An increasing triangle from 1 prints values from the formula i + j - 1. Row i has i consecutive numbers starting at i, separated by spaces.
Remember
Rule: for i from 1 to rows
for j from 1 to i:
print (i + j - 1) + space
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9 ← rows = 5
Same shape as Program 32, different formula. Follows the triangle from 11 in Program 32; next is Program 34 (starts at 0).
Approach
How to Solve It
Start with fixed rows = 5, then generalize with scanf — optionally demo a smaller count for tracing.
Method
Idea
Best for
Fixed formula
Print i + j - 1 for j = 1..i
Learning, interviews, exams
User input rows
Same loops; read rows with scanf
Reusable demos and labs
Pseudocode
Pseudocode
for i from 1 to rows:
for j from 1 to i:
print (i + j - 1) and a space
print newline
Cheat sheet
Goal
Pattern
Walk each row
for (i = 1; i <= rows; i++)
Grow row length
for (j = 1; j <= i; j++)
Print value + space
printf("%d ", i + j - 1);
End the row
printf("\n");
Start at 11 instead
printf("%d ", 9 + i + j); — see Program 32
Printing Numbers vs Starting a New Line
API
Effect
Use for
printf("%d ", i + j - 1)
Stays on the same line
Each value plus a separator space
printf("\n")
Ends the current line
After the inner loop
Print all values on the row without a newline, then end the row once.
Try it
Live Preview
Change the row count and the triangle updates instantly — each row starts at its row number via i + j - 1.
Whole numbers from 1 to 9. Tap a chip or type a value — the preview redraws as you go.
Live result5 rows · 15 values
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9
Trace
Worked Walkthrough — rows = 4
Trace each outer-loop value of i and list the formula results for j = 1..i.
i
Values (i + j - 1)
Printed row
1
1
1
2
2, 3
2 3
3
3, 4, 5
3 4 5
4
4, 5, 6, 7
4 5 6 7
Total values for n rows = n(n + 1)/2 (here 4 × 5 / 2 = 10). That triangular sum is why time is O(n²).
Code
C Programs
Three complete programs: fixed rows, scanf input, and a small demo. Use View Output to reveal sample results.
Example 1 — Fixed rows = 5
Hard-coded row count — classic i + j - 1 formula.
C
#include <stdio.h>
int main(void)
{
int i, j;
for (i = 1; i <= 5; i++)
{
for (j = 1; j <= i; j++)
printf("%d ", i + j - 1);
printf("\n");
}
return 0;
}
Output
1
2 3
3 4 5
4 5 6 7
5 6 7 8 9
How It Works
1. Outer loop grows i.i runs from 1 to 5 — one longer row each pass.
2. Inner loop prints the formula.j runs from 1 to i; each value is i + j - 1 plus a space.
3. Newline.printf("\n") ends the row after the inner loop.
When i = 1 you get 1; when i = 4 you get 4 5 6 7.
Example 2 — User Input Version
Read rows with scanf and clamp to a positive value. Same formula core.
C
#include <stdio.h>
int main(void)
{
int rows;
int i, j;
printf("Enter rows: ");
scanf("%d", &rows);
if (rows < 1) rows = 1;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
printf("%d ", i + j - 1);
printf("\n");
}
return 0;
}
Output (when user enters 3)
Enter rows: 3
1
2 3
3 4 5
How It Works
1. Prompt and clamp. Read rows, then force a minimum of 1 so the outer loop runs.
2. Same nested core. Only the hard-coded 5 becomes the user value.
3. Safer input tip. Unchecked scanf leaves rows uninitialized on bad input. Prefer:
Safer input
if (scanf("%d", &rows) != 1 || rows < 1)
{
printf("Enter a positive whole number of rows.\n");
return 1;
}
Example 3 — Compact rows = 3
Same formula with a smaller row count for quick paper tracing.
C
#include <stdio.h>
int main(void)
{
int rows = 3;
int i, j;
for (i = 1; i <= rows; i++)
{
for (j = 1; j <= i; j++)
printf("%d ", i + j - 1);
printf("\n");
}
return 0;
}
Output
1
2 3
3 4 5
How It Works
1. Same structure. Only rows changes from 5 to 3 — the formula stays i + j - 1.
2. Trace on paper. For i = 2: 2+1-1 = 2, 2+2-1 = 3 → 2 3.
3. Scale up next. Once the small demo is clear, use Examples 1–2 for five rows or user input.
Edge Cases & Pitfalls
Check these before calling the solution done.
wrong formula
Rows start at the wrong number
Using i + j or 9 + i + j shifts every value. Keep i + j - 1 so row i begins at i.
no space
Glued numbers
printf("%d", i + j - 1) without a space merges values like 23. Always append a space in the format string.
\n inside
Broken rows
If printf("\n") sits inside the inner loop, each value lands on its own line. Call the newline only after the row finishes.
j to rows
Rectangle instead of triangle
Looping j to rows instead of i prints the same count every line. Inner bound must be i.
rows = 1
Smallest triangle
Output is a single 1 — a good sanity check.
scanf
Check the return value
If scanf fails, rows may be uninitialized — always test scanf(...) == 1.
Analysis
Time and Space Complexity
Program
Time
Extra space
Fixed / input (Examples 1–2)
O(rows²)
O(1)
Small demo (Example 3)
O(rows²)
O(1)
Total values = n(n + 1)/2 — quadratic in n. For rows = 5 that is 15 printed numbers.
Remember
Key Takeaways
Rule: for each i, print i + j - 1 for j = 1..i.
Row start: when j = 1, the value equals i — each row begins at its row number.
Break the row: call printf("\n") only after the inner loop.
Complexity:O(n²) time from n(n + 1)/2 values; O(1) extra space.
One line: for i from 1 to rows, print i + j - 1 for j = 1..i, then printf("\n").
Frequently Asked Questions
Because when j = 1, the expression i + j - 1 becomes i. Row 4 therefore starts with 4.
j increases by 1, so i + j - 1 increases by 1 as well — producing consecutive numbers on each row.
Program 32 uses 9 + i + j (starts at 11). Program 33 uses i + j - 1 (starts at 1).
Program 33 uses i + j - 1 with i starting at 1. Program 34 uses i + j with i starting at 0.
printf("%d ", i + j - 1) keeps values separated on the same row. printf("\n") ends the row.
Replace 5 with rows in the outer loop bound — see Example 2.
O(n²) for n rows because total prints are 1 + 2 + ... + n = n(n+1)/2.
Check scanf's return value and reject non-positive row counts so the outer loop has a valid range.
🤔
Did you know?
Each printed value is computed as i + j - 1. Row i = 1 prints 1; row i = 4 prints 4, 5, 6, 7 — a left-shifted increasing triangle starting at 1.
