Looking for counter-examples in Collatz-like sequences The following function is very similar to the one involved in Collatz Conjecture
$$ f(n) =
\begin{cases}
3n-1          & \text {if $n$ is odd} \\
\frac{n}{2}   & \text {if $n$ is even} 
\end{cases} ,$$
but counter-examples (to the well-known statement) do exist, meaning that some values of n are such that for any $k$ :
 $$f^{k}(n)≠1\;.$$
For instance, $5, 7, 10, 14, 17, 20, 25, 34, 37, 41, 50, 55, 61, 68, 74, 82, 91, 110, 122, 136, 164, 182, 272$ yield cyclical patterns containing no $1$. But all these values allow to build only two different cyclical patterns:
(5 14 7 20 10) then again 5, etc. 
(17 50 25 74 37 110 55 164 82 41 122 61 182 91 272 136 68 34) then 17 etc.

Or for odd steps counter-examples are:
$$5,7$$
$$17,25,37,55,41,61,91$$
Of course, the sequence starting at,for example,$n=243$ wouldn't be considered as really new since it soon leads to one of these two patterns above.

The question: Is there another similar cyclical pattern?

 A: For even examples, you can take $91\cdot 2^m$ for any $m$. For odd examples, take
$$\frac{91\cdot 2^m+1}{3}$$
for odd $m$.
Of course this is assuming that $91$ is indeed a correct counter example, I did not check that.
A: You can find other counter examples: 5, 7, 10, 14, 17, 20, 25, 34, 37, 41, 50, 55, 61, 68, 74, 82, 91, 110, 122, 136, 164, 182, 272 (brute-force search) but maybe I didn't fully understand what you are asking (maybe you are only interested by odd terms in which case I don't bring any new one)?
Indeed for $n=272$ you get the sequence of terms: 272, 136, 68, 34, 17, 50, 25, 74, 37, 110, 55, 164, 82, 41, 122, 61, 182, 91, 272, etc. where you can see your own $91$ occuring. Actually taking the highest term occuring in your 91-based sequence could have given to you the idea of a higher starting point (since the sequence is cyclycally repeating if I correctly understood your request).
But the fact is that brute-force search doesn't seem to return anything higher than 272.
A: I think an easy example is $93$, one gets the following sequence $278,139,416,208,104,52,26,13,38,19,56,28,14,7,20,10,5,14,7,20,10,5\ldots$
$99$ is also an example, $296,148,74,37,110,55,164,82,41,122,61,182,91,272,136,68,34,17,50,25,74$
As a side remark, if you change $3n-1$ to $3n+1$, then this is a famous conjecture without known counterexamples.
Edit: I wrote a simple program and played with it a little, apparently at least for any $n<100000$, $f^k(n)$ eventually becomes one of your given counter examples...
A: maybe just use the reverse map:
$$\cases{\{{n+1\over 3},2n\} & \text{if }n\equiv 2 \bmod 6 \\{2n}&otherwise}$$
A: I have a counterexample of a type that does not use 3n+1 and produces a proper superset of the odd numbers only. It ends in 1 (but since there are no even numbers, the descent uses other odd ones, including 3). The function is:
$f(n) = \begin{cases} n+ \frac {n+1}{2} & \mbox{if } n+1 \equiv 0 \mbox{ (mod } 4) \\ n-\frac{n-1}{4} & \mbox{if } n-1 \equiv 0 \mbox{ (mod } 8) \\ \frac {n-\frac{n+1}{2}}{2} & \mbox{otherwise} \end{cases}$
Compare for sequence $43$, for example, the odd-even and odd-superset sequences:
43, 130, 65, 196, 98, 49, 148, 74, 37, 112, 56, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
43, 65, 49, 37, 9, 7, 11, 17, 13, 3, 5, 1
