The Pentanacci series is a generalization of the Fibonacci sequence where each term is the sum of the five preceding terms. The first few Pentanacci numbers are as follows – 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464, 912, 1793, 3525, 6930, 13624, 26784, 52656, 103519…..
Nth Term of Pentanacci number is given by:
T(n) = T(n-1) + T(n-2) + T(n-3) + T(n-4) + T(n-5)
with T(0) = T(1) = T(2) = T(3) = 0, T(4) = 1
Find the Nth Pentanacci number
Given a number N. The task is to find the N-th Pentanacci number.
Examples:
Input: N = 7
Output: 2Input: N = 10
Output: 16
Naive Approach: The idea is to follow the recurrence for finding the number and use recursion to solve it.
Recurrence relation:
T(n) = T(n-1) + T(n-2) + T(n-3) + T(n-4) + T(n-5)
Below is the implementation of the above approach:
C++14
// A simple recursive program to print // Nth Pentanacci number #include<bits/stdc++.h>using namespace std; // Recursive function to find the Nth// Pentanacci numberint printpentaRec(int n){ if (n == 0 || n == 1 || n == 2 || n == 3) return 0; else if (n == 4 || n == 5) return 1; else return (printpentaRec(n - 1) + printpentaRec(n - 2) + printpentaRec(n - 3)+ printpentaRec(n - 4)+ printpentaRec(n - 5));} // Function to print the Nth// Pentanacci numbervoid printPenta(int n){ cout << printpentaRec(n) << "\n";} // Driver code int main(){ int n = 10; printPenta(n); return 0;}// This code is contributed by yatinagg |
Java
// A simple recursive program to print // Nth Pentanacci number import java.util.*;class GFG{ // Recursive function to find the Nth// Pentanacci numberstatic int printpentaRec(int n){ if (n == 0 || n == 1 || n == 2 || n == 3) return 0; else if (n == 4 || n == 5) return 1; else return (printpentaRec(n - 1) + printpentaRec(n - 2) + printpentaRec(n - 3) + printpentaRec(n - 4) + printpentaRec(n - 5));} // Function to print the Nth// Pentanacci numberstatic void printPenta(int n){ System.out.print(printpentaRec(n) + "\n");} // Driver code public static void main(String[] args){ int n = 10; printPenta(n); }}// This code is contributed by gauravrajput1 |
Python3
# A simple recursive program to print # Nth Pentanacci number # Recursive program to find the Nth# Pentanacci numberdef printpentaRec(n) : if (n == 0 or n == 1 or\ n == 2 or n == 3): return 0 elif (n == 4 or n == 5): return 1 else : return (printpentaRec(n - 1) + printpentaRec(n - 2) + printpentaRec(n - 3)+ printpentaRec(n - 4)+ printpentaRec(n - 5)) # Function to print the Nth# Pentanacci numberdef printPenta(n) : print(printpentaRec(n)) # Driver code n = 10printPenta(n) |
C#
// A simple recursive program to print // Nth Pentanacci number using System;class GFG{ // Recursive function to find the Nth// Pentanacci numberstatic int printpentaRec(int n){ if (n == 0 || n == 1 || n == 2 || n == 3) return 0; else if (n == 4 || n == 5) return 1; else return (printpentaRec(n - 1) + printpentaRec(n - 2) + printpentaRec(n - 3) + printpentaRec(n - 4) + printpentaRec(n - 5));} // Function to print the Nth// Pentanacci numberstatic void printPenta(int n){ Console.WriteLine(printpentaRec(n));}// Driver codestatic void Main() { int n = 10; printPenta(n); }}// This code is contributed divyeshrabadiya07 |
Javascript
<script> // A simple recursive program to print // Nth Pentanacci number // Recursive function to find the Nth // Pentanacci number function printpentaRec(n) { if (n == 0 || n == 1 || n == 2 || n == 3) return 0; else if (n == 4 || n == 5) return 1; else return (printpentaRec(n - 1) + printpentaRec(n - 2) + printpentaRec(n - 3)+ printpentaRec(n - 4)+ printpentaRec(n - 5)); } // Function to print the Nth // Pentanacci number function printPenta(n) { document.write(printpentaRec(n) + "</br>"); } let n = 10; printPenta(n);// This code is contributed by divyesh072019.</script> |
Output:
16
Efficient Approach: The idea is to use Dynamic Programming to solve this problem. That is memoization of the solution in four variables for the last four terms such that the same subproblem is not computed again and again.
