Multiplibonacci #2

Here are the first few numbers in the Multiplibonacci sequence ${ _{ 2 }{ M } }$ : $1,2,2,{ 2 }^{ 2 },{ 2 }^{ 3 },{ 2 }^{ 5 },{ 2 }^{ 8 }...$ As you might have noticed, the powers in ${ _{ 2 }{ M } }$ are Fibonacci numbers! This means that $\log _{ 2 }{ ({ _{ 2 }{ M }_{ n } }) }$ is always going to be a Fibonacci number.

If you're careful, you might also notice that ${ _{ 2 }{ M } }$ starts with ${ 2 }^{ 0 }$ instead of ${ 2 }^{ 1 }$ . This means that the power in the $n$ th Multiplibonacci number is always the $n$ th Fibonacci number minus 1.

Combining the two facts, we get that $\log _{ 2 }{ ({ _{ 2 }{ M }_{ n } }) }$ is always going to be the $n-1$ st Fibonacci number. Because $n$ is 20 in the problem, we get that $\log _{ 2 }{ ({ _{ 2 }{ M }_{ 20 } }) }$ is the 19th Fibonacci number, which is $\boxed { 4181 }$ .