# Print numbers such that no two consecutive numbers are co-prime and every three consecutive numbers are co-prime

Given an integer N, the task is to print N integers ? 109 such that no two consecutive of these integers are co-prime and every 3 consecutive are co-prime.

Examples:

```Input: N = 3
Output: 6 15 10```
```Input: N = 4
Output: 6 15 35 14```

Approach:

• We can just multiply consecutive primes and for the last number just multiply the gcd(last, last-1) * 2. We do this so that the (n – 1)th number, nth and 1st numbers can also follow the property mentioned in the problem statement.
• Another important part of the problem is the fact that the numbers should be ? 109. If you just multiply consecutive prime numbers, after around 3700 numbers, the value will cross 109. So we need to only use those prime numbers whose product won’t cross 109.
• To do this efficiently, consider a small number of primes, say the first 550 primes, and select them in a way such that on making a product no number gets repeated. We first choose every prime consecutively and then choose the primes with an interval of 2 and then 3 and so on. By doing that, we already make sure that no number gets repeated.

So we will select
5, 11, 17, …
Next time, we can start with 7 and select,
7, 13, 19, …

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `#define limit 1000000000` `#define MAX_PRIME 2000000` `#define MAX 1000000` `#define I_MAX 50000`   `map<``int``, ``int``> mp;`   `int` `b[MAX];` `int` `p[MAX];` `int` `j = 0;` `bool` `prime[MAX_PRIME + 1];`   `// Function to generate Sieve of` `// Eratosthenes` `void` `SieveOfEratosthenes(``int` `n)` `{` `    ``memset``(prime, ``true``, ``sizeof``(prime));`   `    ``for` `(``int` `p = 2; p * p <= n; p++) {`   `        ``// If prime[p] is not changed,` `        ``// then it is a prime` `        ``if` `(prime[p] == ``true``) {` `            ``for` `(``int` `i = p * p; i <= n; i += p)` `                ``prime[i] = ``false``;` `        ``}` `    ``}`   `    ``// Add the prime numbers to the array b` `    ``for` `(``int` `p = 2; p <= n; p++) {` `        ``if` `(prime[p]) {` `            ``b[j++] = p;` `        ``}` `    ``}` `}`   `// Function to return the gcd of a and b` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``if` `(b == 0)` `        ``return` `a;` `    ``return` `gcd(b, a % b);` `}`   `// Function to print the required` `// sequence of integers` `void` `printSeries(``int` `n)` `{` `    ``SieveOfEratosthenes(MAX_PRIME);`   `    ``int` `i, g, k, l, m, d;` `    ``int` `ar[I_MAX + 2];`   `    ``for` `(i = 0; i < j; i++) {` `        ``if` `((b[i] * b[i + 1]) > limit)` `            ``break``;`   `        ``// Including the primes in a series` `        ``// of primes which will be later` `        ``// multiplied` `        ``p[i] = b[i];`   `        ``// This is done to mark a product` `        ``// as existing` `        ``mp[b[i] * b[i + 1]] = 1;` `    ``}`   `    ``// Maximum number of primes that we consider` `    ``d = 550;` `    ``bool` `flag = ``false``;`   `    ``// For different interval` `    ``for` `(k = 2; (k < d - 1) && !flag; k++) {`   `        ``// For different starting index of jump` `        ``for` `(m = 2; (m < d) && !flag; m++) {`   `            ``// For storing the numbers` `            ``for` `(l = m + k; l < d; l += k) {`   `                ``// Checking for occurrence of a` `                ``// product. Also checking for the` `                ``// same prime occurring consecutively` `                ``if` `(((b[l] * b[l + k]) < limit)` `                    ``&& (l + k) < d && p[i - 1] != b[l + k]` `                    ``&& p[i - 1] != b[l]` `                    ``&& mp[b[l] * b[l + k]] != 1) {` `                    ``if` `(mp[p[i - 1] * b[l]] != 1) {`   `                        ``// Including the primes in a` `                        ``// series of primes which will` `                        ``// be later multiplied` `                        ``p[i] = b[l];` `                        ``mp[p[i - 1] * b[l]] = 1;` `                        ``i++;` `                    ``}` `                ``}`   `                ``if` `(i >= I_MAX) {` `                    ``flag = ``true``;` `                    ``break``;` `                ``}` `            ``}` `        ``}` `    ``}`   `    ``for` `(i = 0; i < n; i++)` `        ``ar[i] = p[i] * p[i + 1];`   `    ``for` `(i = 0; i < n - 1; i++)` `        ``cout << ar[i] << ``" "``;`   `    ``g = gcd(ar[n - 1], ar[n - 2]);` `    ``cout << g * 2 << endl;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `n = 4;`   `    ``printSeries(n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG ` `{`   `static` `int` `limit = ``1000000000``;` `static` `int` `MAX_PRIME = ``2000000``;` `static` `int` `MAX = ``1000000``;` `static` `int` `I_MAX = ``50000``;`   `static` `HashMap mp = ``new` `HashMap();`   `static` `int` `[]b = ``new` `int``[MAX];` `static` `int` `[]p = ``new` `int``[MAX];` `static` `int` `j = ``0``;` `static` `boolean` `[]prime = ``new` `boolean``[MAX_PRIME + ``1``];`   `// Function to generate Sieve of` `// Eratosthenes` `static` `void` `SieveOfEratosthenes(``int` `n)` `{` `    ``for``(``int` `i = ``0``; i < MAX_PRIME + ``1``; i++)` `        ``prime[i] = ``true``;`   `    ``for` `(``int` `p = ``2``; p * p <= n; p++)` `    ``{`   `        ``// If prime[p] is not changed,` `        ``// then it is a prime` `        ``if` `(prime[p] == ``true``)` `        ``{` `            ``for` `(``int` `i = p * p; i <= n; i += p)` `                ``prime[i] = ``false``;` `        ``}` `    ``}`   `    ``// Add the prime numbers to the array b` `    ``for` `(``int` `p = ``2``; p <= n; p++) ` `    ``{` `        ``if` `(prime[p]) ` `        ``{` `            ``b[j++] = p;` `        ``}` `    ``}` `}`   `// Function to return the gcd of a and b` `static` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``if` `(b == ``0``)` `        ``return` `a;` `    ``return` `gcd(b, a % b);` `}`   `// Function to print the required` `// sequence of integers` `static` `void` `printSeries(``int` `n)` `{` `    ``SieveOfEratosthenes(MAX_PRIME);`   `    ``int` `i, g, k, l, m, d;` `    ``int` `[]ar = ``new` `int``[I_MAX + ``2``];`   `    ``for` `(i = ``0``; i < j; i++) ` `    ``{` `        ``if` `((b[i] * b[i + ``1``]) > limit)` `            ``break``;`   `        ``// Including the primes in a series` `        ``// of primes which will be later` `        ``// multiplied` `        ``p[i] = b[i];`   `        ``// This is done to mark a product` `        ``// as existing` `        ``mp.put(b[i] * b[i + ``1``], ``1``);` `    ``}`   `    ``// Maximum number of primes that we consider` `    ``d = ``550``;` `    ``boolean` `flag = ``false``;`   `    ``// For different interval` `    ``for` `(k = ``2``; (k < d - ``1``) && !flag; k++)` `    ``{`   `        ``// For different starting index of jump` `        ``for` `(m = ``2``; (m < d) && !flag; m++) ` `        ``{`   `            ``// For storing the numbers` `            ``for` `(l = m + k; l < d; l += k) ` `            ``{`   `                ``// Checking for occurrence of a` `                ``// product. Also checking for the` `                ``// same prime occurring consecutively` `                ``if` `(((b[l] * b[l + k]) < limit) && ` `                      ``mp.containsKey(b[l] * b[l + k]) && ` `                      ``mp.containsKey(p[i - ``1``] * b[l]) && ` `                      ``(l + k) < d && p[i - ``1``] != b[l + k] && ` `                                         ``p[i - ``1``] != b[l] &&` `                             ``mp.get(b[l] * b[l + k]) != ``1``)` `                    ``{` `                    ``if` `(mp.get(p[i - ``1``] * b[l]) != ``1``)` `                    ``{`   `                        ``// Including the primes in a` `                        ``// series of primes which will` `                        ``// be later multiplied` `                        ``p[i] = b[l];` `                        ``mp.put(p[i - ``1``] * b[l], ``1``);` `                        ``i++;` `                    ``}` `                ``}`   `                ``if` `(i >= I_MAX) ` `                ``{` `                    ``flag = ``true``;` `                    ``break``;` `                ``}` `            ``}` `        ``}` `    ``}`   `    ``for` `(i = ``0``; i < n; i++)` `        ``ar[i] = p[i] * p[i + ``1``];`   `    ``for` `(i = ``0``; i < n - ``1``; i++)` `        ``System.out.print(ar[i]+``" "``);`   `    ``g = gcd(ar[n - ``1``], ar[n - ``2``]);` `    ``System.out.print(g * ``2``);` `}`   `// Driver Code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `n = ``4``;` `    ``printSeries(n);` `}` `}`   `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation of ` `# the above approach` `limit ``=` `1000000000` `MAX_PRIME ``=` `2000000` `MAX` `=` `1000000` `I_MAX ``=` `50000`   `mp ``=` `{}`   `b ``=` `[``0``] ``*` `MAX` `p ``=` `[``0``] ``*` `MAX` `j ``=` `0` `prime ``=` `[``True``] ``*` `(MAX_PRIME ``+` `1``)`   `# Function to generate Sieve of` `# Eratosthenes` `def` `SieveOfEratosthenes(n):` `    ``global` `j` `    ``p ``=` `2` `    ``while` `p ``*` `p <``=` `n:`   `        ``# If prime[p] is not changed,` `        ``# then it is a prime` `        ``if` `(prime[p] ``=``=` `True``):` `            ``for` `i ``in` `range``(p ``*` `p, n ``+` `1``, p):` `                ``prime[i] ``=` `False` `        ``p ``+``=` `1`   `    ``# Add the prime numbers to the array b` `    ``for` `p ``in` `range``(``2``, n ``+` `1``):` `        ``if` `(prime[p]):` `            ``b[j] ``=` `p` `            ``j ``+``=` `1`   `# Function to return ` `# the gcd of a and b` `def` `gcd(a, b):`   `    ``if` `(b ``=``=` `0``):` `        ``return` `a` `    ``return` `gcd(b, a ``%` `b)`   `# Function to print the required` `# sequence of integers` `def` `printSeries(n):`   `    ``SieveOfEratosthenes(MAX_PRIME)`   `    ``ar ``=` `[``0``] ``*` `(I_MAX ``+` `2``)`   `    ``for` `i ``in` `range``(j):` `        ``if` `((b[i] ``*` `b[i ``+` `1``]) > limit):` `            ``break`   `        ``# Including the primes in a series` `        ``# of primes which will be later` `        ``# multiplied` `        ``p[i] ``=` `b[i]`   `        ``# This is done to mark a product` `        ``# as existing` `        ``mp[b[i] ``*` `b[i ``+` `1``]] ``=` `1`   `    ``# Maximum number of ` `    ``# primes that we consider` `    ``d ``=` `550` `    ``flag ``=` `False`   `    ``# For different interval` `    ``k ``=` `2` `    ``while` `(k < d ``-` `1``) ``and` `not` `flag:`   `        ``# For different starting ` `        ``# index of jump` `        ``m ``=` `2` `        ``while` `(m < d) ``and` `not` `flag:`   `            ``# For storing the numbers` `            ``for` `l ``in` `range``(m ``+` `k, d, k):`   `                ``# Checking for occurrence of a` `                ``# product. Also checking for the` `                ``# same prime occurring consecutively` `                ``if` `(((b[l] ``*` `b[l ``+` `k]) < limit) ``and` `                    ``(l ``+` `k) < d ``and` `p[i ``-` `1``] !``=` `b[l ``+` `k] ``and` `                     ``p[i ``-` `1``] !``=` `b[l] ``and` `                     ``((b[l] ``*` `b[l ``+` `k] ``in` `mp) ``and` `                     ``mp[b[l] ``*` `b[l ``+` `k]] !``=` `1``)):` `                  `  `                    ``if` `(mp[p[i ``-` `1``] ``*` `b[l]] !``=` `1``):`   `                        ``# Including the primes in a` `                        ``# series of primes which will` `                        ``# be later multiplied` `                        ``p[i] ``=` `b[l]` `                        ``mp[p[i ``-` `1``] ``*` `b[l]] ``=` `1` `                        ``i ``+``=` `1`   `                ``if` `(i >``=` `I_MAX):` `                    ``flag ``=` `True` `                    ``break` `            ``m ``+``=` `1` `        ``k ``+``=` `1`   `    ``for` `i ``in` `range``(n):` `        ``ar[i] ``=` `p[i] ``*` `p[i ``+` `1``]`   `    ``for` `i ``in` `range``(n ``-` `1``):` `        ``print``(ar[i], end ``=` `" "``)`   `    ``g ``=` `gcd(ar[n ``-` `1``], ar[n ``-` `2``])` `    ``print``(g ``*` `2``)`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:` `    ``n ``=` `4` `    ``printSeries(n)`   `# This code is contributed by