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I have been playing around with Fibonacci-type of sequence that involve complex numbers. I have stumbled upon the following sequence, which seemed interesting to me: $$0,1,2i,-3,-4i,5,6i,...$$ so $F_n = 2iF_{n-1} + F_{n-2}$. These look like a sequence of natural numbers (except for $0$) where every other is multiplied by $i$ and the signs change after two sequences.

I understand the algebra behind the above sequence, but I have been wondering whether there is an intuition behind why the sequence looks like a "modified" sequence of natural numbers.

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    $\begingroup$ $(X-i)^2=X^2-(2iX+1)$. $\endgroup$
    – Angina Seng
    Jul 20, 2018 at 14:50
  • $\begingroup$ Isn't this just $n\cdot i^{n-1}$? $\endgroup$
    – mechanodroid
    Jul 20, 2018 at 14:54
  • $\begingroup$ Also related: (1), (2), (3), (4), and so on. $\endgroup$
    – Xander Henderson ♦
    Jul 24, 2018 at 20:27

1 Answer 1

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The characteristic polynomial for your recursion is $$x^2-2ix-1=(x-i)^2$$

Visibly, this has a double root at $x=i$. Thus the general form of the solution to the recursion is $$F_n=Ai^n+Bni^n$$ Using your initial conditions it is easy to specify the solution to your case.

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