Given a positive integer N, find the N-th triacontakaihenagonal number: the N-th polygonal number of a 31-sided polygon, the triacontakaihenagon.
What is a triacontakaihenagonal number?
Polygonal numbers count the dots in a growing family of regular polygons drawn from one shared corner, where each new polygon has one more dot on every side. For a polygon with s sides, the n-th polygonal number is
P(s, n) = ((s − 2)n² − (s − 4)n) / 2
A triacontakaihenagon has 31 sides. Putting s = 31 gives
T(n) = (29n² − 27n) / 2
which produces 1, 31, 90, 178, 295, …
The division is always exact: n(29n − 27) is even, because when n is odd, 29n − 27 is even.
Examples:
Input: N = 2
Output: 31
Explanation: T(2) = (29 × 4 − 27 × 2) / 2 = (116 − 54) / 2 = 31.Input: N = 3
Output: 90
Explanation: T(3) = (261 − 81) / 2 = 90.
Approach
Apply the formula directly. A 64-bit integer type keeps 29n² from overflowing for larger n.
// N-th triacontakaihenagonal (31-gonal) number.
#include <stdio.h>
long long triacontakaihenagonal(long long n)
{
return (29 * n * n - 27 * n) / 2;
}
int main(void)
{
for (long long n = 1; n <= 5; n++) {
printf("T(%lld) = %lld\n", n, triacontakaihenagonal(n));
}
return 0;
}
Output:
T(1) = 1 T(2) = 31 T(3) = 90 T(4) = 178 T(5) = 295
In Python, integers never overflow:
def triacontakaihenagonal(n: int) -> int:
return (29 * n * n - 27 * n) // 2
for n in range(1, 6):
print(f"T({n}) = {triacontakaihenagonal(n)}")
Complexity: O(1) time and O(1) space.
