Perfect totient numbers
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, 40 numbers take around 320 seconds on a Core i5 with 8 GB RAM. <lang C> /*Abhishek Ghosh, 8th December 2018*/
- 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]
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 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)