Chowla numbers
You are encouraged to solve this task according to the task description, using any language you may know.
Chowla numbers are also known as:
- Chowla's function
- chowla numbers
- the chowla function
- the chowla number
- the chowla sequence
The chowla number of n is (as defined by Chowla's function):
- the sum of the divisors of n excluding unity and n
- where n is a positive integer
The sequence is named after Sarvadaman D. S. Chowla, (22 October 1907 ──► 10 December 1995),
a London born Indian American mathematician specializing in number theory.
German mathematician Carl Friedrich Gauss (1777─1855) said:
"Mathematics is the queen of the sciences ─ and number theory is the queen of mathematics".
- Definitions
Chowla numbers can also be expressed as:
chowla(n) = sum of divisors of n excluding unity and n
chowla(n) = sum( divisors(n)) - 1 - n
chowla(n) = sum( properDivisors(n)) - 1
chowla(n) = sum(aliquotDivisors(n)) - 1
chowla(n) = aliquot(n) - 1
chowla(n) = sigma(n) - 1 - n
chowla(n) = sigmaProperDivisiors(n) - 1
chowla(a*b) = a + b, if a and b are distinct primes
if chowla(n) = 0, and n > 1, then n is prime
if chowla(n) = n - 1, and n > 1, then n is a perfect number
- Task
-
- create a chowla function that returns the chowla number for a positive integer n
- Find and display (1 per line) for the 1st 37 integers:
- the integer (the index)
- the chowla number for that integer
- For finding primes, use the chowla function to find values of zero
- Find and display the count of the primes up to 100
- Find and display the count of the primes up to 1,000
- Find and display the count of the primes up to 10,000
- Find and display the count of the primes up to 100,000
- Find and display the count of the primes up to 1,000,000
- Find and display the count of the primes up to 10,000,000
- For finding perfect numbers, use the chowla function to find values of n - 1
- Find and display all perfect numbers up to 35,000,000
- use commas within appropriate numbers
- show all output here
- Related tasks
- See also
-
- the OEIS entry for A48050 Chowla's function.
11l
F chowla(n)
V sum = 0
V i = 2
L i * i <= n
I n % i == 0
sum += i
V j = n I/ i
I i != j
sum += j
i++
R sum
L(n) 1..37
print(‘chowla(’n‘) = ’chowla(n))
V count = 0
V power = 100
L(n) 2..10'000'000
I chowla(n) == 0
count++
I n % power == 0
print(‘There are ’count‘ primes < ’power)
power *= 10
count = 0
V limit = 350'000'000
V k = 2
V kk = 3
L
V p = k * kk
I p > limit
L.break
I chowla(p) == p - 1
print(p‘ is a perfect number’)
count++
k = kk + 1
kk += k
print(‘There are ’count‘ perfect numbers < ’limit)- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There are 25 primes < 100 There are 168 primes < 1000 There are 1229 primes < 10000 There are 9592 primes < 100000 There are 78498 primes < 1000000 There are 664579 primes < 10000000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers < 350000000
Ada
with Ada.Text_IO;
procedure Chowla_Numbers is
function Chowla (N : Positive) return Natural is
Sum : Natural := 0;
I : Positive := 2;
J : Positive;
begin
while I * I <= N loop
if N mod I = 0 then
J := N / I;
Sum := Sum + I + (if I = J then 0 else J);
end if;
I := I + 1;
end loop;
return Sum;
end Chowla;
procedure Put_37_First is
use Ada.Text_IO;
begin
for A in Positive range 1 .. 37 loop
Put_Line ("chowla(" & A'Image & ") = " & Chowla (A)'Image);
end loop;
end Put_37_First;
procedure Put_Prime is
use Ada.Text_IO;
Count : Natural := 0;
Power : Positive := 100;
begin
for N in Positive range 2 .. 10_000_000 loop
if Chowla (N) = 0 then
Count := Count + 1;
end if;
if N mod Power = 0 then
Put_Line ("There is " & Count'Image & " primes < " & Power'Image);
