Generalizing the Lucas numbers $L_n^2+L_n L_{n+1}-L_{n+1}^2=\pm5$ to the tribonacci and tetranacci?

Question

I. Fibonacci

Recall the M relationship $M(x,y) = x^2+xy-y^2$ for Fibonacci numbers from this post,

$$M(F_n,\,F_{n+1})=F_n^2+F_nF_{n+1} -F_{n+1}^2 = \pm1$$

where the sign depends on odd/even $n$. This turns out to have an $n$-nacci analogue.

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II. Tribonacci

Given the alpha relationship,

\begin{align}\alpha(x,y,z) &= \prod_{k=1}^3 \big(x - (r_k - r_k^2) y + r_k z\big)\\[5pt] &= x^3 + 2 y^3 + z^3 - 2 x y z + 2 x^2 y + 2 x y^2 - 2 y z^2 + x^2 z - x z^2 \end{align}

where the $r_k$ are the three roots of $r^3-r^2-r-1=0$. As Niel Sloane pointed out, there are actually four well-studied sequences that are sometimes called "tribonacci", differing only in the three initial values which is either $0$ or $1$,

$A_n = 0, 0, 1; \color{red}{1, 2, 4, 7}, 13, 24,\dots$ (A000073)

$P_n = 0, 1, 0; 1, 2, 3, 6, 11, 20,\dots$ (A001590)

$Q_n = 1, 1, 1; 3, 5, 9, 17, 31, 57,\dots$ (A000213)

$R_n = 1, 1, 0; 2, 3, 5, 10, 18, 33,\dots$ (A081172)

with the first the most well-known. They obey,

\begin{align} \alpha(A_n,\,A_{n+1},\,A_{n+2}) &= \color{red}1\\[5pt] \alpha(P_n,\;P_{n+1},\;P_{n+2}) &= \color{red}2\\[5pt] \alpha(Q_n,\,Q_{n+1},\,Q_{n+2}) &= \color{red}4\\[5pt] \alpha(R_n,\,R_{n+1},\,R_{n+2}) &= \color{red}7\end{align}

Why these match with the red numbers of the first sequence, I do not know.

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III. Tetranacci

Define the delta relationship,

$$\delta(w,x,y,z)=\prod_{k=1}^4 \big(w - (r_k + r_k^2-r_k^3)x - (r_k -r_k^2)y + r_k z\big)$$

where the $r_k$ are now the four roots of $r^4-r^3-r^2-r-1=0$. Expanded out, this is a rather long polynomial with integer coefficients. To compare,

• Fibonacci's $M(x,y)=\pm1$ (one sequence)
• Tribonacci's $\alpha(x,y,z)=1$ (one sequence)
• Tetranacci's $\delta(w,x,y,z)=\pm1$ (five sequences!)

So there is a difference the higher you go. Let,

$T_n = 0, 0, 0, 1; 1, 2, 4, 8, 15,\dots$ (A000078)

$U_n \,= 0, 0, 1, 2; 3, 6, 12, 23,\dots$ (A001630)

$V_n = 1, 1, 0, 1; 3, 5, 9, 18, 35,\dots$ (A251704)

with the first the most well-known. The 4th and 5th sequences, namely $(1, 1, 3, 5; 10, 19,\dots)$ and $(1, 1, 4, 8; 14, 27,\dots)$, do not have OEIS entries yet. Then,

\begin{align} \delta(T_n,\,T_{n+1},\,T_{n+2},\,T_{n+3})\, &= \pm1\\[5pt] \delta(U_n,\,U_{n+1},\,U_{n+2},\,U_{n+3}) &= \pm1\\[5pt] \delta(V_n,\,V_{n+1},\,V_{n+2},\,V_{n+3})\, &= \pm1\end{align}

and the sign depending on whether $n$ is odd or even. Why $\delta = \pm1$ is not exclusive to just one sequence seems unexpected. (And similar for the pentanacci.)

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IV. Question

If we use the three relationships $(M,\, \alpha,\, \delta)$ on related sequences with the same recurrence, then an important invariant appears.

A. Let $r_k$ be the two roots of $r^2-r-1=0$ with discriminant $|d|=5.$ Then,

$$M(L_n,\,L_{n+1}) = L_n^2+L_n L_{n+1}-L_{n+1}^2=\pm5$$

where $L_n = r_1^n+r_2^n = 2,1;3,4,7,11,18,\dots$ (A000032), or the Lucas numbers.

