Padovan Sequence
Last Updated :
24 Jan, 2022
Padovan Sequence similar to Fibonacci sequence with similar recursive structure. The recursive formula is,
P(n) = P(n-2) + P(n-3)
P(0) = P(1) = P(2) = 1
Fibonacci Sequence: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55……
Spiral of squares with side lengths which follow the Fibonacci sequence.
Padovan Sequence: 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37,…..
Spiral of equilateral triangles with side lengths which follow the Padovan sequence.
Examples:
For Padovan Sequence:
P0 = P1 = P2 = 1 ,
P(7) = P(5) + P(4)
= P(3) + P(2) + P(2) + P(1)
= P(2) + P(1) + 1 + 1 + 1
= 1 + 1 + 1 + 1 + 1
= 5
C++
#include<iostream>
using namespace std;
int pad( int n)
{
int pPrevPrev = 1, pPrev = 1, pCurr = 1, pNext = 1;
for ( int i=3; i<=n; i++)
{
pNext = pPrevPrev + pPrev;
pPrevPrev = pPrev;
pPrev = pCurr;
pCurr = pNext;
}
return pNext;
}
int main()
{
int n = 12;
cout << pad(n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static int pad( int n)
{
int []padv= new int [n];
padv[ 0 ]=padv[ 1 ]=padv[ 2 ]= 1 ;
for ( int i = 3 ; i <= n; i++) {
padv[i]=padv[i- 2 ]+padv[i- 3 ];
}
return padv[n- 1 ];
}
public static void main(String args[])
{
int n = 12 ;
System.out.println(pad(n));
}
}
|
Python3
def pad(n):
pPrevPrev, pPrev, pCurr, pNext = 1 , 1 , 1 , 1
for i in range ( 3 , n + 1 ):
pNext = pPrevPrev + pPrev
pPrevPrev = pPrev
pPrev = pCurr
pCurr = pNext
return pNext
print (pad( 12 ))
|
C#
using System;
class GFG {
static int pad( int n)
{
int pPrevPrev = 1, pPrev = 1,
pCurr = 1, pNext = 1;
for ( int i = 3; i <= n; i++) {
pNext = pPrevPrev + pPrev;
pPrevPrev = pPrev;
pPrev = pCurr;
pCurr = pNext;
}
return pNext;
}
public static void Main()
{
int n = 12;
Console.WriteLine(pad(n));
}
}
|
PHP
<?php
function pad( $n )
{
$pPrevPrev = 1; $pPrev = 1;
$pCurr = 1; $pNext = 1;
for ( $i = 3; $i <= $n ; $i ++)
{
$pNext = $pPrevPrev + $pPrev ;
$pPrevPrev = $pPrev ;
$pPrev = $pCurr ;
$pCurr = $pNext ;
}
return $pNext ;
}
$n = 12;
echo (pad( $n ));
?>
|
Javascript
<script>
function pad(n) {
let pPrevPrev = 1;
let pPrev = 1;
let pCurr = 1;
let pNext = 1;
for (let i = 3; i <= n; i++) {
pNext = pPrevPrev + pPrev;
pPrevPrev = pPrev;
pPrev = pCurr;
pCurr = pNext;
}
return pNext;
}
let n = 12;
document.write(pad(n));
</script>
|
Output:
21
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...