer coefficients for all $n?$

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  • $\begingroup$ Is the $\operatorname{gcd}$ in the definition of $u(n)$ a gcd of integer numbers, or polynomials in $q$? $\endgroup$ May 30, 2020 at 6:37
  • $\begingroup$ @Pietro Majer: For arbitrary q it is the polynomial gcd. $\endgroup$ May 30, 2020 at 6:48

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