Find the sum of first N odd Fibonacci numbers

Given a number, N. Find the sum of first N odd Fibonacci numbers.

Note: The answer can be very large so print the answer modulo 10^9+7.

Examples:  

Input : N = 3
Output : 5
Explanation : 1 + 1 + 3
Input : 6
Output : 44
Explanation : 1 + 1 + 3 + 5 + 13 + 21

Approach: Odd Fibonacci series is:  

1, 1, 3, 5, 13, 21, 55, 89......

Prefix sum of odd Fibonacci series is:  

1, 2, 5, 10, 23, 44, 99, 188.....

The formula for the sum of first N odd Fibonacci numbers is: 

a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3) + a(n-4) - a(n-5) for n>5  

Below is the implementation of the above approach: 

C++

// CPP program to Find the sum of
// first N odd Fibonacci numbers
#include <bits/stdc++.h>
using namespace std;

#define mod 1000000007

// Function to calculate sum of first
// N odd Fibonacci numbers
long long sumOddFibonacci(int n)
{
    long long Sum[n + 1];

    // base values
    Sum[0] = 0;
    Sum[1] = 1;
    Sum[2] = 2;
    Sum[3] = 5;
    Sum[4] = 10;
    Sum[5] = 23;

    for (int i = 6; i <= n; i++) {
        Sum[i] = ((Sum[i - 1] + (4 * Sum[i - 2]) % mod -
                  (4 * Sum[i - 3]) % mod + mod) % mod +
                  (Sum[i - 4] - Sum[i - 5] + mod) % mod) % mod;
    }

    return Sum[n];
}

// Driver code
int main()
{
    long long n = 6;

    cout << sumOddFibonacci(n);

    return 0;
}

Java

// Java  program to Find the sum of
// first N odd Fibonacci numbers

import java.io.*;

class GFG {
    static int mod =1000000007;

// Function to calculate sum of first
// N odd Fibonacci numbers
static  int sumOddFibonacci(int n)
{
     int Sum[]=new int[n + 1];

    // base values
    Sum[0] = 0;
    Sum[1] = 1;
    Sum[2] = 2;
    Sum[3] = 5;
    Sum[4] = 10;
    Sum[5] = 23;

    for (int i = 6; i <= n; i++) {
        Sum[i] = ((Sum[i - 1] + (4 * Sum[i - 2]) % mod -
                (4 * Sum[i - 3]) % mod + mod) % mod +
                (Sum[i - 4] - Sum[i - 5] + mod) % mod) % mod;
    }

    return Sum[n];
}

// Driver code

    public static void main (String[] args) {

    int n = 6;
    System.out.println(sumOddFibonacci(n));
    }
//This Code is Contributed by Sachin
}

Python3

# Python3 program to Find the sum of
# first N odd Fibonacci numbers
mod = 1000000007 ;

# Function to calculate sum of
# first N odd Fibonacci numbers
def sumOddFibonacci(n):

    Sum=[0]*(n + 1);

    # base values
    Sum[0] = 0;
    Sum[1] = 1;
    Sum[2] = 2;
    Sum[3] = 5;
    Sum[4] = 10;
    Sum[5] = 23;

    for i in range(6,n+1):
        Sum[i] = ((Sum[i - 1] +
                    (4 * Sum[i - 2]) % mod -
                    (4 * Sum[i - 3]) % mod +
                    mod) % mod + (Sum[i - 4] -
                    Sum[i - 5] + mod) % mod) % mod;

    return Sum[n];

# Driver code
n = 6;
print(sumOddFibonacci(n));

# This code is contributed by mits

C#

// C#  program to Find the sum of
// first N odd Fibonacci numbers

using System;

public class GFG{

static int mod =1000000007;
// Function to calculate sum of first
// N odd Fibonacci numbers
static int sumOddFibonacci(int n)
{
    int []Sum=new int[n + 1];

