Perfect totient numbers: Difference between revisions

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

Related task

Totient function

Also see

the OEIS entry for perfect totient numbers.
mrob list of the first 54

11l

Translation of: Python

<lang 11l>F φ(n)

  R sum((1..n).filter(k -> gcd(@n, k) == 1).map(k -> 1))

F perfect_totient(cnt)

  [Int] r

  L(n0) 1..
     V parts = 0
     V n = n0
     L n != 1
        n = φ(n)
        parts += n
     I parts == n0
        r [+]= n0
        I r.len == cnt
           R r

print(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]

APL

Output:

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

AWK

syntax: GAWK -f PERFECT_TOTIENT_NUMBERS.AWK

BEGIN {

   i = 20
   printf("The first %d perfect totient numbers:\n%s\n",i,perfect_totient(i))
   exit(0)

} function perfect_totient(n, count,m,str,sum,tot) {

   for (m=1; count<n; m++) {
     tot = m
     sum = 0
     while (tot != 1) {
       tot = totient(tot)
       sum += tot
     }
     if (sum == m) {
       str = str m " "
       count++
     }
   }
   return(str)

} function totient(n, i,tot) {

   tot = n
   for (i=2; i*i<=n; i+=2) {
     if (n % i == 0) {
       while (n % i == 0) {
         n /= i
       }
       tot -= tot / i
     }
     if (i == 2) {
       i = 1
     }
   }
   if (n > 1) {
     tot -= tot / n
   }
   return(tot)

} </lang>

Output:

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

<lang BASIC>10 DEFINT A-Z 20 N=3 30 S=N: T=0 40 X=S: S=0 50 FOR I=1 TO X-1 60 A=X: B=I 70 IF B>0 THEN C=A: A=B: B=C MOD B: GOTO 70 80 IF A=1 THEN S=S+1 90 NEXT 100 T=T+S 110 IF S>1 GOTO 40 120 IF T=N THEN PRINT N,: Z=Z+1 130 N=N+2 140 IF Z<20 GOTO 30</lang>

Output:

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

BCPL

<lang bcpl>get "libhdr"

let gcd(a,b) = b=0 -> a, gcd(b, a rem b)

let totient(n) = valof $( let r = 0

   for i=1 to n-1
       if gcd(n,i) = 1 then r := r + 1
   resultis r

$)

let perfect(n) = valof $( let sum = 0 and x = n

   $(  x := totient(x)
       sum := sum + x
   $) repeatuntil x = 1
   resultis sum = n

$)

let start() be $( let seen = 0 and n = 3

   while seen < 20
   $(  if perfect(n)
       $(  writef("%N ", n)
           seen := seen + 1
       $)
       n := n + 2
   $)
   wrch('*N')

$)</lang>

Output:

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

C

Calculates as many perfect Totient numbers as entered on the command line. <lang C>#include<stdlib.h>

include<stdio.h>

long totient(long n){ long tot = n,i;

for(i=2;i*i<=n;i+=2){ if(n%i==0){ while(n%i==0) n/=i; tot-=tot/i; }

if(i==2) i=1; }

if(n>1) tot-=tot/n;

return tot; }

long* perfectTotients(long n){ long *ptList = (long*)malloc(n*sizeof(long)), m,count=0,sum,tot;

for(m=1;count<n;m++){ tot = m; sum = 0;

       while(tot != 1){
           tot = totient(tot);
           sum += tot;
       }
       if(sum == m)

ptList[count++] = m;

return ptList; }

long main(long argC, char* argV[]) { long *ptList,i,n;

if(argC!=2) printf("Usage : %s <number of perfect Totient numbers required>",argV[0]); else{ n = atoi(argV[1]);

ptList = perfectTotients(n);

printf("The first %d perfect Totient numbers are : \n[",n);

for(i=0;i<n;i++) printf(" %d,",ptList[i]); printf("\b]"); }

return 0; } </lang> Output for multiple runs, a is the default executable file name produced by GCC

C:\rossetaCode>a 10
The first 10 perfect Totient numbers are :
[ 3, 9, 15, 27, 39, 81, 111, 183, 243, 255]
C:\rossetaCode>a 20
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]
C:\rossetaCode>a 30
The first 30 perfect Totient numbers are :
[ 3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, 5571, 6561, 8751, 15723, 19683, 36759, 46791, 59049, 65535, 140103, 177147]
C:\rossetaCode>a 40
The first 40 perfect Totient numbers are :
[ 3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, 5571, 6561, 8751, 15723, 19683, 36759, 46791, 59049, 65535, 140103, 177147, 208191, 441027, 531441, 1594323, 4190263, 4782969, 9056583, 14348907, 43046721, 57395631]

C++

<lang cpp>#include <cassert>

include <iostream>
include <vector>

class totient_calculator { public:

   explicit totient_calculator(int max) : totient_(max + 1) {
       for (int i = 1; i <= max; ++i)
           totient_[i] = i;
       for (int i = 2; i <= max; ++i) {
           if (totient_[i] < i)
               continue;
           for (int j = i; j <= max; j += i)
               totient_[j] -= totient_[j] / i;
       }
   }
   int totient(int n) const {
       assert (n >= 1 && n < totient_.size());
       return totient_[n];
   }
   bool is_prime(int n) const {
       return totient(n) == n - 1;
   }

private:

   std::vector<int> totient_;

};

bool perfect_totient_number(const totient_calculator& tc, int n) {

   int sum = 0;
   for (int m = n; m > 1; ) {
       int t = tc.totient(m);
       sum += t;
       m = t;
   }
   return sum == n;

