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# Problem 401: Sum of squares of divisors

The divisors of 6 are 1,2,3 and 6.
The sum of the squares of these numbers is 1+4+9+36=50.

Let sigma2(n) represent the sum of the squares of the divisors of n. Thus sigma2(6)=50.
Let SIGMA2 represent the summatory function of sigma2, that is SIGMA2(n)=sum{sigma2(i)} for i=1 to n.
The first 6 values of SIGMA2 are: 1,6,16,37,63 and 113.

Find SIGMA2(10^15) mod 10^9.

# My Algorithm

Any number i has \lfloor frac{n}{i} \rfloor multiples (integer division).
When calculating SIGMA2(6):

• 1 has \lfloor frac{6}{1} \rfloor = 6 multiples, add 6 * 1^2 = 6 * 1 = 6
• 2 has \lfloor frac{6}{2} \rfloor = 3 multiples, add 3 * 2^2 = 3 * 4 = 24
• 3 has \lfloor frac{6}{3} \rfloor = 2 multiples, add 2 * 3^2 = 2 * 9 = 18
• 4 has \lfloor frac{6}{4} \rfloor = 1 multiples, add 1 * 4^2 = 1 * 16 = 16
• 5 has \lfloor frac{6}{5} \rfloor = 1 multiples, add 1 * 5^2 = 1 * 25 = 25
• 6 has \lfloor frac{6}{6} \rfloor = 1 multiples, add 1 * 6^2 = 1 * 36 = 36
SIGMA2(6) = 6 + 24 + 18 + 16 + 25 + 36 = 113

Unfortunately this requires n iterations, which is no problem for n = 6 but next to impossible for n = 15.
Looking closer at the example reveals that half of the divisors appears only once (4, 5 and 6).
In general, a few number appear very often and the majority just a few times.
Because i appears \lfloor frac{n}{i} \rfloor times, i <= \lfloor frac{n}{i} \rfloor if i <= sqrt{n}.
If i > sqrt{n} then several squares appear the same time. As mentioned before, there are three squares that appear once.

My program applies the basic algorithm for all numbers up to sqrt{n}, which needs sqrt{10^15} approx 3.2 * 10^7 iterations.
Then it determines which squares appear once, twice, etc. going backwards from 10^15 to sqrt{n}.
This can be quite efficient because the sum of a series of squares is (see en.wikipedia.org/wiki/Square_pyramidal_number):
P(n) = sum_{k=1...n}{k^2} = dfrac{n (n + 1) (2n + 1)}{6}

The largest square pyramidal number that appears exactly m times is \lfloor frac{n}{m} \rfloor.
To obtain the sum of all square that appear m times I calculate P(\lfloor frac{n}{m} \rfloor) - P(\lfloor frac{n}{m+1} \rfloor).

pyramidal(n) returns the n-th square pyramidal number (mod 10^9). This function is rather tricky because I hit the limits of C++'s 64 bit integers.
Luckily, G++ has routines for 128 bit integers to work around this issue - but therefore this code doesn't compile with Visual C++.

The code is scattered with mod statement. Maybe a few are redundant.

## Note

Apparently you can implement pyramidal without __int128.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo 6 | ./401

Output:

Note: the original problem's input 1000000000000000 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

# My code

… was written in C++11 and can be compiled with G++, Clang++. You can download it, too.

       #include <iostream>
#include <cmath>

// return sum of 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2 mod modulo
// note: needs GCC's 128 bit integers (not available on Visual C++)
unsigned int pyramidal(unsigned long long n, unsigned int modulo)
{
// inspired by https://stackoverflow.com/questions/39826483/square-pyramidal-number-of-int32-modulo-some-m-using-only-int64-intermediates
unsigned __int128 nn = n;
// solve n * (n + 1) * (2n + 1) / 6
// either nn or nn+1 is even => their product must be divisible by 2
auto x = nn * (nn + 1) / 2;
auto y = 2 * n + 1;
// now equation becomes x * y / 3 (mod modulo)

// is x a multiple of 3 ?
auto div3 = x / 3;
if (div3 * 3 == x)
return ((div3 % modulo) * y) % modulo;

// no, y must be a multiple of 3
return (x % modulo) * (y / 3) % modulo;
}

// count all divisors
unsigned int solve(unsigned long long n, unsigned int modulo)
{
unsigned long long sum = 0;

// for each number i there is a second divisor n/i (integer division)
// for i < sqrt(n) those second divisors are "unique"
// for i > sqrt(n) several i share the same second divisor n/i
unsigned int threshold = sqrt(n);

// look at small divisors
for (unsigned long long i = 1; i <= threshold; i++)
{
// each divisor-th number is a multiple of divisor
auto count  = (n / i) % modulo;
// square (reduce with modulo)
auto square = (i * i) % modulo;

sum += (count * square) % modulo;
}

// apply the same logic in reverse order:
// for each divisor i there is a divisor j = n / i
for (unsigned int j = 1; j <= n / (threshold + 1); j++)
{
auto from = n / j;
auto to   = n / (j + 1);

unsigned long long sumOfConsecutiveSquares = 0;
// note: from > to, therefore decrement instead of increment
/*for (unsigned long long k = from; k > to; k--)
{
auto square = (k * k) % modulo;
sumOfConsecutiveSquares += square;
sumOfConsecutiveSquares %= modulo;
}*/

// same as the for-loop above
auto squareSumFrom = pyramidal(from, modulo); // sum of squares <= from
auto squareSumTo   = pyramidal(to,   modulo); // sum of squares <= to
// add modulo to prevent negative numbers
if (squareSumFrom < squareSumTo)
squareSumFrom += modulo;
sumOfConsecutiveSquares = squareSumFrom - squareSumTo;

// those squares can be found "j" times
sum += (j * sumOfConsecutiveSquares) % modulo;
}

// and again, take care of modulo
return sum % modulo;
}

int main()
{
// 10^9
const auto Modulo = 1000000000;
// 10^15
auto n = 1000000000000000ULL;
std::cin >> n;
std::cout << solve(n, Modulo) << std::endl;
return 0;
}


This solution contains 13 empty lines, 25 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 3.8 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

# Changelog

August 22, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 25% (out of 100%).

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