Chitranayal`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic;             ` `    `  `class` `GFG ` `{`   `static` `int` `limit = 1000000000;` `static` `int` `MAX_PRIME = 2000000;` `static` `int` `MAX = 1000000;` `static` `int` `I_MAX = 50000;`   `static` `Dictionary<``int``,` `                  ``int``> mp = ``new` `Dictionary<``int``,` `                                           ``int``>();`   `static` `int` `[]b = ``new` `int``[MAX];` `static` `int` `[]p = ``new` `int``[MAX];` `static` `int` `j = 0;` `static` `bool` `[]prime = ``new` `bool``[MAX_PRIME + 1];`   `// Function to generate Sieve of` `// Eratosthenes` `static` `void` `SieveOfEratosthenes(``int` `n)` `{` `    ``for``(``int` `i = 0; i < MAX_PRIME + 1; i++)` `        ``prime[i] = ``true``;`   `    ``for` `(``int` `p = 2; p * p <= n; p++)` `    ``{`   `        ``// If prime[p] is not changed,` `        ``// then it is a prime` `        ``if` `(prime[p] == ``true``)` `        ``{` `            ``for` `(``int` `i = p * p; i <= n; i += p)` `                ``prime[i] = ``false``;` `        ``}` `    ``}`   `    ``// Add the prime numbers to the array b` `    ``for` `(``int` `p = 2; p <= n; p++) ` `    ``{` `        ``if` `(prime[p]) ` `        ``{` `            ``b[j++] = p;` `        ``}` `    ``}` `}`   `// Function to return the gcd of a and b` `static` `int` `gcd(``int` `a, ``int` `b)` `{` `    ``if` `(b == 0)` `        ``return` `a;` `    ``return` `gcd(b, a % b);` `}`   `// Function to print the required` `// sequence of integers` `static` `void` `printSeries(``int` `n)` `{` `    ``SieveOfEratosthenes(MAX_PRIME);`   `    ``int` `i, g, k, l, m, d;` `    ``int` `[]ar = ``new` `int``[I_MAX + 2];`   `    ``for` `(i = 0; i < j; i++) ` `    ``{` `        ``if` `((b[i] * b[i + 1]) > limit)` `            ``break``;`   `        ``// Including the primes in a series` `        ``// of primes which will be later` `        ``// multiplied` `        ``p[i] = b[i];`   `        ``// This is done to mark a product` `        ``// as existing` `        ``mp.Add(b[i] * b[i + 1], 1);` `    ``}`   `    ``// Maximum number of primes that we consider` `    ``d = 550;` `    ``bool` `flag = ``false``;`   `    ``// For different interval` `    ``for` `(k = 2; (k < d - 1) && !flag; k++)` `    ``{`   `        ``// For different starting index of jump` `        ``for` `(m = 2; (m < d) && !flag; m++) ` `        ``{`   `            ``// For storing the numbers` `            ``for` `(l = m + k; l < d; l += k) ` `            ``{`   `                ``// Checking for occurrence of a` `                ``// product. Also checking for the` `                ``// same prime occurring consecutively` `                ``if` `(((b[l] * b[l + k]) < limit) && ` `                    ``mp.ContainsKey(b[l] * b[l + k]) && ` `                    ``mp.ContainsKey(p[i - 1] * b[l]) && ` `                    ``(l + k) < d && p[i - 1] != b[l + k] && ` `                                       ``p[i - 1] != b[l] &&` `                            ``mp[b[l] * b[l + k]] != 1)` `                    ``{` `                    ``if` `(mp[p[i - 1] * b[l]] != 1)` `                    ``{`   `                        ``// Including the primes in a` `                        ``// series of primes which will` `                        ``// be later multiplied` `                        ``p[i] = b[l];` `                        ``mp.Add(p[i - 1] * b[l], 1);` `                        ``i++;` `                    ``}` `                ``}`   `                ``if` `(i >= I_MAX) ` `                ``{` `                    ``flag = ``true``;` `                    ``break``;` `                ``}` `            ``}` `        ``}` `    ``}`   `    ``for` `(i = 0; i < n; i++)` `        ``ar[i] = p[i] * p[i + 1];`   `    ``for` `(i = 0; i < n - 1; i++)` `        ``Console.Write(ar[i] + ``" "``);`   `    ``g = gcd(ar[n - 1], ar[n - 2]);` `    ``Console.Write(g * 2);` `}`   `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` `    ``int` `n = 4;` `    ``printSeries(n);` `}` `}`   `// This code is contributed by 29AjayKumar`

