First triangular number whose number of divisors exceeds N
Last Updated :
29 Jul, 2022
Given a number N, find the first triangular number whose number of divisors exceeds N. Triangular numbers are sums of natural numbers, i. e., of the form x*(x+1)/2. First few triangular numbers are 1, 3, 6, 10, 15, 21, 28, ...
Examples:
Input: N = 2
Output: 6
6 is the first triangular number with more than 2 factors.
Input: N = 4
Output: 28
A naive solution is to iterate for every triangular number and count the number of divisors using the Sieve method. At any moment if the number of divisors exceeds the given number N, then we get our answer. If the triangular number which has more than N divisors is X, then the time complexity will be O(X * sqrt(X)) as pre-processing of primes is not possible in case of larger triangular numbers. The naive solution is important to understand in order to solve the problem more efficiently.
An efficient solution will be to use the fact that the triangular number's formula is x*(x+1)/2. The property that we will use is that k and k+1 are coprimes. We know that two co-primes have a distinct set of prime factors. There will be two cases when X is even and odd.
- When X is even, then X/2 and (X+1) will be considered as two numbers whose prime factorisation is to be find out.
- When X is odd, then X and (X+1)/2 will be considered as two numbers whose prime factorisation is to be find out.
Hence the problem has been reduced to the just finding out prime factorization of smaller numbers, which reduces the time complexity significantly. We can reuse the prime factorization for x+1 in the subsequent iterations, and thus factorizing one number in each iteration will do. Iterating till the number of divisors exceeds N and considering the case of even and odd will give us the answer.
Below is the implementation of the above approach.
C++
// C++ efficient program for counting the
// number of numbers <=N having exactly
// 9 divisors
#include <bits/stdc++.h>
using namespace std;
const int MAX = 100000;
// sieve method for prime calculation
bool prime[MAX + 1];
// Function to mark the primes
void sieve()
{
memset(prime, true, sizeof(prime));
// mark the primes
for (int p = 2; p * p < MAX; p++)
if (prime[p] == true)
// mark the factors of prime as non prime
for (int i = p * 2; i < MAX; i += p)
prime[i] = false;
}
// Function for finding no. of divisors
int divCount(int n)
{
// Traversing through all prime numbers
int total = 1;
for (int p = 2; p <= n; p++) {
if (prime[p]) {
// calculate number of divisor
// with formula total div =
// (p1+1) * (p2+1) *.....* (pn+1)
// where n = (a1^p1)*(a2^p2)....
// *(an^pn) ai being prime divisor
// for n and pi are their respective
// power in factorization
int count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
// Function to find the first triangular number
int findNumber(int n)
{
if (n == 1)
return 3;
// initial number
int i = 2;
// initial count of divisors
int count = 0;
// prestore the value
int second = 1;
int first = 1;
// iterate till we get the first triangular number
while (count <= n) {
// even
if (i % 2 == 0) {
// function call to count divisors
first = divCount(i + 1);
// multiply with previous value
count = first * second;
}
// odd step
else {
// function call to count divisors
second = divCount((i + 1) / 2);
// multiply with previous value
count = first * second;
}
i++;
}
return i * (i - 1) / 2;
}
// Driver Code
int main()
{
int n = 4;
// Call the sieve function for prime
sieve();
cout << findNumber(n);
return 0;
}
// Java efficient program for counting the
// number of numbers <=N having exactly
// 9 divisors
public class GFG {
final static int MAX = 100000;
// sieve method for prime calculation
static boolean prime[] = new boolean [MAX + 1];
// Function to mark the primes
static void sieve()
{
for(int i = 0 ; i <= MAX ; i++)
prime[i] = true;
// mark the primes
for (int p = 2; p * p < MAX; p++)
if (prime[p] == true)
// mark the factors of prime as non prime
for (int i = p * 2; i < MAX; i += p)
prime[i] = false;
}
// Function for finding no. of divisors
static int divCount(int n)
{
// Traversing through all prime numbers
int total = 1;
for (int p = 2; p <= n; p++) {
if (prime[p]) {
// calculate number of divisor
// with formula total div =
// (p1+1) * (p2+1) *.....* (pn+1)
// where n = (a1^p1)*(a2^p2)....
