Let $x$ be a variable. Define the following family of sequences (reminiscent of Lucas polynomials) according to the rule: $P_0(x):=0, P_1(x):=1$ and for $n\geq2$ by $$P_n(x)=xP_{n-1}(x)-P_{n-2}(x).$$ Notice that $P_n(2)=n$ for every $n\in\mathbb{Z}_{\geq0}$. Here are a few examples: $$P_2=x, \qquad P_3=x^2-1, \qquad P_4=x^3-2x, \qquad P_5=x^4-3x^2+1.$$

QUESTION 1. Empirical evidence suggest that, for each fixed integers $n, k\geq1$, $$Q_{n,k}(x):=\frac{P_1(x)^{2k-1}+P_2(x)^{2k-1}+\cdots+P_n(x)^{2k-1}}{P_1(x)+P_2(x)+\cdots+P_n(x)} \tag1$$ is a polynomial in $x$.

This is trivial for $k=1$. Is it true for other odd powers $2k-1$?


(1) Specialized values $2k-1=3$ or $5$, etc are still interesting to me.

(2) Even the case of special valuations for $x\in\mathbb{Z}$ are appealing as well, which means (1) becomes a claim on integrality of sequences.

QUESTION 2. Encouraged by the success with QUESTION 1, how about this? $$R_n(x)=\prod_{j=1}^n\frac{P_j(x)^{2k-1}+\cdots+P_n(x)^{2k-1}}{P_j(x)} \tag2$$ is a polynomial in $x$.

  • $\begingroup$ The case $n=2$ is trivial. Seems that $n=3$ says: "if $2k-1\geq1$ is a positive odd integer, then is $(1+x^{2k-1}+(x^2-1)^{2k-1})/(x+x^2)$ a polynomial?" and it's not hard to see that this is true. $\endgroup$
    – znt
    Nov 20, 2016 at 20:29
  • $\begingroup$ It's true for e.g. $2k-1=31$ and $n=10$ so I guess it could well be true in general. $\endgroup$
    – znt
    Nov 20, 2016 at 20:32
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    $\begingroup$ $P_n(2\cos t)=\sin(nt)/\sin t$ is Chebyshev polynomial of the first kind $\endgroup$ Nov 20, 2016 at 20:43
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    $\begingroup$ The Chebyshev polynomial can also be written as $P_n(z+z^{-1}) = (z^{n+1}-z^{-n-1})/(z-z^{-1})$. $\endgroup$ Nov 20, 2016 at 21:37
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    $\begingroup$ @Fedor Petrov They are (normalized) Chebyshev polynomial of the SECOND kind. $\endgroup$ Nov 21, 2016 at 3:05

2 Answers 2


This is response to QUESTION 1.

As Fedor pointed out, we're dealing with the Chebyshev polynomials $P_n(2\cos t)=\sin nt/\sin t$. So we must show that if $$ \sum_{n=1}^N \sin nt = 0 , \quad\quad\quad\quad (1) $$ then also $\sum_{n=1}^N \sin^m nt = 0$ for any odd exponent $m\ge 1$.

We may take $0<t<\pi/2$. Also, the sum in (1) can of course be evaluated, and we find that (1) is equivalent to $$ \cos t/2 = \cos (N+1/2) t . \quad\quad\quad\quad (2) $$ I now claim that if (2) holds for $t$, then it also holds for any multiple of $t$. To see this, we just notice that (2) means that $s=t/2$ satisfies $(2N+1)s = 2\pi M\pm s$, for some $M\ge 1$ and a choice of sign (recall that $0<s<\pi/4$). In other words, (2) requires $s$ to be a rational multiple of $\pi$ with denominator $N$ or $N+1$, and clearly this property is preserved under taking integer multiples.

Now everything is clear: $\sin^m\alpha$ can be written as a linear combination of $\sin j\alpha$, with $j$ odd, and we have just seen that (1) implies that also $\sum_{n=1}^N \sin njt = 0$ for any odd $j$.

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    $\begingroup$ I would rewrite (2) as $2\sin(Nt/2)\sin((N+1)t/2)=0$, this immediately yields that $t=2\pi k/N$ or $t=2\pi k/(N+1)$ with $k\in \mathbb{Z}$. We actually need to check also that all the roots of polynomial in the denominator are simple, but this is also clear as we found $N$ different roots $2\cos (2\pi k/N),2\cos (2\pi k/(N+1))$. $\endgroup$ Nov 20, 2016 at 22:17
  • $\begingroup$ I was not right, by the way:) $\endgroup$ Nov 20, 2016 at 22:17
  • $\begingroup$ @FedorPetrov: Noticed that too :). I'm using that the exponent is odd in the final step, $\sin^{2k} nt$ would not be a linear combination of sines. $\endgroup$ Nov 20, 2016 at 22:22
  • $\begingroup$ You should be able to use that "evaluated" formula to simplify the denominator $P_1\left(x\right) + \cdots + P_n\left(x\right)$ without recourse to trig, right? $\endgroup$ Nov 20, 2016 at 22:27
  • $\begingroup$ @darijgrinberg: Yes, I guess if I use it sufficiently many times, I'll end up with a formula for the quotient. $\endgroup$ Nov 21, 2016 at 0:43

Question 2.

Since $P_n(2\cos t)=\sin(nt)/\sin t$, we get that the roots of $P_n$ are $2\cos(\pi k/n)$, $k=1,\dots,n-1$. That is, a number $\kappa=2\cos(\pi a/b)$, $0<a<b$, gcd$(a,b)=1$, is a root of $P_1\dots P_n$ of multiplicity $[n/b]$. We have to prove that $\kappa$ is a root of the numerator of your fraction of multiplicity at least $[n/b]$. Fix odd integer $r$ and consider the sum $$S_{rmn}(t)=\sum_{j=m}^n \sin(rjt)=-\frac{\sin\frac{n+m}2 rt \sin\frac{n-m-1}2 rt}{\sin \frac{rt}2}.$$ As Christian explained, the $m$-th multiple in the numerator is a linear combination of $(\sin t)^{-1}S_{rmn}(t)$ for different odd $r$. We see that if $n+m$ or $n-m-1$ is divisible by $2b$ (note that both divisibilities are not possible for the same $m$), then $S_{rmn}(\pi a/b)=0$. The total number of such $m$ is the total number of multiples of $2b$ in the set $\{-1,0,1,\dots,2n\}\setminus \{n-1,n\}$. It equals $1+[2n/(2b)]-\delta\geqslant [n/b]$ as desired (here $\delta$ equals 1 if $n-1$ or $n$ is divisible by $2b$, else $\delta=0$.)

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    $\begingroup$ This is quite nice! $\endgroup$ Nov 21, 2016 at 16:13

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