Lucas sequence equivalent for the tribonacci sequence? The Fibonacci and Lucas sequences occur within each other's identities, i.e.
$$F_{2n} = F_{n} * (F_{n-1} + F_{n+1})$$
$$L_{n} = F_{n-1} + F_{n+1}$$
$$F_{2n} = F_{n} * L_{n}$$
The Lucas sequence contains the identity
$$L_{2n} = L_{n}^2 + 2(-1)^n$$
Is there an analog of this relationship in the tribonacci sequence? Specifically, is there a sequence of numbers that are the sums of tribonnaci numbers adjacent to $T_{n}$ which may be similarly squared (cubed?) and offset by a constant that will result in the sum of tribonacci numbers adjacent to $T_{2n}$ (or $T_{3n}$)?
 A: One of the easiest ways to prove the Fibonacci doubling identities is to prove by induction (or combinatorially) that the Fibonacci numbers satisfy
$$\left[ \begin{array}{cc} 1 & 1 \\ 1 & 0 \end{array} \right]^n = \left[ \begin{array}{ccc} F_{n+1} & F_n \\ F_n & F_{n-1} \end{array} \right].$$
(We either need to take $n \ge 1$ or we need to define $F_{-1} = 1$ to make this true for $n = 1$ but either way is fine.) The doubling identities follow from just squaring this matrix. Note that because $L_n = F_{n-1} + F_{n+1}$ (which is the trace of the above matrix) the second doubling identity can just be written $F_{2n} = F_n (F_{n-1} + F_{n+1})$, without mentioning the Lucas numbers at all, which is the way it appears if you square the above matrix.
There's an analogous but more complicated matrix identity for the Tribonacci numbers $T_n$ which goes
$$\left[ \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array} \right]^n = \left[ \begin{array}{ccc} T_{n+2} & T_{n+1} + T_n & T_{n+1} \\ T_{n+1} & T_n + T_{n-1} & T_n \\ T_n & T_{n-1} + T_{n-2} & T_{n-1} \end{array} \right]$$
(defining $T_{-1} = 1$ and $T_{-2} = -1$) and squaring this identity gives doubling formulas for the Tribonacci numbers. You can also cube either of these to get tripling formulas if you like. I don't know how many distinct identities you get but, for example, considering only the top left entry of the square gives
$$T_{2n+2} = T_{n+2}^2 + T_{n+1}(T_{n+1} + T_n) + T_n(T_{n+1})$$
which probably can be rearranged a bunch. Secretly I am thinking about all of this in terms of counting walks on a certain directed graph.
The relationship betwen the Fibonacci numbers and the Lucas numbers happens to be especially nice in ways that are specific to the case of a second-order recurrence relation and don't generalize. You can see this from their explicit formulas $F_n = \frac{\phi^n - \varphi^n}{\phi - \varphi}$ and $L_n = \phi^n + \varphi^n$, which tell us that the identity $F_{2n} = F_n L_n$ follows from the difference of squares identity $a^2 - b^2 = (a - b)(a + b)$. Unfortunately the Tribonacci $T_{3n}$ is not a difference of cubes (although $F_{3n}$ is); $T_n$ has an explicit formula involving a sum of three exponentials, corresponding to the three different roots of the characteristic polynomial $\lambda^3 = \lambda^2 + \lambda + 1$, and I don't know what it is off the top of my head. You can check that $T_5 = 4$ doesn't divide $T_{15} = 1705$ so $\frac{T_{3n}}{T_n}$ isn't an integer in general.
On the other hand, the existence of matrices like these describing the solutions to linear recurrence relations does generalize.