Power := Power * 10;
end if;
end loop;
end Put_Prime;
procedure Put_Perfect is
use Ada.Text_IO;
Count : Natural := 0;
Limit : constant := 350_000_000;
K : Natural := 2;
Kk : Natural := 3;
P : Natural;
begin
loop
P := K * Kk;
exit when P > Limit;
if Chowla (P) = P - 1 then
Put_Line (P'Image & " is a perfect number");
Count := Count + 1;
end if;
K := Kk + 1;
Kk := Kk + K;
end loop;
Put_Line ("There are " & Count'Image & " perfect numbers < " & Limit'Image);
end Put_Perfect;
begin
Put_37_First;
Put_Prime;
Put_Perfect;
end Chowla_Numbers;
- Output:
chowla( 1) = 0 chowla( 2) = 0 chowla( 3) = 0 chowla( 4) = 2 chowla( 5) = 0 chowla( 6) = 5 chowla( 7) = 0 chowla( 8) = 6 chowla( 9) = 3 chowla( 10) = 7 chowla( 11) = 0 chowla( 12) = 15 chowla( 13) = 0 chowla( 14) = 9 chowla( 15) = 8 chowla( 16) = 14 chowla( 17) = 0 chowla( 18) = 20 chowla( 19) = 0 chowla( 20) = 21 chowla( 21) = 10 chowla( 22) = 13 chowla( 23) = 0 chowla( 24) = 35 chowla( 25) = 5 chowla( 26) = 15 chowla( 27) = 12 chowla( 28) = 27 chowla( 29) = 0 chowla( 30) = 41 chowla( 31) = 0 chowla( 32) = 30 chowla( 33) = 14 chowla( 34) = 19 chowla( 35) = 12 chowla( 36) = 54 chowla( 37) = 0 There is 25 primes < 100 There is 168 primes < 1000 There is 1229 primes < 10000 There is 9592 primes < 100000 There is 78498 primes < 1000000 There is 664579 primes < 10000000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers < 350000000
ALGOL 68
BEGIN # find some Chowla numbers ( Chowla n = sum of divisors of n exclusing n and 1 ) #
# returs the Chowla number of n #
PROC chowla = ( INT n )INT:
BEGIN
INT sum := 0;
FOR i FROM 2 WHILE i * i <= n DO
IF n MOD i = 0 THEN
INT j = n OVER i;
sum +:= i + IF i = j THEN 0 ELSE j FI
FI
OD;
sum
END # chowla # ;
FOR n TO 37 DO print( ( "chowla(", whole( n, 0 ), ") = ", whole( chowla( n ), 0 ), newline ) ) OD;
INT count := 0, power := 100;
FOR n FROM 2 TO 10 000 000 DO
IF chowla( n ) = 0 THEN count +:= 1 FI;
IF n MOD power = 0 THEN
print( ( "There are ", whole( count, 0 ), " primes < ", whole( power, 0 ), newline ) );
power *:= 10
FI
OD;
count := 0;
INT limit = 350 000 000;
INT k := 2, kk := 3;
WHILE INT p = k * kk;
p <= limit
DO
IF chowla( p ) = p - 1 THEN
print( ( whole( p, 0 ), " is a perfect number", newline ) );
count +:= 1
FI;
k := kk + 1; kk +:= k
OD;
print( ( "There are ", whole( count, 0 ), " perfect numbers < ", whole( limit, 0 ), newline ) )
END- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There are 25 primes < 100 There are 168 primes < 1000 There are 1229 primes < 10000 There are 9592 primes < 100000 There are 78498 primes < 1000000 There are 664579 primes < 10000000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers < 350000000
Arturo
chowla: function [n]-> sum remove remove factors n 1 n
countPrimesUpTo: function [limit][
count: 1
loop 3.. .step: 2 limit 'x [
if zero? chowla x -> count: count + 1
]
return count
]
loop 1..37 'i -> print [i "=>" chowla i]
print ""
loop [100 1000 10000 100000 1000000 10000000] 'lim [
print ["primes up to" lim "=>" countPrimesUpTo lim]
]
print ""
print "perfect numbers up to 35000000:"
i: 2
while [i < 35000000][
if (chowla i) = i - 1 -> print i
i: i + 2
]
- Output:
1 => 0 2 => 0 3 => 0 4 => 2 5 => 0 6 => 5 7 => 0 8 => 6 9 => 3 10 => 7 11 => 0 12 => 15 13 => 0 14 => 9 15 => 8 16 => 14 17 => 0 18 => 20 19 => 0 20 => 21 21 => 10 22 => 13 23 => 0 24 => 35 25 => 5 26 => 15 27 => 12 28 => 27 29 => 0 30 => 41 31 => 0 32 => 30 33 => 14 34 => 19 35 => 12 36 => 54 37 => 0 primes up to 100 => 25 primes up to 1000 => 168 primes up to 10000 => 1229 primes up to 100000 => 9592 primes up to 1000000 => 78498 primes up to 10000000 => 664579 perfect numbers up to 35000000: 6 28 496 8128 33550336
