Print Number series without using any loop Last Updated : 26 Feb, 2023 Summarize Comments Improve Suggest changes Share Like Article Like Report 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. C++ // C++ program to Print Number // series without using loop #include <iostream> using namespace std; // function print series // using recursion void PrintNumber(int N, int Original, int K, bool flag) { // print the number cout << N << " "; // change flag if number // become negative if (N <= 0) flag = !flag; // base condition for // second_case (Adding K) if (N == Original && !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 (!flag) { PrintNumber(N + K, Original, K, flag); return; } } // driver program int main() { int N = 20, K = 6; PrintNumber(N, N, K, true); return 0; } Java // Java program to Print Number // series without using loop import java.io.*; import java.util.*; class GFG { public static void PrintNumber(int N, int Original, int K, boolean flag) { // print the number System.out.print(N + " "); // change flag if number // become negative if (N <= 0) flag = !flag; // base condition for // second_case (Adding K) if (N == Original && !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 (!flag) { PrintNumber(N + K, Original, K, flag); return; } } public static void main (String[] args) { int N = 20, K = 6; PrintNumber(N, N, K, true); } } // This code is contributed by Mohit Gupta_OMG Python3 # 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 C# // C# program to Print Number // series without using loop using System; public class GFG { // function print series // using recursion static void PrintNumber(int N, int Original, int K, bool flag) { // print the number Console.Write(N + " "); // change flag if number // become negative if (N <= 0) flag = !flag; // base condition for // second_case (Adding K) if (N == Original && !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 (!flag) { PrintNumber(N + K, Original, K, flag); return; } } // driver program static public void Main () { int N = 20, K = 6; PrintNumber(N, N, K, true); } } // This code is contributed by vt_m. PHP <?php // PHP program to Print Number // series without using loop // function print series // using recursion function PrintNumber($N, $Original, $K, $flag) { // print the number echo($N . " "); // change flag if number // become negative if ($N <= 0) $flag = !$flag; // base condition for // second_case (Adding K) if ($N == $Original && !$flag) return; // if flag is true // we subtract value until // number is greater than zero if ($flag == true) { PrintNumber($N - $K, $Original, $K, $flag); return; } // second case (Addition ) if (!$flag) { PrintNumber($N + $K, $Original, $K, $flag); return; } } // Driver Code $N = 20; $K = 6; PrintNumber($N, $N, $K, true); // This code is contributed by Ajit. ?> JavaScript <script> // Javascript program to Print Number // series without using loop // function print series // using recursion function PrintNumber(N, Original, K, flag) { // print the number document.write(N + " "); // change flag if number // become negative if (N <= 0) flag = !flag; // base condition for // second_case (Adding K) if (N == Original && !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 (!flag) { PrintNumber(N + K, Original, K, flag); return; } } // Driver Code let N = 20, K = 6; PrintNumber(N, N, K, true); // This code is contributed by _saurabh_jaiswal </script> Output20 14 8 2 -4 2 8 14 20 Time complexity: O(N/K) Auxiliary space: O(N/K) The extra space is used in recursion call stack. Comment More infoAdvertise with us Next Article Find the nth term of the given series S shubham_singh Follow Improve Article Tags : Mathematical Technical Scripter C++ Programs DSA Practice Tags : Mathematical Similar Reads Find the nth term of the given series Given the first two terms of the series as 1 and 6 and all the elements of the series are 2 less than the mean of the number preceding and succeeding it. The task is to print the nth term of the series. First few terms of the series are: 1, 6, 15, 28, 45, 66, 91, ... Examples: Input: N = 3 Output: 1 3 min read Find Nth term of the series 1, 1, 2, 6, 24... Given a number N. The task is to write a program to find the Nth term in the below series: 1, 1, 2, 6, 24... Examples: Input: 3 Output: 2 For N = 3 Nth term = (N-1)! = 2 Input: 5 Output: 24 Nth term of the series is given by below formula: Nth term = ( N-1)! Below is the required implementation: C++ 3 min read Find Nth term of the series 1, 8, 54, 384... Given a number N. The task is to write a program to find the Nth term in the below series: 1, 8, 54, 384... Examples: Input : 3 Output : 54 For N = 3 Nth term = ( 3*3) * 3! = 54 Input : 2 Output : 8 On observing carefully, the Nth term in the above series can be generalized as: Nth term = ( N*N ) * 4 min read Find Nth term of the series 5, 13, 25, 41, 61... Given a number N. The task is to write a program to find the Nth term in the below series: 5, 13, 25, 41, 61... Examples: Input : 3 Output : 25 For N = 3 Nth term = 3*3 + (3+1)*(3+1) = 25 Input : 5 Output : 61 On observing carefully, the Nth term of the given series can be generalised as: Nth term = 3 min read Find Nth term of the series 0, 6, 0, 12, 0, 90... Given a number N. The task is to write a program to find the Nth term in the below series: 0, 6, 0, 12, 0, 90... Examples: Input : N = 4 Output : 12 For N = 4 Nth term = abs( 4 * ((4-1) * (4-3) * (4-5)) ); = 12 Input : N = 6 Output : 90 We can generalize the Nth term of the above series as: Nth term 3 min read Find Nth term of the series 1, 6, 18, 40, 75, .... Given a number n, the task is to find the n-th term in series 1, 6, 18, 40, 75, ...Example: Input: N = 2 Output: 6 Explanation: 2nd term = (2^2*(2+1))/2 = 6 Input: N = 5 Output: 75 Explanation: 5th term = (5^2*(5+1))/2 = 75 Approach: Nth term = (N^2*(N+1))/2 Implementation of the above approach is g 3 min read Like