Given a number** N**, the task is to find the sum of all the perfect square divisors of numbers from** 1** to **N**. **Examples:**

Input:N = 5Output:9Explanation:N = 5

Perfect square divisors of 1 = 1.

Similarly, perfect square divisors of 2, 3 = 1.

Perfect square divisors of 4 = 1, 4.

Perfect square divisors of 5 = 1 (of course for any prime only 1 will be the perfect square divisor)

So, total sum = 1+1+1+(1+4)+1 = 9.

Input:N = 30Output:126

Input:N = 100Output:910

**Naive Approach:** This approach is based on the approach implemented in this article

The above problem can be solved in **O(N ^{1/k}) **for any

**K**power divisors, where

^{th}**N**is the number up to which we have to find the sum. This is because, in this sum, every number will contribute

**floor(N/p)**or

**int(N/p)**times. Thus, while iterating through these perfect powers, we just need to add

**[p * int(N/p)]**to the sum.

**Time Complexity:** O(√N)

**Efficient Approach:**

- Let us start from start = 2, find the largest range (start to end) for which floor(N/(start
^{2})) = floor(N/(end^{2})) - The contribution of all perfect squares in the interval [start, end] will contribute floor(N/(start
^{2})) times, hence we can do update for this range at once. - Contribution for range [start, end] can be given as:

floor(N/(start^{2}))*(sumUpto(end) – sumUpto(start-1))

- How to find range?

For a given value of start, end can be found by

sqrt(N/K), where K = floor(N/(start^2))

- Now the next range can be found by substituting start = end+1.

**Time complexity:** *O(N ^{1/3})* as

**N/(x**cannot take more than

^{2})**N**different values for a fixed value of

^{1/3}**N**.

Below is the implementation of the above approach:

## C++

`// C++ Program to find the` `// sum of all perfect square` `// divisors of numbers from 1 to N` `#include <bits/stdc++.h>` `using` `namespace` `std;` `#define MOD 1000000007` `#define int unsigned long long` `// Function for finding inverse` `// of a number iteratively` `// Here we will find the inverse` `// of 6, since it appears as` `// denominator in the formula of` `// sum of squares from 1 to N` `int` `inv(` `int` `a)` `{` ` ` `int` `o = 1;` ` ` `for` `(` `int` `p = MOD - 2;` ` ` `p > 0; p >>= 1) {` ` ` `if` `((p & 1) == 1)` ` ` `o = (o * a) % MOD;` ` ` `a = (a * a) % MOD;` ` ` `}` ` ` `return` `o;` `}` `// Store the value of the inverse` `// of 6 once as we don't need to call` `// the function again and again` `int` `inv6 = inv(6);` `// Formula for finding the sum` `// of first n squares` `int` `sumOfSquares(` `int` `n)` `{` ` ` `n %= MOD;` ` ` `return` `(((n * (n + 1))` ` ` `% MOD * (2 * n + 1))` ` ` `% MOD * inv6)` ` ` `% MOD;` `}` `int` `sums(` `int` `n)` `{` ` ` `// No perfect square` ` ` `// exists which is` ` ` `// less than 4` ` ` `if` `(n < 4)` ` ` `return` `0;` ` ` `// Starting from 2, present value` ` ` `// of start is denoted here as` ` ` `// curStart` ` ` `int` `curStart = 2, ans = 0;` ` ` `int` `sqrtN = ` `sqrt` `(n);` ` ` `while` `(curStart <= n / curStart) {` ` ` `int` `V = n / (curStart * curStart);` ` ` `// Finding end of the segment` ` ` `// for which the contribution` ` ` `// will be same` ` ` `int` `end = ` `sqrt` `(n / V);` ` ` `// Using the above mentioned` ` ` `// formula to find ans % MOD` ` ` `ans += (n / (curStart * curStart)` ` ` `% MOD * (sumOfSquares(end)` ` ` `+ MOD` ` ` `- sumOfSquares(curStart - 1)))` ` ` `% MOD;` ` ` `if` `(ans >= MOD)` ` ` `ans -= MOD;` ` ` `// Now for mthe next iteration` ` ` `// start will become end+1` ` ` `curStart = end + 1;` ` ` `}` ` ` `// Finally returning the answer` ` ` `return` `ans;` `}` `// Driver Code` `int32_t main()` `{` ` ` `int` `input[] = { 5 };` ` ` `for` `(` `auto` `x : input) {` ` ` `cout << ` `"sum of all perfect"` ` ` `<< ` `" square divisors from"` ` ` `<< ` `" 1 to "` `<< x` ` ` `<< ` `" is: "` `;` ` ` `// Here we are adding x` ` ` `// because we have not` ` ` `// counted 1 as perfect` ` ` `// squares so if u want to` ` ` `// add it you can just add` ` ` `// that number to the ans` ` ` `cout << x + sums(x) << endl;` ` ` `}` ` ` `return` `0;` `}` |

