How to resolve the algorithm Emirp primes step by step in the F# programming language

Published on 12 May 2024 09:40 PM
#F_Sharp

How to resolve the algorithm Emirp primes step by step in the F# programming language

Table of Contents

Problem Statement

An   emirp   (prime spelled backwards)   are primes that when reversed   (in their decimal representation)   are a different prime. (This rules out palindromic primes.)

In each list, the numbers should be in order. Invoke the (same) program once per task requirement, this will show what limit is used as the upper bound for calculating surplus (regular) primes. The specific method of how to determine if a range or if specific values are to be shown will be left to the programmer.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Emirp primes step by step in the F# programming language

Source code in the fsharp programming language

// Generate emirps. Nigel Galloway: November 19th., 2017
let emirp =
  let rec fN n g = match n with |0->g |_->fN (n/10) (g*10+n%10)
  let     fG n g = n<>g && isPrime g
  primes32() |> Seq.filter (fun n -> fG n (fN n 0))


emirps |> (Seq.take 20) |> Seq.iter (printf "%d ")


emirps |> Seq.skipWhile (fun n->n<7700) |> Seq.takeWhile (fun n->n<=8000) |> Seq.iter (printf "%d ")


printfn "%d" (Seq.item 9999 emirps)


// count # of emirps with n = 2 to 7 digits. Nigel Galloway: August 8th., 2018 
let n=emirp |> Seq.takeWhile(fun n->n<10000000) |> Seq.countBy(fun n->match n with |n when n>999999->7
                                                                                   |n when n> 99999->6
                                                                                   |n when n>  9999->5
                                                                                   |n when n>   999->4
                                                                                   |n when n>    99->3
                                                                                   |_              ->2)
for n,g in n do printfn "%d -> %d" n g

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