$\sqrt{3}$ represented as continued fraction

The numerators should be all $1$. The procedure is to write $x>0$ as $$ x=x_0+y_0 $$ with $0\le y_0<1$. If $y_0=0$, we're done. Otherwise we set $$ \frac{1}{y_0}=x_1+y_1 $$ in the same fashion and go on with the same rule.

In your case, $y_0=\sqrt{3}-1$, so $$ \frac{1}{y_0}=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}= 1+\frac{\sqrt{3}-1}{2} $$ Therefore $x_1=1$ and $y_1=\frac{\sqrt{3}-1}{2}$. Then $$ \frac{1}{y_1}=\frac{2}{\sqrt{3}-1}=\sqrt{3}+1=2+(\sqrt{3}-1) $$ Hence $x_2=2$ and $y_2=\sqrt{3}-1$.

OK, now we'll go on forever with the same numbers. Hence $x_n=1$ for odd $n$ and $x_n=2$ for even $n>0$: $$ \sqrt{3}=[1;1,2,1,2,\dots]=1+ \cfrac{1}{ 1+\cfrac{1}{ 2+\cfrac{1}{ 1+\cfrac{1}{ 2+\ddots } } } } $$

See http://planetmath.org/tableofcontinuedfractionsofsqrtnfor1n102