TITLE: A number $N$ is a $k$-nacci number if and only if ...
QUESTION [10 upvotes]: For $k\ge 2\in\mathbb N$, one can define the $n$-th $k$-nacci number $f_k(n)\ (n=0,1,\cdots)$ as
$$f_k(0)=f_k(1)=\cdots=f_{k}(k-2)=0,\ \ f_{k}(k-1)=1,$$$$f_{k}(n+k)=f_{k}(n)+f_k(n+1)+\cdots+f_{k}(n+k-2)+f_{k}(n+k-1)\ \ \ (n=0,1,2,\cdots)$$
Examples :

$f_2(n)$ (Fibonacci) : $0,1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,\cdots$
$f_3(n)$ (Tribonacci) : $0,0,1,1,2,4,7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136,5768,\cdots$
$f_4(n)$ (Tetranacci) : $0, 0, 0, 1, 1, 2, 4, 8, 15, 29, 56, 108, 208, 401, 773, 1490, 2872, 5536,\cdots$

Here is my question. Let us say "$N$ is $k$-nacci" if a number $N$ is a $k$-nacci number.

Question : For $k\ge 3$, how can we fill in the following blank to make the proposition true?
$$\text{A number $N$ is $k$-nacci if and only if $(\ \ \ \ \ \ \ \ \ \ )$.}$$

Any answer is OK as long as one can prove that the proposition is true. Can anyone help?
For $k=2$, we know that the following proposition is true (see here, here):
$$\text{A number $N$ is Fibonacci if and only if either $5N^2-4$ or $5N^2+4$ is a perfect square.}$$
Added : A user Tito Piezas III asked a related question.

REPLY [4 votes]: Note: Sequences start with $n=0$.

I. $k = 2$

The Lucas numbers,
$$L_n = x_1^n+x_2^n = 2,1,3,4,7,11,18,29,\dots$$
and the Fibonacci numbers,
$$F_n = \frac{x_1^n}{y_1}+\frac{x_2^n}{y_2} = 0,1, 1, 2, 3, 5, 8, 13, 21,\dots$$
where $y_i =2x_i-1$ and the $x_i$ are the roots of $x^2-x-1=0$. Then,
$$x_1^n = \big(\tfrac{1+\sqrt{5}}{2}\big)^n=\tfrac{1}{2}(L_n+F_n\sqrt{5})$$
$$x_2^n = \big(\tfrac{1-\sqrt{5}}{2}\big)^n=\tfrac{1}{2}(L_n-F_n\sqrt{5})$$
Hence,
$$(x_1 x_2)^n = (-1)^n = \tfrac{1}{4}(L_n^2-5F_n^2)$$

II. $k = 3$

(Edited.) Given three sequences with recurrence $s_n = s_{n-1}+s_{n-2}+s_{n-3}$ but different initial values as,
$$\begin{array}{|c|c|c|c|c|c|c|c|c|}
\hline
\text{Name} & \text{Formula} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & OEIS\\
\hline
R_n & x_1^n+x_2^n+x_3^n &3 &1 &3 &7 &11 &21 &39 & 71 & A001644 \\
\hline
S_n &\frac{x_1^n}{y_1}+\frac{x_2^n}{y_2}+\frac{x_3^n}{y_3}&3 &2 &5 &10 &17 &32 &59 &108 &(none)\\
\hline
T_n &\frac{x_1^n}{z_1}+\frac{x_2^n}{z_2}+\frac{x_3^n}{z_3}&0 &1 &1 &2 &4 &7 &13 &24& A000073 \\
\hline
\end{array}$$
$$y_i =\tfrac{1}{19}(x_i^2-7x_i+22)$$ 
$$z_i =-x_i^2+4x_i-1$$ 
and the $x_i$ are the roots of $x^3-x^2-x-1=0$, with $T_n$ as the tribonacci numbers and the real root $T = x_1 \approx 1.83929$ the tribonacci constant. Define,
$$a=\tfrac{1}{3}(19+3\sqrt{33})^{1/3},\quad b=\tfrac{1}{3}(19-3\sqrt{33})^{1/3}$$ 
then we have the nice identity by Pin-Yen Lin,
$$3T^n  = R_{n}+(a+b)S_{n-1}+3(a^2+b^2)T_{n-1}$$
and similar expressions for the complex conjugates $x_2^n$ and $x_3^n$ (after correcting some typos in the paper). Hence it is possible to express,
$$(-x_1x_2x_3)^n = (1)^n = \text{in terms of}\; R_{n},\, S_{n-1},\, T_{n-1}$$
analogous to the $k=2$ case, and get rid of irrationalities. For some reason, Lin didn't bring it to this step so we find it with the help of Mathematica. Since $R_n,\;S_n$ can be expressed as sums of $T_m$, define,
$$a,\;b,\;c = T_{n-1},\;T_{n-2},\;T_{n-3}$$
then we get the diophantine cubic Pell-type equation for the tribonacci numbers,
$$a^3 - 2 a^2 b + 2 b^3 - a^2 c - 2 a b c + 2 b^2 c + a c^2 + 2 b c^2 + c^3=1$$
with an infinite number of integer solutions analogous to the Pell equation for the Fibonacci numbers,
$$p^2-5q^2 = \pm 4.$$

III. $k = 4$

Define sequence A073817,
$$U_n = x_1^n+x_2^n+x_3^n+x_4^n = 4, 1, 3, 7, 15, 26, 51, 99, 191,\dots$$
and the tetranacci numbers,
$$V_n = \frac{x_1^n}{y_1}+\frac{x_2^n}{y_2}+\frac{x_3^n}{y_3}+\frac{x_4^n}{y_4} =0,1, 1, 2, 4, 7, 13, 24, 44, 81,149\dots$$
where $y_i =-x_i^3 + 6x_i - 1$ and the $x_i$ are the roots of $x^4-x^3-x^2-x-1=0$. Then,
$$\text{(insert relationship here)}$$