Here are some results which *suggest* that perhaps what happens is neither very simple nor very random. **But** see the end for a better result.

These are some of the simplest continued fractions and what they lead to. This also tells you what results in simple continued fractions because $r-\frac12$ and $r+\frac12$ have the same continued fraction after the integer part.

The tenth row says that $$r=\frac{1+\sqrt3}{2}=1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{2+\cdots}}}}}$$ and
$$r+\frac{1}{2}=1+\frac{\sqrt3}{2}=1+\cfrac{1}{1+\cfrac{1}{6+\cfrac{1}{2+\cfrac{1}{6+\cfrac{1}{2+\cdots}}}}}$$

Each line is a number $r$ then the continued fraction of $r$ then the continued fraction of $r+\frac12$$$ \begin {array}{ccc} \frac{1+\sqrt5}{2}&[[1]]&[[2,8]]
\\1+\sqrt {2}&[[2]]&[[2],[1,10,1,1]]
\\ \frac{3+\sqrt{13}}{2}&[[3]]&[[3],[1,4,14,4,1,2]]
\\2+\sqrt {5}&[[4]]&[[4],[1,2,1,3]]
\\ \frac{5+\sqrt{29}}{2}&[[5]]&[[5],[1,2,3,1,20,1,3
,2,1,4]]
\\3+\sqrt {10}&[[6]]&[[6],[1,1,1,24,1,1,
1,5]]
\\ \frac{7+\sqrt{53}}{2}&[[7]]&[[7],[1,1,1,3,1
,1,28,1,1,3,1,1,1,6]]
\\4+\sqrt {17}&[[8]]&[[8],[
1,1,1,1,1,7]]
\\ \frac{9+\sqrt{85}}{2}&[[9]]&[[9],[1
,1,1,1,3,2,36,2,3,1,1,1,1,8]]
\\ \frac{1+\sqrt{3}}{2}&[[1,2]]
&[[1,1],[6,2]]
\\ \frac{3+\sqrt{21}}{6}&[ [1,3]]&[[1,
1],[3,4,3,2]]
\\ \frac{1+\sqrt{2}}{2}&[ [1,4]]&[[1,1]
,[2]]
\\ \frac{5+3\sqrt{5}}{10}&[ [1,5]]&[[1,1],[2,26,
2,2]]
\\ \frac{3+\sqrt{15}}{6}&[ [1,6]]&[[1,1],[1,1,4
,1,1,2]]
\\ \frac{7+\sqrt{77}}{14}&[ [1,7]]&[[1,1],[1
,1,2,8,2,1,1,2]]
\\ \frac{2+\sqrt{6}}{4}&[ [1,8]]&[[1
,1],[1,1,1,2]]
\\ \frac{3+\sqrt{13}}{6}&[ [1,9]]&[[1,
1],[1,1,1,42,1,1,1,2]]
\\1+\sqrt {3}&[ [2,1]]&[
[3,4]]
\\ 1+ \frac{\sqrt{15}}{3}&[ [2,3]]&[[2],[1,3,1,3,1,1]]
\\ 1+ \frac{\sqrt{6}}{2}&[ [2,4]]&[[2],[1,2,1,1]]
\\ 1+ \frac{\sqrt{35}}{5}&[ [2,5]]&[[2],[1,2,6,2,1,1]]
\\ 1+ \frac{2\sqrt{3}}{3}&[ [2,6]]&[[2],[1,1,1,8,1,1,1,1
]]
\\ 1+ \frac{3\sqrt{7}}{7}&[ [2,7]]&[[2],[1,1,1,2,1,2,1
,1,1,1]]
\\ 1+ \frac{\sqrt{5}}{2}&[ [2,8]]&[[2],[1]]
\\ 1+ \frac{\sqrt{11}}{3}&[ [2,9]]&[[2],[1,1,1,1,6,1,1,
1,1,1]]\end {array} $$

I also looked at the first few results of the form $[[3,j]]$ which seemed similar. Rational numbers might also be worth a look.

**later**
as Douglas points out. There are patterns which the tables above are just slightly too brief to show.

The continued fraction $[[k]]$ corresponds to $\frac{k+\sqrt{k^2+4}}{2}$ and we have for $k \ge 2$ (and in some cases for $k \ge 1$)

$2k+\sqrt{(2k)^2+1} \hspace{0.5in} [[4k]] \hspace{0.5in} [[4k],[1,1,k-1,1,1,4k-1]]$

$\frac{4k+1+\sqrt{(4k+1)^2+4}}{2} \hspace{0.5in} [[4k+1]] \hspace{0.5in} [[4k+1],[1,1,k-1,1,3,k,16k+4,k,3,1,k-1,1,1,4k]]$

$2k+1+\sqrt{(2k+1)^2+1} \hspace{0.5in} [[4k+2]] \hspace{0.5in} [[4k+2],[1,1,k,16k+8,k,1,1,4k+1]]$

$\frac{4k+3+\sqrt{(4k+3)^2+4}}{2} \hspace{0.5in} [[4k+3]] \hspace{0.5in} [[4k+3],[1,1,k,3,1,k,16k+12,k,1,3,1,1,4k+2 ]]$

Similar things happen for $[[i,k]]$ $i=1,2$ depending on the congruence class $k \mod 4$ for $k$ not too small.