AWK
# syntax: GAWK -f CHOWLA_NUMBERS.AWK
# converted from Go
BEGIN {
for (i=1; i<=37; i++) {
printf("chowla(%2d) = %s\n",i,chowla(i))
}
printf("\nCount of primes up to:\n")
count = 1
limit = 1e7
sieve(limit)
power = 100
for (i=3; i<limit; i+=2) {
if (!c[i]) {
count++
}
if (i == power-1) {
printf("%10s = %s\n",commatize(power),commatize(count))
power *= 10
}
}
printf("\nPerfect numbers:")
count = 0
limit = 35000000
k = 2
kk = 3
while (1) {
if ((p = k * kk) > limit) {
break
}
if (chowla(p) == p-1) {
printf(" %s",commatize(p))
count++
}
k = kk + 1
kk += k
}
printf("\nThere are %d perfect numbers <= %s\n",count,commatize(limit))
exit(0)
}
function chowla(n, i,j,sum) {
if (n < 1 || n != int(n)) {
return sprintf("%s is invalid",n)
}
for (i=2; i*i<=n; i++) {
if (n%i == 0) {
j = n / i
sum += (i == j) ? i : i + j
}
}
return(sum+0)
}
function commatize(x, num) {
if (x < 0) {
return "-" commatize(-x)
}
x = int(x)
num = sprintf("%d.",x)
while (num ~ /^[0-9][0-9][0-9][0-9]/) {
sub(/[0-9][0-9][0-9][,.]/,",&",num)
}
sub(/\.$/,"",num)
return(num)
}
function sieve(limit, i,j) {
for (i=1; i<=limit; i++) {
c[i] = 0
}
for (i=3; i*3<limit; i+=2) {
if (!c[i] && chowla(i) == 0) {
for (j=3*i; j<limit; j+=2*i) {
c[j] = 1
}
}
}
}
- Output:
chowla( 1) = 0
chowla( 2) = 0
chowla( 3) = 0
chowla( 4) = 2
chowla( 5) = 0
chowla( 6) = 5
chowla( 7) = 0
chowla( 8) = 6
chowla( 9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to:
100 = 25
1,000 = 168
10,000 = 1,229
100,000 = 9,592
1,000,000 = 78,498
10,000,000 = 664,579
Perfect numbers: 6 28 496 8,128 33,550,336
There are 5 perfect numbers <= 35,000,000
C
#include <stdio.h>
unsigned chowla(const unsigned n) {
unsigned sum = 0;
for (unsigned i = 2, j; i * i <= n; i ++) if (n % i == 0) sum += i + (i == (j = n / i) ? 0 : j);
return sum;
}
int main(int argc, char const *argv[]) {
unsigned a;
for (unsigned n = 1; n < 38; n ++) printf("chowla(%u) = %u\n", n, chowla(n));
unsigned n, count = 0, power = 100;
for (n = 2; n < 10000001; n ++) {
if (chowla(n) == 0) count ++;
if (n % power == 0) printf("There is %u primes < %u\n", count, power), power *= 10;
}
count = 0;
unsigned limit = 350000000;
unsigned k = 2, kk = 3, p;
for ( ; ; ) {
if ((p = k * kk) > limit) break;
if (chowla(p) == p - 1) {
printf("%d is a perfect number\n", p);
count ++;
}
k = kk + 1; kk += k;
}
printf("There are %u perfect numbers < %u\n", count, limit);
return 0;
}
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There is 25 primes < 100 There is 168 primes < 1000 There is 1229 primes < 10000 There is 9592 primes < 100000 There is 78498 primes < 1000000 There is 664579 primes < 10000000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers < 350000000
C#
using System;
namespace chowla_cs
{
class Program
{
static int chowla(int n)
{
int sum = 0;
for (int i = 2, j; i * i <= n; i++)
if (n % i == 0) sum += i + (i == (j = n / i) ? 0 : j);
return sum;
}
static bool[] sieve(int limit)
{
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
bool[] c = new bool[limit];
for (int i = 3; i * 3 < limit; i += 2)
if (!c[i] && (chowla(i) == 0))
for (int j = 3 * i; j < limit; j += 2 * i)
c[j] = true;
return c;
}
static void Main(string[] args)
{
for (int i = 1; i <= 37; i++)
Console.WriteLine("chowla({0}) = {1}", i, chowla(i));
int count = 1, limit = (int)(1e7), power = 100;
bool[] c = sieve(limit);
for (int i = 3; i < limit; i += 2)
{
if (!c[i]) count++;
if (i == power - 1)
{
Console.WriteLine("Count of primes up to {0,10:n0} = {1:n0}", power, count);
power *= 10;
}
}
count = 0; limit = 35000000;
int k = 2, kk = 3, p;
for (int i = 2; ; i++)
{
if ((p = k * kk) > limit) break;
