For integers $n\geq 1$ let $$H_n=1+\frac{1}{2}+\ldots+\frac{1}{n}$$ the $nth$ harmonic number.
After I've seen the form of the polynomials used by Niven in [1] I wanted to create a puzzle with a new definiton.
Definition. For integers $n\geq 1$, I define the following polynomials $$\operatorname{Niv}_{n}(x):=\frac{1}{n!}x^n(a_n-b_nx)^n,\tag{1}$$ where $a_n$ and $b_n$ are positive integers satisfying $\gcd(a_n,b_n)=1$ and $$H_n=\frac{a_n}{b_n}.\tag{2}$$
I would like to know if such polynomials satisfy some nice recurrence, or some nice ordinary differential equation. Since my definiton was a puzzle (I mean that there are no mathematical reason to define those as I did) isn't required that these have a special features.
Question. Can you set a simple recurrence or a simple ordinary differential equation that satisfy the polynomials $\operatorname{Niv}_{n}(x)$? Only is required an aproach to set a recurrence* or a differential equation. Many thanks.
Isn't required that the order of our equation to be $1$.
Using the definiton one can write $$\left(a_n-b_nx\right)^n\operatorname{Niv}_{n+1}(x)=\frac{x^{n+1}}{n+1}\left(a_{n+1}-b_{n+1}x\right)^{n+1}\operatorname{Niv}_{n}(x).\tag{3}$$ But it does not explode $H_{n+1}=H_n+\frac{1}{n},$ that is $$\frac{a_{n+1}}{b_{n+1}}=\frac{a_n}{b_n}+\frac{1}{n}.\tag{4}$$ Thus I don't know if this recurrence, although it is simple, isn't the best recurrence by capturing the information of the definition. So here is a compromise on the purpose of finding a simple recurrence or well a differential equation, but interesting enough in relation to our definition. Thus you can develop your reasoning to find a simple formula, understanding this word in a flexible way.
*In the case of the recurrence if you prefer it you can write an asymptotic identity (as $n\to\infty$) involving our polynomials.
References:
[1] I. Niven, A simple proof that $\pi$ is irrational, Bull. Amer. Math. Soc. Volume 53, Number 6 (1947).