Gijswijt’s Sequence

Gijswijt’s sequence is a self-describing sequence where the value of each term is equal to the maximum number of repeated blocks of numbers in the sequence preceding the number.

Let us take the ith term of the sequence a(i), Then for this Gijswijt’s Sequence:

where k is the largest natural number such that the word a(1)a(2)..a(n) can be represented as x*(y^k) where the length of y is non zero.



The Gijswijt’s sequence is as follows:

1, 1, 2, 1, 1, 2, 2, 2, 3, 1, ...

It can be proved that every natural number occurs in this sequence at least once but the sequence has very slow growth. The 220th term is 4 whereas the term 5 occurs in the sequence near 10^10^23th term.

Examples:

Input: n = 10
Output: 1, 1, 2, 1, 1, 2, 2, 2, 3, 1

Input: n = 220
Output: 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, …, 3, 2, 1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 1, 2, 1, 1, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 3, 2, 2, 2, 3, 3, 3, 3, 4,

Approach: We have to print the first n terms of the series. We will keep a vector which will store all the previous element of the sequence the next term will be maximum count of repeated blocks of any length. So we will vary the length from 1 to n-1 and then find out the count using a loop.

Example:

  • let the sequence be 1, 1, 2, 1, 1, 2
  • Take values of i from 0 to 5 first pattern contains “2”.
  • 2 is repeated only once from the end of array, next pattern is “1, 2”.
  • 1, 2 is repeated only once from the end of array, the next pattern is “1, 1, 2”.
  • 1, 1, 2 is repeated twice from the end of array so the count is 2.

Below is the implementation of the above approach:

C++

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// C++ program to demonstrate
// Gijswijt's sequence
  
#include <bits/stdc++.h>
using namespace std;
  
// if the sequence is a(1)a(2)a(3)..a(n-1)
// check if the sequence can be represented as
// x*(y^k) find the largest value of k
int find_count(vector<int> ele)
{
  
    // count
    int count = 0;
  
    for (int i = 0; i < ele.size(); i++) {
  
        // pattern of elements of size
        // i from the end of sequence
        vector<int> p;
  
        // count
        int c = 0;
  
        // extract the pattern in a reverse order
        for (int j = ele.size() - 1;
             j >= (ele.size() - 1 - i) && j >= 0;
             j--)
            p.push_back(ele[j]);
  
        int j = ele.size() - 1, k = 0;
  
        // check how many times
        // the pattern is repeated
        while (j >= 0) {
  
            // if the element dosent match
            if (ele[j] != p[k])
                break;
  
            j--;
            k++;
  
            // if the end of pattern is reached
            // set value of k=0 and
            // increase the count
            if (k == p.size()) {
                c++;
                k = 0;
            }
        }
        count = max(count, c);
    }
  
    // return the max count
    return count;
}
  
// print first n terms of
// Gijswijt's sequence
void solve(int n)
{
    // set the count
    int count = 1;
  
    // stoes the element
    vector<int> ele;
  
    // print the first n terms of
    // the sequence
    for (int i = 0; i < n; i++) {
        cout << count << ", ";
  
        // push the element
        ele.push_back(count);
  
        // find the count for next number
        count = find_count(ele);
    }
}
  
// Driver code
int main()
{
    int n = 10;
  
    solve(n);
  
    return 0;
}

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Java

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// Java program to demonstrate
// Gijswijt's sequence
import java.util.*;
  
class GFG 
{
  
    // if the sequence is a(1)a(2)a(3)..a(n-1)
    // check if the sequence can be represented as
    // x*(y^k) find the largest value of k
    static int find_count(Vector<Integer> ele) 
    {
  
        // count
        int count = 0;
  
        for (int i = 0; i < ele.size(); i++) 
        {
  
            // pattern of elements of size
            // i from the end of sequence
            Vector<Integer> p = new Vector<Integer>();
  
            // count
            int c = 0;
  
            // extract the pattern in a reverse order
            for (int j = ele.size() - 1;
                     j >= (ele.size() - 1 - i) && j >= 0;
                     j--) 
            {
                p.add(ele.get(j));
            }
  
            int j = ele.size() - 1, k = 0;
  
            // check how many times
            // the pattern is repeated
            while (j >= 0)
            {
  
                // if the element dosent match
                if (ele.get(j) != p.get(k)) 
                {
                    break;
                }
  
                j--;
                k++;
  