if (chowla(p) == p - 1)
{
Console.WriteLine("{0,10:n0} is a number that is perfect", p);
count++;
}
k = kk + 1; kk += k;
}
Console.WriteLine("There are {0} perfect numbers <= 35,000,000", count);
if (System.Diagnostics.Debugger.IsAttached) Console.ReadKey();
}
}
}
- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to 100 = 25
Count of primes up to 1,000 = 168
Count of primes up to 10,000 = 1,229
Count of primes up to 100,000 = 9,592
Count of primes up to 1,000,000 = 78,498
Count of primes up to 10,000,000 = 664,579
6 is a number that is perfect
28 is a number that is perfect
496 is a number that is perfect
8,128 is a number that is perfect
33,550,336 is a number that is perfect
There are 5 perfect numbers <= 35,000,000
C++
#include <vector>
#include <iostream>
using namespace std;
int chowla(int n)
{
int sum = 0;
for (int i = 2, j; i * i <= n; i++)
if (n % i == 0) sum += i + (i == (j = n / i) ? 0 : j);
return sum;
}
vector<bool> sieve(int limit)
{
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
vector<bool> c(limit);
for (int i = 3; i * 3 < limit; i += 2)
if (!c[i] && (chowla(i) == 0))
for (int j = 3 * i; j < limit; j += 2 * i)
c[j] = true;
return c;
}
int main()
{
cout.imbue(locale(""));
for (int i = 1; i <= 37; i++)
cout << "chowla(" << i << ") = " << chowla(i) << "\n";
int count = 1, limit = (int)(1e7), power = 100;
vector<bool> c = sieve(limit);
for (int i = 3; i < limit; i += 2)
{
if (!c[i]) count++;
if (i == power - 1)
{
cout << "Count of primes up to " << power << " = "<< count <<"\n";
power *= 10;
}
}
count = 0; limit = 35000000;
int k = 2, kk = 3, p;
for (int i = 2; ; i++)
{
if ((p = k * kk) > limit) break;
if (chowla(p) == p - 1)
{
cout << p << " is a number that is perfect\n";
count++;
}
k = kk + 1; kk += k;
}
cout << "There are " << count << " perfect numbers <= 35,000,000\n";
return 0;
}
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1,000 = 168 Count of primes up to 10,000 = 1,229 Count of primes up to 100,000 = 9,592 Count of primes up to 1,000,000 = 78,498 Count of primes up to 10,000,000 = 664,579 6 is a number that is perfect 28 is a number that is perfect 496 is a number that is perfect 8,128 is a number that is perfect 33,550,336 is a number that is perfect There are 5 perfect numbers <= 35,000,000
CLU
% Chowla's function
chowla = proc (n: int) returns (int)
sum: int := 0
i: int := 2
while i*i <= n do
if n//i = 0 then
sum := sum + i
j: int := n/i
if i ~= j then
sum := sum + j
end
end
i := i + 1
end
return(sum)
end chowla
% A number is prime iff chowla(n) is 0
prime = proc (n: int) returns (bool)
return(chowla(n) = 0)
end prime
% A number is perfect iff chowla(n) equals n-1
perfect = proc (n: int) returns (bool)
return(chowla(n) = n-1)
end perfect
start_up = proc ()
LIMIT = 35000000
po: stream := stream$primary_output()
% Show chowla(1) through chowla(37)
for i: int in int$from_to(1, 37) do
stream$putl(po, "chowla(" || int$unparse(i) || ") = "
|| int$unparse(chowla(i)))
end
% Count primes up to powers of 10
pow10: int := 2 % start with 100
primecount: int := 1 % assume 2 is prime, then test only odd numbers
candidate: int := 3
while pow10 <= 7 do
if candidate >= 10**pow10 then
stream$putl(po, "There are "
|| int$unparse(primecount)
|| " primes up to "
|| int$unparse(10**pow10))
pow10 := pow10 + 1
end
if prime(candidate) then primecount := primecount + 1 end
candidate := candidate + 2
end
% Find perfect numbers up to 35 million
perfcount: int := 0
k: int := 2
kk: int := 3
while true do
n: int := k * kk
if n >= LIMIT then break end
if perfect(n) then
perfcount := perfcount + 1
stream$putl(po, int$unparse(n) || " is a perfect number.")