}

int main() {

   totient_calculator tc(10000);
   int count = 0, n = 1;
   std::cout << "First 20 perfect totient numbers:\n";
   for (; count < 20; ++n) {
       if (perfect_totient_number(tc, n))  {
           if (count > 0)
               std::cout << ' ';
           ++count;
           std::cout << n;
       }
   }
   std::cout << '\n';
   return 0;

}</lang>

Output:

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

Cowgol

<lang cowgol>include "cowgol.coh";

sub gcd(a: uint16, b: uint16): (r: uint16) is

   while b != 0 loop
       r := a;
       a := b;
       b := r % b;
   end loop;
   r := a;

end sub;

sub totient(n: uint16): (tot: uint16) is

   var i: uint16 := 1;
   tot := 0;
   while i < n loop
       if gcd(n,i) == 1 then
           tot := tot + 1;
       end if;
       i := i + 1;
   end loop;

end sub;

sub totientSum(n: uint16): (sum: uint16) is

   var x := n;
   sum := 0;
   while x > 1 loop
       x := totient(x);
       sum := sum + x;
   end loop;

end sub;

var seen: uint8 := 0; var n: uint16 := 3; while seen < 20 loop

   if totientSum(n) == n then
       print_i16(n);
       print_char(' ');
       seen := seen + 1;
   end if;
   n := n + 2;

end loop; print_nl();</lang>

Output:

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

Delphi

Library: System.SysUtils

Translation of: Go

<lang Delphi> program Perfect_totient_numbers;

{$APPTYPE CONSOLE}

uses

 System.SysUtils;

function totient(n: Integer): Integer; begin

 var tot := n;
 var i := 2;
 while i * i <= n do
 begin
   if (n mod i) = 0 then
   begin
     while (n mod i) = 0 do
       n := n div i;
     dec(tot, tot div i);
   end;
   if i = 2 then
     i := 1;
   inc(i, 2);
 end;
 if n > 1 then
   dec(tot, tot div n);
 Result := tot;

end;

begin

 var perfect: TArray<Integer>;
 var n := 1;
 while length(perfect) < 20 do
 begin
   var tot := n;
   var sum := 0;
   while tot <> 1 do
   begin
     tot := totient(tot);
     inc(sum, tot);
   end;
   if sum = n then
   begin
     SetLength(perfect, Length(perfect) + 1);
     perfect[High(perfect)] := n;
   end;
   inc(n, 2);
 end;
 writeln('The first 20 perfect totient numbers are:');
 write('[');
 for var e in perfect do
   write(e, ' ');
 writeln(']');
 readln;

end.</lang>

Factor

<lang factor>USING: formatting kernel lists lists.lazy math math.primes.factors ;

perfect? ( n -- ? )

   [ 0 ] dip dup [ dup 2 < ] [ totient tuck [ + ] 2dip ] until
   drop = ;

20 1 lfrom [ perfect? ] lfilter ltake list>array "%[%d, %]\n" printf</lang>

Output:

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

FreeBASIC

Uses the code from the Totient Function example as an include.

<lang freebasic>#include"totient.bas"

dim as uinteger found = 0, curr = 3, sum, toti

while found < 20

   sum = totient(curr)
   toti = sum
   do
       toti = totient(toti)
       sum += toti
   loop while toti <> 1  
   if sum = curr then
       print sum
       found += 1
   end if
   curr += 1

wend</lang>

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]

The following much quicker version uses Euler's product formula rather than repeated invocation of the gcd function to calculate the totient: <lang go>package main

import "fmt"

func totient(n int) int {

   tot := n
   for i := 2; i*i <= n; i += 2 {
       if n%i == 0 {
           for n%i == 0 {
               n /= i
           }
           tot -= tot / i
       }
       if i == 2 {
           i = 1
       }
   }
   if n > 1 {
       tot -= tot / n
   }
   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>

The output is the same as before.

Haskell

<lang haskell>perfectTotients :: [Int] perfectTotients =

 filter ((==) <*> (succ . sum . tail . takeWhile (1 /=) . iterate φ)) [2 ..]

φ :: Int -> Int φ = memoize (\n -> length (filter ((1 ==) . gcd n) [1 .. n]))

memoize :: (Int -> a) -> (Int -> a) memoize f = (!!) (f <$> [0 ..])

main :: IO () main = print $ take 20 perfectTotients</lang>

Output:

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

J

<lang J> Until =: conjunction def 'u^:(0 -: v)^:_' Filter =: (#~`)(`:6) totient =: 5&p: totient_chain =: [: }. (, totient@{:)Until(1={:) ptnQ =: (= ([: +/ totient_chain))&> </lang> With these definitions I've found the first 28 perfect totient numbers

   PTN =: ptnQ Filter >: i.99999
   #PTN
28
   PTN
3 9 15 27 39 81 111 183 243 255 327 363 471 729 2187 2199 3063 4359 4375 5571 6561 8751 15723 19683 36759 46791 59049 65535

