Recursive Collatz Conjecture

Python 2, 70 bytes

f=lambda n,k=0,j=0:n-1and-~f(k*[n/2,n*3+1][n%2]or f(j/99or n,1),k,j+1)

Returns 199 for one hundred or more iterations.

Try it online!

Attache, 40 bytes

`-&3@`#@PeriodicSteps[CollatzSize@Max&1]

Try it online!

This is a new language that I made. I wanted to get around to making a proper infix language, and this is the result: a mathematica knock-off. Hooray?

Explanation

This is a composition of a few functions. These functions are:

• PeriodicSteps[CollatzSize@Max&1] This yields a function which applies its argument until the results contain a duplicate element. This function, CollatzSize@Max&1, is applying CollatzSize to the greater of the input and 1, to avoid the invalid input 0 to CollatSize.
• `# is a quoted operator; when applied monadically in this sense, it obtains the size of its argument
• `-&3 is a bonded function, which bonds the argument 3 to the function `-, which reads as "minus 3". This is because the PeriodicSteps application yields 0s, which need to be accounted for. (It also neatly handles out-of-bounds numbers like 5, which map to -1.)

C (gcc), 75 bytes

i,j;f(n){for(j=0;(i=n)&&j++<100;)for(n=0;i-1;++n)i=i&1?i*3+1:i/2;i=!i*j-1;}

Returns -1 for n>=100 iterations.

Try it online!

C (gcc), 73 bytes

i,j;f(n){for(j=-1;(i=n)&&j++<99;)for(n=0;i-1;++n)i=i&1?i*3+1:i/2;i=!i*j;}

Returns 0 for n>=100 iterations.

Try it online!