## Javascript

 ``

Output:

`6 15 35 14`

Time Complexity: O(MAX_PRIME * I_MAX^2).

The time complexity of the above code is O(MAX_PRIME * I_MAX^2). Here MAX_PRIME is the maximum prime number that we have considered, I_MAX is the maximum number of prime numbers that can be produced and MAX is the maximum number that we have considered.

Space Complexity: O(MAX_PRIME + I_MAX).

The space complexity of the above code is O(MAX_PRIME + I_MAX). Here MAX_PRIME is the maximum prime number that we have considered and I_MAX is the maximum number of prime numbers that can be produced.

Another approach: List all the prime numbers up to 6 million by using the Sieve of Eratosthenes. We know the base condition i.e. N = 3 forms {6, 10, 15}.

So, we use these three values to generate further terms of the sequence.
Like {2, 3, 5}, these primes can not be used to generate sequences because they are already used in {6, 10, 15}. We also can’t use {7, 11}, which we’ll see later.

Now we have a prime list {13, 17, 19, 23, 29, ……}. We take the first prime and multiply it with 6, second with 15, third with 10, again 4th with 6, and so on…

```13 * 6, 17 * 15, 19 * 10, 23 * 6, 29 * 15, ........upto N - 2 terms.
(N - 1)th term = (N - 1)th prime * 7.
Nth term = 7 * 11.
again, first term = first term * 11 to make 1st and last noncoprime.
For example, N = 5
6 * 11 * 13, 15 * 17, 10 * 19, 11 * 19, 7 * 11```

Now we see that we can not use 7 and 11 from the list as these are used to generate the last and second last term.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;` `const` `int` `MAX = 620000;` `int` `prime[MAX];`   `// Function for Sieve of Eratosthenes` `void` `Sieve()` `{` `    ``for` `(``int` `i = 2; i < MAX; i++) {` `        ``if` `(prime[i] == 0) {` `            ``for` `(``int` `j = 2 * i; j < MAX; j += i) {` `                ``prime[j] = 1;` `            ``}` `        ``}` `    ``}` `}`   `// Function to print the required sequence` `void` `printSequence(``int` `n)` `{` `    ``Sieve();` `    ``vector<``int``> v, u;`   `    ``// Store only the required primes` `    ``for` `(``int` `i = 13; i < MAX; i++) {` `        ``if` `(prime[i] == 0) {` `            ``v.push_back(i);` `        ``}` `    ``}` `    ``// Base condition` `    ``if` `(n == 3) {` `        ``cout << 6 << ``" "` `<< 10 << ``" "` `<< 15;` `        ``return``;` `    ``}`   `    ``int` `k;` `    ``for` `(k = 0; k < n - 2; k++) {`   `        ``// First integer in the list` `        ``if` `(k % 3 == 0) {` `            ``u.push_back(v[k] * 6);` `        ``}`   `        ``// Second integer in the list` `        ``else` `if` `(k % 3 == 1) {`   `            ``u.push_back(v[k] * 15);` `        ``}`   `        ``// Third integer in the list` `        ``else` `{` `            ``u.push_back(v[k] * 10);` `        ``}` `    ``}` `    ``k--;`   `    ``// Generate (N - 1)th term` `    ``u.push_back(v[k] * 7);`   `    ``// Generate Nth term` `    ``u.push_back(7 * 11);`   `    ``// Modify first term` `    ``u[0] = u[0] * 11;`   `    ``// Print the sequence` `    ``for` `(``int` `i = 0; i < u.size(); i++) {` `        ``cout << u[i] << ``" "``;` `    ``}` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 4;` `    ``printSequence(n);`   `    ``return` `0;` `}`

