Recursion is basically an idea of connecting any term with the next or previous term of an series. The most famous example of recursion is Fibonacci series \(0,1,1,2,3,5,8,........\) its recursion formula is \(t_{n+2}=t_{n+1}+t_n\) for natural number n.
How many sequences of $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?
$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$
AMC 10B, 2019 Problem 25
Recursion
6 out of 10
challenges and thrills of pre college mathematics
We can deduce, from the given restrictions, that any valid sequence of length $n$ will start with a $0$ followed by either $10$ or $110$. Thus we can define a recursive function $f(n) = f(n-3) + f(n-2)$, where $f(n)$ is the number of valid sequences of length $n$.
This is because for any valid sequence of length $n$, you can append either $10$ or $110$ and the resulting sequence will still satisfy the given conditions.
It is easy to find $f(5) = 1$ with the only possible sequence being $01010$ and $f(6) = 2$ with the only two possible sequences being $011010$ and $010110$ by hand, and then by the recursive formula, we have \(f(19)=65\) so option C is correct option.
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