I am trying to make a function that allows a user to input a number and the result will be a list containing Fibonacci numbers up to the input and one above if the input is not in the series. For example, input of 4 will return [0, 1, 1, 2, 3, 5] but input of 3 would return [0, 1, 1, 2, 3]. I have managed to do this using the function below :

    def fibonacci(n):series = [0]if (n == 0):passelse:series.append(1)if (n == 1):passelse:while(series[len(series)-1] < n):newValue = series[len(series)-1] + series[len(series)-2]series.append(newValue)print(series)

However, I would now like to be able to do this recursively, any ideas?

A possible solution is the following

def fibo_up_to(n):if n < 2:return [1,1]else:L = fibo_up_to(n-1)if L[-1] < n:L.append(L[-1] + L[-2])return L

the idea is that to return the list of all fibonacci numbers less than n you can ask for the list of those less than n-1 and then possibly add just one element. This works from 2 on if we define the first two numbers being [1, 1]. Using [0, 1] instead creates a problem for 2 because a single next element is not enough.

This code is not inefficient on time (fibonacci is a double recursion, this is a simple recursion), but uses a lot of stack space.