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# Sum of numbers from 1 to N which are in Lucas Sequence

• Last Updated : 11 May, 2021

Given a number N. The task is to find the sum of numbers from 1 to N, which are present in the Lucas Sequence.

The Lucas numbers are in the following integer sequence:

`2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123 ......`

Examples:

```Input :  N = 10
Output : 17

Input : N = 5
Output : 10```

Approach:

• Loop through all the Lucas numbers which are less than the given value N.
• Initialize a sum variable with 0.
• Keep on adding these lucas numbers to get the required sum.

Below is the implementation of the above approach:

## C++

 `// C++ program to find sum of numbers from``// 1 to N which are in Lucas Sequence``#include ``using` `namespace` `std;` `// Function to return the``// required sum``int` `LucasSum(``int` `N)``{``    ``// Generate lucas number and keep on``    ``// adding them``    ``int` `sum = 0;``    ``int` `a = 2, b = 1, c;` `    ``sum += a;` `    ``while` `(b <= N) {` `        ``sum += b;``        ``int` `c = a + b;``        ``a = b;``        ``b = c;``    ``}` `    ``return` `sum;``}` `// Driver code``int` `main()``{``    ``int` `N = 20;``    ``cout << LucasSum(N);``    ``return` `0;``}`

## Java

 `// java program to find sum of numbers from``// 1 to N which are in Lucas Sequence``class` `GFG``{` `// Function to return the``// required sum``static` `int` `LucasSum(``int` `N)``{``    ``// Generate lucas number and keep on``    ``// adding them``    ``int` `sum = ``0``;``    ``int` `a = ``2``, b = ``1``, c;` `    ``sum += a;` `    ``while` `(b <= N) {` `        ``sum += b;``        ``c = a + b;``        ``a = b;``        ``b = c;``    ``}` `    ``return` `sum;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``20``;``    ``System.out.println(LucasSum(N));``    ` `}``// This code is contributed by princiraj1992``}`

## Python3

 `# Python3 program to find Sum of``# numbers from 1 to N which are``# in Lucas Sequence` `# Function to return the``# required Sum``def` `LucasSum(N):``    ` `    ``# Generate lucas number and``    ``# keep on adding them``    ``Sum` `=` `0``    ``a ``=` `2``    ``b ``=` `1``    ``c ``=` `0` `    ``Sum` `+``=` `a` `    ``while` `(b <``=` `N):` `        ``Sum` `+``=` `b``        ``c ``=` `a ``+` `b``        ``a ``=` `b``        ``b ``=` `c` `    ``return` `Sum` `# Driver code``N ``=` `20``print``(LucasSum(N))` `# This code is contributed``# by mohit kumar`

## C#

 `// C# program to find sum of numbers from``// 1 to N which are in Lucas Sequence``using` `System;` `class` `GFG``{` `// Function to return the``// required sum``static` `int` `LucasSum(``int` `N)``{``    ``// Generate lucas number and keep on``    ``// adding them``    ``int` `sum = 0;``    ``int` `a = 2, b = 1, c;` `    ``sum += a;` `    ``while` `(b <= N)``    ``{` `        ``sum += b;``        ``c = a + b;``        ``a = b;``        ``b = c;``    ``}` `    ``return` `sum;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 20;``    ``Console.WriteLine(LucasSum(N));``}``}` `// This code contributed by Rajput-Ji`

## PHP

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## Javascript

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Output:

`46`

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