Question about a recursively defined ordered triple sequence

Question

Let $(x_n, y_n, z_n)$ be the sequence of ordered triples recursively defined by $(x_0, y_0, z_0)=(0,1,2)$ and for any $n$ in the Naturals, $(x_{n+1}, y_{n+1}, z_{n+1})=(x_n-y_n, y_n-z_n, z_n-x_n)$ and let $S_n=(x_n)^2+(y_n)^2+(z_n)^2$.

What are the values of $a$, if any, for which $(S_{a+5}-S_{a+4}+S_{a+3}+S_{a+1})^{1/7}$ is a natural number?

What is the minimum value of $b$, if any, for which $\displaystyle{\sum_{k=0}^b \frac{2^k}{S_k}}>0.199$?

Answer 1

Let’s first calculate some first terms. Notice that

• $(x_1,y_1,z_1)=(-1,-1,2)$
• $(x_2,y_2,z_2)=(0,-3,3)$
• $(x_3,y_3,z_3)=(3,-6,3)$
• $(x_4,y_4,z_4)=(9,-9,0)$
• $(x_5,y_5,z_5)=(18,-9,-9)$
• $\ldots$

With this, we can notice pretty quickly by induction that $$(x_{n+2},y_{n+2},z_{n+2})=(-3y_n,-3z_n,-3x_n),$$ and that in particular, $$S_{n}=2\cdot3^n.$$ The value you want to calculate is therefore, quite conveniently, $$\left(S_{a+5}-S_{a+4}+S_{a+2}+S_{a+1}\right)^{\frac{1}{7}}$$ $$=\left(2\cdot3^{a+1}\cdot\left(3^4-3^3+3^2+3^0\right)\right)^{\frac{1}{7}}$$ $$=\left(2^7\cdot3^{a+1}\right)^{\frac{1}{7}}$$ $$=2\cdot3^{\frac{a+1}{7}}.$$ This will be an integer if and only if $\boxed{a\equiv6\pmod{7}}$.