Showing that the tribonacci sequence has relatively prime terms

Question

Let $f_k$ denote the Fibonacci numbers. It turns out that $gcd(f_k,f_{k+1})=1$ for all $k\ge 1$ and I understand this proof. If one defines the tribonacci numbers by $t_1=t_2=t_3=1$ and $t_{k+3}=t_{k+2}+t_{k+1}+t_k$, can one similarly prove that $gcd(t_k,t_{k+1}),t_{k+2})=1$ for all $k$?

Answer 1

As darij grinberg says in the comments, the property does hold for the tribonacci numbers as you define them.

To see this, we show by induction that if at any point in the sequence, $\gcd(t_k,t_{k+1},t_{k+2})=1$, then it will always be true that $\gcd(t_k,t_{k+1},t_{k+2})=1$ for all $k$. Since $\gcd(1,1,1)=1$, the base case is satisfied for the tribonacci numbers.

Suppose $\gcd(t_{k-1},t_k,t_{k+1})=1$. Then $\gcd(t_k,t_{k+1},t_{k+2})=\gcd(t_k,t_{k+1},t_{k-1}+t_k+t_{k+1})$. If $p$ is any number that divides $t_k$, $t_{k+1}$, and $t_{k-1}+t_k+t_{k+1}$, then it also divides $(t_{k-1}+t_k+t_{k+1}) - t_k - t_{k+1} = t_{k-1}$. Then $p$ divides $\gcd(t_{k-1},t_k,t_{k+1})=1$, so $\gcd(t_k,t_{k+1},t_{k+2})=1$.

Answer 2

We have $$ \begin{pmatrix}t_{n+3} \\ t_{n+2} \\ t_{n+1} \end{pmatrix} =\begin{pmatrix}1&1&1\\1&0&0\\0&1&0\end{pmatrix} \begin{pmatrix}t_{n+2} \\ t_{n+1} \\ t_{n} \end{pmatrix} $$ The matrix has determinant $1$ and so is invertible over the integers.

Therefore, the common divisors of $t_{n+3}, t_{n+2}, t_{n+1}$ are exactly the same as the common divisors of $t_{n+2}, t_{n+1}, t_{n}$.