rename statics/euler and sync b93 code with main repo

www/statics/euler/Euler_Problem-072_explanation.md

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+First we need the number of proper reduced fractions for a denominator `d`.
+This is the amount of numbers `d<n` with `gcd(d,n) == 1`.
+Coincidentally this is again Eulers Phi-Function.
+
+Now we need to find the sum over all phi values from `1` to `1 000 000`.
+
+First we get all the prime numbers from `2` to `limit/2` (with an [Sieve of Eratosthenes](https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes)).
+Then we iterate through all possible prime combinations smaller than the limit (practically this is prime factorization).
+While we do this we calculate on the side the phi value of these numbers (with the prime factorization this is trivial) and sum these values together.
+
+For all values bigger than `LIMIT/2` where we haven't got a factorization we use `phi(p) = p-1`.
+Because these *must* be prime. *(see [Eulers totient function](https://en.wikipedia.org/wiki/Euler's_totient_function))*
+
+The rest is a bit of clever optimization, so this all does not take too long.
+
+ - First we assume every number is prime and calculate the sum of the phi function of all these primes.
+   Then when we find a number (that is not a prime) we subtract the old (wrong) value from the result and add the new (correct) value.
+ - We abort our loops early when we run into a number `x > limit`. Because every other factor will only increase `x`.
+ - The value of phi is always calculated from the last phi value. So we don't need to totally recalculate it in every step.
+
+After looking a little bit around this is not ***the** fastest solution to this problem.
+But it's the one I found and I think it's reasonable fast.