Does the Apéry-like sequence $A_n=(n!)^2\sum_{k=0}^n { \rho \choose k}^2 { \rho+n-k \choose n-k}, \rho=e^{2 \pi i/3}$ change signs infinitely often?

Question

This is an integer sequence OEIS sequence A217703.

It satisfies an order 3 recurrence which is the constant term $A_n=u_n(0)$ of a three term recurrent sequence of polynomial defined by

$$u_0(x)=1,u_1(x)=x, u_{n+1}(x)=(x+2n(n+1))u_n(x)-n^4u_{n-1}(x).$$

It was conjectured that $A_n<0$ if and only if $2 \le n \le 57$ and $n >2177$ but it turns positive at $n=81935$ and negative again at $n=3082582$, and it should be the case that it changes signs infinitely often.

The very interesting explicit formula for $A_n=(n!)^2\sum_{k=0}^n { \rho \choose k}^2 { \rho+n-k \choose n-k}, \rho=e^{2 \pi i/3}$ with complex binomial comes from a formula in Z. W. Sun's original posting A sequence of irreducible polynomials

$$u_n(x^2+x+1)= (n!)^2\sum_{k=0}^n { x \choose k}^2 {x+n-k \choose n-k}.$$

However the explicit formula does not seem to be of any help for showing infinitely many sign changes.

Reasons suggesting infinitely many sign changes

It can be shown that a 3-term recurring polynomial sequence

$p_0(x)=b_0, p_1(x)=x+a_0, p_{n+1}(x)=(x+a_n)p_n(x)-b_np_{n-1}(x),$

with all $a_j$ real and $b_j>0$ have only real and distinct roots which strictly interlace, ie. there is exactly one root of $p_n(x)$ strictly between adjacent roots of $p_{n+1}(x)$.

An interesting consequence of such interlacing sequence is that is that the roots are aligned. If we let

$r_{n,1}>,...,>r_{n,n}$

be the roots of $p_n(x)$, then for fixed $k \ge 1$, the $k$th largest root $r_{n,k}$ is strictly increasing and the $k$th smallest root $r_{n,n-k+1}$ is decreasing, so that we must have

$ lim_{n \rightarrow \infty} r_{n,k}=\alpha_k \in (-\infty,\infty],$

$ lim_{n \rightarrow \infty} r_{n,n-k+1}=\beta_k\in [-\infty,\infty) $

It could happen that all the $\alpha_k$ are the same , also for $\beta_k$ as is the case for the classical orthogonal polynomials.

Since $sign(u_n(0))=(-1)^m$, where $m$ is the number of positive roots of $u_n(x)$. We can explain the sign changes by there is only one positive root for $1<n<58$, $r_{n,2}$ become also positive for $n=58$ and also $r_{n,3}>0$ for $n \ge 2178$, $r_{n,4}>0$ for $n \ge 81935$, $r_{n,5}>0$ for $n \ge 3082582$ etc. There are infinitely many sign changes because all $r_{n,k}$ becomes eventually positive.

It could happen that all $\alpha_k$ are distinct (it seems these are the only two possibilities) as in the special case

$a_n=an,b_n=bn$

for $a>0,b>0$.

In this case all $\beta_k=-\infty$ but $\alpha_k=b/a-a(k-1)$ so there are exactly $\lceil b/a^2 \rceil$ sign changes.

Answer 1

I see two ways to prove this.

Method 1: Generalities about orthogonal polynomials.

In standard notation, your polynomials are $$u_n(x)=S_n\left(\frac 34-x;\frac 12,\frac12,\frac 12\right).$$ This is clear by comparing the recurrence relation. If you want to compare the explicit expressions, you need to apply a ${}_3F_2$-transformation. The polynomials $S_n$ are orthogonal on the real line, with weight $$\frac{|\Gamma(1/2+i x)|^6}{|\Gamma(2ix)|^2}=8\pi^2\frac{x\sinh(\pi x)}{\cosh^2(\pi x)}.$$ The corresponding moment problem is determinate, that is, this weight function is unique.

For quite general systems of orthogonal polynomials, one can in principle obtain a weight function (the weight function in the determinate case) from the zeroes, see e.g. G. Freud, Orthogonal Polynomials, Chapter 2. If $r_{n,k}$ are the zeroes of $p_n$, then one can approximate the orthogonality measure by a quadrature formula $$\tag{1}\sum_{k=1}^n\lambda_{n,k}f(r_{n,k}).$$ Then there is at least a subsequence $n=n_m$ so that this converges to an integral $$\tag{2}\int f(x)\,d\rho(x)$$ as $m\rightarrow\infty$, and $p_n$ are orthogonal with respect to $d\rho$.

In the case at hand, it follows that the zeroes $(r_{n,k})_{n,k}$ are dense in $\mathbb R$. If there is an interval $I$ without zeroes and we take $f$ as the characteristic function of that interval, then (1) vanishes but (2) doesn't. In particular, the $k$-th largest zero $r_{n,k}\rightarrow\infty$. As you mention, it then follows from interlacing of zeroes that $u_n(0)$ changes sign infinitely often. To repeat the argument, since $u_n(x)\rightarrow\infty$ as $x\rightarrow\infty$ and has only simple zeroes, the sign of $u_n(0)$ is the parity of the number $m_n$ of positive zeroes. We have seen that $m_n\rightarrow\infty$ and, by interlacing, $m_{n+1}\in\{m_n,m_n+1\}$. So $m_n$ has to change parity infinitely often. Here I ignored the possibilty that $u_n(0)=0$ (I don't know if this can happen). But in that case it follows from interlacing that $m_{n-1}$ and $m_{n+1}$ have distinct parity.

Method 2: Asymptotics

I believe that one can prove an asymptotic formula like $$u_n(0)\sim C_n\big(\cos(A\log n+B)+\mathcal O(1/n)\big),$$ where $C_n>0$ and $A$, $B$ are some elementary constants. See J. A. Wilson, Asymptotics for the ${}_4F_3$ polynomials, J. Approx. Theory 66 (1991), 58-71. The polynomials considered there are more general. This would again show that there are infinitely many sign changes, and that the time between them increases exponentially.