Java

<lang Java> import java.util.ArrayList; import java.util.List;

public class PerfectTotientNumbers {

   public static void main(String[] args) {
       computePhi();
       int n = 20;
       System.out.printf("The first %d perfect totient numbers:%n%s%n", n, perfectTotient(n));
   }
   
   private static final List<Integer> perfectTotient(int n) {
       int test = 2;
       List<Integer> results = new ArrayList<Integer>();
       for ( int i = 0 ; i < n ; test++ ) {
           int phiLoop = test;
           int sum = 0;
           do {
               phiLoop = phi[phiLoop];
               sum += phiLoop;
           } while ( phiLoop > 1);
           if ( sum == test ) {
               i++;
               results.add(test);
           }
       }
       return results;
   }

   private static final int max = 100000;
   private static final int[] phi = new int[max+1];

   private static final void computePhi() {
       for ( int i = 1 ; i <= max ; i++ ) {
           phi[i] = i;
       }
       for ( int i = 2 ; i <= max ; i++ ) {
           if (phi[i] < i) continue;
           for ( int j = i ; j <= max ; j += i ) {
               phi[j] -= phi[j] / i;
           }
       }
   }

} </lang>

Output:

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

JavaScript

<lang javascript>(() => {

   'use strict';

   // main :: IO ()
   const main = () =>
       showLog(
           take(20, perfectTotients())
       );

   // perfectTotients :: Generator [Int]
   function* perfectTotients() {
       const
           phi = memoized(
               n => length(
                   filter(
                       k => 1 === gcd(n, k),
                       enumFromTo(1, n)
                   )
               )
           ),
           imperfect = n => n !== sum(
               tail(iterateUntil(
                   x => 1 === x,
                   phi,
                   n
               ))
           );
       let ys = dropWhileGen(imperfect, enumFrom(1))
       while (true) {
           yield ys.next().value - 1;
           ys = dropWhileGen(imperfect, ys)
       }
   }

   // GENERIC FUNCTIONS ----------------------------

   // abs :: Num -> Num
   const abs = Math.abs;

   // dropWhileGen :: (a -> Bool) -> Gen [a] -> [a]
   const dropWhileGen = (p, xs) => {
       let
           nxt = xs.next(),
           v = nxt.value;
       while (!nxt.done && p(v)) {
           nxt = xs.next();
           v = nxt.value;
       }
       return xs;
   };

   // enumFrom :: Int -> [Int]
   function* enumFrom(x) {
       let v = x;
       while (true) {
           yield v;
           v = 1 + v;
       }
   }

   // enumFromTo :: Int -> Int -> [Int]
   const enumFromTo = (m, n) =>
       m <= n ? iterateUntil(
           x => n <= x,
           x => 1 + x,
           m
       ) : [];

   // filter :: (a -> Bool) -> [a] -> [a]
   const filter = (f, xs) => xs.filter(f);

   // gcd :: Int -> Int -> Int
   const gcd = (x, y) => {
       const
           _gcd = (a, b) => (0 === b ? a : _gcd(b, a % b)),
           abs = Math.abs;
       return _gcd(abs(x), abs(y));
   };

   // iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]
   const iterateUntil = (p, f, x) => {
       const vs = [x];
       let h = x;
       while (!p(h))(h = f(h), vs.push(h));
       return vs;
   };

   // Returns Infinity over objects without finite length.
   // This enables zip and zipWith to choose the shorter
   // argument when one is non-finite, like cycle, repeat etc

   // length :: [a] -> Int
   const length = xs =>
       (Array.isArray(xs) || 'string' === typeof xs) ? (
           xs.length
       ) : Infinity;

   // memoized :: (a -> b) -> (a -> b)
   const memoized = f => {
       const dctMemo = {};
       return x => {
           const v = dctMemo[x];
           return undefined !== v ? v : (dctMemo[x] = f(x));
       };
   };

   // showLog :: a -> IO ()
   const showLog = (...args) =>
       console.log(
           args
           .map(JSON.stringify)
           .join(' -> ')
       );

   // sum :: [Num] -> Num
   const sum = xs => xs.reduce((a, x) => a + x, 0);

   // tail :: [a] -> [a]
   const tail = xs => 0 < xs.length ? xs.slice(1) : [];

   // take :: Int -> [a] -> [a]
   // take :: Int -> String -> String
   const take = (n, xs) =>
       'GeneratorFunction' !== xs.constructor.constructor.name ? (
           xs.slice(0, n)
       ) : [].concat.apply([], Array.from({
           length: n
       }, () => {
           const x = xs.next();
           return x.done ? [] : [x.value];
       }));

   // MAIN ---
   main();

})();</lang>

Output:

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

jq

Adapted from Julia

Works with: jq

Works with gojq, the Go implementation of jq

One small point of interest in the following implementation is the way the cacheing of totient values is accomplished using a helper function (`cachephi`). <lang jq>

jq optimizes the recursive call of _gcd in the following:

def gcd(a;b):

 def _gcd:
   if .[1] != 0 then [.[1], .[0] % .[1]] | _gcd else .[0] end;
 [a,b] | _gcd ;

def count(s): reduce s as $x (0; .+1);

A perfect totient number is an integer that is equal to the sum of its iterated totients.
aka Euler's phi function

def totient:

 . as $n
 | count( range(0; .) | select( gcd($n; .) == 1) );

input: the cache
output: the updated cache

def cachephi($n):

 ($n|tostring) as $s
 | if (has($s)|not) then .[$s] = ($n|totient) else . end ;

Emit the stream of perfect totients

def perfect_totients:

 . as $n
 | foreach range(1; infinite) as $i ({cache: {}};
       .tot = $i
       | .tsum = 0
       | until( .tot == 1;

.tot as $tot

            | .cache |= cachephi($tot)
            | .tot = .cache[$tot|tostring]
            | .tsum += .tot);
       if .tsum == $i then $i else empty end );

