Perfect totient numbers

Generate and show here, the first twenty Perfect totient numbers.

Perfect totient numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Go

<lang go>package main

import "fmt"

func gcd(n, k int) int {

   if n < k || k < 1 {
       panic("Need n >= k and k >= 1")
   }

   s := 1
   for n&1 == 0 && k&1 == 0 {
       n >>= 1
       k >>= 1
       s <<= 1
   }

   t := n
   if n&1 != 0 {
       t = -k
   }
   for t != 0 {
       for t&1 == 0 {
           t >>= 1
       }
       if t > 0 {
           n = t
       } else {
           k = -t
       }
       t = n - k
   }
   return n * s

}

func totient(n int) int {

   tot := 0
   for k := 1; k <= n; k++ {
       if gcd(n, k) == 1 {
           tot++
       }
   }
   return tot

}

func main() {

   var perfect []int
   for n := 1; len(perfect) < 20; n += 2 {
       tot := n
       sum := 0
       for tot != 1 {
           tot = totient(tot)
           sum += tot
       }
       if sum == n {
           perfect = append(perfect, n)
       }
   }
   fmt.Println("The first 20 perfect totient numbers are:")
   fmt.Println(perfect)

}</lang>

Output:

The first 20 perfect totient numbers are:
[3 9 15 27 39 81 111 183 243 255 327 363 471 729 2187 2199 3063 4359 4375 5571]

Python

<lang python>from math import gcd from functools import lru_cache from itertools import islice, count

@lru_cache(maxsize=None) def φ(n):

   return sum(1 for k in range(1, n + 1) if gcd(n, k) == 1)

def perfect_totient():

   for n0 in count(1):
       parts, n = 0, n0
       while n != 1:
           n = φ(n)
           parts += n
       if parts == n0:
           yield n0
       

if __name__ == '__main__':

   print(list(islice(perfect_totient(), 20)))</lang>

Output:

[3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, 5571]