It is well known that the n-anacci (higher degree Fibonacci, that is Tribonacci and so on) numbers can be computed in closed form from roots of polynomials in the way Eric Weisstein at Mathworld describes it:
http://mathworld.wolfram.com/Fibonaccin-StepNumber.html
Consider the Fibonacci like linear recurrence with only difference that it has positive coefficients $\frac{1}{a}$ and $\frac{1}{b}$ like this:
$$F(n)=\frac{1}{a}*F(n-1)+\frac{1}{b}*F(n-2)$$
Let:
$$x=\lim_{n\to \infty } \, -\frac{F(n)}{F(n-1)}$$
Prove or disprove that $x$ is a solution to:
$$-a + bx + abx^2$$
when $a \geq 1$ and $b \geq 1$.
Do the same in the case of the cubic equation with the Tribonacci like recurrence:
$$T(n)=\frac{1}{a}*T(n-1)+\frac{1}{b}*T(n-2)+\frac{1}{c}*T(n-3)$$
Let:
$$x=\lim_{n\to \infty } \, -\frac{T(n)}{T(n-1)}$$
Prove or disprove that $x$ is a solution to:
$$ab^2 - abcx + b^2cx^2 + ab^2cx^3$$
when $a \geq 1$, $b \geq 1$ and $c \geq 1$.
(*Mathematica*)
(*quadratic equation start*)
Clear[a, b, c, nn, x]
nn = 200;
a = Random[];
b = Random[];
LinearRecurrence[{1/a, 1/b}, {1, 1}, nn];
x = -N[Ratios[%], 30][[nn - 1]];
(-a + b*x + a*b*x^2)
(*quadratic equation end*)
(*cubic equation start*)
Clear[a, b, c, nn, x]
nn = 200;
a = Random[];
b = Random[];
c = Random[];
LinearRecurrence[{1/a, 1/b, 1/c}, {1, 1, 1}, nn];
x = -N[Ratios[%], 30][[nn - 1]];
(a*b^2 - a*b*c*x + b^2*c*x^2 + a*b^2*c*x^3)
(*cubic equation end*)
The output in both cases, quadratic and cubic, is zero. Therefore $x$ appears to be a root.
Also if you can continue the pattern to tetranacci numbers and further, I would be happy.
Any solution to a linear recursion with constant coefficients $$ 0=x_{n+p}+a_{p-1}x_{n+p-1}+…+a_1x_{n+1}+a_0x_n $$ can be written as $$ x_n=c_1q_1^n+c_2q_2^n+…+ c_pq_p^n $$ where the number $q_k$ are the roots of $$ 0=q^p+a_{p-1}q^{p-1}+…+a_1q+a_0 $$ Assuming that $c_1\ne 0$, $q_1$ is the root with the largest absolute value and all others have a smaller absolute value, the fraction $$ \frac{x_{n+1}}{x_n} =q_1·\frac{c_1+c_2(q_2/q_1)^{n+1}+…+c_p(q_p/q_1)^{n+1}}{c_1+c_2(q_2/q_1)^{n}+…+c_p(q_p/q_1)^{n}} $$ has indeed the proposed limit.