Below is the implementation of the above approach:
C++14
// C++14 implementation to print // Nth Pentanacci numbers. #include<bits/stdc++.h>using namespace std; // Function to print Nth // Pentanacci number void printpenta(int n){ if (n < 0) return; // Initialize first five // numbers to base cases int first = 0; int second = 0; int third = 0; int fourth = 0; int fifth = 1; // Declare a current variable int curr = 0; if (n == 0 || n == 1 || n == 2 || n == 3) cout << first << "\n"; else if (n == 5) cout << fifth << "\n"; else { // Loop to add previous five numbers // for each number starting from 5 // and then assign first, second, // third, fourth to second, third, fourth // and curr to fifth respectively for(int i = 5; i < n; i++) { curr = first + second + third + fourth + fifth; first = second; second = third; third = fourth; fourth = fifth; fifth = curr; } cout << curr << "\n"; }} // Driver code int main(){ int n = 10; printpenta(n); return 0;}// This code is contributed by yatinagg |
Java
// Java implementation to print // Nth Pentanacci numbers. import java.util.*;class GFG{ // Function to print Nth // Pentanacci number static void printpenta(int n){ if (n < 0) return; // Initialize first five // numbers to base cases int first = 0; int second = 0; int third = 0; int fourth = 0; int fifth = 1; // Declare a current variable int curr = 0; if (n == 0 || n == 1 || n == 2 || n == 3) System.out.print(first + "\n"); else if (n == 5) System.out.print(fifth + "\n"); else { // Loop to add previous five numbers // for each number starting from 5 // and then assign first, second, // third, fourth to second, third, // fourth and curr to fifth respectively for(int i = 5; i < n; i++) { curr = first + second + third + fourth + fifth; first = second; second = third; third = fourth; fourth = fifth; fifth = curr; } System.out.print(curr + "\n"); }} // Driver code public static void main(String[] args){ int n = 10; printpenta(n); }}// This code is contributed by Princi Singh |
Python3
# Python3 implementation to print # Nth Pentanacci numbers. # Function to print Nth # Pentanacci number def printpenta(n) : if (n < 0): return # Initialize first five # numbers to base cases first = 0 second = 0 third = 0 fourth = 0 fifth = 1 # declare a current variable curr = 0 if (n == 0 or n == 1 or\ n == 2 or n == 3): print(first) elif (n == 5): print(fifth) else: # Loop to add previous five numbers # for each number starting from 5 # and then assign first, second, # third, fourth to second, third, fourth # and curr to fifth respectively for i in range(5, n): curr = first + second +\ third + fourth + fifth first = second second = third third = fourth fourth = fifth fifth = curr print(curr) # Driver code n = 10printpenta(n) |
C#
// C# implementation to print // Nth Pentanacci numbers. using System;class GFG{ // Function to print Nth // Pentanacci number static void printpenta(int n){ if (n < 0) return; // Initialize first five // numbers to base cases int first = 0; int second = 0; int third = 0; int fourth = 0; int fifth = 1; // Declare a current variable int curr = 0; if (n == 0 || n == 1 || n == 2 || n == 3) Console.Write(first + "\n"); else if (n == 5) Console.Write(fifth + "\n"); else { // Loop to add previous five numbers // for each number starting from 5 // and then assign first, second, // third, fourth to second, third, // fourth and curr to fifth respectively for(int i = 5; i < n; i++) { curr = first + second + third + fourth + fifth; first = second; second = third; third = fourth; fourth = fifth; fifth = curr; } Console.Write(curr + "\n"); }}// Driver code public static void Main(String[] args){ int n = 10; printpenta(n); }}// This code is contributed by shikhasingrajput |
Javascript
<script> // Javascript implementation to print // Nth Pentanacci numbers. // Function to print Nth // Pentanacci number function printpenta(n) { if (n < 0) return; // Initialize first five // numbers to base cases let first = 0; let second = 0; let third = 0; let fourth = 0; let fifth = 1; // Declare a current variable let curr = 0; if (n == 0 || n == 1 || n == 2 || n == 3) document.write(first + "</br>"); else if (n == 5) document.write(fifth + "</br>"); else { // Loop to add previous five numbers // for each number starting from 5 // and then assign first, second, // third, fourth to second, third, // fourth and curr to fifth respectively for(let i = 5; i < n; i++) { curr = first + second + third + fourth + fifth; first = second; second = third; third = fourth; fourth = fifth; fifth = curr; } document.write(curr + "</br>"); } } let n = 10; printpenta(n); </script> |
Output:
16
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