There are $2^n$ equations for $T^n(x)$, and so there are $2^n$ fixed points. However, some of the fixed points are in the same cycle. So for $T^3(x)$, the fixed points are $\left\{0,\frac29,\frac27,\frac49,\frac47,\frac69=\frac23,\frac67,\frac89\right\}$, and the cycles for $T^3(x)$ are $\left\{\left\{0\right\},\left\{\frac29,\frac49,\frac89\right\},\left\{\frac27,\frac47,\frac67\right\},\left\{\frac69=\frac23\right\}\right\}$. So there are 4 period-3 orbits for $T(x)$, but only 2 with a prime period of 3, and thus only 2 3-cycles.
**EDIT**:
Prove that the tent map has exactly nine 6-cycles.
$$T(x)=\left\{\begin{matrix}2x \qquad \ \ \ \ \ 0\le x\le\frac12 \\ 2-2x \quad \frac12< x\le1\end{matrix}\right.$$
$T^n(x)$ has $2^n$ piecewise functions, $n \in N$ by the definition of an iterating function and by $T(x)$.
$T^n(x)$ has $2^n$ fixed points where $T^n(x)=x$. By (1).
Iff $T^n(x)=x$, then $x$ is an element of a period-$n$ orbit of $T(x)$. By definition of orbit.
The fixed points of $T^n(x)$ comprise all elements of the period-$n$ orbits of $T(x)$. By (3).
Given $m<n$ where $m,n \in N$, iff $\frac mn \in N$, then $m$ is a factor of $n$. The factors of $n$ are $\{m_1,m_2,...,m_k\}$ where $k \in N$. By the definition of factor.
$T^n(x)=T^{m_i}(T^{m_j}(x))$, where $m_i \cdot m_j=n$ and $i,j \in N$. By iterating function.
The fixed points of $T^{m_k}(x)$ are also fixed points of $T^n(x)$. By (6).
Given $s$ unique fixed points of $T^{m_k}(x)$, there are $2^n-s$ fixed points of $T^n(x)$ which are not fixed points of any $T^{m_k}(x)$. By (7) and (2).
The fixed points of $T^n(x)$ which are not fixed points of $T^{m_k}(x)$ are all the elements of period-$n$ orbits of $T(x)$ with prime period $n$. By (7) and (3).
There are $n$ unique elements to each $n$-cycle of $T(x)$. By definition of $n$-cycle.
There are exactly $\frac {2^n-s}{n}$ $n$-cycles of $T(x)$. By (10).
For $T^6(x)$, there are $\frac {2^6-s}{6}$ 6-cycles of $T(x)$. By (11).
$s$=number of unique fixed points of $\{T^1(x),T^2(x),T^3(x)\}=2+2+6=10$. By (8) and (5).
There are $\frac {2^6-10}{6}$ 6-cycles of $T(x)$. $$\frac {2^6-10}{6}=\frac {64-10}{6}=\frac {54}{6}=9$$ By (13),(11),(12).
There are exactly nine 6-cycles of $T(x)$. By (14).
Hope that's more comprehensive and helpful.
Actually, after some manipulations, you can use the master theorem! Let us see how. First, let us prove the following lemma:
Lemma: The function $T$ is non-decreasing, i.e. $T(n) \leq T(n+1)$ for all $n \in \mathbb{N}$.
Proof. By strong induction on $n \in \mathbb{N}$.
Base cases: For all $0 \leq n \leq 6$, one has $T(n) = 1 = T(n+1)$. Moreover, $T(7) = 1 < 2\,T(0) + 8^{\pi/2} = T(8)$, as $\big\lfloor \frac{8}{\sqrt{2}} \big\rfloor =5$.
Inductive step: Let $n > 7$. The strong induction hypothesis is $T(k) \leq T(k+1)$ for all $0 \leq k < n$. The goal is to prove that $T(n) \leq T(n+1)$.
By definition,
\begin{align}
T(n) &= 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2}
&
T(n+1) &= 2\,T\big(\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2}\,.