"The first 20 perfect totient numbers:", limit(20; perfect_totients)</lang>

Output:

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

Julia

<lang julia>using Primes

eulerphi(n) = (r = one(n); for (p,k) in factor(abs(n)) r *= p^(k-1)*(p-1) end; r)

const phicache = Dict{Int, Int}()

cachedphi(n) = (if !haskey(phicache, n) phicache[n] = eulerphi(n) end; phicache[n])

function perfecttotientseries(n)

   perfect = Vector{Int}()
   i = 1
   while length(perfect) < n
       tot = i
       tsum = 0
       while tot != 1
           tot = cachedphi(tot)
           tsum += tot
       end
       if tsum == i
           push!(perfect, i)
       end
       i += 1
   end
   perfect

end

println("The first 20 perfect totient numbers are: $(perfecttotientseries(20))") println("The first 40 perfect totient numbers are: $(perfecttotientseries(40))")

</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]
The first 40 perfect totient numbers are: [3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, 5571, 6561, 8751, 15723, 19683, 36759, 46791, 59049, 65535, 140103, 177147, 208191, 441027, 531441, 1594323, 4190263, 4782969, 9056583, 14348907, 43046721, 57395631]

Kotlin

Translation of: Go

<lang scala>// Version 1.3.21

fun totient(n: Int): Int {

   var tot = n
   var nn = n
   var i = 2
   while (i * i <= nn) {
       if (nn % i == 0) {
           while (nn % i == 0) nn /= i
           tot -= tot / i
       }
       if (i == 2) i = 1
       i += 2
   }
   if (nn > 1) tot -= tot / nn
   return tot

}

fun main() {

   val perfect = mutableListOf<Int>()
   var n = 1
   while (perfect.size < 20) {
       var tot = n
       var sum = 0
       while (tot != 1) {
           tot = totient(tot)
           sum += tot
       }
       if (sum == n) perfect.add(n)
       n += 2
   }
   println("The first 20 perfect totient numbers are:")
   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]

MAD

<lang MAD> NORMAL MODE IS INTEGER

           INTERNAL FUNCTION(Y,Z)
           ENTRY TO GCD.
           A = Y
           B = Z

LOOP WHENEVER A.E.B, FUNCTION RETURN A

           WHENEVER A.G.B, A = A-B 
           WHENEVER A.L.B, B = B-A
           TRANSFER TO LOOP
           END OF FUNCTION
           
           INTERNAL FUNCTION(C)
           ENTRY TO TOTENT.
           E = 0
           THROUGH LOOP, FOR D=1, 1, D.GE.C

LOOP WHENEVER GCD.(C,D).E.1, E = E+1

           FUNCTION RETURN E
           END OF FUNCTION
           
           INTERNAL FUNCTION(G)
           ENTRY TO PERFCT.
           H = G
           I = 0

LOOP H = TOTENT.(H)

           I = I+H
           WHENEVER H.G.1, TRANSFER TO LOOP
           FUNCTION RETURN I.E.G
           END OF FUNCTION
           
           SEEN = 0
           THROUGH LOOP, FOR N=3, 2, SEEN.GE.20
           WHENEVER PERFCT.(N)
               SEEN = SEEN+1
               PRINT FORMAT FMT,N
           END OF CONDITIONAL

LOOP CONTINUE

           VECTOR VALUES FMT = $I9*$
           END OF PROGRAM </lang>

Output:

Maple

<lang Maple>iterated_totient := proc(n::posint, total)

if NumberTheory:-Totient(n) = 1 then
  return total + 1;
else
  return iterated_totient(NumberTheory:-Totient(n), total + NumberTheory:-Totient(n));
end if;

end proc:

isPerfect := n -> evalb(iterated_totient(n, 0) = n):

count := 0: num_list := []: for i while count < 20 do

if isPerfectTotient(i) then
 num_list := [op(num_list), i];
 count := count + 1;
end if;

end do; num_list;</lang>

Output:

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

Mathematica / Wolfram Language

<lang Mathematica>ClearAll[PerfectTotientNumberQ] PerfectTotientNumberQ[n_Integer] := Total[Rest[Most[FixedPointList[EulerPhi, n]]]] == n res = {}; i = 0; While[Length[res] < 20,

i++;
If[PerfectTotientNumberQ[i], AppendTo[res, i]]
]

res</lang>

Output:

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

Nim

<lang nim>import strformat

func totient(n: int): int =

 var tot = n
 var nn = n
 var i = 2
 while i * i <= nn:
   if nn mod i == 0:
     while nn mod i == 0:
       nn = nn div i
     dec tot, tot div i
   if i == 2:
     i = 1
   inc i, 2
 if nn > 1:
   dec tot, tot div nn
 tot

var n = 1 var num = 0 echo "The first 20 perfect totient numbers are:" while num < 20:

 var tot = n
 var sum = 0
 while tot != 1:
   tot = totient(tot)
   inc sum, tot
 if sum == n:
   write(stdout, fmt"{n} ")
   inc num
 inc n, 2

write(stdout, "\n")</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

Pascal

I am using a really big array to calculate the Totient of every number up to 1.162.261.467, the 46.te perfect totient number. ( I can only test up to 1.5e9 before I get - out of memory ( 6.5 GB ) ). I'm doing this, by using only prime numbers to calculate the Totientnumbers. After that I sum up the totient numbers Tot[i] := Tot[i]+Tot[Tot[i]]; Tot[Tot[i]] is always < Tot[i], so it is already calculated. So I needn't calculations going trough so whole array ending up in Tot[2].
With limit 57395631 it takes "real 0m2,025s "
The c-program takes "real 3m12,481s"
A test with using floating point/SSE is by 2 seconds faster for 46.th perfect totient number, with the coming new Version of Freepascal 3.2.0 <lang pascal>program Perftotient; {$IFdef FPC}