    // base values
    Sum[0] = 0;
    Sum[1] = 1;
    Sum[2] = 2;
    Sum[3] = 5;
    Sum[4] = 10;
    Sum[5] = 23;

    for (int i = 6; i <= n; i++) {
        Sum[i] = ((Sum[i - 1] + (4 * Sum[i - 2]) % mod -
                (4 * Sum[i - 3]) % mod + mod) % mod +
                (Sum[i - 4] - Sum[i - 5] + mod) % mod) % mod;
    }

    return Sum[n];
}

// Driver code



    static public void Main (){
        int n = 6;
    Console.WriteLine(sumOddFibonacci(n));
    }
//This Code is Contributed by Sachin
}

Javascript

<script>
// javascript  program to Find the sum of
// first N odd Fibonacci numbers

    var mod = 1000000007;

    // Function to calculate sum of first
    // N odd Fibonacci numbers
    function sumOddFibonacci(n) {
        var Sum = Array(n + 1).fill(0);

        // base values
        Sum[0] = 0;
        Sum[1] = 1;
        Sum[2] = 2;
        Sum[3] = 5;
        Sum[4] = 10;
        Sum[5] = 23;

        for (i = 6; i <= n; i++) {
            Sum[i] = ((Sum[i - 1] + (4 * Sum[i - 2]) % mod - (4 * Sum[i - 3]) % mod + mod) % mod
                    + (Sum[i - 4] - Sum[i - 5] + mod) % mod) % mod;
        }

        return Sum[n];
    }

    // Driver code



        var n = 6;
        document.write(sumOddFibonacci(n));

// This code contributed by umadevi9616
</script>

PHP

<?php
// PHP program to Find the sum of
// first N odd Fibonacci numbers
$mod = 1000000007 ;

// Function to calculate sum of
// first N odd Fibonacci numbers
function sumOddFibonacci($n)
{
    global $mod;
    $Sum[$n + 1] = array();

    // base values
    $Sum[0] = 0;
    $Sum[1] = 1;
    $Sum[2] = 2;
    $Sum[3] = 5;
    $Sum[4] = 10;
    $Sum[5] = 23;

    for ($i = 6; $i <= $n; $i++)
    {
        $Sum[$i] = (($Sum[$i - 1] +
                    (4 * $Sum[$i - 2]) % $mod -
                    (4 * $Sum[$i - 3]) % $mod +
                    $mod) % $mod + ($Sum[$i - 4] -
                    $Sum[$i - 5] + $mod) % $mod) % $mod;
    }

    return $Sum[$n];
}

// Driver code
$n = 6;
echo sumOddFibonacci($n);

// This code is contributed by jit_t
?>

Output

44

Time Complexity: O(n), Auxiliary Space: O(n)

Efficient approach : Space optimization O(1)

In previous approach we the current value Sum[i] is only depend upon the previous 6 values i.e. Sum[i-1], Sum[i-2], …. Sum[i-5]  So to optimize the space we can keep track of previous and current values by the help of three variables Sum0, Sum1 , …. ,  Sum5 which will reduce the space complexity from O(N) to O(1).  

Implementation Steps:

• Create variables Sum0, Sum1 , …. ,  Sum5 to keep track of previous values of Sum[].
• Initialize base case.
• Create a variable Sum to store current value.
• Iterate over subproblem using loop and update Sum.
• After every iteration update Sum0, Sum1 , …. ,  Sum5 for further iterations.
• At last return Sum.

Implementation:

C++

// CPP program to Find the sum of
// first N odd Fibonacci numbers
#include <bits/stdc++.h>
using namespace std;

#define mod 1000000007

// Function to calculate sum of first
// N odd Fibonacci numbers
long long sumOddFibonacci(int n)
{
//     long long Sum[n + 1];

    // base values
    int Sum0 = 0;
    int Sum1 = 1;
    int Sum2 = 2;
    int Sum3 = 5;
    int Sum4 = 10;
    int Sum5 = 23;

    int Sum;

    for (int i = 6; i <= n; i++) {
        Sum = ((Sum5 + (4 * Sum4) % mod -
                (4 * Sum3) % mod + mod) % mod +
                (Sum2 - Sum1 + mod) % mod) % mod;

