Odd part of $n-1$ and primes

Question

Using $n=11$ as an example:

Step 1 : 11 - 1 =  10. Get the odd part of  10, which is 5    
Step 2 : 11 - 5 =   6. Get the odd part of   6, which is 3
Step 3 : 11 - 3 =   8. Get the odd part of   8, which is 1

Continuing this operation (with $11-1$) repeats the same steps as above. There are three consecutive odd numbers $1,3,5$ from $1$ to $(n-1)/2$ in the cycle, so the number $11$ has a "full counter cycle".

Is there any counterexample that a number has a "full counter cycle" but isn't prime?

Answer 1

Let $n$ be a positive integer, and define the sequence $(c_i)_{i\geq0}$ inductively by $$c_0=n-1,\qquad c_{i+1}=\operatorname{odd}(n-c_i).$$ I understand your question to be:

Is there a composite $n$ such that every odd number less that $\frac{n-1}{2}$ occurs in the sequence $(c_i)_{i\geq0}$?

Note that $\gcd(n,c_0)=1$, and that $$\gcd(n,c_{i+1})=\gcd(n,\operatorname{odd}(n-c_i))\mid\gcd(n,n-c_i)=\gcd(n,c_i).$$ Hence $\gcd(n,c_i)=1$ for all $i\geq0$. Hence no divisor of $n$ other than $1$ occurs in the sequence $(c_i)_{i\geq0}$. So no counterexample exists with $n$ composite and odd.

A counterexample with $n$ composite and even is easy; take $n=4$.