 {$MODE DELPHI} {$CodeAlign proc=32,loop=1}

{$IFEND} uses

 sysutils;

const

 cLimit = 57395631;//177147;//4190263;//57395631;//1162261467;//

//global var

 TotientList : array of LongWord;
 Sieve : Array of byte;
 SolList : array of LongWord;
 T1,T0 : INt64;

procedure SieveInit(svLimit:NativeUint); var

 pSieve:pByte;
 i,j,pr :NativeUint;

Begin

 svlimit := (svLimit+1) DIV 2;
 setlength(sieve,svlimit+1);
 pSieve := @Sieve[0];
 For i := 1 to svlimit do
 Begin
   IF pSieve[i]= 0 then
   Begin
     pr := 2*i+1;
     j := (sqr(pr)-1) DIV 2;
     IF  j> svlimit then
       BREAK;
     repeat
       pSieve[j]:= 1;
       inc(j,pr);
     until j> svlimit;
   end;
 end;
 pr := 0;
 j := 0;
 For i := 1 to svlimit do
 Begin
   IF pSieve[i]= 0 then
   Begin
     pSieve[j] := i-pr;
     inc(j);
     pr := i;
   end;
 end;
 setlength(sieve,j);

end;

procedure TotientInit(len: NativeUint); var

 pTotLst : pLongWord;
 pSieve  : pByte;
 test : double;
 i: NativeInt;
 p,j,k,svLimit : NativeUint;

Begin

 SieveInit(len);
 T0:= GetTickCount64;
 setlength(TotientList,len+12);
 pTotLst := @TotientList[0];

//Fill totient with simple start values for odd and even numbers //and multiples of 3

 j := 1;
 k := 1;// k == j DIV 2
 p := 1;// p == j div 3;
 repeat
   pTotLst[j] := j;//1
   pTotLst[j+1] := k;//2 j DIV 2; //2
   inc(k);
   inc(j,2);
   pTotLst[j] := j-p;//3
   inc(p);
   pTotLst[j+1] := k;//4  j div 2
   inc(k);
   inc(j,2);
   pTotLst[j] := j;//5
   pTotLst[j+1] := p;//6   j DIV 3 <=  (div 2) * 2 DIV/3
   inc(j,2);
   inc(p);
   inc(k);
 until j>len+6;

//correct values of totient by prime factors

 svLimit := High(sieve);
 p := 3;// starting after 3
 pSieve := @Sieve[svLimit+1];
 i := -svlimit;
 repeat
   p := p+2*pSieve[i];
   j := p;

// Test := (1-1/p);

   while j <= cLimit do
   Begin

// pTotLst[j] := trunc(pTotLst[j]*Test);

     k:= pTotLst[j];
     pTotLst[j]:= k-(k DIV p);
     inc(j,p);
   end;
   inc(i);
 until i=0;

 T1:= GetTickCount64;
 writeln('totient calculated in ',T1-T0,' ms');
 setlength(sieve,0);

end;

function GetPerfectTotient(len: NativeUint):NativeUint; var

 pTotLst : pLongWord;
 i,sum: NativeUint;

Begin

 T0:= GetTickCount64;
 pTotLst := @TotientList[0];
 setlength(SolList,100);
 result := 0;
 For i := 3 to Len do
 Begin
   sum := pTotLst[i];
   pTotLst[i] := sum+pTotLst[sum];
 end;
 //Check for solution ( IF ) in seperate loop ,reduces time consuption ~ 12% for this function
 For i := 3 to Len do
   IF pTotLst[i] =i then
   Begin
     SolList[result] := i;
     inc(result);
   end;

 T1:= GetTickCount64;
 setlength(SolList,result);
 writeln('calculated totientsum in ',T1-T0,' ms');
 writeln('found ',result,' perfect totient numbers');

end;

var

 j,k : NativeUint;

Begin

 TotientInit(climit);
 GetPerfectTotient(climit);
 k := 0;
 For j := 0 to High(Sollist) do
 Begin
   inc(k);
   if k > 4 then
   Begin
     writeln(Sollist[j]);
     k := 0;
   end
   else
     write(Sollist[j],',');
 end;

end.</lang>

OutPut

compiled with fpc 3.0.4 -O3 "Perftotient.pas"
totient calculated in 32484 ms
calculated totientsum in 8244 ms
found 46 perfect totient numbers
3,9,15,27,39
81,111,183,243,255
327,363,471,729,2187
2199,3063,4359,4375,5571
6561,8751,15723,19683,36759
46791,59049,65535,140103,177147
208191,441027,531441,1594323,4190263
4782969,9056583,14348907,43046721,57395631
129140163,172186887,236923383,387420489,918330183
1162261467,
real  0m47,690s
*
found 40 perfect totient numbers
...
real  0m2,025s

Perl

Library: ntheory

<lang perl>use ntheory qw(euler_phi);

sub phi_iter {

   my($p) = @_;
   euler_phi($p) + ($p == 2 ? 0 : phi_iter(euler_phi($p)));

}

my @perfect; for (my $p = 2; @perfect < 20 ; ++$p) {

   push @perfect, $p if $p == phi_iter($p);

}

printf "The first twenty perfect totient numbers:\n%s\n", join ' ', @perfect;</lang>

