# Program to print numeric pattern | Set – 2

• Difficulty Level : Medium
• Last Updated : 22 Sep, 2021

Given a number as ‘num’, and Number of lines as ‘num_of_lines’ where ‘num’ implies the starting number from which the pattern has to be started and ‘num_of_lines’ implies the number of lines that have to be printed. Now, according to the above information, print a pattern as given below.
Examples:

```Input: num = 7, num_of_lines = 4
Output: 7
14 15
28 29 30 31
56 57 58 59 60 61 62 63

Input: num = 3, num_of_lines = 3
Output: 3
6 7
12 13 14 15```

Observations:

• Elements of the first column are the multiple of the previous element in that column.
• No. of elements in each row is the twice of the no. of elements in the previous row.
• Also, to generate next element in the row, add 1 to previous element of that row.

Approach:
So, start a loop from 0 to num_of_lines-1, to take care of the number of rows to be printed and another loop inside the first loop, from 0 till limit-1, the limit will be initialized to 1, and its value is increased exponentially. Now inside the loop, just increase the number by 1 to print the next number of that row.

## C++

 `// C++ program to print the``// given numeric pattern``#include ``using` `namespace` `std;` `// Function to print th epattern``void` `printPattern (``int` `num, ``int` `numOfLines )``{``    ``int` `n = num, num2, x = 1, limit = 1;` `    ``// No. of rows to be printed``    ``for` `(``int` `i = 0; i < numOfLines; i++) {``        ` `        ``// No. of elements to be printed in each row``        ``for` `(``int` `j = 0; j < limit; j++) {``            ``if` `(j == 0)``                ``num2 = num;` `            ``// Print the element``            ``cout << num2++ << ``" "``;``        ``}``        ``num *= 2;``        ``limit = num / n;``        ``cout << endl;``    ``}``}` `// Drivers code``int` `main()``{` `    ``int` `num = 3;``    ``int` `numOfLines = 3;` `    ``printPattern(num,  numOfLines);``    ` `    ``return` `0;``}`

## Java

 `// Java program to print the``// given numeric pattern``class` `solution_1``{``    ` `// Function to print``// the pattern``static` `void` `printPattern (``int` `num,``                          ``int` `numOfLines)``{``    ``int` `n = num, num2 = ``0``,``        ``x = ``1``, limit = ``1``;` `    ``// No. of rows to``    ``// be printed``    ``for` `(``int` `i = ``0``;``             ``i < numOfLines; i++)``    ``{``        ` `        ``// No. of elements to be``        ``// printed in each row``        ``for` `(``int` `j = ``0``; j < limit; j++)``        ``{``            ``if` `(j == ``0``)``                ``num2 = num;` `            ``// Print the element``            ``System.out.print(num2++ + ``" "``);``        ``}``        ` `        ``num *= ``2``;``        ``limit = num / n;``        ``System.out.println();``    ``}``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `num = ``3``;``    ``int` `numOfLines = ``3``;` `    ``printPattern(num, numOfLines);``}``}` `// This code is contributed``// by Arnab Kundu`

## Python 3

 `# Python 3 program to print``# the given numeric pattern` `# Function to print th epattern``def` `printPattern (num, numOfLines ):` `    ``n ``=` `num``    ``limit ``=` `1` `    ``# No. of rows to be printed``    ``for` `i ``in` `range``(``0``, numOfLines):``        ` `        ``# No. of elements to be``        ``# printed in each row``        ``for` `j ``in` `range``(limit):``            ``if` `j ``=``=` `0``:``                ``num2 ``=` `num` `            ``# Print the element``            ``print``(num2, end ``=` `" "``)``            ``num2 ``+``=` `1``    ` `        ``num ``*``=` `2``        ``limit ``=` `num ``/``/` `n``        ``print``()``        ` `# Driver code``if` `__name__ ``=``=` `"__main__"``:` `    ``num ``=` `3``    ``numOfLines ``=` `3` `    ``printPattern(num, numOfLines)` `# This code is contributed``# by ChitraNayal`

## C#

 `// C# program to print the``// given numeric pattern``using` `System;``class` `GFG``{``    ` `// Function to print``// the pattern``static` `void` `printPattern(``int` `num,``                         ``int` `numOfLines)``{``    ``int` `n = num, num2 = 0,``        ``limit = 1;` `    ``// No. of rows to``    ``// be printed``    ``for` `(``int` `i = 0;``             ``i < numOfLines; i++)``    ``{``        ` `        ``// No. of elements to be``        ``// printed in each row``        ``for` `(``int` `j = 0; j < limit; j++)``        ``{``            ``if` `(j == 0)``                ``num2 = num;` `            ``// Print the element``            ``Console.Write(num2++ + ``" "``);``        ``}``        ` `        ``num *= 2;``        ``limit = num / n;``        ``Console.Write(``"\n"``);``    ``}``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `num = 3;``    ``int` `numOfLines = 3;` `    ``printPattern(num, numOfLines);``}``}` `// This code is contributed by Smitha`

## PHP

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## Javascript

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Output:
```3
6 7
12 13 14 15```

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