Find the sum of first N odd Fibonacci numbers

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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     
} 

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
?>

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>

Output:

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;
}

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))

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));
    }
}

Output:

Time Complexity: O(n)

Auxiliary Space: O(1)

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