Perfect totient numbers: Difference between revisions
Alextretyak (talk | contribs) Added 11l |
MaiconSoft (talk | contribs) m Added Delphi reference to Pascal code |
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3 9 15 27 39 81 111 183 243 255 327 363 471 729 2187 2199 3063 4359 4375 5571 |
3 9 15 27 39 81 111 183 243 255 327 363 471 729 2187 2199 3063 4359 4375 5571 |
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</pre> |
</pre> |
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=={{header|Delphi}}== |
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{{libheader| System.SysUtils}} |
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{{Trans|Go}} |
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<lang Delphi> |
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program Perfect_totient_numbers; |
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{$APPTYPE CONSOLE} |
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uses |
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System.SysUtils; |
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function totient(n: Integer): Integer; |
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begin |
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var tot := n; |
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var i := 2; |
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while i * i <= n do |
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begin |
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if (n mod i) = 0 then |
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begin |
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while (n mod i) = 0 do |
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n := n div i; |
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dec(tot, tot div i); |
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end; |
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if i = 2 then |
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i := 1; |
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inc(i, 2); |
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end; |
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if n > 1 then |
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dec(tot, tot div n); |
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Result := tot; |
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end; |
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begin |
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var perfect: TArray<Integer>; |
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var n := 1; |
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while length(perfect) < 20 do |
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begin |
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var tot := n; |
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var sum := 0; |
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while tot <> 1 do |
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begin |
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tot := totient(tot); |
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inc(sum, tot); |
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end; |
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if sum = n then |
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begin |
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SetLength(perfect, Length(perfect) + 1); |
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perfect[High(perfect)] := n; |
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end; |
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inc(n, 2); |
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end; |
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writeln('The first 20 perfect totient numbers are:'); |
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write('['); |
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for var e in perfect do |
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write(e, ' '); |
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writeln(']'); |
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readln; |
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end.</lang> |
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=={{header|Factor}}== |
=={{header|Factor}}== |
Revision as of 00:35, 17 March 2021
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.
- mrob list of the first 54
11l
<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]
AWK
<lang 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
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
Delphi
<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]
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
<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]
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]
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
<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
<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
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]
Racket
<lang 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)
<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 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 1st 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
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)