// *(an^pn) ai being prime divisor
// for n and pi are their respective
// power in factorization
int count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
// Function to find the first triangular number
static int findNumber(int n)
{
if (n == 1)
return 3;
// initial number
int i = 2;
// initial count of divisors
int count = 0;
// prestore the value
int second = 1;
int first = 1;
// iterate till we get the first triangular number
while (count <= n) {
// even
if (i % 2 == 0) {
// function call to count divisors
first = divCount(i + 1);
// multiply with previous value
count = first * second;
}
// odd step
else {
// function call to count divisors
second = divCount((i + 1) / 2);
// multiply with previous value
count = first * second;
}
i++;
}
return i * (i - 1) / 2;
}
public static void main(String args[])
{
int n = 4;
// Call the sieve function for prime
sieve();
System.out.println(findNumber(n));
}
// This Code is contributed by ANKITRAI1
}
# Python 3 efficient program for counting the
# number of numbers <=N having exactly
# 9 divisors
from math import sqrt
MAX = 100000
prime = [ True for i in range(MAX + 1)]
# Function to mark the primes
def sieve():
# mark the primes
k = int(sqrt(MAX))
for p in range(2,k,1):
if (prime[p] == True):
# mark the factors of prime as non prime
for i in range(p * 2,MAX,p):
prime[i] = False
# Function for finding no. of divisors
def divCount(n):
# Traversing through all prime numbers
total = 1
for p in range(2,n+1,1):
if (prime[p]):
# calculate number of divisor
# with formula total div =
# (p1+1) * (p2+1) *.....* (pn+1)
# where n = (a1^p1)*(a2^p2)....
# *(an^pn) ai being prime divisor
# for n and pi are their respective
# power in factorization
count = 0
if (n % p == 0):
while (n % p == 0):
n = n / p
count += 1
total = total * (count + 1)
return total
# Function to find the first triangular number
def findNumber(n):
if (n == 1):
return 3
# initial number
i = 2
# initial count of divisors
count = 0
# prestore the value
second = 1
first = 1
# iterate till we get the first triangular number
while (count <= n):
# even
if (i % 2 == 0):
# function call to count divisors
first = divCount(i + 1)
# multiply with previous value
count = first * second
# odd step
else:
# function call to count divisors
second = divCount(int((i + 1) / 2))
# multiply with previous value
count = first * second
i += 1
return i * (i - 1) / 2
# Driver Code
if __name__ == '__main__':
n = 4
# Call the sieve function for prime
sieve()
print(int(findNumber(n)))
# This code is contributed by
# Surendra_Gangwar
// C# efficient program for counting the
// number of numbers <=N having exactly
// 9 divisors
using System;
public class GFG {
static int MAX = 100000;
// sieve method for prime calculation
static bool[] prime = new bool [MAX + 1];
// Function to mark the primes
static void sieve()
{
for(int i = 0 ; i <= MAX ; i++)
prime[i] = true;
// mark the primes
for (int p = 2; p * p < MAX; p++)
if (prime[p] == true)
// mark the factors of prime as non prime
for (int i = p * 2; i < MAX; i += p)
prime[i] = false;
}
// Function for finding no. of divisors
static int divCount(int n)
{
// Traversing through all prime numbers
int total = 1;
for (int p = 2; p <= n; p++) {
if (prime[p]) {
// calculate number of divisor
// with formula total div =
// (p1+1) * (p2+1) *.....* (pn+1)
// where n = (a1^p1)*(a2^p2)....
