Diophantine reciprocals II

Problem 110

In the following equation $x$, $y$, and $n$ are positive integers.

$$\frac 1 x + \frac 1 y = \frac 1 n$$

It can be verified that when $n = 1260$ there are $113$ distinct solutions and this is the least value of $n$ for which the total number of distinct solutions exceeds one hundred.

What is the least value of $n$ for which the number of distinct solutions exceeds four million?

NOTE: This problem is a much more difficult version of problem 108 and as it is well beyond the limitations of a brute force approach it requires a clever implementation.



In [7]:

    
open Euler

let p110() =
    let n = 1260
    let k = 12 // started at 15 and went down until answer repeated
    let step = Math.Primes |> Seq.take k |> Seq.fold ( * ) 1L
    let solutions (n:int64) = (Math.DivisorSigma(2, n) + 1) / 2
    let rec answer guess =
        if solutions guess < 4000000 then 
            answer (guess + step)
        else guess
    answer step

Euler.Timer(p110)









    Out[7]:





val it : int64 * float = (9350130049860600L, 0.11362)



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