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Given a n, check it is the Stella Octangula number or not. A number of the form n(2n^{2} - 1)  - quicklatex - Stella Octangula Number – GeeksforGeeks where n is a whole number(0, 1, 2, 3, 4, …) is called . First few Octangula numbers are 0, 1, 14, 51, 4, 245, 426, 679, 16, 1449, 1990, …

Stella octangula numbers which are perfect squares are 1 and 9653449.

Given a number x, check if it is Stella octangula.

Examples:

Input: x = 51
Output: Yes
For n = 3, the value of expression
n(2n2 - 1) is 51

Input: n = 53
Output: No

A simple solution is to run a loop starting from n = 0. For every n, check if n(2n2 – 1) is equal to x. We run the loop while value of n(2n2 – 1) is smaller than or equal to x.

An efficient solution is to use Unbounded Binary Search. We first find a value of n such that n(2n2 – 1) is greater than x using repeated doubling. Then we apply Binary Search.

// Program to check if a number is Stella
// Octangula Number
#include <bits/stdc++.h>
using namespace std;

// Returns value of n*(2*n*n - 1)
int f(int n) {
   return n*(2*n*n - 1);
}

// Finds if a value of f(n) is equl to x
// where n is in interval [low..high]
bool binarySearch(int low, int high, int x)
{
    while (low <= high) {
        long long mid = (low + high) / 2;

        if (f(mid) < x)
            low = mid + 1;
        else if (f(mid) > x)
            high = mid - 1;
        else
            return true;
    }
    return false;
}

// Returns true if x isStella Octangula Number.
// Else returns false.
bool isStellaOctangula(int x)
{
    if (x == 0)
      return true;

    // Find 'high' for binary search by
    // repeated doubling
    int i = 1;
    while (f(i) < x)
        i = i*2;

    // If condition is satisfied for a
    // power of 2.
    if (f(i) == x)
        return true;

    //  Call binary search
    return binarySearch(i/2, i, x);
}

// driver code
int main()
{
    int n = 51;
    if (isStellaOctangula(n))
        cout << "Yes";
    else
        cout << "No";

    return 0;
}


- avatar - Stella Octangula Number &#8211; GeeksforGeeks

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