Problem - Givens Two number N and K, our task is to subtract a number K from N until number(N) is greater than zero, once the N becomes negative or zero then we start adding K until that number become the original number(N). Note : Not allow to use any loop. Examples :

Input : N = 15 K = 5  
Output : 15 10 5 0 1 5 10 15

Input : N = 20 K = 6
Output : 20 14 8 2 -4 2 8 14 20 

Explanation - We can do it using recursion idea is that we call the function again and again until N is greater than zero (in every function call we subtract N by K). Once the number becomes negative or zero we start adding K in every function call until the number becomes the original number. Here we use a single function for both addition and subtraction but to switch between addition or subtraction function we used a Boolean flag. 

# Python program to Print Number
# series without using loop

def PrintNumber(N, Original, K, flag):
    #print the number
    print(N, end = " ")
    
    # change flag if number
    # become negative
    
    if (N <= 0):
        if(flag==0):
            flag = 1
        else:
            flag = 0
        
    # base condition for
    # second_case (Adding K)
    
    if (N == Original and (not(flag))):
        return
    
    # if flag is true
    # we subtract value until
    # number is greater than zero
    
    if (flag == True):
        PrintNumber(N - K, Original, K, flag)
        return
    
    # second case (Addition )
    if (not(flag)):
        PrintNumber(N + K, Original, K, flag);
        return
    
N = 20
K = 6
PrintNumber(N, N, K, True)

# This code is contributed by Mohit Gupta_OMG

Output :

20 14 8 2 -4 2 8 14 20 

The time complexity is O(N/K), where N is the input number and K is the step value.

The auxiliary space is O(N/K)

Approach Name: Recursive Mathematical Approach using Function Calls.

Steps:

1. Define a function print_series(N, K) that takes two parameters, N and K.
2. Check if N is less than or equal to zero, if yes, print N and return.
3. Print N and call the print_series(N-K, K) function recursively.
4. Print N again.

def print_series(N, K):
    if N <= 0:
        print(N, end=" ")
        return
    print(N, end=" ")
    print_series(N-K, K)
    print(N, end=" ")

N = 20
K = 6
print_series(N, K)

20 14 8 2 -4 2 8 14 20 

 It has a time complexity of O(log(N/K)) and an auxiliary space of O(log(N/K)).

Please refer complete article on Print Number series without using any loop for more details!