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28

Score 100 8605 I used an algorithm that starts with one solution and repeatedly tries to split a prime \$p\$ in the solution into two other primes \$q_ 1\$ and \$q_ 2\$ that satisfy \$\frac1{p-1} = \frac1{q_1-1}+\frac1{q_2-1}\$. It is known (and can be quickly checked) that the positive integer solutions to \$\frac1n = \frac1x + \frac1y\$ are in one-to-one ...


13

Score 263 385 425 426 with only primes < 1.000.000 (was: non-competitive, now it is; score can be increased by running the program longer) I followed the same path as Wheat Wizard: iteratively search for primes in the solution that can be replaced with a longer list of primes with the same result. I wrote that Python program that does exactly this. It ...


11

Score 32 34 36 {5, 7, 11, 13, 17, 23, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 101, 113, 131, 137, 151, 211, 229, 241, 281, 313, 379, 401, 433, 457, 491, 521, 571, 601, 25117, 293362609} This is an improvement of Arnauld's answer. I just noticed that \$ \dfrac{1}{19-1}=\dfrac{1}{73-1}+\dfrac{1}{61-1}+\dfrac{1}{41-1} \$ But 41 and 61 were already used in ...


7

Score 22 Just to get the ball rolling. {5,7,11,13,17,19,23,31,37,41,43,47,53,59,61,67,71,79,137,491,25117,293362609} Compute the fraction I suspect that the sequence can be made arbitrary large, but my code is currently too messy and inefficient for anything significantly better than that.


1

This non-answer expands on the modifications to @GregMartin's answer referenced in this comment there. The allDivisors to goodFactorizations block can be sped up noticeably by not asking Divisors to factor the square of a number (and also a few other changes). The @GregMartin's original code: divMethod[p_] := Module[{}, allDivisors = Divisors[(p - 1)^2]...


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