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# Check if the n-th term is odd or even in a Fibonacci like sequence

Consider a sequence a0, a1, …, an, where ai = ai-1 + ai-2. Given a0, a1, and a positive integer n. The task is to find whether an is odd or even.
Note that the given sequence is like Fibonacci with the difference that the first two terms can be anything instead of 0 or 1.
Examples :

```Input : a0 = 2, a1 = 4, n =3
Output : Even
a2 = 6, a3 = 10
And a3 is even.

Input : a0 = 1, a1 = 9, n = 2
Output : Even```

Method 1: The idea is to find the sequence using the array and check if nth element is even or odd.

## C++

 `// CPP Program to check if the nth is odd or even in a``// sequence where each term is sum of previous two term``#include ``using` `namespace` `std;``#define MAX 100` `// Return if the nth term is even or odd.``bool` `findNature(``int` `a, ``int` `b, ``int` `n)``{``    ``int` `seq[MAX] = { 0 };` `    ``seq = a;``    ``seq = b;` `    ``for` `(``int` `i = 2; i <= n; i++)``        ``seq[i] = seq[i - 1] + seq[i - 2];` `    ``// Return true if odd``    ``return` `(seq[n] & 1);``}` `// Driven Program``int` `main()``{``    ``int` `a = 2, b = 4;``    ``int` `n = 3;` `    ``(findNature(a, b, n) ? (cout << ``"Odd"``                                 ``<< ``" "``)``                         ``: (cout << ``"Even"``                                 ``<< ``" "``));` `    ``return` `0;``}`

## Java

 `// Java Program to check if``// the nth is odd or even``// in a sequence where each``// term is sum of previous``// two term` `// Return if the nth``// term is even or odd.``class` `GFG``{``public` `static` `int` `findNature(``int` `a,``                             ``int` `b, ``int` `n)``{``    ``int``[] seq = ``new` `int``[``100``];` `    ``seq[``0``] = a;``    ``seq[``1``] = b;` `    ``for` `(``int` `i = ``2``; i <= n; i++)``        ``seq[i] = seq[i - ``1``] + seq[i - ``2``];` `    ``// Return true if odd``    ``if``((seq[n] & ``1``) != ``0``)``    ``return` `1``;``    ``else``    ``return` `0``;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `a = ``2``, b = ``4``;``    ``int` `n = ``3``;``    ``if``(findNature(a, b, n) == ``1``)``    ``System.out.println(``"Odd "``);``    ``else``    ``System.out.println(``"Even "``);``}``}` `// This code is contributed``// by mits`

## Python3

 `# Python3 Program to check if``# the nth is odd or even in a``# sequence where each term is``# sum of previous two term``MAX` `=` `100``;` `# Return if the nth``# term is even or odd.``def` `findNature(a, b, n):``    ``seq ``=` `[``0``] ``*` `MAX``;``    ``seq[``0``] ``=` `a;``    ``seq[``1``] ``=` `b;` `    ``for` `i ``in` `range``(``2``, n ``+` `1``):``        ``seq[i] ``=` `seq[i ``-` `1``] ``+` `seq[i ``-` `2``];` `    ``# Return true if odd``    ``return` `(seq[n] & ``1``);` `# Driver Code``a ``=` `2``;``b ``=` `4``;``n ``=` `3``;` `if``(findNature(a, b, n)):``    ``print``(``"Odd"``);``else``:``    ``print``(``"Even"``);` `# This code is contributed by mits`

## C#

 `// C# Program to check if the``// nth is odd or even in a``// sequence where each term``// is sum of previous two term``using` `System;` `// Return if the nth term``// is even or odd.``class` `GFG``{``public` `static` `int` `findNature(``int` `a,``                             ``int` `b, ``int` `n)``{``    ``int``[] seq = ``new` `int``;` `    ``seq = a;``    ``seq = b;` `    ``for` `(``int` `i = 2; i <= n; i++)``        ``seq[i] = seq[i - 1] +``                 ``seq[i - 2];` `    ``// Return true if odd``    ``if``((seq[n] & 1)!=0)``    ``return` `1;``    ``else``    ``return` `0;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `a = 2, b = 4;``    ``int` `n = 3;``    ``if``(findNature(a, b, n) == 1)``    ``Console.Write(``"Odd "``);``    ``else``    ``Console.Write(``"Even "``);``}``}` `// This code is contributed``// by mits`

## PHP

 ``

## Javascript

 ``

Output:

`Even`

Method 2 (efficient) :
Observe, the nature (odd or even) of the nth term depends on the previous terms, and the nature of the previous term depends on their previous terms and finally depends on the initial value i.e a0 and a1
So, we have four possible scenarios for a0 and a1
Case 1: When a0 an a1 is even
In this case each of the value in the sequence will be even only.
Case 2: When a0 an a1 is odd
In this case, observe a2 is even, a3 is odd, a4 is odd and so on. So, we can say ai is even if i is of form 3*k – 1, else odd.
Case 3: When a0 is even and a1 is odd
In this case, observe a2 is odd, a3 is even, a4 and a5 is odd, a6 is even and so on. So, we can say, ai is even if i is multiple of 3, else odd
Case 4: When a0 is odd and a1 is even
In this case, observe a2 and a3 is odd, a4 is even, a5 and a6 is odd, a7 is even and so on. So, we can say, ai is even if i is of the form 3*k + 1, k >= 0, else odd.
