Interesting Problem : To Find nth NON-FIBONACCI NUMBER

First Few Non-Fibonacci nos. are : 4,6,7,9,10,11,12..........................

This is very similar to generating fibonacci numbers.Initialize your loop with the 3 initial fibonacci numbers a=1,b=1,c=2.Now , whenever this loop iterates, this tuple moves on to the next tuple of fibonacci numbers.(a=1,b=2,c=3).At every iteration, the number of non-fibonacci numbers jumped over is c-b+1.Keep adding this count as long as it is less than N. When this count exceeds N, it means that we just crossed the Nth Non-Fibonacci number on the numberLine. So if we move back the required numbers (the number by which count exceeds N), we will get the answer.

Complexity: O(log n)

              a=1 b=2 c=3

              while(n>0)

              {

               a=b;

               b=c;

               c=a+b;

               n-=(c-b-1);

              }

              n+=(c-b-1);

              ans=b+n;