C Program To Print 1361015.. Number Series Up To N

This article will guide you through creating a C program to generate and print a specific number series: 1, 3, 6, 10, 15... up to a user-defined limit N. You will learn how to identify the pattern and implement an iterative solution.

Problem Statement

The challenge is to generate a sequence of numbers where each subsequent number is the sum of the previous number and an increasing integer. Starting with 1, the series progresses by adding 2, then 3, then 4, and so on. The program needs to accept an integer N and print all numbers in this series that are less than or equal to N.

Example

For an input N = 20, the expected output of the series would be: 1, 3, 6, 10, 15

Background & Knowledge Prerequisites

To understand this article, you should have a basic grasp of:

• Variables: Declaring and using integer variables in C.
• Loops: for or while loops for repetitive tasks.
• Input/Output: Using printf() for printing and scanf() for reading user input.
• Arithmetic Operators: Basic addition and assignment.

Use Cases or Case Studies

This type of series pattern, often called triangular numbers (minus one adjustment here), appears in various computational and mathematical contexts:

• Combinatorics: Counting combinations or arrangements where the number of choices increases sequentially.
• Algorithmic Puzzles: As a building block for more complex number theory or sequence generation problems.
• Mathematical Modeling: Representing cumulative growth or patterns where the rate of change itself grows linearly.

Solution Approaches

For generating this series, a simple iterative approach is highly effective. We can maintain the current number in the series and an increment value, updating both in each step of a loop.

Approach 1: Iterative Summation

This approach uses a loop to generate each number in the series by adding an incrementally increasing value to the previous number.

• Summary: Initialize the first term as 1 and an incrementer as 2. In each iteration, print the current term, update the current term by adding the incrementer, and then increment the incrementer itself.

// Generate Series 1,3,6,10,15...
#include <stdio.h>

int main() {
    // Step 1: Declare variables for the limit, current series term, and incrementer
    int N;             // User-defined limit
    int current_term = 1; // Starts with the first term of the series
    int increment = 2; // The value to add to the current_term, starts with 2 (1 + 2 = 3)

    // Step 2: Prompt the user to enter the limit N
    printf("Enter the upper limit (N): ");
    scanf("%d", &N);

    printf("Series up to %d: ", N);

    // Step 3: Loop to generate and print the series
    // Continue as long as the current_term does not exceed N
    while (current_term <= N) {
        // Step 3a: Print the current term
        printf("%d", current_term);

        // Step 3b: Add a comma and space if it's not the last term to be printed
        // This check prevents an extra comma at the end
        if (current_term + increment <= N) {
            printf(", ");
        }
        
        // Step 3c: Calculate the next term
        current_term += increment; // Add the current increment to get the next term

        // Step 3d: Increment the incrementer for the next iteration
        increment++; // The increment itself increases by 1 each time (2, then 3, then 4...)
    }
    printf("\\n"); // Print a newline at the end for clean output

    return 0;
}

Run

• Sample Output:

Enter the upper limit (N): 20
Series up to 20: 1, 3, 6, 10, 15

Enter the upper limit (N): 30
Series up to 30: 1, 3, 6, 10, 15, 21, 28

• Stepwise Explanation:

1. Initialization: We declare N for the limit, current_term is set to 1 (the first number in the series), and increment is set to 2. This increment is what we'll add to current_term to get the *next* series number.
2. User Input: The program prompts the user to enter the maximum value N for the series.
3. Looping: A while loop continues as long as current_term is less than or equal to N.
4. Printing: Inside the loop, current_term is printed. A conditional printf(", ") ensures commas separate numbers without an trailing comma.
5. Updating current_term: The current_term is updated by adding the increment to it. For example, if current_term is 1 and increment is 2, the next current_term becomes 3.
6. Updating increment: The increment itself is then increased by 1. So, after increment was 2, it becomes 3 for the next iteration. This ensures the additions are 2, then 3, then 4, and so on.
7. Termination: The loop exits when current_term exceeds N.

Conclusion

Generating the series 1, 3, 6, 10, 15... up to a given limit N in C is straightforward using an iterative approach. By identifying the pattern of an increasing increment, we can efficiently compute and print each term of the series with a simple loop. This method is clear, concise, and easy to understand for beginners.

Summary

• The series starts with 1.
• Each subsequent term is generated by adding an incrementing value (2, then 3, then 4, etc.) to the previous term.
• A while loop effectively iterates, calculating and printing terms until the specified limit N is reached.
• Separate variables are used for the current_term and the increment value, both updated within the loop.