Given a positive integer n, the task is to print the n’th non Fibonacci number. The Fibonacci numbers are defined as:

Fib(0) = 0 Fib(1) = 1 for n >1, Fib(n) = Fib(n-1) + Fib(n-2)

First few Fibonacci numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ……..

**Examples :**

Input : n = 2 Output : 6 Input : n = 5 Output : 10

A** naive solution** is to find find Fibonacci numbers and then print first n numbers not present in the found Fibonacci numbers.

A **better solution **is to use the formula of Fibonacci numbers and keep adding of gap between two consecutive Fibonacci numbers. Value of sum of gap is count of non-fibonacci numbers seen so far. Below is implementation of above idea.

## C++

`// C++ program to find n'th Fibonacci number ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Returns n'th Non-Fibonacci number ` `int` `nonFibonacci(` `int` `n) ` `{ ` ` ` `// curr is to keep track of current fibonacci ` ` ` `// number, prev is previous, prevPrev is ` ` ` `// previous of previous. ` ` ` `int` `prevPrev=1, prev=2, curr=3; ` ` ` ` ` `// While count of non-fibonacci numbers ` ` ` `// doesn't become negative or zero ` ` ` `while` `(n > 0) ` ` ` `{ ` ` ` `// Simple Fibonacci number logic ` ` ` `prevPrev = prev; ` ` ` `prev = curr; ` ` ` `curr = prevPrev + prev; ` ` ` ` ` `// (curr - prev - 1) is count of ` ` ` `// non-Fibonacci numbers between curr ` ` ` `// and prev. ` ` ` `n = n - (curr - prev - 1); ` ` ` `} ` ` ` ` ` `// n might be negative now. Make sure it ` ` ` `// becomes positive by removing last added ` ` ` `// gap. ` ` ` `n = n + (curr - prev - 1); ` ` ` ` ` `// n must be now positive and less than or equal ` ` ` `// to gap between current and previous, i.e., ` ` ` `// (curr - prev - 1); ` ` ` ` ` `// Now add the positive n to previous Fibonacci ` ` ` `// number to find the n'th non-fibonacci. ` ` ` `return` `prev + n; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `cout << nonFibonacci(5); ` ` ` `return` `0; ` `} ` |

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## Java

`// Java program to find ` `// n'th Fibonacci number ` `import` `java.io.*; ` ` ` `class` `GFG ` `{ ` ` ` `// Returns n'th Non- ` ` ` `// Fibonacci number ` ` ` `static` `int` `nonFibonacci(` `int` `n) ` ` ` `{ ` ` ` ` ` `// curr is to keep track of ` ` ` `// current fibonacci number, ` ` ` `// prev is previous, prevPrev ` ` ` `// is previous of previous. ` ` ` `int` `prevPrev = ` `1` `, prev = ` `2` `, curr = ` `3` `; ` ` ` ` ` `// While count of non-fibonacci ` ` ` `// numbers doesn't become ` ` ` `// negative or zero ` ` ` `while` `(n > ` `0` `) ` ` ` `{ ` ` ` `// Simple Fibonacci number logic ` ` ` `prevPrev = prev; ` ` ` `prev = curr; ` ` ` `curr = prevPrev + prev; ` ` ` ` ` `// (curr - prev - 1) is count ` ` ` `// of non-Fibonacci numbers ` ` ` `// between curr and prev. ` ` ` `n = n - (curr - prev - ` `1` `); ` ` ` `} ` ` ` ` ` `// n might be negative now. Make ` ` ` `// sure it becomes positive by ` ` ` `// removing last added gap. ` ` ` `n = n + (curr - prev - ` `1` `); ` ` ` ` ` `// n must be now positive and less ` ` ` `// than or equal to gap between ` ` ` `// current and previous, i.e., ` ` ` `// (curr - prev - 1); ` ` ` ` ` `// Now add the positive n to ` ` ` `// previous Fibonacci number ` ` ` `// to find the n'th non-fibonacci. ` ` ` `return` `prev + n; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main (String args[]) ` ` ` `{ ` ` ` `System.out.println(nonFibonacci(` `5` `)); ` ` ` `} ` `} ` ` ` `// This code is contributed by aj_36 ` |