B. Let $r_k$ be the three roots of $r^3-r^2-r-1=0$ with discriminant $|d|=44.$ Then,

$$\alpha(S_n,\,S_{n+1},\,S_{n+2}) = 44$$

where $S_n = r_1^n+r_2^n+r_3^n = 3, 1, 3; 7, 11, 21, 39,\dots$ (A001644).

C. Let $r_k$ be the four roots of $r^4-r^3-r^2-r-1=0$ with discriminant $|d|=563.$ Then,

$$\delta(W_n,\,W_{n+1},\,W_{n+2},\,W_{n+3}) = \pm563$$

where $W_n = r_1^n+r_2^n+r_3^n+r_4^n = 4, 1, 3, 7; 15, 26, 51,\dots$ (A073817).

Question: In general, given a relationship like $(M,\, \alpha,\, \delta)$ where the input are $n$ consecutive terms of an $n$-nacci sequence, it seems the first $n$ terms determine the invariant output. But why is it for these three (and presumably for similar sequences $r_1^n+r_2^n+\dots+r_k^n$) the output, of all numbers, is the discriminant?

Answer 1

Here's the $n = 3$ case with $C=1$. (I already assumed that the characteristic equation is monic, since otherwise fixing $C$ is irrelevant.)
The $n=2$ case can be done similarly and the algebra is simple enough, so I'm not showing it.
I believe that this approach generalizes, but I am not that willing to check the algebra. (See details at the end.)

We substitute in $ x = 1 + 1 + 1, y = r_1 + r_2 + r_3$, $z = r_1 ^2 + r_2^2 + r_3^2$, and normalize the degree by multiplying with $r_1 r_2 r_3 =1$ as needed. The $k=1$ term of the product $N(x, y, z)$ can be written as

$$ (1+1+1) r_1r_2r_3 - ( (r_1+r_2+r_3) \cdot r_1 - r_1^2) ( r_1 + r_2 + r_3) + r_1(r_1^2 + r_2^2 + r_3^2)\\ = r_1 ( r_1 - r_2)(r_1 - r_3).$$

Hence, it follows that $ N(x, y, z) = (r_1r_2r_3) (r_1 - r_2)^2 ( r_2-r_3)^2 (r_3-r_1)^2$, which is the discriminant of the polynomial.

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If we had substituted in $ x = r_1 + r_2 + r_3$, $y = r_1 ^2 + r_2^2 + r_3^2$, $z = r_1^3 + r_2^3 + r_3^3$, then we get

$$ (r_1 + r_2 + r_3) r_1r_2r_3 - ( (r_1+r_2+r_3) \cdot r_1 - r_1^2) ( r_1 ^2 + r_2^2 + r_3^2) + r_1(r_1^3 + r_2^3 + r_3^3)\\ = r_1^2 ( r_1 - r_2)(r_1 - r_3).$$

This strongly suggests some identity is in play.

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Scratch work for generalization:

Consider the characteristic polynomial $ r^n - \sum_{i=0}^{n-1} a_i r^i = 0 $ with $a_0 = (- 1)^{n}$ and roots $r_i$.

Consider the polynomial $$N(x_1, x_2, \ldots x_n) = \prod_k [ (x_1 + ( r_k^{n-1} - \sum_{i=0}^{n-2} a_i r_k^i) x_2 + ( r_k^{n-2} - \sum_{i=0}^{n-3} a_i r_k^i) x_3 + ( r_k^{n-3} - \sum_{i=0}^{n-4} a_i r_k^i) x_4 + (r_k) x_n ]. $$

Let $ L_n = \sum r_i ^n$.

Claim:

$L_0 \prod r_i + ( r_k^{n-1} - \sum_{i=0}^{n-2} a_i r_k^i) L_1 + ( r_k^{n-2} - \sum_{i=0}^{n-3} a_i r_k^i) L_2 + ( r_k^{n-3} - \sum_{i=0}^{n-4} a_i r_k^i) L_3 + (r_k) L_{n-1} = r_k \prod_{j \neq k} (r_k - r_j)$.

Proof idea: Consider the LHS fully expanded in terms of $r_i$, and treat it as a polynomial in $r_k$ namely $f(r_k)$,

• Show that for $ k \neq j$, $f(r_j) = 0$. I don't see why this is true as yet , possibly some telescoping with Newton's identities?
• Show that $f(0) = 0 $. Obvious.
• Show that the coefficient of $r_k^n $ is 1: $(r_k) L_{n-1}$ contributes 1, and each of the rest contribute 0.
• Thus, applying the Remainder Factor Theorem, the LHS factors as $r_k \prod_{j \neq k} (r_k - r_j)$.