\end{align}
According to the properties of the floor function, $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + \big\lfloor \frac{1}{\sqrt{2}} \big\rfloor + 1 = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + 1$, and $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + \big\lfloor \frac{1}{\sqrt{2}} \big\rfloor \leq \big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor$, since $\sqrt{2} > 1$.
Therefore, there are only two cases:
- either $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor$ and then $T(n) \leq 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2} = T(n+1)$, where the inequality holds because from $\frac{\pi}{2} > 0$ it follows that $n^\frac{\pi}{2} < (n+1)^\frac{\pi}{2}$ ;
- or $\big\lfloor \frac{n+1}{\sqrt{2}} \big\rfloor = \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor + 1$; we can apply the strong induction hypothesis to $T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 \big)$ because $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5 < n$ (indeed, $n + 5 > n = \lfloor n \rfloor \geq \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor $ since $\lfloor \cdot \rfloor$ is non-decreasing), so $T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 \big)\leq T \big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor- 5 + 1\big) = T \big(\big\lfloor \frac{n + 1}{\sqrt{2}} \big\rfloor- 5 \big)$ and hence $T(n) \leq 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + (n+1)^\frac{\pi}{2} = T(n+1)$. $\qquad\square$
As $T$ is non-decreasing by the lemma above (and $\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5 < \big\lfloor \frac{n}{\sqrt{2}} \big\rfloor \leq \frac{n}{\sqrt{2}}$), for $n > 7$ one has $T(n) = 2\,T\big(\big\lfloor \frac{n}{\sqrt{2}} \big\rfloor - 5\big) + n^\frac{\pi}{2} \leq 2\,T\big(\frac{n}{\sqrt{2}}\big) + n^\frac{\pi}{2}$. Therefore, if we set
\begin{align}
S(n) =
\begin{cases}
1 &\text{if } n = 0 \\
2\,S\big(\frac{n}{\sqrt{2}}\big) + n^\frac{\pi}{2} & \text{otherwise}
\end{cases}
\end{align}
then $T(n) \leq S(n)$ for all $n \in \mathbb{N}$ and so, for any function $g$, $S(n) \in O(g(n))$ implies $T(n) \in O(g(n))$, i.e. the fact that $S$ grows asymptotically no faster than $g$ implies that $T$ grows asymptotically no faster than $g$.
The point is that we can use the master theorem to find a $g$ such that $S(n) \in O(g(n))$. Using the same notations as in Wikipedia article:
\begin{align}
a &= 2 & b&= \sqrt{2} & c_\text{crit} &= \log_\sqrt{2} 2 = 2 & f(n) &= n^{\pi/2}
\end{align}
thus, $S(n) \in O(n^2)$ by the master theorem (since $\pi/2 < 2 = c_\text{crit}$), and hence $T(n) \in O(n^2)$.
Best Answer
We have a recurrence that is ostensibly over the integers. But if $n$ is an integer, then $\sqrt{n}$ is not necessarily an integer. One way of getting an informal answer is to imagine that we start with an integer $n$ of the form $2^{2^m}$. Then $\sqrt{n}=2^{2^{m-1}}$. (To find the square root, we divide the exponent by $2$.)
So going up from $2^{2^0}$ to $2^{2^m}$, we have incremented $T$ by $m$. If we start with $T(2)=0$, then $$T\left(2^{2^m}\right)=m.$$ If $T(2)$ has some other value $a$, then $T\left(2^{2^m}\right)=a+m$. Note that $m=\log_2\left(\log_2\left(2^{2^m}\right)\right).$ So for this value of $n$, and $T(2)=0$, we have $T(n)=\log_2(\log_2(n))$. If $T(2)=a$, just add $a$.
Remark: The above calculation can be carried out in basically the same way if we assume that $T(n)=T(\lfloor\sqrt{n}\rfloor)+1$. The conclusion is that for large $n$, $T(n)$ behaves like $\lg\lg n$. And $\lg\lg n$ grows glacially slowly.