Check if a Number is a Tetracontagon Number

Given a positive integer N, find out whether N is a tetracontagonal number (also called a tetracontagon number): one of the numbers produced by 40-sided polygons.

What is a tetracontagonal number?

Polygonal (figurate) numbers count the dots needed to draw a growing family of regular polygons that share one corner, where each new polygon has one more dot on every side. A tetracontagon has 40 sides, so the tetracontagonal numbers are the polygonal numbers with s = 40.

For a polygon with s sides, the k-th polygonal number is

P(s, k) = ((s − 2)k² − (s − 4)k) / 2

With s = 40 this becomes

T(k) = (38k² − 36k) / 2 = 19k² − 18k

so the sequence starts 1, 40, 117, 232, 385, …

Examples:

Input: N = 40
Output: Yes
Explanation: T(2) = 19 × 4 − 18 × 2 = 40.

Input: N = 35
Output: No
Explanation: T(1) = 1 and T(2) = 40, and no term lies between them.

Approach

N is tetracontagonal exactly when 19k² − 18k = N has a positive whole-number solution k. Solving this quadratic for k gives

k = (18 + √(76N + 324)) / 38

So two checks are enough, both with whole numbers only:

1. 76N + 324 is a perfect square, say d².
2. 18 + d is divisible by 38. Then k = (18 + d) / 38.

Working with integers avoids the rounding errors that a floating-point square root can cause for large N.

// Check whether N is a tetracontagonal (40-gonal) number.
#include <cmath>
#include <iostream>
using namespace std;

bool isTetracontagonal(long long n)
{
    if (n < 1) return false;
    long long disc = 76 * n + 324;          // must be a perfect square
    long long d = llround(sqrt((long double)disc));
    while (d * d > disc) --d;               // correct any rounding
    while ((d + 1) * (d + 1) <= disc) ++d;
    return d * d == disc && (18 + d) % 38 == 0;
}

int main()
{
    for (long long n : {40LL, 35LL, 117LL, 385LL}) {
        cout << n << (isTetracontagonal(n) ? " is" : " is not")
             << " a tetracontagonal number\n";
    }
    return 0;
}

Output:

40 is a tetracontagonal number
35 is not a tetracontagonal number
117 is a tetracontagonal number
385 is a tetracontagonal number

The same check in Python, where math.isqrt gives an exact integer square root:

import math

def is_tetracontagonal(n: int) -> bool:
    if n < 1:
        return False
    disc = 76 * n + 324
    d = math.isqrt(disc)
    return d * d == disc and (18 + d) % 38 == 0

for n in (40, 35, 117, 385):
    print(n, "is" if is_tetracontagonal(n) else "is not", "a tetracontagonal number")

Complexity: O(1) arithmetic per number (one integer square root) and O(1) extra space.