This question is part of Question 456683 in MO which has not been answered.

Define polynomials $$(u_n(x))=(1,0,-x^2,x^2 -1,x^4-x^2,-2x^4+3x^2, -x^6+3x^4-3x^2+1,\dots)$$ by their generating function $$\sum_{n \geq 0}u_n(x)z^n=\frac{1+z}{1+z+x^2 z^2+z^3+z^4}.$$ Then $$u_n(x)=\sum_{0\leq 2j\leq n}(-1)^{n-j}a_{j,n-2j}x^{2j}$$ with $$\sum_{n \geq 0}a_{j,n}z^n=\frac{1}{(1-z)^j (1-z^3)^{j+1}}$$ because $ \sum_{n,j}(-1)^{n-j} a_{j,n-2j}x^{2j}z^n=\sum_j(-1)^j (x z)^{2j}\sum_n(-1)^n a_{j,n-2j}z^{n-2j} =\sum_j \frac{(x z)^{2j}}{(1+z)^j (1+z^3) ^{j+1}}=\frac{1+z}{1+z+x^2 z^2 +z^3+z^4}.$

Computations suggest that

$$ u_n(x) u_{n+6}(x)- u_{n+3}(x)^2=x^2 (u_{n+1}(x) u_{n+5}(x)- u_{n+3}(x)^2).$$

More generally let $r,s$ be positive integers and $$h_n(x)=u_n(x)u_{n+r+s}(x)-u_{n+r}(x)u_{n+s}(x)$$. Then $h_n(x)$ satisfies the recursion $$h_{n+5}(x)-h_n(x)=(x^2-1)(h_{n+4}(x)+ h_{n+3}(x)- h_{n+2}(x)- h_{n+1}(x)).$$

Is there a simple method to prove these conjectures?

  • $\begingroup$ Does this recurrence hold for $h_n(x) = u_n(x) u_{n+r}(x)$? $\endgroup$ Nov 26, 2023 at 16:47
  • $\begingroup$ Judging from my own answer, no, $u_n(x) u_{n+r}(x)$ should be a recurrence of order $9$ rather than $5$. $\endgroup$ Nov 26, 2023 at 18:58

2 Answers 2


Here's the proof that $$h_n(x) = (x^2-1)(h_{n-1}(x)+h_{n-2}(x)-h_{n-3}(x)-h_{n-4}(x))+h_{n-5}(x)$$ for $h_n(x) = u_n(x) u_{n+r+s}(x) - u_{n+r}(x) u_{n+s}(x)$, as defined above:

Generally, the way it is defined, it holds that $$ -u_n(x) = u_{n-1}(x)+x^2 u_{n-2}(x) + u_{n-3}(x) + u_{n-4}(x). $$ In other words, $u_n(x)$ is a linear recurrence with a characteristic polynomial $$ p(z) = z^4+z^3+x^2z^2+z+1. $$ What it means is, assume $$ p(z) = (z-z_1)(z-z_2)(z-z_3)(z-z_4), $$ then, assuming that the roots are distinct, $u_n(x)$ can be represented as $$ u_n(x) = a_1(x) z_1(x)^n + \dots + a_4(x)z_4(x)^n. $$

Using this result, we can express $h_n(x)$ defined above as $$ h_n(x) = \sum\limits_{i=1}^4 \sum\limits_{j=1}^4 a_i a_j\left(z_i^n z_j^{n+r+s}-z_i^{n+r} z_j^{n+s}\right). $$

To understand why the expression above results into a linear recurrence of small degree, we should note that for $x \in (\sqrt{6}, +\infty)$, if $r$ is one of the roots, then the full set of roots is $\{r,r^*,\frac{1}{r},\frac{1}{r^*}\}$, where $r^*$ is complex conjugate of $r$. This stems from the following facts:

  1. For $x > 0$, there are no real roots of $p(z)$;
  2. $p(z) = p(z)^*$, meaning that for any root $r$, its conjugate $r^*$ must also be a root;
  3. $p(z) = z^4 p(\frac{1}{z})$, meaning that for any root $r$, its inverse $\frac{1}{r}$ must also be a root.
  4. Generally, $x^2 = z_1 z_2 + z_1 z_3 + z_1 z_4 + z_2 z_3 + z_2 z_4 + z_3 z_4$, meaning that for $x > \sqrt{6}$ one of the roots must be not on the unit circle, hence having distinct conjugate and inverse.

