# Sum of Perrin Numbers

Given a number positive number n, find value of P0 + P1 + P2 + …. + Pn where pi indicates i’th Perrin number. First few Perrin numbers are 3, 0, 2, 3, 2, 5, 5, 7…….

Examples:

```Input  : 4
Output : 8
Explanation : 3 + 0 + 2 + 3

Input  : 6
Output : 15
Explanation : 3 + 0 + 2 + 3 + 2 + 5
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

In previous post, we have introduced Perrin Numbers. In mathematical terms, the sequence p(n) of Perrin numbers is defined by the recurrence relation

``` P(n) = P(n-2) + P(n-3) for n > 2,

with initial values
P(0) = 3, P(1) = 0, P(2) = 2.
```

Method 1 (Using Recursive Formula of n’th Perrin Number)
We can simply add numbers using above recursive formula of n’th Perrin Number.

## C++

 `// C++ program to calculate sum of Perrin Numbers ` `#include ` `using` `namespace` `std; ` ` `  `// function for sum of first n Perrin number. ` `int` `calSum(``int` `n) ` `{ ` `    ``int` `a = 3, b = 0, c = 2; ` `    ``if` `(n == 0) ``// n=0 ` `        ``return` `3; ` `    ``if` `(n == 1) ``// n=1 ` `        ``return` `3; ` `    ``if` `(n == 2) ``// n=2 ` `        ``return` `5; ` ` `  `    ``// calculate k=5 sum of three previous step. ` `    ``int` `sum = 5; ` ` `  `    ``// Sum remaining numbers ` `    ``while` `(n > 2) { ` `        ``int` `d = a + b; ``// calculate next term ` `        ``sum += d; ` `        ``a = b; ` `        ``b = c; ` `        ``c = d; ` `        ``n--; ` `    ``} ` ` `  `    ``return` `sum; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `n = 9; ` `    ``cout << calSum(n); ` `    ``return` `0; ` `} `

## Java

 `// Java program to calculate ` `// sum of Perrin Numbers ` `import` `java.lang.*; ` ` `  `class` `GFG { ` ` `  `    ``// function for sum of first n Perrin number. ` `    ``static` `int` `calSum(``int` `n) ` `    ``{ ` ` `  `        ``int` `a = ``3``, b = ``0``, c = ``2``; ` `        ``if` `(n == ``0``) ``// n=0 ` `            ``return` `3``; ` `        ``if` `(n == ``1``) ``// n=1 ` `            ``return` `3``; ` `        ``if` `(n == ``2``) ``// n=2 ` `            ``return` `5``; ` ` `  `        ``// calculate k=5 sum of three previous step. ` `        ``int` `sum = ``5``; ` ` `  `        ``// Sum remaining numbers ` `        ``while` `(n > ``2``) { ` ` `  `            ``// calculate next term ` `            ``int` `d = a + b; ` `            ``sum += d; ` `            ``a = b; ` `            ``b = c; ` `            ``c = d; ` `            ``n--; ` `        ``} ` ` `  `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``9``; ` `        ``System.out.print(calSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to calculate ` `# sum of Perrin Numbers ` ` `  `# function for sum of first ` `# n Perrin number. ` `def` `calSum(n): ` ` `  `    ``a ``=` `3` `    ``b ``=` `0` `    ``c ``=` `2` ` `  `    ``if` `(n ``=``=` `0``):  ``# n = 0 ` `        ``return` `3` `    ``if` `(n ``=``=` `1``):  ``# n = 1 ` `        ``return` `3` `    ``if` `(n ``=``=` `2``):  ``# n = 2 ` `        ``return` `5` `  `  `    ``# calculate k = 5 sum of ` `    ``# three previous step. ` `    ``sum` `=` `5` `  `  `    ``# Sum remaining numbers ` `    ``while` `(n > ``2``): ` ` `  `        ``# calculate next term ` `        ``d ``=` `a ``+` `b ` `        ``sum` `=` `sum` `+` `d ` `        ``a ``=` `b ` `        ``b ``=` `c ` `        ``c ``=` `d ` `        ``n ``=` `n``-``1` `     `  `    ``return` `sum` ` `  `# Driver code ` ` `  `n ``=` `9` `print``(calSum(n)) ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program to calculate ` `// sum of Perrin Numbers ` `using` `System; ` ` `  `class` `GFG { ` ` `  `    ``// function for sum of first n Perrin number. ` `    ``static` `int` `calSum(``int` `n) ` `    ``{ ` ` `  `        ``int` `a = 3, b = 0, c = 2; ` ` `  `        ``if` `(n == 0) ``// n=0 ` `            ``return` `3; ` `        ``if` `(n == 1) ``// n=1 ` `            ``return` `3; ` `        ``if` `(n == 2) ``// n=2 ` `            ``return` `5; ` ` `  `        ``// calculate k=5 sum of three ` `        ``// previous step. ` `        ``int` `sum = 5; ` ` `  `        ``// Sum remaining numbers ` `        ``while` `(n > 2) { ` ` `  `            ``// calculate next term ` `            ``int` `d = a + b; ` `            ``sum += d; ` `            ``a = b; ` `            ``b = c; ` `            ``c = d; ` `            ``n--; ` `        ``} ` ` `  `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `Main() ` `    ``{ ` ` `  `        ``int` `n = 9; ` ` `  `        ``Console.WriteLine(calSum(n)); ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` 2) ` `    ``{ ` `         `  `        ``// calculate next term ` `        ``\$d` `= ``\$a` `+ ``\$b``;  ` `         `  `        ``\$sum` `+= ``\$d``; ` `        ``\$a` `= ``\$b``; ` `        ``\$b` `= ``\$c``; ` `        ``\$c` `= ``\$d``; ` `        ``\$n``--; ` `    ``} ` ` `  `    ``return` `\$sum``; ` `} ` ` `  `    ``// Driver code ` `    ``\$n` `= 9; ` `    ``echo` `calSum(``\$n``); ` `     `  `// This code is contributed by ajit. ` `?> `

Output:

```49
```

Method 2 (Using Direct Formula)
The idea is to find relationship between the sum of Perrin numbers and n’th Perrin number.

```p(i) refers to the i’th perrin number.
S(i) refers to sum of perrin numbers till p(i),

We can rewrite the relation P(n) = P(n-2) + P(n-3)
as below :
P(n-3)    = P(n)  -  P(n-2)

Similarly,
P(n-4)    = P(n-1)  -  P(n-3)
P(n-5)    = P(n-2)  -  P(n-4)
.          .           .
.          .             .
.          .             .
P(1)      = P(4)    -  P(2)
P(0)      = P(3)    -  P(1)
-------------------------------
Adding all the equations, on left side, we have
{(n) + P(n-1) - P(1) - P(2) which is S(n-3).
Therefore,
S(n-3) = P(n) + P(n-1) - P(1) - P(2)
S(n-3) = P(n) + P(n-1) - 2
S(n)   = P(n+3) + P(n+2) - 2
```

In order to find S(n), we can simply calculate the (n+3)’th and (n+2) Perrin number and subtract 2 from the result.

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Improved By : jit_t

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