Last Digit of the Sum of Fibonacci Numbers

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I am trying to find the last digit of sum of Fibonacci Series. I calculate the sum as F(n+2) - 1. The below code is working fine but it is slow for large numbers (e.g 99999). How can I optimize this?

n = int(input())

def last_digit(n):
    a, b = 0, 1
    for i in range(n+2):
        a, b = b, a + b
    return (a-1) % 10

print(last_digit(n))

Look at this table: http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html notice that fib(60) last digit is 0 and fib(61) last digit is 1, that is same as fib(0) and fib(1), thus starting at 60 last digits starts to repeat, so you can calculate last digit for fib(n%60) rather than fib(n). For example last digit is same for fib(115) and fib(55) and equal to 5.

Sum of Fibonacci Numbers, SOLVED IT. Works on all range of inputs. It works on the following algorithm. The idea is to notice that the last digits of fibonacci numbers also occur in  Task: Given an integer n, find the last digit of the nth Fibonacci number F(n) (that is, F(n) mod 10). Input Format: The input consists of a single integer n . Constraints: 0 ≤ n ≤ 10 ^7.


The series of final digits of Fibonacci numbers repeats with a cycle of 60. Therefore, you can optimize the calculation of the sum of n terms to F((n+2) % 60) - 1. Also, to stay in the integer range, you can keep only the last digit of each term:

def last_digit(n):
    a, b = 0, 1
    for i in range((n + 2) % 60):
        a, b = b, (a + b) % 10
    return 9 if a == 0 else a - 1

print([last_digit(n) for n in range(1, 11)])

Output:

[1, 2, 4, 7, 2, 0, 3, 4, 8, 3]

Fibonacci Sum of Large Numbers(Only Last Digit to be Printed , Now we have to calculate the last digit of the (full or partial) sum of Fibonacci numbers. Also these problems are presented in the the edX  If you write out a sequence of Fibonacci numbers, you can see that the last digits repeat every 60 numbers. The 61st Fibonacci number is 2504730781961. The 62nd is 4052739537881. Since these end in 1 and 1, the 63rd Fibonacci number must end in 2, etc. and so the pattern starts over.


Here's code optimized specifically for the last digit:

def fib(n):
    a, b = 0, 1
    r = 1
    if n < 1:
        return 0
    for i in range(n - 1):
        a, b = b, (a + b)%10
        r += b
        r %= 10
    return r

It works by getting only the last digit of the next term and adding that to the result. It then gets the last digit of the result and sets it to itself. It repeats until it gets to the term number and returns a one-digit number :D

Fun Fact: Try the above function on 99. Returns 0. What about 999? 0. 9999? 0. Continue this :D

Last Digit of the Sum of Fibonacci Numbers, Instead of summing Fn-1 + Fn-2, we just sum the last digit of Fn-1 and last digit of Fn-2. Advanced Problem 6: Huge Fibonacci Number modulo m. Last Digit of the Sum of Fibonacci Numbers Given an integer 𝑛, find the last digit of the sum 𝐹0 + 𝐹1 + · · · + 𝐹𝑛. Considering that n could be as big as 10^14, the naive solution of summing up all the Fibonacci numbers as long as we calculate them is leading too slowly to the result.


Try to use the Pisano Period property. If you want to compute the last digit, Pisano Period of 10 will be 60. Knowing this, you can have a function similar to:

def fibonacci_sum(n):

    pisano = 60

    if n < 2: return n

    n %= pisano

    fib_arr = [1,1]
    for _ in range(n):
        fib_arr.append((fib_arr[-1] + fib_arr[-2]) % 10)

    return (fib_arr[-1] - 1) % 10

For more information refer to saveriogzz Github CS_Curriculum repo.

Cheers!

Fibonacci Numbers Lines Definition and Uses, The Last Digit of a Large Fibonacci Number — aadimator i grows the ith iteration of the loop computes the sum of longer and longer numbers. Here's code optimized specifically for the last digit: def fib(n): a, b = 0, 1 r = 1 if n < 1: return 0 for i in range(n - 1): a, b = b, (a + b)%10 r += b r %= 10 return r It works by getting only the last digit of the next term and adding that to the result.


Here is a simple C++ program

//outputs last digit of ( sum of fib number till n)
#include<iostream>
using namespace std;

int64_t fib_sum_digit(int64_t n)
{
    int fl[60] ={0};
    fl[0] = 0;
    fl[1] = 1;
   // int64_t sum60 = 1;

    for(int i = 2 ; i<60 ; i++)
    {
        fl[i] = (fl[i-1] +fl[i-2])%10 ; 
        //sum60 += fl[i]; 
    }


    int64_t sum = 0;
    // sum += (sum60*(n/60));               ///sum60%10 always  = 0 ;

    for(int i = 1; i<=(n%60); i++ )
    {
        sum += (fl[i]);
        //cout<<i<<","<<sum<<"->";         ///debug
    }

return sum%10;
}

int main()
{
    int64_t n;
    cin>>n;

    int64_t ans = fib_sum_digit(n);
    cout<<ans;

    return 0;
}

What is the Fibonacci Sequence (aka Fibonacci Series)?, 1 Patterns in the Fibonacci Numbers. 1.1 The Final Digits; 1.2 Digit Sums. 1.2.1 A new research question for you to try. 1.3 calculator Fibonacci Number Digit  Fibonacci numbers tend to get quite big - as a very naive lower bound, the number at least doubles after every two iterations. This gets hard to use quite quickly - according to the naive lower bound, the number will exceed 2 64 after at most the 129-th term.


Fibonacci Sequence, Since the Fibonacci numbers are determined by a two-term recurrence, and since the last digit of a sum is determined by the sum of the last  Given a positive integer N. The task is to find the sum of squares of all Fibonacci numbers up to N-th fibonacci number. That is, f 0 2 + f 1 2 + f 2 2 +..+f n 2 where f i indicates i-th fibonacci number.


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