Output:

The first twenty Perfect totient numbers:
3 9 15 27 39 81 111 183 243 255 327 363 471 729 2187 2199 3063 4359 4375 5571

Phix

Translation of: Go

<lang Phix>function totient(integer n)

   integer tot = n, i = 2
   while i*i<=n do
       if mod(n,i)=0 then
           while true do
               n /= i
               if mod(n,i)!=0 then exit end if
           end while
           tot -= tot/i
       end if
       i += iff(i=2?1:2)
   end while
   if n>1 then
       tot -= tot/n
   end if
   return tot

end function

sequence perfect = {} integer n = 1 while length(perfect)<20 do

   integer tot = n,
           tsum = 0
   while tot!=1 do
       tot = totient(tot)
       tsum += tot
   end while
   if tsum=n then
       perfect &= n
   end if
   n += 2

end while printf(1,"The first 20 perfect totient numbers are:\n") ?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}

PicoLisp

<lang PicoLisp>(gc 16) (de gcd (A B)

  (until (=0 B)
     (let M (% A B)
        (setq A B B M) ) )
  (abs A) )

(de totient (N)

  (let C 0
     (for I N
        (and (=1 (gcd N I)) (inc 'C)) )
     C ) )

(de totients (NIL)

  (let (C 0  N 1)
     (while (> 20 C)
        (let (Cur N  S 0)
           (while (> Cur 1)
              (inc 'S (setq Cur (totient Cur))) )
           (when (= S N)
              (inc 'C)
              (prin N " ")
              (flush) )
           (inc 'N 2) ) )
     (prinl) ) )

(totients)</lang>

Output:

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

PL/I

<lang pli>perfectTotient: procedure options(main);

   gcd: procedure(aa, bb) returns(fixed);
       declare (aa, bb, a, b, c) fixed;
       a = aa; 
       b = bb;
       do while(b ^= 0);
           c = a;
           a = b;
           b = mod(c, b);
       end;
       return(a);
   end gcd;
   
   totient: procedure(n) returns(fixed);
       declare (i, n, s) fixed;
       s = 0;
       do i=1 to n-1;
           if gcd(n,i) = 1 then s = s+1;
       end;
       return(s);
   end totient;
   
   perfect: procedure(n) returns(bit);
       declare (n, x, sum) fixed;
       sum = 0;
       x = n;
       do while(x > 1);
           x = totient(x);
           sum = sum + x;
       end;
       return(sum = n);
   end perfect;
   
   declare (n, seen) fixed;
   seen = 0;
   do n=3 repeat(n+2) while(seen<20);
       if perfect(n) then do;
           put edit(n) (F(5));
           seen = seen+1;
           if mod(seen,10) = 0 then put skip;
       end;
   end;

end perfectTotient;</lang>

Output:

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

PL/M

<lang plm>100H: BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS; EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT; PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;

PRINT$NUMBER: PROCEDURE (N);

   DECLARE S (7) BYTE INITIAL ('..... $');
   DECLARE (N, P) ADDRESS, C BASED P BYTE;
   P = .S(5);

DIGIT:

   P = P - 1;
   C = N MOD 10 + '0';
   N = N / 10;
   IF N > 0 THEN GO TO DIGIT;
   CALL PRINT(P);

END PRINT$NUMBER;

GCD: PROCEDURE (A, B) ADDRESS;

   DECLARE (A, B, C) ADDRESS;
   DO WHILE B <> 0;
       C = A;
       A = B;
       B = C MOD B;
   END;
   RETURN A;

END GCD;

TOTIENT: PROCEDURE (N) ADDRESS;

   DECLARE (I, N, S) ADDRESS;
   S = 0;
   DO I=1 TO N-1;
       IF GCD(N,I) = 1 THEN S = S+1;
   END;
   RETURN S;

END TOTIENT;

PERFECT: PROCEDURE (N) BYTE;

   DECLARE (N, X, SUM) ADDRESS;
   X = N;
   SUM = 0;
   DO WHILE X > 1;
       X = TOTIENT(X);
       SUM = SUM + X;
   END;
   RETURN SUM = N;

END PERFECT;

DECLARE N ADDRESS, SEEN BYTE; SEEN = 0; N = 3; DO WHILE SEEN < 20;

   IF PERFECT(N) THEN DO;
       CALL PRINT$NUMBER(N);
       SEEN = SEEN+1;
   END;
   N = N+2;

END; CALL EXIT; EOF</lang>

Output:

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]

Alternatively, by composition of generic functions:

<lang python>Perfect totient numbers

from functools import lru_cache from itertools import count, islice from math import gcd import operator

perfectTotients :: () -> [Int]

def perfectTotients():

   An unbounded sequence of perfect totients.
      OEIS A082897
   
   def p(x):
       return x == 1 + sum(
           iterateUntil(eq(1))(
               phi
           )(x)[1:]
       )
   return filter(p, count(2))

@lru_cache(maxsize=None) def phi(n):

   Euler's totient function.
      The count of integers up to n which
      are relatively prime to n.
   
   return len([
       x for x in enumFromTo(1)(n)
       if 1 == gcd(n, x)
   ])

TEST ----------------------------------------------------
main :: IO ()

def main():

   First twenty perfect totient numbers
   print(
       take(20)(
           perfectTotients()
       )
   )

GENERIC -------------------------------------------------

curry :: ((a, b) -> c) -> a -> b -> c

def curry(f):

   A curried function derived
      from an uncurried function.
   
