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Jacobsthal and Jacobsthal-Lucas Numbers
Jacobsthal Numbers
Lucas sequence 𝑈𝑛(𝑃,𝑄) where P = 1 and Q = -2 are called Jacobsthal numbers. The recurrence relation for Jacobsthal numbers is,
$$\mathrm{𝐽_𝑛 = 0\: 𝑓𝑜𝑟 \: 𝑛 = 0}$$
$$\mathrm{𝐽_𝑛 = 1\: 𝑓𝑜𝑟 \: 𝑛 = 1}$$
$$\mathrm{𝐽_𝑛 = 𝐽_𝑛−1 + 2𝐽_{𝑛−2}\: 𝑓𝑜𝑟 \: 𝑛 > 1}$$
Following are the Jacobsthal numbers −
0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, ….
Jacobsthal-Lucas Numbers
Complementary Lucas sequence $\mathrm{𝑉_𝑛(𝑃,𝑄)}$ where P = 1 and Q = -2 are called JacobsthalLucas numbers. The recurrence relation for Jacobsthal-Lucas numbers is,
$\mathrm{𝐽_𝑛}$ = 2 𝑓𝑜𝑟 𝑛 = 0
$\mathrm{𝐽_𝑛}$ = 1 𝑓𝑜𝑟 𝑛 = 1
$\mathrm{𝐽_𝑛 = 𝐽_{𝑛−1} + 2𝐽_{𝑛−2}\: 𝑓𝑜𝑟 \: 𝑛 > 1}$
Following are the Jacobsthal-Lucas numbers −
2, 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025, 2047, 4097, ….
Problem Statement
Given an integer, N. Find the Nth Jacobsthal and Jacobsthal-Lucas number.
Example 1
Input: 2
Output: Jacobsthal = 1 Jacobsthal-Lucas = 5
Explanation
Jacobsthal Numbers - 0, 1, 1
Jacobsthal-Lucas Numbers - 2, 1, 5
Example 2
Input: 12
Output: Jacobsthal = 1365 Jacobsthal-Lucas = 4097
Explanation
Jacobsthal Numbers − 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365
Jacobsthal-Lucas Numbers − 2, 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025, 2047, 4097
Approach 1: Recursive Approach
Using the recurrence relation of Jacobsthal and Jacobsthal-Lucas numbers, we can form a recursive relation as, $\mathrm{𝐽_𝑛 = 𝐽_{𝑛−1} + 2𝐽_{𝑛−2}}$
Pseudocode
procedure jacobsthalNumber (n) if n == 0 ans = 0 end if if n == 1 ans = 1 end if ans = jacobsthalNumber(n-1) + 2*jacobsthalNumber(n-2) end procedure procedure jacobsthalLucasNumber (n) if n == 0 ans = 2 end if if n == 1 ans = 1 end if ans = jacobsthalLucasNumber(n-1) + 2*jacobsthalLucasNumber(n-2) end procedure
Example
In the following program, we have created separate functions for jacobsthal and jacobsthal-lucas numbers and used their recurrence relation to recursively calculate the nth number.
#include<bits/stdc++.h> using namespace std; // Recursive function for finding the nth Jacobsthal Number int jacobsthalNumber (int n){ // Defining the first case in recurrence relation if (n == 0) { return 0; } // Defining the second case in recurrence relation if (n == 1) { return 1; } // Defining the third case in recurrence relation when n>1 return jacobsthalNumber(n-1) + 2*jacobsthalNumber(n-2); } // Recursive function for finding the nth Jacobsthal-Lucas Number int jacobsthalLucasNumber (int n){ // Defining the first case in recurrence relation if (n == 0) { return 2; } // Defining the second case in recurrence relation if (n == 1) { return 1; } // Defining the third case in recurrence relation when n>1 return jacobsthalLucasNumber(n-1) + 2*jacobsthalLucasNumber(n-2); } int main(){ int N = 7; cout << N << "th Jacobsthal number = " << jacobsthalNumber(N) << endl; cout << N << "th Jacobsthal-Lucas number = " << jacobsthalLucasNumber(N) << endl; }
Output
7th Jacobsthal number = 43 7th Jacobsthal-Lucas number = 127
Time Complexity − O(2n) due to the recursive tree formed.
Space Complexity − O(1) + Recursive stack space.
Approach 2: Memoization
Memoization is helpful in reducing time complexity as it helps compute the function for one input only once by storing the results.
