Revisions to 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?

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edited Nov 8 at 13:41

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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 this arethese 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.

Is there an explicit formula for $p_n(x)$ in this case ?

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 this 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.

Is there an explicit formula for $p_n(x)$ in this case ?

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.

added 6 characters in body

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edited Nov 8 at 0:53

CHUAKS

1.4k
7
18

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 this 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=0$$\beta_k=-\infty$ but $\alpha_k=b/a-a(k-1)$ so there are exactly $\lceil b/a^2 \rceil$ sign changes.

Is there an explicit formula for $p_n(x)$ in this case ?

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 this 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 $\beta_k=0$ but $\alpha_k=b/a-a(k-1)$ so there are exactly $\lceil b/a^2 \rceil$ sign changes.

Is there an explicit formula for $p_n(x)$ in this case ?

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 this 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.

Is there an explicit formula for $p_n(x)$ in this case ?

added 1661 characters in body

Source Link

edited Nov 7 at 18:44

CHUAKS

1.4k
7
18

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 again 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

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

This looks like Apéry's sequence and it also satisfies similar congruence $p\mid A_{p+1}$.

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 this 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 $\beta_k=0$ but $\alpha_k=b/a-a(k-1)$ so there are exactly $\lceil b/a^2 \rceil$ sign changes.

Is there an explicit formula for $p_n(x)$ in this case ?

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 again at $n=81935$ 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

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

This looks like Apéry's sequence and it also satisfies similar congruence $p\mid A_{p+1}$.

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

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.