Recurrence relationship pleased by $\lfloor(1+\sqrt{3})^n\rfloor$

This is an adhere to up to a question I had actually asked previously concerning a straight reappearance partnership satsified by $\lfloor(1+\sqrt{5})^n\rfloor$. I screwed up there, and also I in fact suggested to inquire about $L(n)=\lfloor(1+\sqrt{3})^n\rfloor$.

Adhering to Douglas' pointer I have actually established that the values (at the very least the first 1000) please the adhering to reappearance :

$L(2n+5)=8L(2n+3)-4L(2n+1)$

The inquiry is just how do I confirm something similar to this. I can confirm the reappearance for the values inside the flooring function, yet flooring function as a whole does not commute with enhancement and also reproduction.

Clearly, it is very easy to show

$(1+\sqrt{3})^{2n+5}=8(1+\sqrt{3})^{2n+3}-4(1+\sqrt{3})^{2n+1}$

yet I am not exactly sure just how to confirm the reappearance from below.

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2019-12-02 02:54:32
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Answers: 1

Prove the very same relationship for $1-\sqrt 3$, and also examine that the resulting powers are all smaller sized than 1, all adverse, and also when included in the very same powers of $1+\sqrt 3$ you wind up with an integer : For all $k=1,2,\dots$, $$(1+\sqrt 3)^k+(1-\sqrt 3)^k $$ is an integer.

There are numerous means of examining this reality. As an example, by induction. Or by utilizing the Binomial theory.

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2019-12-03 04:25:53
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