Let $n$ be an odd positive integer, Let $o=\operatorname{ord}\_n 2$ be the order of 2 modulo $n$ and $m$ the period of $1/n, k$ is number of distinct odd residues contained in set $\{2^1,2^2,...,2^{n−1}\}$ modulo $n$.


If $o,m$ and $k$ are even power of 2 and $k$ divide $n-1$, then $n$ is item in the sequence $17, 257, 641, 65537, …$.


It seems all known items in the sequence are Fermat factors, How come? 



Question on odd positive integers and Fermat factors
*Hmm, unfortunately you deleted instructive portions of your OP, but let me answer based on that earlier version which included explicitela the iterations scheme.* 


You give a game of iteration on numbers, which I'd like to rewrite for some given positive odd number $z$ and an initial $a=1$ :  

$$ b = {z+a\over 2^A } \tag 1
$$
where $A$ is taken such that $b$ is odd again. 


You iterate that formula until this cycles:
$$ c = {z+b\over 2^B } \qquad d = {z+c\over 2^C } \qquad \cdots \qquad a = {z+h\over 2^H } \tag 2
$$
Let us denote the number of iterates with $N$ and the sum of exponents $S=A+B+C+\cdots+H$ .  

You wonder now about the observation that some numbers $z$ satisfy multiple conditions: that $N$ is a power of $2$, that $S$ is a power of $2$, (and about the period in decimal expansion of $1/z$ but which I shall not discuss here). 




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Let us look at some simple to detect properties. 


* **What numbers $z$ have a cycle of $N=1$ ?**  

this means $b=a$ and constructing a formula gives (here is $S=A$)
$$\begin{array}{rll} a &= { z+a\over 2^S } \\
 a2^S &= z+a \\
 z &=a(2^S-1) 
 \end{array} \tag {3.1}
$$
$\qquad \qquad $ Assuming $ a=1 $ gives the set of solutions
$$ \begin{array}{rll}
 z &\in \{1,3,7,15,\ldots\} \\
 A=S &\in \{1,2,3,4,\ldots\}
 \end{array} \tag {3.2} $$
* **What numbers $z$ have a cycle of $N=2$ ?**  

this means $c=a$ in definition (2) and a formula gives (here $S=A+B$)
$$\begin{array}{rll} b &= { z+a\over 2^A } && a &= { z+b\over 2^B } \\
 a &= { z+{ z+a\over 2^A }\over 2^B } & 
 = { z2^A+ z+a \over 2^{A+B} } \\
 a2^S -a &= z2^A+ z \\
 z &= a{ 2^S-1 \over 2^A+1 } & \left(= a{ (2^S-1)(2^{A}-1) \over 2^{2A}-1 }\right)
 \end{array} \tag {4.1}
$$
$\qquad \qquad $ Assuming $ a=1 $ gives the set of solutions
$$ \Tiny \begin{array}{rr|ll}
z & S & A & B \\ \hline
1 & 2 & 1 & 1 \\
5 & 4 & 1 & 3 \\
21 & 6 & 1 & 5 \\
 \vdots & \vdots & \vdots & \vdots \\ \hline
3 & 4 & 2 & 2 \\
51 & 8 & 2 & 6 \\
 \vdots & \vdots & \vdots & \vdots \\ \hline
7 & 6 & 3 & 3 \\
455 & 12 & 3 & 9 \\
 \vdots & \vdots & \vdots & \vdots \\ \hline
15 & 8 & 4 & 4 \\
3855 & 16 & 4 & 12 \\
986895 & 24 & 4 & 20 \\
 \vdots & \vdots & \vdots & \vdots \\ \hline
31 & 10 & 5 & 5 \\
31775 & 20 & 5 & 15 \\
32537631 & 30 & 5 & 25 \\
 \vdots & \vdots & \vdots & \vdots \\ \hline
63 & 12 & 6 & 6 \\
258111 & 24 & 6 & 18 \\
1057222719 & 36 & 6 & 30 \\
 \vdots & \vdots & \vdots & \vdots \\
\end{array} \tag {4.2} $$
This table displays easy to recognize patterns for setting up formulae for $z$ and $S,A,B$ for the $N=2$ case. One shot is surely  

$$ S=2A\cdot k \qquad \implies z={2^{2kA}-1\over2^A+1}=
{(2^{2kA}-1)(2^{A}-1)\over 2^{2A}-1}$$  

and the left parenthese is always divisible by the denominator.  

For a better example: just to look at your question of $N=2^j$ && $ S=2^k$ for positive $j \le k$ we can extract cases for $j=1$, $N=2^j=2$ first  

$$ \Tiny \begin{array}{rr|ll}
z & S & A & B \\ \hline
1 & 2 & 1 & 1 \\
5 & 4 & 1 & 3 \\
3 & 4 & 2 & 2 \\
85 & 8 & 1 & 7 \\
51 & 8 & 2 & 6 \\
15 & 8 & 4 & 4 \\
21845 & 16 & 1 & 15 \\
13107 & 16 & 2 & 14 \\
3855 & 16 & 4 & 12 \\
255 & 16 & 8 & 8 \\
1431655765 & 32 & 1 & 31 \\
858993459 & 32 & 2 & 30 \\
252645135 & 32 & 4 & 28 \\
16711935 & 32 & 8 & 24 \\
65535 & 32 & 16 & 16 \\
6148914691236517205 & 64 & 1 & 63 \\
3689348814741910323 & 64 & 2 & 62 \\
1085102592571150095 & 64 & 4 & 60 \\
71777214294589695 & 64 & 8 & 56 \\
281470681808895 & 64 & 16 & 48 \\
4294967295 & 64 & 32 & 32 \\
 \vdots & \vdots & \vdots & \vdots \\ 
\end{array} \tag {4.3} $$  

As by (4.1) with $a=1$ and $S=2^k$ we have analytically  

$$\begin{array}{rll} 
 z &= { 2^{2^k}-1 \over 2^A+1 } \\
 &= { (2^{2^{k-1}}+1)(2^{2^{k-2}}+1)...(2^1+1)(2^1-1)\over 2^{A}+1 }
 &= { 3 \cdot 5 \cdot 17 \cdot 257 \cdot \ldots \cdot(2^{2^{k-1}}+1)\over 2^{A}+1 }
 \end{array} \tag {4.4}
$$  

Because the parentheses in the numerator have no common factor, we can draw conclusions of possible $A$ in the denominator to make the fraction an integer value. (Empirically we can observe that all $A$ are perfect powers of $2$ and it is surely not difficult to prove this)




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This can analoguously be extended to larger cycle-lengthes $N>2$ . It might furtherly be useful, to introduce a notation which combines the set of exponents $A,B,C,...$ with the resulting integer value $z$ in a fully compacted form, say
$$ z = T(A\_1,A\_2,A\_3,...,A\_N) \tag 5$$
and look, for instance, at solutions $z$ for typical patterns of the exponents, like $z=T(1,1,1,...,1)$ or $z=T(1,1,...1,A\_N)$ with $A\_N\gt 1$ and $S=N-1+A$ to develop short formulae for the integer solutions. 


I'm not going deeper into this nice game, but leave it to you as a suggestion how to approach analytical answers to your questions.