                // if the end of pattern is reached
                // set value of k=0 and
                // increase the count
                if (k == p.size()) 
                {
                    c++;
                    k = 0;
                }
            }
            count = Math.max(count, c);
        }
  
        // return the max count
        return count;
    }
  
    // print first n terms of
    // Gijswijt's sequence
    static void solve(int n) 
    {
          
        // set the count
        int count = 1;
  
        // stoes the element
        Vector<Integer> ele = new Vector<Integer>();
  
        // print the first n terms of
        // the sequence
        for (int i = 0; i < n; i++) 
        {
            System.out.print(count + ", ");
  
            // push the element
            ele.add(count);
  
            // find the count for next number
            count = find_count(ele);
        }
    }
  
    // Driver code
    public static void main(String[] args) 
    {
        int n = 10;
  
        solve(n);
    }
  
// This code is contributed by PrinciRaj1992

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Python3

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# Python3 program to demonstrate
# Gijswijt's sequence
  
# if the sequence is a(1)a(2)a(3)..a(n-1)
# check if the sequence can be represented as
# x*(y^k) find the largest value of k
def find_count(ele):
  
    # count
    count = 0
  
    for i in range(len(ele)):
  
        # pattern of elements of size
        # i from the end of sequence
        p = []
  
        # count
        c = 0
        j = len(ele) - 1
  
        # extract the pattern in a reverse order
        while j >= (len(ele) - 1 - i) and j >= 0:
            p.append(ele[j])
            j -= 1
  
        j = len(ele) - 1
        k = 0
  
        # check how many times
        # the pattern is repeated
        while j >= 0:
  
            # if the element dosent match
            if ele[j] != p[k]:
                break
  
            j -= 1
            k += 1
  
            # if the end of pattern is reached
            # set value of k=0 and
            # increase the count
            if k == len(p):
                c += 1
                k = 0
  
        count = max(count, c)
  
    # return the max count
    return count
  
# print first n terms of
# Gijswijt's sequence
def solve(n):
  
    # set the count
    count = 1
  
    # stoes the element
    ele = []
  
    # print the first n terms of
    # the sequence
    for i in range(n):
        print(count, end = " ")
  
        # push the element
        ele.append(count)
  
        # find the count for next number
        count = find_count(ele)
  
# Driver Code
if __name__ == "__main__":
    n = 10
  
    solve(n)
  
# This code is contributed by
# sanjeev2552

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C#

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// C# program to demonstrate
// Gijswijt's sequence
using System;
using System.Collections.Generic;
      
class GFG 
{
  
    // if the sequence is a(1)a(2)a(3)..a(n-1)
    // check if the sequence can be represented as
    // x*(y^k) find the largest value of k
    static int find_count(List<int> ele) 
    {
  
        // count
        int count = 0;
  
        for (int i = 0; i < ele.Count; i++) 
        {
  
            // pattern of elements of size
            // i from the end of sequence
            List<int> p = new List<int>();
  
            // count
            int c = 0, j;
  
            // extract the pattern in a reverse order
            for (j = ele.Count - 1;
                 j >= (ele.Count - 1 - i) && j >= 0;
                 j--) 
            {
                p.Add(ele[j]);
            }
            j = ele.Count - 1;
            int k = 0;
  
            // check how many times
            // the pattern is repeated
            while (j >= 0)
            {
  
                // if the element doesn't match
                if (ele[j] != p[k]) 
                {
                    break;
                }
  
                j--;
                k++;
  
                // if the end of pattern is reached
                // set value of k=0 and
                // increase the count
                if (k == p.Count) 
                {
                    c++;
                    k = 0;
                }
            }
            count = Math.Max(count, c);
        }
  
        // return the max count
        return count;
    }
  
    // print first n terms of
    // Gijswijt's sequence
    static void solve(int n) 
    {
          
        // set the count
        int count = 1;
  
        // stoes the element
        List<int> ele = new List<int>();
  
        // print the first n terms of
        // the sequence
        for (int i = 0; i < n; i++) 
        {
            Console.Write(count + ", ");
  
            // push the element
            ele.Add(count);
  
            // find the count for next number
            count = find_count(ele);
        }
    }
  
    // Driver code
    public static void Main(String[] args) 
    {
        int n = 10;
  
        solve(n);
    }
}
  
// This code is contributed by Rajput-Ji

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Output:

1, 1, 2, 1, 1, 2, 2, 2, 3, 1, 


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Second year Department of Information Technology Jadavpur University

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