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## Java

`// Java program to find the` `// sum of all perfect square` `// divisors of numbers from 1 to N` `import` `java.util.*;` `class` `GFG{` `static` `final` `int` `MOD = ` `7` `;` `// Function for finding inverse` `// of a number iteratively` `// Here we will find the inverse` `// of 6, since it appears as` `// denominator in the formula of` `// sum of squares from 1 to N` `static` `int` `inv(` `int` `a)` `{` ` ` `int` `o = ` `1` `;` ` ` `for` `(` `int` `p = MOD - ` `2` `;` ` ` `p > ` `0` `; p >>= ` `1` `) ` ` ` `{` ` ` `if` `((p & ` `1` `) == ` `1` `)` ` ` `o = (o * a) % MOD;` ` ` ` ` `a = (a * a) % MOD;` ` ` `}` ` ` `return` `o;` `}` `// Store the value of the inverse` `// of 6 once as we don't need to call` `// the function again and again` `static` `int` `inv6 = inv(` `6` `);` `// Formula for finding the sum` `// of first n squares` `static` `int` `sumOfSquares(` `int` `n)` `{` ` ` `n %= MOD;` ` ` `return` `(((n * (n + ` `1` `)) % ` ` ` `MOD * (` `2` `* n + ` `1` `)) % ` ` ` `MOD * inv6) % MOD;` `}` `static` `int` `sums(` `int` `n)` `{` ` ` ` ` `// No perfect square` ` ` `// exists which is` ` ` `// less than 4` ` ` `if` `(n < ` `4` `)` ` ` `return` `0` `;` ` ` `// Starting from 2, present value` ` ` `// of start is denoted here as` ` ` `// curStart` ` ` `int` `curStart = ` `2` `, ans = ` `0` `;` ` ` `int` `sqrtN = (` `int` `)Math.sqrt(n);` ` ` `while` `(curStart <= n / curStart) ` ` ` `{` ` ` `int` `V = n / (curStart * curStart);` ` ` `// Finding end of the segment` ` ` `// for which the contribution` ` ` `// will be same` ` ` `int` `end = (` `int` `)Math.sqrt(n / V);` ` ` `// Using the above mentioned` ` ` `// formula to find ans % MOD` ` ` `ans += (n / (curStart * curStart) % ` ` ` `MOD * (sumOfSquares(end) + MOD - ` ` ` `sumOfSquares(curStart - ` `1` `))) % MOD;` ` ` `if` `(ans >= MOD)` ` ` `ans -= MOD;` ` ` `// Now for mthe next iteration` ` ` `// start will become end+1` ` ` `curStart = end + ` `1` `;` ` ` `}` ` ` `// Finally returning the answer` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `input[] = {` `5` `};` ` ` `for` `(` `int` `x : input)` ` ` `{` ` ` `System.out.print(` `"sum of all perfect "` `+ ` ` ` `"square divisors from "` `+` ` ` `"1 to "` `+ x + ` `" is: "` `);` ` ` `// Here we are adding x` ` ` `// because we have not` ` ` `// counted 1 as perfect` ` ` `// squares so if u want to` ` ` `// add it you can just add` ` ` `// that number to the ans` ` ` `System.out.print(x + sums(x) + ` `"\n"` `);` ` ` `}` `}` `}` `// This code is contributed by Amit Katiyar` |

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## Python3

`# Python3 program to find the ` `# sum of all perfect square ` `# divisors of numbers from 1 to N ` `from` `math ` `import` `*` `MOD ` `=` `1000000007` `# Function for finding inverse ` `# of a number iteratively ` `# Here we will find the inverse ` `# of 6, since it appears as ` `# denominator in the formula of ` `# sum of squares from 1 to N ` `def` `inv (a):` ` ` `o ` `=` `1` ` ` `p ` `=` `MOD ` `-` `2` ` ` `while` `(p > ` `0` `):` ` ` `if` `(p ` `%` `2` `=` `=` `1` `):` ` ` `o ` `=` `(o ` `*` `a) ` `%` `MOD` ` ` `a ` `=` `(a ` `*` `a) ` `%` `MOD` ` ` `p >>` `=` `1` ` ` `return` `o ` `# Store the value of the inverse ` `# of 6 once as we don't need to call ` `# the function again and again ` `inv6 ` `=` `inv(` `6` `)` `# Formula for finding the sum ` `# of first n squares` `def` `sumOfSquares (n):` ` ` `n ` `%` `=` `MOD` ` ` `return` `(((n ` `*` `(n ` `+` `1` `)) ` `%` ` ` `MOD ` `*` `(` `2` `*` `n ` `+` `1` `)) ` `%` ` ` `MOD ` `*` `inv6) ` `%` `MOD` `def` `sums (n):` ` ` `# No perfect square exists which` ` ` `# is less than 4` ` ` `if` `(n < ` `4` `):` ` ` `return` `0` ` ` `# Starting from 2, present value` ` ` `# of start is denoted here as curStart` ` ` `curStart ` `=` `2` ` ` `ans ` `=` `0` ` ` `sqrtN ` `=` `int` `(sqrt(n))` ` ` `while` `(curStart <` `=` `n ` `/` `/` `curStart):` ` ` `V ` `=` `n ` `/` `/` `(curStart ` `*` `curStart)` ` ` `# Finding end of the segment for ` ` ` `# which the contribution will be same` ` ` `end ` `=` `int` `(sqrt(n ` `/` `/` `V))` ` ` `# Using the above mentioned` ` ` `# formula to find ans % MOD` ` ` `ans ` `+` `=` `((n ` `/` `/` `(curStart ` `*` `curStart) ` `%` ` ` `MOD ` `*` `(sumOfSquares(end) ` `+` ` ` `MOD ` `-` `sumOfSquares(curStart ` `-` `1` `))) ` `%` `MOD)` ` ` `if` `(ans >` `=` `MOD):` ` ` `ans ` `-` `=` `MOD` ` ` `# Now for mthe next iteration` ` ` `# start will become end+1` ` ` `curStart ` `=` `end ` `+` `1` ` ` `# Finally return the answer` ` ` `return` `ans ` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `Input` `=` `[` `5` `]` ` ` `for` `x ` `in` `Input` `:` ` ` `print` `(` `"sum of all perfect "` `\` ` ` `"square "` `, end ` `=` `'')` ` ` `print` `(` `"divisors from 1 to"` `, x,` ` ` `"is: "` `, end ` `=` `'')` ` ` `# Here we are adding x because we have` ` ` `# not counted 1 as perfect squares so if u ` ` ` `# want to add it you can just add that ` ` ` `# number to the ans` ` ` `print` `(x ` `+` `sums(x))` `# This code is contributed by himanshu77` |