Below is the implementation of this approach:

## C++

 `// CPP Program to check if the nth is odd or even in a``// sequence where each term is sum of previous two term``#include ``using` `namespace` `std;` `// Return if the nth term is even or odd.``bool` `findNature(``int` `a, ``int` `b, ``int` `n)``{``    ``if` `(n == 0)``        ``return` `(a & 1);` `    ``if` `(n == 1)``        ``return` `(b & 1);` `    ``// If a is even``    ``if` `(!(a & 1)) {` `        ``// If b is even``        ``if` `(!(b & 1))``            ``return` `false``;``        ` `        ``// If b is odd``        ``else``            ``return` `(n % 3 != 0);``    ``}` `    ``// If a is odd``    ``else` `{``        ``// If b is odd``        ``if` `(!(b & 1))``            ``return` `((n - 1) % 3 != 0);` `        ``// If b is even``        ``else``            ``return` `((n + 1) % 3 != 0);``    ``}``}` `// Driven Program``int` `main()``{``    ``int` `a = 2, b = 4;``    ``int` `n = 3;` `    ``(findNature(a, b, n) ? (cout << ``"Odd"``                                 ``<< ``" "``)``                         ``: (cout << ``"Even"``                                 ``<< ``" "``));` `    ``return` `0;``}`

## Java

 `// Java Program to check if the nth is odd or even in a``// sequence where each term is sum of previous two term` `class` `GFG{``// Return if the nth term is even or odd.``static` `boolean` `findNature(``int` `a, ``int` `b, ``int` `n)``{``    ``if` `(n == ``0``)``        ``return` `(a & ``1``)==``1``?``true``:``false``;` `    ``if` `(n == ``1``)``        ``return` `(b & ``1``)==``1``?``true``:``false``;` `    ``// If a is even``    ``if` `((a & ``1``)==``0``) {` `        ``// If b is even``        ``if` `((b & ``1``)==``0``)``            ``return` `false``;``        ` `        ``// If b is odd``        ``else``            ``return` `(n % ``3` `!= ``0``);``    ``}` `    ``// If a is odd``    ``else` `{``        ``// If b is odd``        ``if` `((b & ``1``)==``0``)``            ``return` `((n - ``1``) % ``3` `!= ``0``);` `        ``// If b is even``        ``else``            ``return` `((n + ``1``) % ``3` `!= ``0``);``    ``}``}` `// Driven Program``public` `static` `void` `main(String[] args)``{``    ``int` `a = ``2``, b = ``4``;``    ``int` `n = ``3``;` `    ``if``(findNature(a, b, n))``    ``System.out.println(``"Odd"``);``    ``else``    ``System.out.println(``"Even"``);` `}``}``// This Code is contributed by mits`

## Python3

 `# Python3 Program to check if the``# nth is odd or even in a``# sequence where each term is``# sum of previous two term` `# Return if the nth``# term is even or odd.``def` `findNature(a, b, n):``    ``if` `(n ``=``=` `0``):``        ``return` `(a & ``1``);` `    ``if` `(n ``=``=` `1``):``        ``return` `(b & ``1``);` `    ``# If a is even``    ``if` `((a & ``1``) ``=``=` `0``):` `        ``# If b is even``        ``if` `((b & ``1``) ``=``=` `0``):``            ``return` `False``;``        ` `        ``# If b is odd``        ``else``:``            ``return` `True` `if``(n ``%` `3` `!``=` `0``) ``else` `False``;` `    ``# If a is odd``    ``else``:``        ``# If b is odd``        ``if` `((b & ``1``) ``=``=` `0``):``            ``return` `True` `if``((n ``-` `1``) ``%` `3` `!``=` `0``) ``else` `False``;` `        ``# If b is even``        ``else``:``            ``return` `True` `if``((n ``+` `1``) ``%` `3` `!``=` `0``) ``else` `False``;` `# Driver Code``a ``=` `2``;``b ``=` `4``;``n ``=` `3``;` `if` `(findNature(a, b, n) ``=``=` `True``):``    ``print``(``"Odd"``, end ``=` `" "``);``else``:``    ``print``(``"Even"``, end ``=` `" "``);``    ` `# This code is contributed by mits`

## C#

 `// C# Program to check if the nth is odd or even in a``// sequence where each term is sum of previous two term` `class` `GFG{``// Return if the nth term is even or odd.``static` `bool` `findNature(``int` `a, ``int` `b, ``int` `n)``{``    ``if` `(n == 0)``        ``return` `(a & 1)==1?``true``:``false``;` `    ``if` `(n == 1)``        ``return` `(b & 1)==1?``true``:``false``;` `    ``// If a is even``    ``if` `((a & 1)==0) {` `        ``// If b is even``        ``if` `((b & 1)==0)``            ``return` `false``;``        ` `        ``// If b is odd``        ``else``            ``return` `(n % 3 != 0);``    ``}` `    ``// If a is odd``    ``else` `{``        ``// If b is odd``        ``if` `((b & 1)==0)``            ``return` `((n - 1) % 3 != 0);` `        ``// If b is even``        ``else``            ``return` `((n + 1) % 3 != 0);``    ``}``}` `// Driven Program``static` `void` `Main()``{``    ``int` `a = 2, b = 4;``    ``int` `n = 3;` `    ``if``(findNature(a, b, n))``    ``System.Console.WriteLine(``"Odd"``);``    ``else``    ``System.Console.WriteLine(``"Even"``);` `}``}``// This Code is contributed by mits`

## PHP

 ``

## Javascript

 ``

Output :

`Even`

Time Complexity: O(1)

Auxiliary Space: O(1)

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