Nth Fibonacci Number

public class FibanocciSeries {
 public static void main(String[] args) {
  // 0 1 1 2 3 5 8 13.....
  // 0 1 2 3 4 5 6 7.....
  int n = getNthFibanocciNumber_Rec(4);
  System.out.println(n);// recursively calculated value

  n = fib(4);
  System.out.println(n);// fibanocci number without recursion
 }

 public static int fib(int n) {
  int a = 0, b = 1, c = 1;
  while (n > 1) {
   c = a + b;
   a = b;
   b = c;
   n--;
  }
  return c;
 }

 public static int getNthFibanocciNumber_Rec(int n) {
  if (n == 0)
   return 0;
  if (n == 1 || n == 2)
   return 1;
  else
   return getNthFibanocciNumber_Rec(n - 1)
     + getNthFibanocciNumber_Rec(n - 2);
 }
}

Refer this for O(logN) solution :
 http://tech-queries.blogspot.com/2010/09/nth-fibbonacci-number-in-ologn.html

Checking given number is fibanocci or not: A positive integer ω is a Fibonacci number if and only if one of 5ω2 + 4 and 5ω2 - 4 is a perfect square.

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) ;
}

Finding Nth Fibonacci number using Dynamic Programming : 

        static int[] fib = new int[10];
 //Top down approach(memorization)
 static int getFibanocci(int n) {
  if(n<=1){
   fib[n]=n;
  }    
  else if(fib[n]==0){
   fib[n] = getFibanocci(n-1)+getFibanocci(n-2); 
  }
  return fib[n];
 }
 
 //bottom up approach(tabulation)
 static int getFibanocciNumber_BottomUp(int n){
  int[] a = new int[n+1];
  a[1]=1;
  for(int i=2;i<=n;i++){
   a[i] = a[i-1]+a[i-2];
  }
  return a[n];
 }