<< problem 291 - Panaitopol Primes Zeckendorf Representation - problem 297 >>

# Problem 293: Pseudo-Fortunate Numbers

An even positive integer N will be called admissible, if it is a power of 2 or its distinct prime factors are consecutive primes.
The first twelve admissible numbers are 2,4,6,8,12,16,18,24,30,32,36,48.

If N is admissible, the smallest integer M > 1 such that N+M is prime, will be called the pseudo-Fortunate number for N.

For example, N=630 is admissible since it is even and its distinct prime factors are the consecutive primes 2,3,5 and 7.
The next prime number after 631 is 641; hence, the pseudo-Fortunate number for 630 is M=11.
It can also be seen that the pseudo-Fortunate number for 16 is 3.

Find the sum of all distinct pseudo-Fortunate numbers for admissible numbers N less than 10^9.

# My Algorithm

The problem description starts with "an even positive integer [...]" and continues with "[...] its distinct prime factors are consecutive primes.".
Therefore any admissible number must be a multiple of 2 and its prime factors must start with 2, 3, 5, 7, ... without any gap.
Its general form is: 2^a * 3^b * 5^c * ... where a exponent is only zero if all following exponents are zero, too.
The highest relevant base (prime) is 23 because 2^1 * 3^1 * 5^1 * 7^1 * 9^1 * 11^1 * 13^1 * 17^1 * 19^1 * 23^1 = 223092870 < 10^9.
Extending the series with 29^1 would exceed 10^9.

In step 1 of my algorithm I create all such admissible numbers up to 10^9 in ascending order:

• put a 2 in the automatically sorted todo container, its highest prime factor is 2 (index 0 of factors)
• then repeat until todo is empty:
• pick the lowest number from todo and store it in admissible
• multiply the same number by all values of factor[0 ... index+1] (those are the prime factors) and store them in todo (if < 10^9)
In step 2 I run a simple prime sieve (trial division) to find all prime numbers up to sqrt{10^9}.
Step 3 needs these prime numbers to run a primality test on each admissible number's right neighbors.
Admissible numbers are always even, therefore only the numbers +3, +5, +7, +9, ... have to be checked.

## Note

There are a few optimization tricks:

• some admissible numbers are smaller than the prime numbers generated in step 2 → simple linear search in primes
• for this specific range, not all primes in step 2 are actually needed → it's sufficient to have all primes up to sqrt{2.8 * 10^8} → 30% faster

# 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 50 | ./293

Output:

Note: the original problem's input 1000000000 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 <map>
#include <set>

int main()
{
unsigned int limit = 1000000000;
std::cin >> limit;

// ---------- step 1: find all admissible numbers ----------

// generate all numbers 2^a * 3^b * 5^c * 7^d * 11^e ...
const unsigned int NumFactors = 9;
const unsigned int factors[NumFactors] = { 2,3,5,7,11,13,17,19,23 };
// product of all factors is just under 10^9, including the next prime (29) would be bigger than 10^9

// [number] => [index of its highest prime factor]
std::map<unsigned int, unsigned char> todo = { { 2, 0 } };
while (!todo.empty())
{
// pick smallest number and remove from to-do list
auto current = *(todo.begin());
todo.erase(todo.begin());

auto number   = current.first;
auto maxPrime = current.second;

// multiply by all already used prime numbers and the next one
for (unsigned char i = 0; i <= maxPrime + 1; i++) // can start with i = maxPrime, too
{
// only up to 9 primes are relevant
if (i == NumFactors)
break;

// put on to-do list if below 10^9
auto next = number * (unsigned long long)factors[i];
if (next < limit)
todo[next] = std::max(maxPrime, i);
}
}
// note: only 6656 admissible numbers

// ---------- step 2: generate prime numbers up to sqrt(10^9) ----------

// find at least one prime factor for each number below 10^9
// => all up to sqrt(10^9) which is about 31622
// and a few more to be on the safe side ... not needed in this case, but safety first :-)
auto primeLimit = limit + 1000;
// but I can cheat and reduce primeLimit as long as I get the correct result
// => makes the program about 30% faster
//primeLimit = 271623361;

std::vector<unsigned int> primes = { 2 };
// simply trial-division prime generator
for (unsigned int i = 3; i*i <= primeLimit; i += 2)
{
bool isPrime = true;

// test against all prime numbers we have so far (in ascending order)
for (auto x : primes)
{
// prime is too large to be a divisor
if (x*x > i)
break;

// divisible => not prime
if (i % x == 0)
{
isPrime = false;
break;
}
}

// yes, we have a prime
if (isPrime)
primes.push_back(i);
}
// only 1911 out of the 3401 primes are actually needed in step 3

// ---------- step 3: look for each admissible number's fortunate number ----------

// std::set automatically filters duplicates
std::set<unsigned int> fortunate;
size_t pos = 0;
{
// every admissible number is even and m > 1
// => therefore the first candidate is at m = 3
const unsigned int MinDistance = 3;

// fast check for small primes
if (x <= primes.back() - MinDistance)
{
while (primes[pos] < x + MinDistance)
pos++;

auto distance = primes[pos] - x;
// might be a duplicate but std::set makes sure that all values are unique
fortunate.insert(distance);
continue;
}

// every prime is odd (except for 2)
// => I look at every second number
for (unsigned int m = MinDistance; ; m += 2)
{
auto candidate = x + m;

// is it a prime ?
bool isPrime = true;

// trial division
for (auto p : primes) // actually I could skip p = 2 because all candidates are odd
if (candidate % p == 0)
{
// it's a composite number
isPrime = false;
break;
}

// no, keep going ...
if (!isPrime)
continue;

// yes, a prime
auto distance = candidate - x;
// might be a duplicate but std::set ensures that all values are unique
fortunate.insert(distance);
break;
}
}

unsigned int sum = 0;
for (auto x : fortunate)
sum += x;

// display result
std::cout << sum << std::endl;
return 0;
}


This solution contains 26 empty lines, 40 comments and 4 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 0.10 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

July 26, 2017 submitted solution

# Difficulty

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

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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
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