   return lambda x: lambda y: f(x, y)

enumFromTo :: Int -> Int -> [Int]

def enumFromTo(m):

   Enumeration of integer values [m..n]
   return lambda n: range(m, 1 + n)

eq (==) :: Eq a => a -> a -> Bool

eq = curry(operator.eq) True if a and b are comparable and a equals b.

iterateUntil :: (a -> Bool) -> (a -> a) -> a -> [a]

def iterateUntil(p):

   A list of the results of repeated
      applications of f, until p matches.
   
   def go(f, x):
       vs = []
       v = x
       while True:
           if p(v):
               break
           vs.append(v)
           v = f(v)
       return vs

   return lambda f: lambda x: go(f, x)

take :: Int -> [a] -> [a]
take :: Int -> String -> String

def take(n):

   The prefix of xs of length n,
      or xs itself if n > length xs.
   
   return lambda xs: (
       xs[0:n]
       if isinstance(xs, (list, tuple))
       else list(islice(xs, n))
   )

MAIN ---

if __name__ == '__main__':

   main()</lang>

Output:

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

Quackery

totient is defined at Totient function#Quackery.

<lang Quackery> [ 0 over

   [ dup 1 != while 
     totient
     dup dip +
     again ]  
   drop = ]                is perfecttotient ( n --> b )    

 [ [] 1
   [ dup perfecttotient if
       [ dup dip join ]
     2 +
   over size 20 = 
   until ] drop ]          is task            (   -->  )

</lang>

Output:

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

Racket

lang racket

(require math/number-theory)

(define (tot n)

 (match n
   [1 0]
   [n (define t (totient n))
      (+ t (tot t))]))

(define (perfect? n)

 (= n (tot n)))

(define-values (ns i)

 (for/fold ([ns '()] [i 0])
           ([n (in-naturals 1)]
            #:break (= i 20)
            #:when (perfect? n))
   (values (cons n ns) (+ i 1))))

(reverse ns) </lang>

Raku

(formerly Perl 6)

Works with: Rakudo version 2018.11

<lang perl6>use Prime::Factor;

my \𝜑 = lazy 0, |(1..*).hyper.map: -> \t { t * [*] t.&prime-factors.squish.map: 1 - 1/* } my \𝜑𝜑 = Nil, |(3, *+2 … *).grep: -> \p { p == sum 𝜑[p], { 𝜑[$_] } … 1 };

put "The first twenty Perfect totient numbers:\n", 𝜑𝜑[1..20];</lang>

Output:

The first twenty Perfect totient numbers:
3 9 15 27 39 81 111 183 243 255 327 363 471 729 2187 2199 3063 4359 4375 5571

REXX

unoptimized

<lang rexx>/*REXX program calculates and displays the first N perfect totient numbers. */ parse arg N . /*obtain optional argument from the CL.*/ if N== | N=="," then N= 20 /*Not specified? Then use the default.*/ @.= . /*memoization array of totient numbers.*/ p= 0 /*the count of perfect " " */ $= /*list of the " " " */

   do j=3  by 2  until p==N;   s= phi(j)        /*obtain totient number for a number.  */
   a= s                                         /* [↓]  search for a perfect totient #.*/
                               do until a==1;           a= phi(a);            s= s + a
                               end   /*until*/
   if s\==j  then iterate                       /*Is  J  not a perfect totient number? */
   p= p + 1                                     /*bump count of perfect totient numbers*/
   $= $ j                                       /*add to perfect totient numbers list. */
   end   /*j*/

say 'The first ' N " perfect totient numbers:" /*display the header to the terminal. */ say strip($) /* " " list. " " " */ exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ gcd: parse arg x,y; do until y==0; parse value x//y y with y x; end; return x /*──────────────────────────────────────────────────────────────────────────────────────*/ phi: procedure expose @.; parse arg z; if @.z\==. then return @.z /*was found before?*/

    #= z==1;         do m=1  for z-1;   if gcd(m, z)==1  then #= # + 1;    end  /*m*/
    @.z= #;   return #                                              /*use memoization. */</lang>

output when using the default input of : 20

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

optimized

This REXX version is over twice as fast as the unoptimized version.

It takes advantage of the fact that all known perfect totient numbers less than 3²² have at least one of these factors: 3, 5, or 7

(3²² = 31,381,059,609). <lang rexx>/*REXX program calculates and displays the first N perfect totient numbers. */ parse arg N . /*obtain optional argument from the CL.*/ if N== | N=="," then N= 20 /*Not specified? Then use the default.*/ @.= . /*memoization array of totient numbers.*/ p= 0 /*the count of perfect " " */ $= /*list of the " " " */

    do j=3  by 2  until p==N                    /*obtain the totient number for index J*/
    if j//3\==0   then  if j//5\==0   then  if j//7\==0   then iterate
    s= phi(j);  a= s                            /* [↑]  J  must have 1 of these factors*/
                              do until a==1;  if @.a==.  then a= phi(a);    else a= @.a
                                              s= s + a
                              end   /*until*/
    if s\==j  then iterate                      /*Is  J  not a perfect totient number? */
    p= p + 1                                    /*bump count of perfect totient numbers*/
    $= $ j                                      /*add to perfect totient numbers list. */
    end   /*j*/

say 'The first ' N " perfect totient numbers:" /*display the header to the terminal. */ say strip($) /* " " list. " " " */ exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ gcd: parse arg x,y; do until y==0; parse value x//y y with y x; end; return x /*──────────────────────────────────────────────────────────────────────────────────────*/ phi: procedure expose @.; parse arg z; if @.z\==. then return @.z /*was found before?*/

    #= z==1;         do m=1  for z-1;   if gcd(m, z)==1  then #= # + 1;    end  /*m*/
    @.z= #;   return #                                              /*use memoization. */</lang>

output is identical to the 1^st REXX version.

