Perfect totient numbers

From Rosetta Code
Task
Perfect totient numbers
You are encouraged to solve this task according to the task description, using any language you may know.

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


Related task


Also see


AWK[edit]

 
# 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)
}
 
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

C[edit]

Calculates as many perfect Totient numbers as entered on the command line.

#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;
}
 

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]

Factor[edit]

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

Go[edit]

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)
}
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:

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)
}

The output is the same as before.

Haskell[edit]

import Data.Bool (bool)
 
perfectTotients :: [Int]
perfectTotients =
[2 ..] >>=
((bool [] . return) <*>
((==) <*> (succ . sum . tail . takeWhile (1 /=) . iterate φ)))
 
φ :: Int -> Int
φ = memoize (\n -> length (filter ((1 ==) . gcd n) [1 .. n]))
 
memoize :: (Int -> a) -> (Int -> a)
memoize f = (map f [0 ..] !!)
 
main :: IO ()
main = print $ take 20 perfectTotients
Output:
[3,9,15,27,39,81,111,183,243,255,327,363,471,729,2187,2199,3063,4359,4375,5571]

J[edit]

 
Until =: conjunction def 'u^:(0 -: v)^:_'
Filter =: (#~`)(`:6)
totient =: 5&p:
totient_chain =: [: }. (, [email protected]{:)Until(1={:)
ptnQ =: (= ([: +/ totient_chain))&>
 

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

JavaScript[edit]

(() => {
'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();
})();
Output:
[3,9,15,27,39,81,111,183,243,255,327,363,471,729,2187,2199,3063,4359,4375,5571]

Julia[edit]

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))")
 
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[edit]

Translation of: Go
// 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)
}
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[edit]

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

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.
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[edit]

Library: ntheory
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;
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

Perl 6[edit]

Works with: Rakudo version 2018.11
my \𝜑  = Nil, |(1..*).hyper.map: -> $t { +(^$t).grep: * gcd $t == 1 };
my \𝜑𝜑 = Nil, |(2..*).grep: -> $p { $p == sum 𝜑[$p], { 𝜑[$_] }1 };
 
put "The first twenty Perfect totient numbers:\n", 𝜑𝜑[1..20];
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[edit]

Translation of: Go
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
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[edit]

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

Python[edit]

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

REXX[edit]

unoptimized[edit]

/*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 /*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. */
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[edit]

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 one of these factors:   3,   5,   or   7

(322   =   31,381,059,609).

/*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 /*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. */
output   is identical to the 1st REXX version.


Ruby[edit]

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

Scala[edit]

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.

//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
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[edit]

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

Swift[edit]

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)))
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[edit]

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;
})
}
perfectTotientW().walk(20).println();
Output:
L(3,9,15,27,39,81,111,183,243,255,327,363,471,729,2187,2199,3063,4359,4375,5571)