Playing with Wolfram a bit, we can also notice that the actual transition point from roots on unit circle to the set of $4$ roots as described above is even earlier, at $x > \frac{3}{2}$ rather than $x > \sqrt{6}$. This is probably due to the fact that at $x=\frac{3}{2}$ we have $p(z) = (z^2+\frac{z}{2}+1)^2$, hence the bifurcation.

Knowing that the roots are $\{r, r^*, \frac{1}{r}, \frac{1}{r^*}\}$, we may notice that all cases of $z_i = z_j$ will cancel each other out, thus only the following possible products $z_i z_j$ will survive: $$ z_i z_j \in \left\{1,\|r\|,\frac{r^2}{\|r\|}, \frac{(r^*)^2}{\|r\|},\frac{1}{\|r\|}\right\}. $$ So, while the initial recurrence corresponded to $4$ distinct roots, the new one will correspond to $5$ distinct roots. Since you provided the resulting recurrence in your answer, my guess would be that if the initial characteristic polynomial is $$ p(z) = (z-r)(z-r^*)\left(z-\frac{1}{r}\right)\left(z-\frac{1}{r^*}\right) = x^4+x^3+x^2z^2+z+1, $$ then the final one is $$ q(z) = (z-1)(z-\|r\|)\left(z-\frac{r^2}{\|r\|}\right)\left(z-\frac{(r^*)^2}{\|r\|}\right)\left(z-\frac{1}{\|r\|}\right), $$ and from your resulting formula it seems that, in fact, $$ q(z) = z^5 - (x^2-1)(z^4+z^3-z^2-z) - 1. $$

Well, so far we only somewhat proved this result for $x \in (\sqrt{6},+\infty)$, but with generating functions it's quite typical that the results that are true for some continuous range of $x$ are also true for $x$ as a formal variable, so I hope this should do it.

UPD. This can also be shown symbolically with sympy:

from sympy import *

x, z = symbols('x z')

p = z**4 + z**3 + x**2 * z**2 + z + 1
r = list(roots(p, z))

q = (prod(z-r[i]*r[j] for i in range(4) for j in range(i))).expand()
expand(factor(q) / gcd(q, q.diff(z))) # remove multiple roots

Yields $q(z) = - x^{2} z^{4} - x^{2} z^{3} + x^{2} z^{2} + x^{2} z + z^{5} + z^{4} + z^{3} - z^{2} - z - 1$, which is, indeed $$ q(z) = z^5 - (x^2-1)(z^4+z^3-z^2-z) - 1, $$ thus proving that the recurrence for $h_n(x)$ is, indeed $$ h_n(x) = (x^2-1)(h_{n-1}(x)+h_{n-2}(x)-h_{n-3}(x)-h_{n-4}(x))+h_{n-5}(x). $$

  • $\begingroup$ @ Oleksandr Kulkov: This explains why these expressions have a simpler recurrence. $\endgroup$ Nov 27, 2023 at 13:38
  • $\begingroup$ Also note the update for more conclusive proof that $q(z)$ is exactly as I assumed. $\endgroup$ Nov 27, 2023 at 13:57

The question is about a sequence of polynomials in $\,x\,$ whose generating function is

$$ \sum_{n \geq 0}u_nz^n := \sum_{n \geq 0}u_n(x)z^n = \frac{1+z}{1+z+x^2 z^2+z^3+z^4}. \tag1 $$

