Hey yes, my formula is wrong! Thanks! Let me translate your labeling of the Fibonacci graph to a loop signature.

\\(f \rightarrow a_1=1\\)

\\(h°g \rightarrow a_2=2\\)

I had proposed to count paths like this example with length 4:

\\(2\cdot a_1+1\cdot a_2\\)

While in fact these paths are all different:

\\(a_1a_1a_2\\)

\\(a_1a_2a_1\\)

\\(a_2a_1a_1\\)

I still like naming loops. The formula I gave simply isn't _enough_! For each such sum I'd also have to count the permutations which look different. Counting them shouldn't be too hard, but armed with such an ability there might be a better way to count them alltogether. Or even general paths with arbitrary start and end...

If it weren't a bit late for me, I might find out how to count these permutations right away, and then perhaps nevertheless try to correct my formula that way.