Pseudocode −
Initialize two arrays j[N+1] and jL[N+1] with -1 procedure jacobsthalNumber (n) if n == 0 ans = 0 end if if n == 1 ans = 1 end if if j[n] == -1 j[n] = jacobsthalNumber(n-1) + 2*jacobsthalNumber(n-2) end if ans = j[n] end procedure procedure jacobsthalLucasNumber (n) if n == 0 ans = 2 end if if n == 1 ans = 1 end if if jL[n] == -1 jL[n] = jacobsthalLucasNumber(n-1) + 2*jacobsthalLucasNumber(n-2) end if ans = jL[n] end procedure
Example
In this program, we use memoization to reduce the time complexity of the previous approach.
#include<bits/stdc++.h> using namespace std; int N = 12; // initializing memoization table vector<int> j(N+1, -1), jL(N+1, -1); // Memoized function for finding the nth Jacobsthal Number int jacobsthalNumber (int n){ // Defining the first case in recurrence relation if (n == 0) { return 0; } // Defining the second case in recurrence relation if (n == 1) { return 1; } // Checking if nth jacobsthal number is already calculated or not if(j[n] != -1) { return j[n]; } else { return j[n] = jacobsthalNumber(n-1) + 2*jacobsthalNumber(n-2); } } // Memoized function for finding the nth Jacobsthal-Lucas Number int jacobsthalLucasNumber (int n){ // Defining the first case in recurrence relation if (n == 0) { return 2; } // Defining the second case in recurrence relation if (n == 1) { return 1; } // Checking if nth jacobsthal-lucas number is already calculated or not if(jL[n] != -1) { return jL[n]; } else { return jL[n] = jacobsthalLucasNumber(n-1) + 2*jacobsthalLucasNumber(n-2); } } // Driver code int main(){ cout << N << "th Jacobsthal number = " << jacobsthalNumber(N) << endl; cout << N << "th Jacobsthal-Lucas number = " << jacobsthalLucasNumber(N) << endl; }
Output
12th Jacobsthal number = 1365 12th Jacobsthal-Lucas number = 4097
Time Complexity − O(N) as repetitive recursive calls are not calculated.
Space Complexity − O(N) (because of the space used for the memoization table) + recursive stack space.
Approach 3: Dynamic Programming
Pseudocode −
procedure jacobsthal (n) initialize dp table jacobsthalDP[n+1] jacobsthalDP[0] = 0 jacobsthalDP[1] = 1 for i = 2 to n jacobsthalDP[i] = jacobsthalDP[i-1] + 2*jacobsthalDP[i-2] end for ans = jacobsthalDP[n] end procedure procedure jacobsthalLucas (n) initialize dp table jacobsthalLucasDP[n+1] jacobsthalLucasDP[0] = 2 jacobsthalLucasDP[1] = 1 for i = 2 to n jacobsthalLucasDP[i] = jacobsthalLucasDP[i-1] + 2*jacobsthalLucasDP[i-2] end for ans = jacobsthalLucasDP[n] end procedure
Example
In the following program, dynamic programming approach is used.
#include<bits/stdc++.h> using namespace std; // Function for finding the nth Jacobsthal Number int jacobsthalNumber (int n){ int jacobsthalDP[n+1]; // Defining base condition jacobsthalDP[0] = 0; jacobsthalDP[1] = 1; for (int i = 2 ; i <= n ; i++){ jacobsthalDP[i] = jacobsthalDP[i-1] + 2*jacobsthalDP[i-2]; } return jacobsthalDP[n]; } // Function for finding the nth Jacobsthal-Lucas Number int jacobsthalLucasNumber (int n){ int jacobsthalLucasDP[n+1]; // Defining base condition jacobsthalLucasDP[0] = 2; jacobsthalLucasDP[1] = 1; for (int i = 2 ; i <= n ; i++){ jacobsthalLucasDP[i] = jacobsthalLucasDP[i-1] + 2*jacobsthalLucasDP[i-2]; } return jacobsthalLucasDP[n]; } int main(){ int N = 5; cout << N << "th Jacobsthal number = " << jacobsthalNumber(N) << endl; cout << N << "th Jacobsthal-Lucas number = " << jacobsthalLucasNumber(N) << endl; }
Output
5th Jacobsthal number = 11 5th Jacobsthal-Lucas number = 31
Time Complexity − O(n) as for loop is only executed once in each function
Space Complexity − O(n) for the dp table. In this approach, no extra stack recursive space is used.
Conclusion
In conclusion, Jacobsthal numbers are obtained from the Lucas sequence and Jacobsthal-Lucas numbers are obtained from the complementary Lucas sequence. In order to find the nth Jacobsthal and Jacobsthal-Lucas number, we can move from a recursive to a dynamic programming approach with minimum time and space complexity.