This immediately implies that the sequence satisfies a linear recurrence with coefficients that are polynomials in $\,x.\,$ It is easy to verify using only the linear recurrence the following results for all integer $\,n\,$ where $\,y := x^2-1\,$

\begin{align*} u_{n+4} &= -u_n - u_{n+1} - x^2u_{n+2} - u_{n+3}, \\ u_{n+5} &= u_n +y\,u_{n+2} -y\,u_{n+3}, \\ u_{n+6} &= y\,u_n + x^2u_{n+1} +x^2y\,u_{n+2} + 2y\,u_{n+3}, \\ u_{n+7} &= -2y\,u_n -y\,u_{n+1} +x^2(3-2x^2)u_{n+2} +y(y-1)u_{n+3}. \tag2 \end{align*}

Define the sequence of polynomials

$$ g_n := (u_n u_{n+6} -u_{n+3} ^2) - x^2(u_{n+1} u_{n+5} - u_{n+3} ^2). \tag3 $$

Use the results in equation $(2)$ to verify that

$$ g_{n+1} = g_n = (u_n+u_{n+3})^2 + x^2(u_{n+1}u_{n+3} - u_{n+1}u_{n+2} + u_nu_{n+2}). \tag4 $$

Use induction to conclude that $\,g_n = g_0 := y\,I_1\,$ for all integer $\,n.\,$

From equation $(1)$ the first few values of the $\,u\,$ sequence are

$$ u_{-1} = 0,\;\; u_0 = 1,\;\; u_1 = 0,\;\; u_2 = -x^2,\;\; u_3 = x^2-1. \tag5 $$

Verify that $\,g_0 = I_1 = 0\,$ for these initial values of $\,u.\,$

This proves the first conjecture

$$ u_n(x)u_{n+6}(x)-u_{n+3}(x)^2 = x^2(u_{n+1}(x)u_{n+5}(x) - u_{n+3}(x)^2). \tag6 $$

Define the sequence

$$ f_n := (u_{n+1} + u_{n+2})^2 + y\,u_{n+1}u_{n+2} + u_nu_{n+1} +(u_{n+2}-u_n)u_{n+3}. \tag7 $$

Similar to equation $(4)$ verify that $\,f_{n+1} = f_n\,$ and conclude that $\,f_n = f_0 := I_2\,$ for all integer $\,n.\,$ Verify that $\,f_0=I_2=1\,$ For the initial values of $\,u\,$ in equation $(5)$.

Verify the general recurrence result

\begin{align*} I_1u_{n+4}u_{n-4}=\; & (2I_1+x^2I_2)u_{n+3}u_{n-3} + ((y^2-2)I_1 - x^4I_2)u_{n+2}u_{n-2} +\\ & (1-y)I_1u_{n+1}u_{n-1} + y((1-y)I_1 + x^2I_2)u_nu_n. \tag8 \end{align*}

Verify that equation $(8)$ is equivalent to equation $(6)$ if $\,I_1=0\,$ and $\,I_2=1\,$ which is implied by equation $(5)$.

Note that so far, all the algebra required can easily be done by hand.

Given any two integers $\,r,s\,$ define the sequence

$$ h_n := u_n u_{n+r+s} -u_{n+r} u_{n+s} . \tag9 $$

It is conjectured that the sequence $\,h_n\,$ satisfies the recursion

$$ h_{n+5}-h_n = y(h_{n+4} + h_{n+3} - h_{n+2} - h_{n+1}). \tag{10} $$

This is surprisingly simple to prove in great generality. Suppose that for a fixed positive integer $\,m\,$ and constants $\,c_k\,$ the sequence $\,u\,$ satisfies the linear recursion

$$ u_n := \sum_{k=1}^m c_k u_{n-k}. \tag{11} $$

In the general case, where the characteristic polynomial of the recursion has no repeated roots, there exists constants $\,a_k\,$ and roots $\,r_k\,$ such that

$$ u_n = \sum_{k=1}^m a_k r_k{}^n. \tag{12} $$

That is, the sequence $\,u\,$ is a linear combination of powers of the roots $\,r.\,$ Suppose $\,v\,$ and $\,w\,$ are any two sequences with the same roots. Then their product sequence is a linear combination of powers of the product of two roots. That is, there exists constants $\,b_{i,j}\,$ which only depend on the roots $\,r_k\,$ such that

