Super Sequence

Algebra Level 4

A sequence, $S_{n}$ is defined such that $S_{1}=1, S_{2}=2$ and $S_{n}=S_{n-1} \times S_{n-2}$ for all $n>2$ where $n$ is a positive integer.

Which of these are the correct form for $S_{n}$ true for all values of $n>2$ ?

Notations:

$f(n)$ denotes the $n^\text{th}$ Fibonacci number , where $f(0) = 0, f(1) = 1$ and $f(n) = f(n-1) + f(n-2)$ for $n=2,3,4,\ldots$ .
$!$ is the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

2^{n!}

f(n+1)

2^{n-1}

2^{f(n-1)}

2^{f(n)}

2 solutions

William Whitehouse
Feb 25, 2017

Taking logs base 2 of the statements given shows that $log_{2}(S_{1})=0$ , $log_{2}(S_{2})=1$ and $log_{2}(S_{n})=log_{2}(S_{n-1})+log_{2}(S_{n-2})$ .

Calculating the third term using this formula shows that $log_{2}(S_{3})=1$ and from this point on the log to base two of the sequence is the fibonacci numbers. The first fibonacci number appears at n=2 so the offset is 1. In other words $log_{2}(S_{n})=f(n-1)$ and so $S_{n}=2^{f(n-1)}$

Chew-Seong Cheong
Mar 1, 2017

Note that:

$\begin{aligned} S_1 & = 1 = 2^0 = 2^{f(0)} \\ S_2 & = 2 = 2^1 = 2^{f(1)} \\ S_3 & = S_1S_2 = 2^{f(0)+f(1)} = 2^{f(2)} \\ S_4 & = S_2S_3 = 2^{f(1)+f(2)} = 2^{f(3)} \\ ... & = \ ... \\ S_n & = S_{n-2}S_{n-1} = \boxed{2^{f(n-1)}} \end{aligned}$

Super Sequence

2 solutions

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