Revision 189752e7-da9c-48ae-918f-476b66cd9a11 - Code Golf Stack Exchange

#><>, <strike>27 26</strike> 23 bytes

 \ln;
 \::2%:@5*1+2,*+:2=?

Like the other ><> answers, this builds the sequence on the stack. Once the sequence reaches 2, the size of the stack is the number of steps taken.

Thanks to @Hohmannfan, saved 3 bytes by a very clever method of computing the next value directly. The formula used to calculate the next value in the sequence is:

![f(n)=n\cdot\frac{5(n\bmod2)+1}{2}+(n\bmod2)](http://mathurl.com/hvjobv2.png)

The fraction maps even numbers to 0.5, and odd numbers to 3. Multiplying by `n` and adding `n%2` completes the calculation - no need to choose the next value at all!

**Edit 2:** Here's the pre-@Hohmannfan version:

 \ln;
 \:::3*1+@2,@2%?$~:2=?

The trick here is that both `3n+1` and `n/2` are computed at each step in the sequence, and the one to be dropped from the sequence is chosen afterwards. This means that the code doesn't need to branch until 1 is reached, and the calculation of the sequence can live on one line of code.

**Edit:** Golfed off another character after realising that the only positive integer that can lead to 1 is 2. As the output of the program doesn't matter for input < 2, the sequence generation can end when 2 is reached, leaving the stack size being the exact number of steps required.

Previouser version:

 \~ln;
 \:::3*1+@2,@2%?$~:1=?