Test if a number is fibonacci

Question

I know how to make the list of the Fibonacci numbers, but i don\'t know how can i test if a given number belongs to the fibonacci list - one way that comes in mind is genera

Answer 1

Positive integer ω is a Fibonacci number

If and only if one of 5ω² + 4 and 5ω² - 4 is a perfect square

from The (Fabulous) FIBONACCI Numbers by Alfred Posamentier and Ingmar Lehmann

bool isFibonacci(int  w)
{
       double X1 = 5 * Math.Pow(w, 2) + 4;
       double X2 = 5 * Math.Pow(w, 2) - 4;

       long X1_sqrt = (long)Math.Sqrt(X1);
       long X2_sqrt = (long)Math.Sqrt(X2);   

       return (X1_sqrt*X1_sqrt == X1) || (X2_sqrt*X2_sqrt == X2) ;
}

I copied it from this source

---

Snippet that prints Fibonacci numbers between 1k and 10k.

for (int i = 1000; i < 10000; i++)
{
         if (isFibonacci(i))
              Console.Write(" "+i);
}

OMG There are only FOUR!!!

With other method

from math import *

phi = 1.61803399
sqrt5 = sqrt(5)

def F(n):
    return int((phi**n - (1-phi)**n) /sqrt5)

def isFibonacci(z):
    return F(int(floor(log(sqrt5*z,phi)+0.5))) == z

print [i for i in range(1000,10000) if isFibonacci(i)]

Answer 2

A very nice test is that N is a Fibonacci number if and only if 5 N^2 + 4 or 5N^2 – 4 is a square number. For ideas on how to efficiently test that a number is square refer to the SO discussion.

Hope this helps

Answer 3

This is my solution I'm not sure if it benchmarks. I hope this helps!

def is_fibonacci?(i)
  a,b=0,1
    until b >= i
        a,b=b,a+b
        return true if b == i
    end
end

what a,b=b,a+b is doing

 0, 1 = 1, 0 +1
 1, 1 = 1, 1 + 1
 1, 2 = 2, 1 + 2
 2, 3 = 3, 2 + 3

fib1 = fib2
fib2 = fib1 + fib2

Answer 4

#include <stdio.h>
#include <math.h>

int main()
{
int number_entered, x, y;

printf("Please enter a number.\n");
scanf("%d", &number_entered);
x = y = 5 * number_entered^2 + 4;        /*Test if 5N^2 + 4 is a square number.*/
x = sqrt(x);
x = x^2;
if (x == y)
{
        printf("That number is in the Fibonacci sequence.\n");
    }
x = y = 5 * number_entered^2 - 4;        /*Test if 5N^2 - 4 is a square number.*/
x = sqrt(x);
x = x^2;
if (x == y)
{
    printf("That number is in the Fibonacci sequence.\n");
}
else
{
    printf("That number isn't in the Fibonacci sequence.\n");
}
return 0;
}

Will this work?

Answer 5

A positive integer ω is a Fibonacci number if and only if either 5ω² + 4 or 5ω² - 4 is a perfect square.

See The Fabulous Fibonacci Numbers for more.

Answer 6

Based on earlier answers by me and psmears, I've written this C# code.

It goes through the steps slowly, and it can clearly be reduced and optimized:

// Input: T: number to test.
// Output: idx: index of the number in the Fibonacci sequence.
//    eg: idx for 8 is 6. (0, 1, 1, 2, 3, 5, 8)
// Return value: True if Fibonacci, False otherwise.
static bool IsFib(long T, out int idx)
{
    double root5 = Math.Sqrt(5);
    double PSI = (1 + root5) / 2;

    // For reference, IsFib(72723460248141) should show it is the 68th Fibonacci number

    double a;

    a = T*root5;
    a = Math.Log(a) / Math.Log(PSI);
    a += 0.5;
    a = Math.Floor(a);
    idx = (Int32)a;

    long u = (long)Math.Floor(Math.Pow(PSI, a)/root5 + 0.5);

    if (u == T)
    {
        return true;
    }
    else
    {
        idx = 0;
        return false;
    }
}

Testing reveals this works for the first 69 Fibonacci numbers, but breaks down for the 70th.

F(69) = 117,669,030,460,994 - Works
F(70) = 190,392,490,709,135 - Fails

In all, unless you're using a BigInt library of some kind, it is probably better to have a simple lookup table of the Fibonacci Numbers and check that, rather than run an algorithm.

A list of the first 300 Numbers is readily available online.

But this code does outline a workable algorithm, provided you have enough precision, and don't overflow your number representation system.