Increasing Shifted Number Triangle in C++

Beginner
⏱️ 5 min read
📚 Updated: Aug 2025
🎯 2 Code Examples
Nested Loops
Increasing shifted number triangle output (1, 2 3, 3 4 5, 4 5 6 7, 5 6 7 8 9) in C++ programming

What You’ll Learn

How to print an increasing shifted number triangle in C++ where each row begins with the row number and continues sequentially.

The key idea is to print i + j - 1 while the inner loop runs from 1 to i.

⭐ Pattern Output

For rows = 5, the pattern looks like this:

Output
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9
1

Complete C++ Program

Fixed five rows: the inner loop prints i values, and the expression i + j - 1 generates consecutive numbers on each row.

C++
#include <iostream>
using namespace std;

int main() {
    int rows = 5;
    int i, j;

    for (i = 1; i <= rows; i++) {
        for (j = 1; j <= i; j++) {
            if (j > 1) cout << ' ';
            cout << (i + j - 1);
        }
        cout << "\n";
    }

    return 0;
}

🧠 How It Works

1

Choose the row count

int rows = 5; sets how many lines the triangle prints.

Setup
2

Outer loop (rows)

for (i = 1; i <= rows; i++) controls the current row number i.

Row control
3

Inner loop (print i values)

for (j = 1; j <= i; j++) prints exactly i numbers on row i.

Column control
4

Compute the value

Printing i + j - 1 makes each row start at i and increase by 1 as j grows.

Value rule
=

Shifted consecutive numbers

Total printed numbers are 1+2+…+n = n(n+1)/2, so time complexity is O(n²) for n rows.

2

Variation — User Input Version

Let the user decide the number of rows at runtime using cin:

C++
#include <iostream>
using namespace std;

int main() {
    int rows;
    int i, j;

    cout << "Enter the number of rows: ";
    cin >> rows;

    for (i = 1; i <= rows; i++) {
        for (j = 1; j <= i; j++) {
            if (j > 1) cout << ' ';
            cout << (i + j - 1);
        }
        cout << "\n";
    }

    return 0;
}

💡 Tips for Enhancement

Try These

  • Validate input (e.g., check cin.fail()) before using rows
  • Right-align the triangle by printing leading spaces before each row
  • Start each row from a different base (e.g., base + i + j - 1)
  • Format with fixed width using setw() for neat columns
  • Try the decreasing-width version by printing j = 1..(rows-i+1)

Avoid

  • Hard-coding 5 in multiple places (use rows)
  • Printing extra spaces at the end of each line (use a separator rule)
  • Forgetting the newline after each row
  • Using endl inside loops (it flushes output and slows printing)

Key Takeaways

1

The outer loop selects the row number i.

2

The inner loop prints exactly i numbers on row i.

3

The expression i + j - 1 creates the shifted consecutive sequence per row.

4

Work is triangular: n(n+1)/2, which gives O(n²) time.

❓ Frequently Asked Questions

Because row i prints i + 1 - 1 first, which equals i. So row 1 starts at 1, row 2 starts at 2, row 3 starts at 3, and so on.
Yes. Add a base: print base + i + j - 1. If base = 10, the first row starts at 11 instead of 1.
Simply omit printing the separator space. For example, remove the if (j > 1) cout << ' '; line.
O(n²) for n rows: total prints are 1+2+…+n = n(n+1)/2.

Explore More C++ Number Patterns!

Keep practicing nested loops by building triangles, pyramids, and shifting sequences.

All Number Patterns →
Did you know?

The formula i + j - 1 is a classic way to build diagonals of consecutive numbers in grids and triangles. It shows up often in pattern problems because it naturally increases by 1 as you move right.

About the author

Mari Selvan M P
Mari Selvan M P 🔗

Developer, cloud engineer, and technical writer

  • Experience 12 years building web and cloud systems
  • Focus Full Stack Development, AWS, and Developer Education

I write practical tutorials so students and working developers can learn by doing—from databases and APIs to deployment on AWS.

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