I didn't know that was a requirement. OK then

   from itertools import cycle
   
   fizz = cycle(["","","fizz"])
   buzz = cycle(["","","","","buzz"])
   for idx, z in enumerate(zip(fizz,buzz)):
      print(f"{idx}: {z[0]}{z[1]}")

Happy now?

This doesn't look right either. You should really look up what FizzBuzz is. The output should be like:

  1
  2
  fizz
  4
  buzz
  fizz

You can't just print the number for every line. Check this out - https://rosettacode.org/wiki/FizzBuzz or just run the program in the OP which prints correct output.

I should really look it up because it’s such an important thing to get canonically correct.

It's not important to look it up. I only suggested it because you posted two programs and neither solve the problem people are discussing here. If you really want to solve a boring version of the problem, by all means you should. I'm nobody to tell you otherwise.

All I'm saying is that if you water the problem down into something trivial and boring, the solution will be trivial and boring too. No surprise there. The other answers here are more complicated because they solve the canonical FizzBuzz problem, not the boring version of it that is trivial to solve without conditionals.

it's more that fizzbuzz i think is one of the - if not the - simplest way to see if people understand requirements. some of the comments here remind me of thedailywtf.com comment sections.

being correct is good and being confused is bad