## Java

 `// Java implementation of the approach` `import` `java.util.*;`   `class` `GFG` `{` `    ``static` `int` `MAX = ``620000``;` `    ``static` `int``[] prime = ``new` `int``[MAX];`   `    ``// Function for Sieve of Eratosthenes ` `    ``static` `void` `Sieve()` `    ``{` `        ``for` `(``int` `i = ``2``; i < MAX; i++) ` `        ``{` `            ``if` `(prime[i] == ``0``) ` `            ``{` `                ``for` `(``int` `j = ``2` `* i; ` `                         ``j < MAX; j += i) ` `                ``{` `                    ``prime[j] = ``1``;` `                ``}` `            ``}` `        ``}` `    ``}`   `    ``// Function to print the required sequence ` `    ``static` `void` `printSequence(``int` `n) ` `    ``{` `        ``Sieve();` `        ``Vector v = ``new` `Vector();` `        ``Vector u = ``new` `Vector();`   `        ``// Store only the required primes ` `        ``for` `(``int` `i = ``13``; i < MAX; i++)` `        ``{` `            ``if` `(prime[i] == ``0``) ` `            ``{` `                ``v.add(i);` `            ``}` `        ``}` `        `  `        ``// Base condition ` `        ``if` `(n == ``3``) ` `        ``{` `            ``System.out.print(``6` `+ ``" "` `+ ``10` `+ ``" "` `+ ``15``);` `            ``return``;` `        ``}`   `        ``int` `k;` `        ``for` `(k = ``0``; k < n - ``2``; k++)` `        ``{`   `            ``// First integer in the list ` `            ``if` `(k % ``3` `== ``0``)` `            ``{` `                ``u.add(v.get(k) * ``6``);` `            ``} ` `            `  `            ``// Second integer in the list ` `            ``else` `if` `(k % ``3` `== ``1``) ` `            ``{`   `                ``u.add(v.get(k) * ``15``);` `            ``}` `            `  `            ``// Third integer in the list ` `            ``else` `            ``{` `                ``u.add(v.get(k) * ``10``);` `            ``}` `        ``}` `        ``k--;`   `        ``// Generate (N - 1)th term ` `        ``u.add(v.get(k) * ``7``);`   `        ``// Generate Nth term ` `        ``u.add(``7` `* ``11``);`   `        ``// Modify first term ` `        ``u.set(``0``, u.get(``0``) * ``11``);`   `        ``// Print the sequence ` `        ``for` `(``int` `i = ``0``; i < u.size(); i++)` `        ``{` `            ``System.out.print(u.get(i) + ``" "``);` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `main(String[] args)` `    ``{` `        ``int` `n = ``4``;` `        ``printSequence(n);` `    ``}` `} `   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program for the above approach` `MAX` `=` `620000` `prime ``=` `[``0``] ``*` `MAX`   `# Function for Sieve of Eratosthenes` `def` `Sieve():`   `    ``for` `i ``in` `range``(``2``, ``MAX``):` `        ``if` `(prime[i] ``=``=` `0``):` `            ``for` `j ``in` `range``(``2` `*` `i, ``MAX``, i):` `                ``prime[j] ``=` `1`   `# Function to print the required sequence` `def` `printSequence (n):`   `    ``Sieve()` `    ``v ``=` `[]` `    ``u ``=` `[]`   `    ``# Store only the required primes` `    ``for` `i ``in` `range``(``13``, ``MAX``):` `        ``if` `(prime[i] ``=``=` `0``):` `            ``v.append(i)`   `    ``# Base condition` `    ``if` `(n ``=``=` `3``):` `        ``print``(``6``, ``10``, ``15``)` `        ``return`   `    ``k ``=` `0` `    ``for` `k ``in` `range``(n ``-` `2``):`   `        ``# First integer in the list` `        ``if` `(k ``%` `3` `=``=` `0``):` `            ``u.append(v[k] ``*` `6``)`   `        ``# Second integer in the list` `        ``elif` `(k ``%` `3` `=``=` `1``):` `            ``u.append(v[k] ``*` `15``)`   `        ``# Third integer in the list` `        ``else``:` `            ``u.append(v[k] ``*` `10``)` `    `  `    ``# Generate (N - 1)th term` `    ``u.append(v[k] ``*` `7``)`   `    ``# Generate Nth term` `    ``u.append(``7` `*` `11``)`   `    ``# Modify first term` `    ``u[``0``] ``=` `u[``0``] ``*` `11`   `    ``# Print the sequence` `    ``print``(``*``u)`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:`   `    ``n ``=` `4` `    ``printSequence(n)`   `# This code is contributed by himanshu77`

## C#

 `// C# implementation of the approach` `using` `System;` `using` `System.Collections.Generic; `   `class` `GFG` `{` `    ``static` `int` `MAX = 620000;` `    ``static` `int``[] prime = ``new` `int``[MAX];`   `    ``// Function for Sieve of Eratosthenes ` `    ``static` `void` `Sieve()` `    ``{` `        ``for` `(``int` `i = 2; i < MAX; i++) ` `        ``{` `            ``if` `(prime[i] == 0) ` `            ``{` `                ``for` `(``int` `j = 2 * i; ` `                        ``j < MAX; j += i) ` `                ``{` `                    ``prime[j] = 1;` `                ``}` `            ``}` `        ``}` `    ``}`   `    ``// Function to print the required sequence ` `    ``static` `void` `printSequence(``int` `n) ` `    ``{` `        ``Sieve();` `        ``List<``int``> v = ``new` `List<``int``>();` `        ``List<``int``> u = ``new` `List<``int``>();`   `        ``// Store only the required primes ` `        ``for` `(``int` `i = 13; i < MAX; i++)` `        ``{` `            ``if` `(prime[i] == 0) ` `            ``{` `                ``v.Add(i);` `            ``}` `        ``}` `        `  `        ``// Base condition ` `        ``if` `(n == 3) ` `        ``{` `            ``Console.Write(6 + ``" "` `+ 10 + ``" "` `+ 15);` `            ``return``;` `        ``}`   `        ``int` `k;` `        ``for` `(k = 0; k < n - 2; k++)` `        ``{`   `            ``// First integer in the list ` `            ``if` `(k % 3 == 0)` `            ``{` `                ``u.Add(v[k] * 6);` `            ``} ` `            `  `            ``// Second integer in the list ` `            ``else` `if` `(k % 3 == 1) ` `            ``{`   `                ``u.Add(v[k] * 15);` `            ``}` `            `  `            ``// Third integer in the list ` `            ``else` `            ``{` `                ``u.Add(v[k] * 10);` `            ``}` `        ``}` `        ``k--;`   `        ``// Generate (N - 1)th term ` `        ``u.Add(v[k] * 7);`   `        ``// Generate Nth term ` `        ``u.Add(7 * 11);`   `        ``// Modify first term ` `        ``u[0] = u[0] * 11;`   `        ``// Print the sequence ` `        ``for` `(``int` `i = 0; i < u.Count; i++)` `        ``{` `            ``Console.Write(u[i] + ``" "``);` `        ``}` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main(String[] args)` `    ``{` `        ``int` `n = 4;` `        ``printSequence(n);` `    ``}` `}`   `// This code is contributed by Princi Singh`

## Javascript

 ``

Output:

`858 255 119 77`

Time Complexity : O(n*log(n))

Space Complexity: O(n)

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