Consider the following sequence of polynomials
$$p_n(x) = x^n + \binom{n}{1} a x^{n-1}+ \binom{n}{2} a(a+h) x^{n-2} + \cdots + a(a+h) \cdots ( a+ (n-1) h) = \sum_{k=0}^n \binom{n}{k} a^{(k)} x^{n-k} $$
where $a^{(k)} = a(a+h) \cdots (a+(k-1) h)$ for $k \ge 1$, $a^{(0)} = 1$.
I want to show the following formula for the discriminant of $p_n(x)$
$$D(P_n(x)) = \epsilon(n) H(n) h^{\binom{n-1}{2}} a^{n-1} (a+h)^{n-2} \cdots (a+(n-2) h) $$ where $\epsilon(n)$ has period $4$, starting with $1$, $1$, $-1$, $-1$, and $H(n)$ is the hyperfactorial.
Notes:
We have the equalities $p_n'(x) = n p_{n-1}(x)$ for $n\ge 0$ ( so $(p_n(x))_n$ is an Appell sequence)
We can reduce to the case $h=1$. However, it is useful to see the variation with respect to $h$. If $h=0$ we get $p_n(x) = (x+a)^n$.
The sequence $(p_n(x))$ is associated to the power series in $(1- h D)^{-\frac{a}{h}}$ that is $$(1- h D)^{-\frac{a}{h}} (x^n) = p_n(x)$$ when $D= \frac{d}{dx}$.
Why I am interested in the discriminant of $p_n(x)$? I am trying to show that if $a$, $h> 0$ then $p_n(x)$ has no real roots if $n$ even, and has exactly one root if $n$ odd. Knowing the determinant would allow ( using continuity) to reduce to the case $a=h=1$, which is known ( the partial sums of the exponential have $0$ or $1$ real root). Why I care about the sign of the polynomial $p_n(x)$? It has to do with derivatives of functions of the form $\frac{e^x}{x^a}$ ( being convex and so on).
The formula checks for all values tested.
There is a formula of the form
$$ p_{n, a+b}(x+y) = \sum_{k=0}^n\binom{n}{k} p_{k,a}(x) p_{n-k, b}(y)$$ although it's not clear how that would help.
- In general for a polynomial of the form $$x^n + a_1 x^{n-1} + a_1 a_2 x^{n-2} + a_1 a_2 a_3 x^{n-3} + \cdots + a_1 a_2 \cdots a_n$$ the discriminant is an expression in $a_1$, $\ldots$, $a_n$ divisible by $a_1^{n-1} a_2^{n-2} \cdots $, an expected thing since we do get lots of roots $=0$ if some of the $a_i$ is $0$. However, in general the other factor does not look clear.
Any feedback would be appreciated!
$\bf{Added:}$ I've found this paper which, together with the papers in its bibliography, give a lot of info, probably will also take care of the above. The paper is
Resultants and discriminants of Chebyshev and related polynomials, by Dilcher and Stolarsky
Especially interesting in the bibliography is the paper by Cygankova (1962). If anybody cares to calculate ( with a CAS), the discriminants of many series of polynomials have a predictable form ( proving it may be more difficult).