Must a certain continued fraction have "small" partial quotients? I have reformulated the original question, which appears at the bottom, it a way that seems more likely to produce a reference.
New version: Let $\Delta$ be a positive nonsquare integer congruent to $0$ or $1$ modulo $4$.  If $\Delta$ is even, expand $\frac{\sqrt{\Delta}}{2}$ in a simple continued fraction.  If $\Delta$ is odd, expand $\frac{\sqrt{\Delta}+1}{2}$ in a simple continued fraction.  Let $m$ be the greatest integer with $m \equiv \Delta \pmod{2}$ and $m^2 < \Delta$. Is the following claim true?
Claim: The continued fraction expansion consists of $\left\lceil \frac{m}{2} \right\rceil$ followed by a purely periodic sequence.  The repeating block has final quotient $m$, but all other quotients do not exceed $\frac{m}{2}$.  
What surprises me is the bound on the other quotients.  When $\mathbb{Q}(\sqrt{\Delta})$ has large fundamental unit, the repeating block will either need to be long or contain some big quotients.  It seems something is forcing the quotients to stay small, putting everything into the length instead.
Is it known if the claim always holds?  Is there a proof somewhere?  
** Original post follows: **  Let $(t,u)$ be the fundamental solution to the Pellian equation (when solvable):  $t^2 - \Delta u^2 = -4$.  Otherwise, let $(t,u)$ be the fundamental solution to $t^2 - \Delta u^2 = 4$.  Data shows that these solutions can fluctuate quite wildly, and some can be "very large" in some sense compared to $\Delta$.  
Set $r = \frac{t-u}{2u}$ if $\Delta$ is odd and $r = \frac{t}{2u}$ if $\Delta$ is even.  Expand $r$ in the simple continued fraction with an odd number of quotients if $t^2 - \Delta u^2 = -4$ was solvable and with an even number of partial quotients otherwise.  Using the theorem of Hermite and Serret, it can be shown that after dropping the first partial quotient, the remaining ones form a symmetric sequence.  What surprises me is the following observation, which I have found to hold numerically for small discriminants:
Observation:
Let $m$ be the greatest integer with $m \equiv \Delta \pmod{2}$ and $m^2 < \Delta$.  Then the partial quotients of the continued fraction expansion of $r$ do not exceed $m/2$.  
An example:  when $\Delta = 193$, the equation $t^2 - 193 u^2 = -4$ is solvable with fundamental solution $(t,u) = (3528264,253970)$.  Also, $m = 13$, and $r = 1637147/253970$.  The continued fraction expansion has partial quotients $(6,2,4,6,1,2,1,1,1,1,2,1,6,4,2)$, which are all less than or equal to $m/2$.
The observation surprised me because when $t$ and $u$ swing up to large values,   $r$ can be quite large.  There seems to be something forcing the continued fraction toward having great length rather than large partial quotients.  
Is it known if the observation always holds?  Is there a proof somewhere?  
 A: Let $\epsilon = \Delta \mod 2 \in \{0 ; 1\}$ and $t = \frac 12 (\sqrt \Delta +\epsilon)$. $t$ is a quadratic algebraic number and it satisfies the equation $t^2 - \epsilon t - \frac 14(\Delta-\epsilon)$.
Consider $t$ together with his conjugate $\overline t < 0$ (because $\Delta>1$) and look at what happens during the continued fraction process.
If $(t,\overline t$) are solutions to equations of the form $ax^2+bx+c=0$ with $ac =a^2t\overline t<0$ and $b^2-4ac = a^2(t-\overline t)^2 = \Delta$,
then when you shift them by $k = \lfloor t \rfloor$, this has the effect of replacing $b$ with $b+2ka$ and $c$ with $ak^2+bk+c$. But since $t$ stays positive, and $\overline t$ gets even more negative, $t\overline t$ stays negative and so the new $ac$ is still negative ; and of course $a$ doesn't change and so neither does $\Delta$.
If $t<1$ and you take their reciprocal, this simply switches $a$ with $c$, and you still have $ac<0$ and $b^2-4ac = \Delta$. Also you can multiply $a,b,c$ by $-1$ at this point so we can only look at equations with $a>0$ and $c<0$.
Therefore during this whole process the triplets $(a,b,c)$ you "visit" all satisfy $a>0, c<0$ and $b^2 = \Delta + 4ac < \Delta$, so you can only visit a finite number of them before getting into a loop at some point.
Now, more importantly you are asking about the largest shift operation you can do. Obviously, $k = \lfloor t \rfloor \le t \le t - \overline{t}$, and since $(t - \overline t)^2 = \Delta/a^2$, the largest shifts are bound by $\sqrt \Delta / a $.
The worst case is when $a=1$ and your starting points are right in the middle of this family (they are at $b= -\epsilon $), so the shift length there should be around $\sqrt D/2$.
Eventually when you finish the loop, you end up at one of its extremities where $b$ is the biggest negative integer possible, that is $x^2-mx-\frac 14(\Delta-m^2)$ where $m$ is the largest positive integer such that $m^2 < \Delta$ and $4$ divides $\Delta-m^2$ (so this coincides with your $m$). When you do this shift completely, the length is $\lfloor t \rfloor = \lfloor \frac 12 (m+\sqrt \Delta) \rfloor = m$ (because $m \le \sqrt \Delta < m+2$).
For the sake of completeness, it takes $(m-\epsilon)/2$ steps to get from $b=-m$ to $b=-\epsilon$ (your starting point), which leaves a length of $(m+\epsilon)/2 = \lceil \frac m2 \rceil$ for the starting point.
The next worst case is when $a=2$, specifically when $b$ is the biggest negative possible, so for the roots of $2x^2-nx - \frac 18(\Delta-n^2)$ where $n$ is the largest positive integer less than $\sqrt D$ such that $8$ divides $\Delta-n^2$.
Here, $k= \lfloor t \rfloor = \lfloor \frac 14(n+\sqrt \Delta) \rfloor$.
If $\Delta = 1 \mod 8$ then $n=m$ and so you get $k=\lfloor \frac 12 \lfloor \frac 12(m+\sqrt \Delta) \rfloor \rfloor = \lfloor \frac m2 \rfloor$.
If $\Delta = 0 \mod 4$ then $n=m$ or $n=m-2$ (whichever is a multiple of $4$). In the second case, $k = \lfloor \frac {m-1}2 \rfloor = \lfloor \frac m 2 \rfloor - 1$.
If $\Delta = 5 \mod 8$ then you cannot reach a triplet with $a=2$ anyway.
