The cubic relationship of three consecutive terms of tribonacci, Padovan, and Narayana sequences?

Question

I. Fibonacci and Lucas numbers

$$F_n = 1,1,2,3,5,8,13,\dots\\[7pt] L_n = 2, 1, 3, 4, 7, 11, 18\dots$$

There is a relationship between two consecutive terms such that,

$$M(x,y) = x^2+xy-y^2$$ $$M(F_n,\,F_{n+1}) = \pm1$$ $$M(L_n,\,L_{n+1}) = \pm5$$

with the sign depending on whether $n$ is odd or even. Of course, $d=5$ is prominent in the golden ratio and Fibonacci numbers. It turns out this may have a cubic counterpart.

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II. Tribonacci and Trucas numbers

$A_n = 1, 1, 2, 4, 7, 13, 24, 44,\dots$ (A000073)

$B_n = 1, 3, 4, 8, 15, 27, 50, 92,\dots$ (A007486)

and the latter is quaintly called the Trucas numbers, in analogy to the Lucas numbers above. Define,

$$\alpha(x,y,z) = x^3 + 2 y^3 + z^3 - 2 x y z + 2 x^2 y + 2 x y^2 - 2 y z^2 + x^2 z - x z^2 $$

then we conjecture three consecutive terms obey,

$$\alpha(A_n,\,A_{n+1},\,A_{n+2}) = 1$$ $$\;\alpha(B_n,\,B_{n+1},\,B_{n+2})\, =\, 11$$

where solving for the tribonacci constant in $P_1(r)=r^3-r^2-r-1=0\,$ needs $d=11$. Note that just like the cubic Pell equation, then $\alpha$ factors into linear forms in $(x,y,z)$ over a cubic extension,

$$\alpha(x,y,z) = (x - r_1 y + r_1^2 y + r_1 z)\,(x - r_2 y + r_2^2 y + r_2 z)\,(x - r_3 y + r_3^2 y + r_3 z)$$

where the $r_k$ are the three roots of $P_1(r)=0$.

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III. Padovan and Perrin sequence

$C_n = 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12,\dots$ (A000931)

$D_n = 3, 0, 2, 3, 2, 5, 5, 7, 10, 12, 17\dots$ (A001608)

Define,

$$\beta(x,y,z) = x^3 + y^3 + z^3 - 3 x y z + x y^2 - y^2 z + 2 x^2 y - x z^2 $$

then,

$$\beta(C_n,\,C_{n+1},\,C_{n+2}) = 1$$ $$\;\beta(D_n,\,D_{n+1},\,D_{n+2})\, =\, 23$$

where solving for the plastic constant in $P_2(r)=r^3-r-1=0\,$ needs $d=23$. Similarly, given the three roots $r_k$ of $P_2(r)=0$, then $\beta$ factors as,

$$\beta(x,y,z) = (x + r_1^2 y + r_1 z)\,(x + r_2^2 y + r_2 z)\,(x + r_3^2 y + r_3 z)$$

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IV. Narayana and companion sequence

$E_n = 1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28,\dots$ (A000930)

$F_n = 1, 1, 4, 5, 6, 10, 15, 21, 31, 46,\dots$ (A001609)

Define,

$$\gamma(x,y,z) = x^3 + y^3 + z^3 - 3 x y z + x y^2 + y^2 z + x^2 z - 2 y z^2$$

then,

$$\gamma(E_n,\,E_{n+1},\,E_{n+2}) = 1$$ $$\;\gamma(F_n,\,F_{n+1},\,F_{n+2})\, =\, 31$$

where solving for the supergolden ratio in $P_3(r)=r^3-r^2-1=0\,$ needs $d=31$. Given the three roots $r_k$ of $P_3(r)=0$, then,

$$\gamma(x,y,z) = (x - r_1 y + r_1^2 y + r_1 z)\,(x - r_2 y + r_2^2 y + r_2 z)\,(x - r_3 y + r_3^2 y + r_3 z)$$

So $\alpha$ and $\gamma$ have the same factorization but different $P_n(r)=0$.

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V. Question

Q: So do the six sequences obey the relationships $(\alpha,\,\beta,\,\gamma)$ and the conjectured values?

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P.S. For those curious how $(\alpha,\,\beta,\,\gamma)$ were found, I simply used three large enough consecutive terms, say the $A_{100},\,A_{101},\,A_{102}$ terms, and used the Integer Relations subroutine of Mathematica to find a relation between the monomials $x^3,x^2y,xyz,\dots$ A result with single-digit coefficients usually is the correct one.

Answer 1

The conjectures are all true. They just need a lot of algebra to prove.

Let's focus on the relation

$$\alpha(x,y,z) = x^3 + 2 y^3 + z^3 - 2 x y z + 2 x^2 y + 2 x y^2 - 2 y z^2 + x^2 z - x z^2 \tag{1}$$

If $(x,y,z)$ is any block of three consecutive Lucas numbers, then the next such block including overlapping is given by

$(y,z,x+y+z).$

So if we can prove that

$\alpha(y,z,x+y+z)=\alpha(x,y,z),$

then we have by induction that $\alpha(A_n,A_{n+1},A_{n+2})=\alpha(1,1,2)$

and the right side is then computed as $1$.

