Why am I getting this error from Dave Beazly's programme?

[Pedroski55]

This is from Dave Beazly's Generators: The Final Frontier here. Part 3, the example inline_recursive.py

When I run inline_recursive.py in Idle, by copying and pasting the parts into Idle, it works and prints lots of:

print('Tick:', n)

But I get this error when I run inline_recursive.py in bash, right after the first tick is printed:

pedro@pedro-HP:~/myPython/yield/tutorial2014$ ./inline_recursive.py
Tick: 0

Error:
exception calling callback for <Future at 0x75e01f9e0be0 state=finished returned NoneType>
Traceback (most recent call last):
  File "/usr/lib/python3.10/concurrent/futures/_base.py", line 342, in _invoke_callbacks
    callback(self)
  File "/home/pedro/myPython/yield/tutorial2014/./inline_recursive.py", line 26, in _wakeup
    self.step(None, exc)
  File "/home/pedro/myPython/yield/tutorial2014/./inline_recursive.py", line 14, in step
    fut = self._gen.throw(exc)
  File "/home/pedro/myPython/yield/tutorial2014/./inline_recursive.py", line 24, in _wakeup
    self.step(result, None)
  File "/home/pedro/myPython/yield/tutorial2014/./inline_recursive.py", line 16, in step
    fut = self._gen.send(value)
  File "/home/pedro/myPython/yield/tutorial2014/./inline_recursive.py", line 44, in recursive
    Task(recursive(n+1)).step()
  File "/home/pedro/myPython/yield/tutorial2014/./inline_recursive.py", line 16, in step
    fut = self._gen.send(value)
  File "/home/pedro/myPython/yield/tutorial2014/./inline_recursive.py", line 42, in recursive
    yield pool.submit(time.sleep, 0.001)
  File "/usr/lib/python3.10/concurrent/futures/thread.py", line 169, in submit
    raise RuntimeError('cannot schedule new futures after '
RuntimeError: cannot schedule new futures after interpreter shutdown

Like I said, it works in the Idle shell.

#! /usr/bin/python3

# inline_recursive.py
#
# Bizarre inline recursive example

class Task:
    def __init__(self, gen):
        self._gen = gen

    def step(self, value=None, exc=None):
        try:
            if exc:
                fut = self._gen.throw(exc)
            else:
                fut = self._gen.send(value)
            fut.add_done_callback(self._wakeup)
        except StopIteration as exc:
            pass

    def _wakeup(self, fut):
        try:
            result = fut.result()
            self.step(result, None)
        except Exception as exc:
            self.step(None, exc)

# Example
if __name__ == '__main__':
    from concurrent.futures import ThreadPoolExecutor
    import time

    pool = ThreadPoolExecutor(max_workers=8)

    """ Error
    File "/usr/lib/python3.10/concurrent/futures/thread.py", line 169, in submit
        raise RuntimeError('cannot schedule new futures after '
    RuntimeError: cannot schedule new futures after interpreter shutdown
    """
    def recursive(n):
        # this submit causes a problem
        yield pool.submit(time.sleep, 0.001)
        print('Tick:', n)
        Task(recursive(n+1)).step()

    Task(recursive(0)).step()

I found, on the other hand, examples that work with .ProcessPoolExecutor(), like below, will not work in Idle, but work in bash!

#! /usr/bin/python3
import concurrent.futures

def worker(task):
    result = task * 2
    print(f"Task {task}: Result = {result}\n")
    return result

if __name__ == "__main__":
    tasks = [1, 2, 3, 4, 5]
    # will not work in the shell

    with concurrent.futures.ProcessPoolExecutor() as executor:
        results = executor.map(worker, tasks)
        print("Results:", list(results))

Any tips please?