Ring

<lang ring> perfect = [] n = 1 while len(perfect)<20

     totnt = n
     tsum = 0
     while totnt!=1 
           totnt = totient(totnt)
           tsum = tsum + totnt
     end
     if tsum=n 
        add(perfect,n)
     ok
     n = n + 2

end see "The first 20 perfect totient numbers are:" + nl showarray(perfect)

func totient n

    totnt = n
    i = 2
    while i*i <= n
          if n%i=0 
             while true
                   n = n/i
                   if n%i!=0 
                      exit
                   ok
             end 
             totnt = totnt - totnt/i
          ok
          if i=2
             i = i + 1
          else
             i = i + 2
          ok
    end
   if n>1 
      totnt = totnt - totnt/n
   ok
   return totnt

func showArray array

    txt = ""
    see "["
    for n = 1 to len(array)
        txt = txt + array[n] + ","
    next
    txt = left(txt,len(txt)-1)
    txt = txt + "]"
    see txt

</lang>

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]

Ruby

<lang ruby>require "prime"

class Integer

 def φ
   prime_division.inject(1) {|res, (pr, exp)| res *= (pr-1) * pr**(exp-1) } 
 end

 def perfect_totient?
   f, sum = self, 0
   until f == 1 do
     f = f.φ
     sum += f
   end
   self == sum
 end

end

puts (1..).lazy.select(&:perfect_totient?).first(20).join(", ") </lang>

Output:

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

Scala

In this example we define a function which determines whether or not a number is a perfect totient number, then use it to construct a lazily evaluated list which contains all perfect totient numbers. Calculating the first n perfect totient numbers only requires taking the first n elements from the list. <lang scala>//List of perfect totients def isPerfectTotient(num: Int): Boolean = LazyList.iterate(totient(num))(totient).takeWhile(_ != 1).foldLeft(0L)(_+_) + 1 == num def perfectTotients: LazyList[Int] = LazyList.from(3).filter(isPerfectTotient)

//Totient Function @tailrec def scrub(f: Long, num: Long): Long = if(num%f == 0) scrub(f, num/f) else num def totient(num: Long): Long = LazyList.iterate((num, 2: Long, num)){case (ac, i, n) => if(n%i == 0) (ac*(i - 1)/i, i + 1, scrub(i, n)) else (ac, i + 1, n)}.dropWhile(_._3 != 1).head._1</lang>

Output:

scala> perfectTotients.take(20).mkString(", ")
res1: String = 3, 9, 15, 27, 39, 81, 111, 183, 243, 255, 327, 363, 471, 729, 2187, 2199, 3063, 4359, 4375, 5571

Sidef

<lang ruby>func perfect_totient({.<=1}, sum=0) { sum } func perfect_totient( n, sum=0) { __FUNC__(var(t = n.euler_phi), sum + t) }

say (1..Inf -> lazy.grep {|n| perfect_totient(n) == n }.first(20))</lang>

Output:

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

Swift

<lang swift>public func totient(n: Int) -> Int {

 var n = n
 var i = 2
 var tot = n

 while i * i <= n {
   if n % i == 0 {
     while n % i == 0 {
       n /= i
     }

     tot -= tot / i
   }

   if i == 2 {
     i = 1
   }

   i += 2
 }

 if n > 1 {
   tot -= tot / n
 }

 return tot

}

public struct PerfectTotients: Sequence, IteratorProtocol {

 private var m = 1

 public init() { }

 public mutating func next() -> Int? {
   while true {
     defer {
       m += 1
     }

     var tot = m
     var sum = 0

     while tot != 1 {
       tot = totient(n: tot)
       sum += tot
     }

     if sum == m {
       return m
     }
   }
 }

}

print("The first 20 perfect totient numbers are:") print(Array(PerfectTotients().prefix(20)))</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]

Wren

Translation of: Go

The version using Euler's product formula. <lang ecmascript>var totient = Fn.new { |n|

   var tot = n
   var i = 2
   while (i*i <= n) {
       if (n%i == 0) {
           while(n%i == 0) n = (n/i).floor
           tot = tot - (tot/i).floor
       }
       if (i == 2) i = 1
       i = i + 2
   }
   if (n > 1) tot = tot - (tot/n).floor
   return tot

}

var perfect = [] var n = 1 while (perfect.count < 20) {

   var tot = n
   var sum = 0
   while (tot != 1) {
       tot = totient.call(tot)
       sum = sum + tot
   }
   if (sum == n) perfect.add(n)
   n = n + 2

} System.print("The first 20 perfect totient numbers are:") System.print(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]

zkl

<lang zkl>var totients=List.createLong(10_000,0); // cache fcn totient(n){ if(phi:=totients[n]) return(phi);

  totients[n]=[1..n].reduce('wrap(p,k){ p + (n.gcd(k)==1) })

} fcn perfectTotientW{ // -->iterator

  (1).walker(*).tweak(fcn(z){
     parts,n := 0,z;
     while(n!=1){ parts+=( n=totient(n) ) }
     if(parts==z) z else Void.Skip;
  })

}</lang> <lang zkl>perfectTotientW().walk(20).println();</lang>

Output:

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