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## Python

`# Python program to find n'th ` `# Fibonacci number ` ` ` `# Returns n'th Non-Fibonacci ` `# number ` `def` `nonFibonacci(n): ` ` ` ` ` `# curr is to keep track of ` ` ` `# current fibonacci number, ` ` ` `# prev is previous, prevPrev ` ` ` `# is previous of previous. ` ` ` `prevPrev ` `=` `1` ` ` `prev ` `=` `2` ` ` `curr ` `=` `3` ` ` ` ` `# While count of non-fibonacci ` ` ` `# numbers doesn't become ` ` ` `# negative or zero ` ` ` `while` `n > ` `0` `: ` ` ` `prevPrev ` `=` `prev ` ` ` `prev ` `=` `curr ` ` ` `curr ` `=` `prevPrev ` `+` `prev ` ` ` ` ` `# (curr - prev - 1) is ` ` ` `# count of non-Fibonacci ` ` ` `# numbers between curr ` ` ` `# and prev. ` ` ` `n ` `=` `n ` `-` `(curr ` `-` `prev ` `-` `1` `) ` ` ` ` ` `# n might be negative now. ` ` ` `# Make sure it becomes positive ` ` ` `# by removing last added gap. ` ` ` `n ` `=` `n ` `+` `(curr ` `-` `prev ` `-` `1` `) ` ` ` ` ` `# n must be now positive and ` ` ` `# less than or equal to gap ` ` ` `# between current and previous, ` ` ` `# i.e., (curr - prev - 1) ` ` ` ` ` `# Now add the positive n to ` ` ` `# previous Fibonacci number to ` ` ` `# find the n'th non-fibonacci. ` ` ` `return` `prev ` `+` `n ` ` ` `# Driver code ` `print` `(nonFibonacci(` `5` `)) ` ` ` `# This code is contributed by anuj_67. ` |

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## C#

`// C# program to find ` `// n'th Fibonacci number ` `using` `System; ` ` ` `class` `GFG ` `{ ` ` ` `// Returns n'th Non- ` ` ` `// Fibonacci number ` ` ` `static` `int` `nonFibonacci (` `int` `n) ` ` ` `{ ` ` ` ` ` `// curr is to keep track of ` ` ` `// current fibonacci number, ` ` ` `// prev is previous, prevPrev ` ` ` `// is previous of previous. ` ` ` `int` `prevPrev = 1, prev = 2, curr = 3; ` ` ` ` ` `// While count of non-fibonacci ` ` ` `// numbers doesn't become ` ` ` `// negative or zero ` ` ` `while` `(n > 0) ` ` ` `{ ` ` ` `// Simple Fibonacci number logic ` ` ` `prevPrev = prev; ` ` ` `prev = curr; ` ` ` `curr = prevPrev + prev; ` ` ` ` ` `// (curr - prev - 1) is count ` ` ` `// of non-Fibonacci numbers ` ` ` `// between curr and prev. ` ` ` `n = n - (curr - prev - 1); ` ` ` `} ` ` ` ` ` `// n might be negative now. Make ` ` ` `// sure it becomes positive by ` ` ` `// removing last added gap. ` ` ` `n = n + (curr - prev - 1); ` ` ` ` ` `// n must be now positive and less ` ` ` `// than or equal to gap between ` ` ` `// current and previous, i.e., ` ` ` `// (curr - prev - 1); ` ` ` ` ` `// Now add the positive n to ` ` ` `// previous Fibonacci number ` ` ` `// to find the n'th non-fibonacci. ` ` ` `return` `prev + n; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main () ` ` ` `{ ` ` ` `Console.WriteLine (nonFibonacci(5)); ` ` ` `} ` `} ` ` ` `//This code is contributed by aj_36 ` |

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## PHP

`<?php ` `// PHP program to find ` `// n'th Fibonacci number ` ` ` `// Returns n'th Non- ` `// Fibonacci number ` `function` `nonFibonacci(` `$n` `) ` `{ ` ` ` `// curr is to keep track of ` ` ` `// current fibonacci number, ` ` ` `// prev is previous, prevPrev ` ` ` `// is previous of previous. ` ` ` `$prevPrev` `= 1; ` ` ` `$prev` `= 2; ` ` ` `$curr` `= 3; ` ` ` ` ` `// While count of non-fibonacci ` ` ` `// numbers doesn't become ` ` ` `// negative or zero ` ` ` `while` `(` `$n` `> 0) ` ` ` `{ ` ` ` `// Simple Fibonacci ` ` ` `// number logic ` ` ` `$prevPrev` `= ` `$prev` `; ` ` ` `$prev` `= ` `$curr` `; ` ` ` `$curr` `= ` `$prevPrev` `+ ` `$prev` `; ` ` ` ` ` `// (curr - prev - 1) is count ` ` ` `// of non-Fibonacci numbers ` ` ` `// between curr and prev. ` ` ` `$n` `= ` `$n` `- (` `$curr` `- ` `$prev` `- 1); ` ` ` `} ` ` ` ` ` `// n might be negative now. Make ` ` ` `// sure it becomes positive by ` ` ` `// removing last added gap. ` ` ` `$n` `= ` `$n` `+ (` `$curr` `- ` `$prev` `- 1); ` ` ` ` ` `// n must be now positive and ` ` ` `// less than or equal to gap ` ` ` `// between current and previous, ` ` ` `// i.e., (curr - prev - 1); ` ` ` ` ` `// Now add the positive n to ` ` ` `// previous Fibonacci number ` ` ` `// to find the n'th non-fibonacci. ` ` ` `return` `$prev` `+ ` `$n` `; ` `} ` ` ` `// Driver code ` `echo` `nonFibonacci(5); ` ` ` `// This code is contributed by m_kit ` `?> ` |

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**Output :**

10

**Time Complexity : **O(n)

**Auxiliary Space : **O(1)

The above problem and solution are contributed by **Hemang Sarkar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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