Is there a formula for the pattern $1,1,1,0,1,0,1,0,1,0\ldots$? Or a mathematical principle? Starting from $n= 0$:
$1,1,1,0,1,0,1,0,1,0\ldots$
The $0,1,0,1$ pattern keeps on going. Is there a formula for this? 
 A: Try this one:
\begin{align}
f(n)&=
1-\operatorname{sgn}(n-1)\cdot(n\bmod2)
\end{align}
Edit
Here 
\begin{align}
\operatorname{sgn}(n)&=
\begin{cases}
-1,\quad n<0\\
\phantom{-}0,\quad n=0\\
\phantom{-}1,\quad n>0
\end{cases}
.
\end{align}
The function $f$ works as follows: 
for all even $n\ge0$
we have $(n\bmod2)=0$, hence $f(n)=1-\operatorname{sgn}(n-1)\cdot0=1$;
for $n=1$ the term $\operatorname{sgn}(n-1)=0$, hence $f(n)=1-0\cdot(n\bmod2)=1$;
and for all odd $n>1$, 
 $f(n)=1-\operatorname{sgn}(n-1)\cdot(n\bmod2)=1-1\cdot1=0$.
Edit2
More exotic function that produces the same output as $f$ for $x=0,1,2,\dots$:

\begin{align}
g(x)&=
1+\sin \left( \tfrac\pi2\,\cos(\tfrac\pi2\,x) 
-\arctan(x-1) 
-\tfrac12\,\arcsin\left({\tfrac{2\,(x-1)}{{x}^{2}-2\,x+2}}\right)  
\right)
.
\end{align}
A: orlp's answer, ${1099\over9900} =0.111010101…$, uses decimal arithmetic to produce the desired sequence (as digits of a number).  A similar result, but using base 2, is as follows:
$$0.111010101…_2 = {1\over4}+0.101010101…_2= {1\over4}+ {1\over2}\sum\limits_{j=0}^\infty({1\over4})^j ={1\over4}+ {1\over2}\times {4\over3}= {11\over12}$$
That is, the binary representation of base-10 fraction ${11\over12}$ is $0.111010101…_2$.
A: The Online Encyclopedia of Integer Sequences has a few interesting results, including http://oeis.org/A266591:
"Middle column of the "Rule 37" elementary cellular automaton starting with a single ON (black) cell."
A: $$\frac{1099}{9900} = 0.111010101\dots$$
A: The formula of the sequence is
$$a_n=\begin{cases}
0, &\text{ if }n \text{ is odd and }n \neq 1 \\
1,&\text{ if }n \text{ is even or } n=1 \end{cases}.
$$
That's all. There is no simpler formula. You could replace "$\text{if }n \text{ is even or } n=1$" by "$\text{otherwise}$", if you prefer that notation. Trying to write this without a case split makes things unnecessary complicated. Interpreting the sequence as digits of a number, as some answers do, only makes sense if you know that the sequence actually represents digits of a number. The pattern is simply too basic to have a deeper meaning.
A: You could define it this way: $$n_k=\begin{cases}1, &\text{ if }k=0,1\\ \frac{1}{2}\left((-1)^k+1)\right),&\text{ if }k\ge 2\end{cases}.$$ Or if you want a completely closed formula, you could use $$n_k=\frac{1}{2}\left(1-\frac{1}{2}\big(1-(-1)^{k!}\big)\right)\big((-1)^k+1\big)+\frac{1}{2}\big(1-(-1)^{k!}\big).$$ 
I actually figured out a better formula. Try $$n_k=\frac{1}{2}\left(1+(-1)^k\right)+\left(1-\frac{1}{2}\left(1+(-1)^{k\cdot k!}\right)\right).$$ Note that the first term describes the sequence $1 , 0, 1, 0, \dots$ while the second term describes the sequence $0,1,0,0,0,0,\dots$. So, adding gives $1,1,1,0,1,0,1,0,\dots$.
