Total kilometers you can write with N pencils

There are n pencils, each having l length. Each can write 4 kilometers. After writing 4 kilometers it has l/4 length. One can join 4 pencils which are having l/4 length and can make 1 pencil. One can’t make pencil of pieces if remaining pieces are 3 or 2 or 1 in number but one can include these remaining pieces whenever needed.
Write a relation independent of l,length of given pencil, for how much one can write from n pencils.

Examples :

Input : 4
Output : 20

Approach:
Suppose We use 3 pencils that will 12 and generate 3 used pencils, now if the remaining pencils are greater than zero at least 1 unused pencil can be used with those 3 unused to write 4 and that will generate 1 more unused pencil. This will keep repeating.

if(n-3 >= 1){
 f(n) = f(n-3) + 12 + 4
} else{
 f(n) = 4*n
}

Above relation can be optimized in O(1)
n - 3 + x <= 3     //x is max no of time we can subtract 3 without n-3 <= 3 

x  >  (n-3)/3
ie. n/3 - 1 if it divides exactly else n/3

//C++ program to find relation independent of l,length of given pencil, for how much one can write from n pencils. 
#include<bits/stdc++.h>
using namespace std;

//Function to find no of pencils
int count(int n){
    int x = (n/3) - 1;
    if(n%3){
      x++;
    }
   return (4*x + 4*n);
}

//Driver function
int main(){
   int n = 5;
  cout<<count(n)<<endl;
}

Output :

Recursively it can be written as:

int count(int n){
  if(n<4){
    return (n*4);
   } else{
     return (16+count(n-3));
   } 
}

Examples :

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