            //assigning values for further iteration
            Sum0 = Sum1;
            Sum1 = Sum2;
            Sum2 = Sum3;
            Sum3 = Sum4;
            Sum4 = Sum5;
            Sum5 = Sum;
    }

      // return final answer
    return Sum;
}

// Driver code
int main()
{
    long long n = 6;

      // function call
    cout << sumOddFibonacci(n);

    return 0;
}

Java

import java.util.*;

public class Main {

    // Function to calculate sum of first
    // N odd Fibonacci numbers
    static long sumOddFibonacci(int n)
    {
        int mod = 1000000007;

        // base values
        int Sum0 = 0;
        int Sum1 = 1;
        int Sum2 = 2;
        int Sum3 = 5;
        int Sum4 = 10;
        int Sum5 = 23;

        int Sum;

        for (int i = 6; i <= n; i++) {
            Sum = ((Sum5 + (4 * Sum4) % mod
                    - (4 * Sum3) % mod + mod)
                       % mod
                   + (Sum2 - Sum1 + mod) % mod)
                  % mod;

            // assigning values for further iteration
            Sum0 = Sum1;
            Sum1 = Sum2;
            Sum2 = Sum3;
            Sum3 = Sum4;
            Sum4 = Sum5;
            Sum5 = Sum;
        }

        // return final answer
        return Sum5;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int N = 6;
        System.out.println(sumOddFibonacci(N));
    }
}

Python

MOD = 1000000007

# Function to calculate sum of first
# N odd Fibonacci numbers
def sumOddFibonacci(n):
    # base values
    Sum0 = 0
    Sum1 = 1
    Sum2 = 2
    Sum3 = 5
    Sum4 = 10
    Sum5 = 23

    for i in range(6, n+1):
        Sum = ((Sum5 + (4 * Sum4) % MOD -
                (4 * Sum3) % MOD + MOD) % MOD +
                (Sum2 - Sum1 + MOD) % MOD) % MOD

        # assigning values for further iteration
        Sum0 = Sum1
        Sum1 = Sum2
        Sum2 = Sum3
        Sum3 = Sum4
        Sum4 = Sum5
        Sum5 = Sum

    # return final answer
    return Sum

# Driver code
n = 6

# function call
print(sumOddFibonacci(n))

C#

using System;

class GFG
{
    const int mod = 1000000007;

    // Function to calculate the sum of the first N odd Fibonacci numbers
    static long SumOddFibonacci(int n)
    {
        long sum1 = 1;
        long sum2 = 2;
        long sum3 = 5;
        long sum4 = 10;
        long sum5 = 23;
        long sum = 0;

        for (int i = 6; i <= n; i++)
        {
            sum = ((sum5 + (4 * sum4) % mod - (4 * sum3) % mod + mod) % mod
                + (sum2 - sum1 + mod) % mod) % mod;

            // Shift values for further iterations
            sum1 = sum2;
            sum2 = sum3;
            sum3 = sum4;
            sum4 = sum5;
            sum5 = sum;
        }

        // Return the final answer
        return sum;
    }

    // Driver code
    static void Main()
    {
        long n = 6;

        // Function call
        long result = SumOddFibonacci((int)n);
        Console.WriteLine(result);
    }
}

Javascript

function sumOddFibonacci(n) {
    const mod = 1000000007;

    // Initialize base values
    let Sum0 = 0;
    let Sum1 = 1;
    let Sum2 = 2;
    let Sum3 = 5;
    let Sum4 = 10;
    let Sum5 = 23;

    let Sum;

    for (let i = 6; i <= n; i++) {
        Sum = (((Sum5 + (4 * Sum4) % mod - (4 * Sum3) % mod + mod) % mod + (Sum2 - Sum1 + mod) % mod) % mod);

        // Assign values for further iteration
        Sum0 = Sum1;
        Sum1 = Sum2;
        Sum2 = Sum3;
        Sum3 = Sum4;
        Sum4 = Sum5;
        Sum5 = Sum;
    }

    // Return the final answer
    return Sum5;
}

// Driver Code
const N = 6;
console.log(sumOddFibonacci(N));

Output:

44

Time Complexity: O(n)

Auxiliary Space: O(1)