How to generate function n times in python and count digits?

Question

I want to generate the fibonacci function N times based on user input. (I am thinking this as a while loop or an if statement like if N in range). Also there is second user input defined as Y. Y represents the amount of digits of the repeated function and I want a count of how many numbers generated have Y amount of digits.

Below is my non complete code:

N = int(input("Enter N: "))
Y = int(input("Enter Y: "))


def fibonacci(n): 
   if n <= 1:
     return n
   else:
     return fibonacci(n-2) + fibonacci(n-1)

nterms = 10

# check if the number of terms is valid
if nterms <= 0:
   print("Please enter a positive integer")
else:
   print("Fibonacci sequence:")
   for i in range(nterms):
       print(fibonacci(i))

Thanks in advance

Please give an example of your expected input and output for the functionality you are planning to create. — r4bb1t, Nov 16 '19 at 08:42
@manavhs13 please comment if an answer did not help you, or mark the answer as solved if it solved your problem. — jkazan, Nov 17 '19 at 00:59
**What happens** when you try using this code? **How is that different** from what is supposed to happen? — Karl Knechtel, Mar 10 '23 at 09:21

jkazan · Accepted Answer · 2019-11-17T15:03:52.730

Try this (read comments in the code below):

from math import sqrt

def fibonacci(n):
    return int(((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5))) # WA algorithm

# Handle N from user input
while True:
    N = int(input("Enter N: "))
    Y = int(input("Enter Y: "))
    if N < 1 or Y < 1:
        print("Plese enter a positive integers for N and Y")
    else:
        break

count = 0 # Counter of the instances of fibonacci numbers containing Y digits
for n in range(N):
    fib = fibonacci(n) # Calls fibonacci
    print(fib) # Print Fibonacci number
    if len(str(fib)) == Y: # Number of digits in the Fibonacci number
        count += 1 # Count up if number of digits = Y

print("\nFibonacci numbers with {} digits occured {} times" .format(Y, count))

The algorithm for calculating Fibonacci numbers comes from Wolfram Alpha

EDIT: In order to save time to find results of multiple Ys, you can keep statistics in a dictionary like so:

from math import sqrt

# Handle N from user input
while True:
    N = int(input("Enter N: "))
    if N < 1:
        print("Plese enter a positive integers for N and Y")
    else:
        break

size = {} # Dictionary with number of occurences of lengths of Fibonacci numbers
count = 0 # Counter of the instances of fibonacci numbers containing Y digits
for n in range(N):
    fib = int(((1+sqrt(5))**n-(1-sqrt(5))**n)/(2**n*sqrt(5))) # WA algorithm
    print(fib) # Print Fibonacci number
    s = str(len(str(fib)))
    if s not in size:
        size[s] = 1
    else:
        size[s] += 1

print()
for key, val in size.items():
    print("{}\toccured\t{} times" .format(key, val))

This will yield an output like:

Enter N: 13
0
1
1
2
3
5
8
13
21
34
55
89
144

1   occured 7 times
2   occured 5 times
3   occured 1 times

Τhanks for the code it is very useful. However, I need to generate the sequence N times. N is not the number of terms but how many times the sequence is generated. — manavhs13, Nov 17 '19 at 12:13
Thanks, but this one does not print the sequence N times. Also, an addition to prompt the user for n: the amount of terms required — manavhs13, Nov 17 '19 at 13:16
In your example, you run the amount of terms depending on the loop. To my surprise you never use `N` as I commented on the question, but you use `nterms` and then you run `fibonacci(0), fibonacci(1), ..., fibonacci(nterms)`, i.e. the amount of terms should be `n` in the loop over `N` elements I suppose. Anyway, I updated the code to print the sequence. — jkazan, Nov 17 '19 at 14:54

How to generate function n times in python and count digits?

1 Answers1