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# Second Fibonacci Sequence problem?

I changed the Fibonacci sequence so that the first two terms were 2. I only saw 4 powers of two in this sequence (2, 2, 4, and 16). Is there a fifth power of two in this pattern, and is there an infinite number of powers of two in this pattern?

Note by Phillip Williams
3 years, 1 month ago

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## Comments

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I got this formula where $$f_{0}=2$$ and $$f_{1}=2$$: $\dfrac{5+\sqrt{5}}{5}\left(\dfrac{1+\sqrt{5}}{2}\right)^n+\dfrac{5-\sqrt{5}}{5}\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ where $$n$$ represents the $$nth$$ term of the sequence.

- 3 years, 1 month ago

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Cool formula... How did you solve for that formula?

- 3 years, 1 month ago

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Since it follows the Fibonacci recurrence relation, the general form would be $a\left(\dfrac{1+\sqrt{5}}{2}\right)^n + b\left(\dfrac{1-\sqrt{5}}{2}\right)^n$ To get $$a$$ and $$b$$, you need to solve these two equations: $a+b=c_{0}$$a\left(\frac{1+\sqrt{5}}{2}\right)+b\left(\frac{1-\sqrt{5}}{2}\right)=c_{1}$ where $$c_{0}$$ is the first term and $$c_{1}$$ is the second term

- 3 years, 1 month ago

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I have no idea. I've only reached 2692538.

- 3 years, 1 month ago

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