$$ h_n:=v_n w_n = \sum_{1\le i\le j\le m} b_{i,j}(r_ir_j)^n. \tag{13} $$

This implies that there exists constants $\,d_k\,$ which depend on the roots $\,r_k\,$such that there is a linear recursion for $\,h\,$

$$ h_n = \sum_{k=1}^M d_k h_{n-k}. \tag{14} $$

where $\,M\le m(m+1)/2.\,$ Note that this same recursion holds for any linear combination of products of two sequences with the same recursion in equation $(11)$. Thus, in particular, the recursion in equation $(14)$ holds for the sequence in equation $(9)$.

For the particular sequence $\,u_n\,$ defined in equation $(1)$ it suffices to show that $\,h_n := u_n^2\,$ satisfies the recursion in equation $(10)$ using the same techniques used to prove the first conjecture and this implies that the sequence $\,h\,$ in equation $(9)$ also satisfies the same recursion. Note that the characteristic polynomial for $\,u\,$ is palindromic which leads to the simple form of the recursion in equation $(10)$.

This proves the second conjecture.

I used Wolfram Mathematica extensively to study the natural generalization of the original sequence of polynomials $\,u_n.\,$ Here is a summary of the results I found.

Define the sequence $\,u_n\,$ by the recursion with constants $\,c_2,c_3$

$$ u_n = -(c_2 u_{n-1} + c_3 u_{n-2} + c_2 u_{n-3} + u_{n-4}). \tag{15} $$

Define the sequence $\,h_n\,$ as in equation $(9)$. Then it satisfies the recursion

$$ h_{n+5} - h_n = (1-c_3)(h_{n+4} - h_{n+1}) + (c_2^2-c_3)(h_{n+3} - h_{n+2}) \tag{16} $$

which is the generalization of equation $(10)$

Notice the general Somos-8 sequence identity

$$ v_0 u_{n+4}u_{n-4} = -(v_1 u_{n+3}u_{n-3} + v_2 u_{n+2}u_{n-2} + v_3 u_{n+1}u_{n-1} + v_4 u_{n}u_{n}) \tag{17} $$

where the constants $\,v_0,v_1,v_2,v_3,v_4\,$ depend on $\,c_2,c_3\,$ and the initial values $\,u_0,u_1,u_2,u_3.\,$ This is the generalization of equation $(6)$.

  • $\begingroup$ @ Somos: Very nice! You first express $u_{n+4},\dots, u_{n+7}$ as linear expressions in $u_n, \dots, u_{n+3}.$ It suffices to verify this for small $n.$ Substituting this in $g_n(x)$ and $g_{n+1}(x)$ gives the same formula $F(u_n,u_{n+1},u_{n+2},u_{n+3})$ which by induction gives $g_n(x)=0.$ $\endgroup$ Nov 27, 2023 at 13:27
  • $\begingroup$ @JohannCigler Thanks, but almost correct. Expressing $\,u_{n+4},\dots,u_{n+7}\,$ as a linear combination in $\,u_n,\dots,u_{n+3}\,$ does not need verification for small $\,n\,$, but comes directly from the recursion $\,u_{n+4} = -u_n - u_{n+1} - x^2u_{n+2} - u_{n+3},\,$ which comes from equation $(1)$. I am still working on the 2nd conjecture. $\endgroup$
    – Somos
    Nov 27, 2023 at 15:01
  • $\begingroup$ Note that the order of the recurrence in $(10)$ is smaller than the one in $(14)$, and they're, generally, different recursions. E.g. $h_n = u_n u_{n+r}$ might not adhere to the one in $(10)$, but it will always adhere to the one in $(14)$. $\endgroup$ Nov 29, 2023 at 18:00
  • $\begingroup$ @OleksandrKulkov Good point! I forgot that little detail. I will add more explanation. $\endgroup$
    – Somos
    Nov 29, 2023 at 18:02

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