# Tag Info

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.

3

Python 2.7, 341 301 253 bytes, optimized for speed lambda x,y:(x==0and g(y))or(y==0and g(x))or(x*y and p(x*x+y*y)) def p(n,r=[2]):a=lambda n:r+range(r[-1],int(n**.5)+1);r+=[i for i in a(n)if all(i%j for j in a(i))]if n>r[-1]**2else[];return all(n%i for i in r if i*i<n) g=lambda x:abs(x)%4>2and p(abs(x)) Try it online! #pRimes. need at least one ...

1

Ruby -rprime, 65 60 80 bytes Didn't notice the "can't use isPrime" rule... ->a,b{r=->n{(2...n).all?{|i|n%i>0}};c=(a+b).abs;r[a*a+b*b]||a*b==0&&r[c]&&c%4>2} Try it online!

1

Mathematica, 149 Characters If[a==0,#[[3]]&&Mod[Abs@b,4]==3,If[b==0,#[[2]]&&Mod[Abs@a,4]==3,#[[1]]]]&[(q=#;Total[Boole@IntegerQ[q/#]&/@Range@q]<3&&q!=0)&/@{a^2+b^2,Abs@a,Abs@b}] The code doesn't use any standard prime number features of mathematica, instead it counts the number of integers in the list {n/1,n/2,...,n/n}...

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