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
-
- the OEIS entry for perfect totient numbers.
C
Calculates as many number of 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]
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 }
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.
Perl
<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
Perl 6
<lang perl6>my \π = Nil, |(1..*).hyper.map: -> $t { +(^$t).grep: * gcd $t == 1 }; my \ππ = Nil, |(2..*).grep: -> $p { $p == sum π[$p], { π[$_] } β¦ 1 };
put "The first twenty Perfect totient numbers:\n", ππ[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
Python
<lang python>from math import gcd from functools import lru_cache from itertools import islice, count
@lru_cache(maxsize=None) def Ο(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 = Ο(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]
REXX
<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.*/ p= 0 /*the count of perfect totient numbers.*/ @.=. /*memoization array of totient numbers.*/ $= /*list of the perfect totient numbers. */
do j=3 by 2 until p==N; s= phi(j) /*obtain totient number for a number. */ a= s /* [β] 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. */ /*ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ*/ gcd: parse arg x,y; do until y==0; parse value x//y y with y x; end /*until*/
return x
/*ββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββββ*/ 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
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]
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)