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This is a somewhat different proof of Joro's claim that $\lfloor(\phi^6)^n\rfloor$ is odd for all $n$, where $\phi$ is the golden ratio. Maybe it will shed some additional light.

Let $F_n$ be the Fibonacci sequence $0,1,1,2,\ldots$. Let $c=\phi^6$ and let $d=\bar{\phi}^6$, where $\bar{\phi}$ is the $\mathbb{Q}$-conjugate of $\phi$. The equation $\phi^n=F_{n-1}+\phi F_n$ is easily verified by induction, and it implies that $$c^n=F_{6n-1}+\phi F_{6n}.$$ Taking $\mathbb{Q}$-conjugates, we obtain $$d^n=F_{6n-1}+\bar{\phi} F_{6n}.$$

Adding the two equations, $$c^n+d^n=2F_{6n-1}+F_{6n}.$$

Now it is well known that $F_n$ is a divisibility sequence, meaning that if $m\mid n$ then $F_n\mid F_m$. Since $F_6$ is even, so is $F_{6n}$. Therefore $2F_{6n-1}+F_{6n}$ is an even integer. On the other hand, $d$ is between zero and 1, therefore $c^n=2F_{6n-1}+F_{6n}-d^n$ has floor an odd integer, as required.

This is a somewhat different proof of Joro's claim that $\lfloor(\phi^6)^n\rfloor$ is odd for all $n$, where $\phi$ is the golden ratio. Maybe it will shed some additional light.

Let $F_n$ be the Fibonacci sequence $0,1,1,2,\ldots$. Let $c=\phi^6$ and let $d=\bar{\phi}^6$, where $\bar{\phi}$ is the $\mathbb{Q}$-conjugate of $\phi$. The equation $\phi^n=F_{n-1}+\phi F_n$ is easily verified by induction, and it implies that $$c^n=F_{6n-1}+\phi F_{6n}.$$ Taking $\mathbb{Q}$-conjugates, we obtain $$d^n=F_{6n-1}+\bar{\phi} F_{6n}.$$

Adding the two equations, $$c^n+d^n=2F_{6n-1}+F_{6n}.$$

Now it is well known that $F_n$ is a divisibility sequence, meaning that if $m\mid n$ then $F_n\mid F_m$. Since $F_6$ is even, so is $F_{6n}$. Therefore $2F_{6n-1}+F_{6n}$ is an even integer. On the other hand, $d$ is between 0 and 1, therefore $c^n=2F_{6n-1}+F_{6n}-d^n$ has floor an odd integer, as required.

This is a somewhat different proof of Joro's claim that $\lfloor(\phi^6)^n\rfloor$ is odd for all $n$, where $\phi$ is the golden ratio. Maybe it will shed some additional light.

Let $F_n$ be the Fibonacci sequence $0,1,1,2,\ldots$. Let $c=\phi^6$ and let $d=\bar{\phi}^6$, where $\bar{\phi}$ is the $\mathbb{Q}$-conjugate of $\phi$. The equation $\phi^n=F_{n-1}+\phi F_n$ is easily verified by induction, and it implies that $$c^n=F_{6n-1}+\phi F_{6n}.$$ Taking $\mathbb{Q}$-conjugates, we obtain $$d^n=F_{6n-1}+\bar{\phi} F_{6n}.$$

Adding the two equations, $$c^n+d^n=2F_{6n-1}-F_{6n}.$$

Now it is well known that $F_n$ is a divisibility sequence, meaning that if $m\mid n$ then $F_n\mid F_m$. Since $F_6$ is even, so is $F_{6n}$. Therefore $2F_{6n-1}-F_{6n}$ is an even integer. On the other hand, $d$ is between zero and 1, therefore $c^n=2F_{6n-1}-F_{6n}-d^n$ has floor an odd integer, as required.

This is a somewhat different proof of Joro's claim that $\lfloor(\phi^6)^n\rfloor$ is odd for all $n$, where $\phi$ is the golden ratio. Maybe it will shed some additional light.

Let $F_n$ be the Fibonacci sequence $0,1,1,2,\ldots$. Let $c=\phi^6$ and let $d=\bar{\phi}^6$, where $\bar{\phi}$ is the $\mathbb{Q}$-conjugate of $\phi$. The equation $\phi^n=F_{n-1}+\phi F_n$ is easily verified by induction, and it implies that $$c^n=F_{6n-1}+\phi F_{6n}.$$ Taking $\mathbb{Q}$-conjugates, we obtain $$d^n=F_{6n-1}+\bar{\phi} F_{6n}.$$

Adding the two equations, $$c^n+d^n=2F_{6n-1}+F_{6n}.$$

Now it is well known that $F_n$ is a divisibility sequence, meaning that if $m\mid n$ then $F_n\mid F_m$. Since $F_6$ is even, so is $F_{6n}$. Therefore $2F_{6n-1}+F_{6n}$ is an even integer. On the other hand, $d$ is between zero and 1, therefore $c^n=2F_{6n-1}+F_{6n}-d^n$ has floor an odd integer, as required.

This is a somewhat different proof of Joro's claim that $\lfloor(\phi^6)^n\rfloor$ is odd for all $n$, where $\phi$ is the golden ratio. Maybe it will shed some additional light.

Let $F_n$ be the Fibonacci sequence $0,1,1,2,\ldots$. Let $c=\phi^6$ and let $d=\bar{\phi}^6$, where $\bar{\phi}$ is the $\mathbb{Q}$-conjugate of $\phi$. The equation $\phi^n=F_{n-1}+\phi F_n$ is easily verified by induction, and it implies that $$c^n=F_{6n-1}+\phi F_{6n}.$$ Taking $\mathbb{Q}$-conjugates, we obtain $$d^n=F_{6n-1}+\bar{\phi} F_{6n}.$$

Adding the two equations, $$c^n+d^n=2F_{6n-1}-F_{6n}.$$

Now it is well known that $F_n$ is a divisibility sequence, meaning that if $m\mid n$ then $F_n\mid F_m$. Since $F_6$ is even, so is $F_{6n}$. Therefore $2F_{6n-1}-F_{6n}$ is an even integer. On the other hand, $d$ is between zero and 1, therefore $c_n=2F_{6n-1}-F_{6n}-d^n$ has floor an odd integer, as required.

This is a somewhat different proof of Joro's claim that $\lfloor(\phi^6)^n\rfloor$ is odd for all $n$, where $\phi$ is the golden ratio. Maybe it will shed some additional light.

Let $F_n$ be the Fibonacci sequence $0,1,1,2,\ldots$. Let $c=\phi^6$ and let $d=\bar{\phi}^6$, where $\bar{\phi}$ is the $\mathbb{Q}$-conjugate of $\phi$. The equation $\phi^n=F_{n-1}+\phi F_n$ is easily verified by induction, and it implies that $$c^n=F_{6n-1}+\phi F_{6n}.$$ Taking $\mathbb{Q}$-conjugates, we obtain $$d^n=F_{6n-1}+\bar{\phi} F_{6n}.$$

Adding the two equations, $$c^n+d^n=2F_{6n-1}-F_{6n}.$$

Now it is well known that $F_n$ is a divisibility sequence, meaning that if $m\mid n$ then $F_n\mid F_m$. Since $F_6$ is even, so is $F_{6n}$. Therefore $2F_{6n-1}-F_{6n}$ is an even integer. On the other hand, $d$ is between zero and 1, therefore $c^n=2F_{6n-1}-F_{6n}-d^n$ has floor an odd integer, as required.