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?

#FibonacciNumbers #Fibonacci

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

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

Phillip Williams - 6 years, 6 months ago

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.

Marc Vince Casimiro - 6 years, 6 months ago

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

Phillip Williams - 6 years, 6 months ago

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

Marc Vince Casimiro - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$