Restrictions:
- First and the last digit has to be
1
.- Contain no two consecutive
1's
and- Contain no three consecutive
0's
From my observation first two and last two digits will be fixed as 10
and 01
, because it can't be 11… or …11. So we got n-4
places left to fill. Apart from the first 2 and last 2 places if we add one more space then that space can only have 1
, not 0
. If we add one more place then there are two possible way (100101, 101001
). So that's what I got
For
n=6 2 ways
n=7 2 ways
n=8 3 ways
n=9 4 ways
I don't know if I am doing it right or not and what's the easy way to find the number of ways for this scenario rather than counting.
I would appreciate some guidance for a general rule of a binary string of length n
that do not contain 3 consecutive 0's
and 2 consecutive 1's
while starting and ending with 1
Best Answer
Ignoring the $1$ at the front, any such sequence can be broken into blocks of $01$ and $001$. Let $k$ be the number of blocks of $01$. Then the number of blocks of $001$ will be $(n-1-2k)/3$, which is only possible when this number is an integer. This means there are $(n-1-2k)/3+k=(n+k-1)/3$ blocks total. We must choose $k$ of these blocks to be $01$, which can be done in $\binom{(n+k-1)/3}k$ ways. Therefore, the number of sequences is $$ \sum_{\substack{k=0}}^{\lfloor (n-1)/2\rfloor}\binom{\frac{n+k-1}3}{k} $$ where the summation only ranges over $k$ for which $n+k-1$ is an integer multiple of $3$.
It is also possible to get a "closed form" by solving the linear recurrence $$ a_n=a_{n-2}+a_{n-3},\\ a_2=0,\\ a_1=1,\\ a_0=0. $$ To solve this, see for example:
https://en.wikipedia.org/wiki/Recurrence_relation#Roots_of_the_characteristic_polynomial.