So we first write down

$$\alpha(y,z,x+y+z) = y^3 + 2 z^3 + (x+y+z)^3 - 2 y z (x+y+z) + 2 y^2 z + 2 y z^2 - 2 z (x+y+z)^2 + y^2 (x+y+z) - y (x+y+z)^2 \tag{2}$$

and then brace ourselves for the algebraic explosion that will occur when we expand all the products containing $x+y+z$. It will be convenient to first render the trinomial powers

$(x+y+z)^2=x^2+2xy+2xz+y^2+2yz+z^2$

$(x+y+z)^3=x^3+3x^2y+3xz^2+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^3$

With these in hand, we set out yo "divide and conquer". We focus in turn on the ciefficients of $x^3$, then $x^2y$, etc in (2) and compare the results to (1). For instance, the only terms in (2) that contain terms with $x^2y$ are

$(x+y+z)^3-y(x+y+z)^2,$

as all other terms contain factors that cannot be fitted into $x^2y$. Using the trnomial power expansions given earlier we find that the coefficient if $x^2y$ in (2) is $3-1=2$, which matches (1).

All ten degree-three combinations of $x,y,z$ give such a match, proving tge claim.

Answer 2

I will show that $ \alpha (x, y, z) = \alpha (y, z, x+y+z) $ similar to Oscar's solution, but without the tedious expansion.
Furthermore, I will show how to generalize to (3-term) linear recurrences and determine the $ \alpha$ polynomials explicitly. You might prefer to skip to this section directly.

Observe that

1. As OP mentioned, $\alpha(x,y,z) = \prod_i (x - r_i y + r_i^2 y + r_i z)$, where $r_i$ are the roots of $r^3 - r^2 - r - 1 = 0 $.
2. $r_1 r_2 r_3 = 1$ by Vieta's formula
3. $\frac{ 1 + r_i} { r_i} = r_i^2 - r_i$. Easiest way to show this is to multiply by the denominator to get $r_i ^3 - r_i^2 - r_i - 1 = 0 $.
4. $ (y - r_i z + r_i^2 z + r_i (x+y+z)) = r_i ( x + \frac{ 1 + r_i}{r_i} y + r_i z) = r_i(x - r_i y + r_i^2 y + r_i z) $
5. Hence $\alpha(y, z, x+y+z) = \prod (y - r_i z + r_i^2 z + r_i (x+y+z)) = r_1r_2r_3 \prod (x - r_i y + r_i^2 y + r_i z) = \alpha(x, y, z) $

We apply the same steps to $ \beta$ to prove that $ \beta(x, y, z) = \beta(y, z, x+y)$.

1. As OP mentioned, $\beta(x,y,z) = \prod (x + r_i^2 y + r_i z),$ where $r_i$ are the roots of $r^3 -r - 1 = 0 $
2. $r_1 r_2 r_3 = 1$ by Vieta's formula.
3. $\frac{1 + r_i}{r_i} = r_i^2$. Again, multiply by the denominator.
4. $(y + r_i^2 z + r_i(x+y) ) = r_i ( x + \frac{1+r_i}{r_i} y + r_iz) = r_i ( x + r_i^2 y + r_iz)$.
5. Hence $ \beta(y, z, x+y) = \beta(x, y, z)$.

We leave $ \gamma$ to the reader, since the generalization below explains why these work.

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Generalization: We show how to generalize this to any (3-term) linear recurrence relation that satisfies $ r^3 - Ar^2 - Br - C = 0$ with roots $r_i$.
We want to determine functions $M(r), N(r)$ such that $$y + M(r)z + N(r) (Az + By + Cx) = r ( x + M(r)y + N(r) z).$$ Calculating the coefficients of $x, y, z$, we want

$$CN(r) - r = 0, \quad 1+N(r)B-rM(r) = 0, \quad M(r) +AN(r) - rN(r) = 0. $$

The first two equations gives us $N(r) = \frac{r}{C}$, $M(r) = \frac{ Br+C}{Cr}$, which substituting into the last gives us $r^3 - Ar^2 - Br - C = 0 $ which is satisfied by the roots $r_i$. Note that we can express $ M(r)$ as a polynomial $ \frac{r^2 - Ar}{C}$, which is what OP had in the factorization. Thus with

$$ \alpha (x,y, z ) = \prod (x +\frac{ Br_i+C}{Cr_i}y + \frac{r_i}{C} z ),$$ we will get $\alpha(y, z, Az + By + Cx) =\prod r_i(x +\frac{ Br_i+C}{Cr_i}y + \frac{r_i}{C} z ) = C \alpha(x, y, z)$.
OP's sequences satisfy $C=1$, hence $\alpha(x, y, z)$ is an invariant.