// *(an^pn) ai being prime divisor
// for n and pi are their respective
// power in factorization
int count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
// Function to find the first triangular number
static int findNumber(int n)
{
if (n == 1)
return 3;
// initial number
int i = 2;
// initial count of divisors
int count = 0;
// prestore the value
int second = 1;
int first = 1;
// iterate till we get the first triangular number
while (count <= n) {
// even
if (i % 2 == 0) {
// function call to count divisors
first = divCount(i + 1);
// multiply with previous value
count = first * second;
}
// odd step
else {
// function call to count divisors
second = divCount((i + 1) / 2);
// multiply with previous value
count = first * second;
}
i++;
}
return i * (i - 1) / 2;
}
public static void Main()
{
int n = 4;
// Call the sieve function for prime
sieve();
Console.Write(findNumber(n));
}
}
<?php
// PHP efficient program for counting the
// number of numbers <=N having exactly
// 9 divisors
$MAX = 10000;
// sieve method for $prime calculation
$prime = array_fill(0, $MAX + 1, true);
// Function to mark the primes
function sieve()
{
global $prime;
global $MAX;
// mark the primes
for ($p = 2; $p * $p < $MAX; $p++)
if ($prime[$p] == true)
// mark the factors of prime
// as non prime
for ($i = $p * 2;
$i < $MAX; $i += $p)
$prime[$i] = false;
}
// Function for finding no. of divisors
function divCount($n)
{
global $prime;
// Traversing through all prime numbers
$total = 1;
for ($p = 2; $p <= $n; $p++)
{
if ($prime[$p])
{
// calculate number of divisor
// with formula $total div =
// (p1+1) * (p2+1) *.....* (pn+1)
// where $n = (a1^p1)*(a2^p2)....
// *(an^pn) ai being $prime divisor
// for $n and pi are their respective
// power in factorization
$count = 0;
if ($n % $p == 0)
{
while ($n % $p == 0)
{
$n = $n / $p;
$count++;
}
$total = $total * ($count + 1);
}
}
}
return $total;
}
// Function to find the first
// triangular number
function findNumber($n)
{
if ($n == 1)
return 3;
// initial number
$i = 2;
// initial count of divisors
$count = 0;
// prestore the value
$second = 1;
$first = 1;
// iterate till we get the
// first triangular number
while ($count <= $n)
{
// even
if ($i % 2 == 0)
{
// function call to $count divisors
$first = divCount($i + 1);
// multiply with previous value
$count = $first * $second;
}
// odd step
else
{
// function call to $count divisors
$second = divCount(($i + 1) / 2);
// multiply with previous value
$count = $first * $second;
}
$i++;
}
return $i * ($i - 1) / 2;
}
// Driver Code
$n = 4;
// Call the sieve function for prime
sieve();
echo findNumber($n);
// This code is contributed by ihritik
?>
<script>
// javascript efficient program for counting the
// number of numbers <=N having exactly
// 9 divisors
const MAX = 100000;
// sieve method for prime calculation
let prime = new Array(MAX + 1).fill(0);
// Function to mark the primes
function sieve()
{
for (i = 0; i <= MAX; i++)
prime[i] = true;
// mark the primes
for (p = 2; p * p < MAX; p++)
if (prime[p] == true)
// mark the factors of prime as non prime
for (i = p * 2; i < MAX; i += p)
prime[i] = false;
}
// Function for finding no. of divisors
function divCount(n)
{
// Traversing through all prime numbers
var total = 1;
for (p = 2; p <= n; p++) {
if (prime[p]) {
// calculate number of divisor
// with formula total div =
// (p1+1) * (p2+1) *.....* (pn+1)
// where n = (a1^p1)*(a2^p2)....
// *(an^pn) ai being prime divisor
// for n and pi are their respective
// power in factorization
var count = 0;
if (n % p == 0) {
while (n % p == 0) {
n = n / p;
count++;
}
total = total * (count + 1);
}
}
}
return total;
}
// Function to find the first triangular number
function findNumber(n) {
if (n == 1)
return 3;
// initial number
var i = 2;
// initial count of divisors
var count = 0;
// prestore the value
var second = 1;
var first = 1;
// iterate till we get the first triangular number
while (count <= n) {
// even
if (i % 2 == 0) {
// function call to count divisors
first = divCount(i + 1);
// multiply with previous value
count = first * second;
}
// odd step
else {
// function call to count divisors
second = divCount((i + 1) / 2);
// multiply with previous value
count = first * second;
}
i++;
}
return i * (i - 1) / 2;
}
var n = 4;
// Call the sieve function for prime
sieve();
document.write(findNumber(n));
// This code contributed by Rajput-Ji
</script>
Time Complexity: O(N*logN),
- Sieve of eratosthenes will cost O(N*log(logN)) time, but
- we are using nested loops where the outer loop traverses N times and the inner loop traverses logN times as in every traversal we are decrementing by floor division of factor of n.
Auxiliary Space: O(105), as we are using extra space for primer array.
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