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# Problem 581: 47-smooth triangular numbers

A number is p-smooth if it has no prime factors larger than p.
Let T be the sequence of triangular numbers, ie T(n)=n(n+1)/2.
Find the sum of all indices n such that T(n) is 47-smooth.

# My Algorithm

If a triangular number T(n) = n (n+1) / 2 is 47-smooth then n and n+1 must be 47-smooth, too.
In order to solve the problem I have to find all consecutive 47-smooth numbers n and n+1.

After having solved problem 563, I suddenly realized that a similar approach works for this problem, too:

• generate all 47-smooth numbers 2^a * 3^b * 5^c * 7^d * 11^e * 13^f * 17^g * 19^h * 23^i * 29^j * 31^k * 37^l * 41^m * 43^n * 47^o
• store them in a min-heap and always retrieve the smallest one
• if it's the successor of the value previously picked from the min-heap then another triangular number T(n-1) was found (same as T(last))
This algorithm is simple, fast, and ... doesn't know when to terminate:
I have no clue what the largest 47-smooth triangular number is.
Within less than 3 seconds my program extracts more than 1500 triangular 47-smooth numbers - and then it seems there are no more such numbers.

After a while I aborted the search and hoped for the best ... and I was right, there are no more numbers.
In the code below you find two manually added constants:
• the program "knows" that it must stop after it analyzed the 10 millionth 47-smooth number (see iterations)
• the program "knows" that there are no solutions beyond 1111111111111 (see if (todo < 1111111111111ULL) next.push(todo))
I added these constants to make sure the program finishes and shows the correct result (with a reasonable memory consumption).
There are slightly more optimal values but I decided to go for "nice looking" ones.

## Alternative Approaches

I never heard of Størmer's theorem (see en.wikipedia.org/wiki/Størmer's theorem) but it can be used to solve this problem.
Moreover, OEIS A117581 contains the limits for various primes - the 15th entry on the list is 1109496723126 (47 is the 15th prime).

# 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):
Note: Enter n to find all n-smooth triangular numbers (n <= 47)

This is equivalent to
echo 3 | ./581

Output:

Note: the original problem's input 47 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++, Visual C++. You can download it, too.

       #include <iostream>
#include <vector>
#include <deque>
#include <queue>
#include <functional>

int main()
{
unsigned int limit = 47;
std::cin >> limit;

// primes <= 47 in descending order
std::vector<unsigned int> primes = { 47,43,41,37,31,29,23,19,17,13,11,7,5,3,2 };
// remove large prime depending on user input
while (!primes.empty() && primes.front() > limit)
primes.erase(primes.begin());

// min-heap
std::priority_queue<unsigned long long, std::vector<unsigned long long>, std::greater<unsigned long long>> next;
// "seed" value
next.push(1);

// will store the final result
unsigned long long sum  = 0;

// 47-smooth number from previous iteration
unsigned long long last = 1; // must not be zero, any other value is fine

// I saw in my first runs that no solutions were found beyond the 10 millionth 47-smooth number
for (unsigned int iteration = 0; iteration < 10000000; iteration++)
{
// fetch next 47-smooth number
auto current = next.top();
next.pop();

// two consecutive 47-smooth numbers ? => T(last) is 47-smooth, too
if (last + 1 == current)
{
sum += last;
//std::cout << "T(" << last << "), sum=" << sum << " @ iteration " << iterations << " " << next.size() << std::endl;
}

// remember for next iteration
last = current;

// find further 47-smooth numbers
for (auto p : primes)
{
auto todo = current * p;

// heuristic: ignore numbers beyond the largest 47-smooth T(n)
if (todo < 1111111111111ULL)
next.push(todo);

// any prime smaller than the largest prime yields a number already found in "next"
if (current % p == 0)
break;
}
}

std::cout << sum << std::endl;
return 0;
}


This solution contains 12 empty lines, 14 comments and 5 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 2.7 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 68 MByte.

(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

October 12, 2017 submitted solution

# Difficulty

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

# Heatmap

Please click on a problem's number to open my solution to that problem:

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published the flashing problem is the one I solved most recently

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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
I scored 13,486 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

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