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## C#

`// C# program to find the` `// sum of all perfect square` `// divisors of numbers from 1 to N` `using` `System;` `class` `GFG{` `static` `readonly` `int` `MOD = 7;` `// Function for finding inverse` `// of a number iteratively` `// Here we will find the inverse` `// of 6, since it appears as` `// denominator in the formula of` `// sum of squares from 1 to N` `static` `int` `inv(` `int` `a)` `{` ` ` `int` `o = 1;` ` ` `for` `(` `int` `p = MOD - 2;` ` ` `p > 0; p >>= 1) ` ` ` `{` ` ` `if` `((p & 1) == 1)` ` ` `o = (o * a) % MOD;` ` ` `a = (a * a) % MOD;` ` ` `}` ` ` ` ` `return` `o;` `}` `// Store the value of the inverse` `// of 6 once as we don't need to call` `// the function again and again` `static` `int` `inv6 = inv(6);` `// Formula for finding the sum` `// of first n squares` `static` `int` `sumOfSquares(` `int` `n)` `{` ` ` `n %= MOD;` ` ` `return` `(((n * (n + 1)) % ` ` ` `MOD * (2 * n + 1)) % ` ` ` `MOD * inv6) % MOD;` `}` `static` `int` `sums(` `int` `n)` `{` ` ` `// No perfect square` ` ` `// exists which is` ` ` `// less than 4` ` ` `if` `(n < 4)` ` ` `return` `0;` ` ` `// Starting from 2, present ` ` ` `// value of start is denoted ` ` ` `// here as curStart` ` ` `int` `curStart = 2, ans = 0;` ` ` `int` `sqrtN = (` `int` `)Math.Sqrt(n);` ` ` `while` `(curStart <= n / curStart) ` ` ` `{` ` ` `int` `V = n / (curStart * curStart);` ` ` `// Finding end of the segment` ` ` `// for which the contribution` ` ` `// will be same` ` ` `int` `end = (` `int` `)Math.Sqrt(n / V);` ` ` `// Using the above mentioned` ` ` `// formula to find ans % MOD` ` ` `ans += (n / (curStart * curStart) % ` ` ` `MOD * (sumOfSquares(end) + MOD - ` ` ` `sumOfSquares(curStart - ` ` ` `1))) % MOD;` ` ` `if` `(ans >= MOD)` ` ` `ans -= MOD;` ` ` `// Now for mthe next iteration` ` ` `// start will become end+1` ` ` `curStart = end + 1;` ` ` `}` ` ` `// Finally returning ` ` ` `// the answer` ` ` `return` `ans;` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[]input = {5};` ` ` `foreach` `(` `int` `x ` `in` `input)` ` ` `{` ` ` `Console.Write(` `"sum of all perfect "` `+ ` ` ` `"square divisors from "` `+` ` ` `"1 to "` `+ x + ` `" is: "` `);` ` ` `// Here we are adding x` ` ` `// because we have not` ` ` `// counted 1 as perfect` ` ` `// squares so if u want to` ` ` `// add it you can just add` ` ` `// that number to the ans` ` ` `Console.Write(x + sums(x) + ` `"\n"` `);` ` ` `}` `}` `}` `// This code is contributed by Rajput-Ji` |

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**Output:**

sum of all perfect square divisors from 1 to 5 is: 9

**Time Complexity: **O(N^{1/3})