end
k := kk + 1
kk := kk + k
end
stream$putl(po, "There are " || int$unparse(perfcount) ||
" perfect numbers < 35,000,000.")
end start_up- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There are 25 primes up to 100 There are 168 primes up to 1000 There are 1229 primes up to 10000 There are 9592 primes up to 100000 There are 78498 primes up to 1000000 There are 664579 primes up to 10000000 6 is a perfect number. 28 is a perfect number. 496 is a perfect number. 8128 is a perfect number. 33550336 is a perfect number. There are 5 perfect numbers < 35,000,000.
Cowgol
include "cowgol.coh";
sub chowla(n: uint32): (sum: uint32) is
sum := 0;
var i: uint32 := 2;
while i*i <= n loop
if n % i == 0 then
sum := sum + i;
var j := n / i;
if i != j then
sum := sum + j;
end if;
end if;
i := i + 1;
end loop;
end sub;
var n: uint32 := 1;
while n <= 37 loop
print("chowla(");
print_i32(n);
print(") = ");
print_i32(chowla(n));
print("\n");
n := n + 1;
end loop;
n := 2;
var power: uint32 := 100;
var count: uint32 := 0;
while n <= 10000000 loop
if chowla(n) == 0 then
count := count + 1;
end if;
if n % power == 0 then
print("There are ");
print_i32(count);
print(" primes < ");
print_i32(power);
print_nl();
power := power * 10;
end if;
n := n + 1;
end loop;
count := 0;
const LIMIT := 35000000;
var k: uint32 := 2;
var kk: uint32 := 3;
loop
n := k * kk;
if n > LIMIT then break; end if;
if chowla(n) == n-1 then
print_i32(n);
print(" is a perfect number.\n");
count := count + 1;
end if;
k := kk + 1;
kk := kk + k;
end loop;
print("There are ");
print_i32(count);
print(" perfect numbers < ");
print_i32(LIMIT);
print_nl();- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There are 25 primes < 100 There are 168 primes < 1000 There are 1229 primes < 10000 There are 9592 primes < 100000 There are 78498 primes < 1000000 There are 664579 primes < 10000000 6 is a perfect number. 28 is a perfect number. 496 is a perfect number. 8128 is a perfect number. 33550336 is a perfect number. There are 5 perfect numbers < 35000000
D
import std.stdio;
int chowla(int n) {
int sum;
for (int i = 2, j; i * i <= n; ++i) {
if (n % i == 0) {
sum += i + (i == (j = n / i) ? 0 : j);
}
}
return sum;
}
bool[] sieve(int limit) {
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
auto c = new bool[limit];
for (int i = 3; i * 3 < limit; i += 2) {
if (!c[i] && (chowla(i) == 0)) {
for (int j = 3 * i; j < limit; j += 2 * i) {
c[j] = true;
}
}
}
return c;
}
void main() {
foreach (i; 1..38) {
writefln("chowla(%d) = %d", i, chowla(i));
}
int count = 1;
int limit = cast(int)1e7;
int power = 100;
bool[] c = sieve(limit);
for (int i = 3; i < limit; i += 2) {
if (!c[i]) {
count++;
}
if (i == power - 1) {
writefln("Count of primes up to %10d = %d", power, count);
power *= 10;
}
}
count = 0;
limit = 350_000_000;
int k = 2;
int kk = 3;
int p;
for (int i = 2; ; ++i) {
p = k * kk;
if (p > limit) {
break;
}
if (chowla(p) == p - 1) {
writefln("%10d is a number that is perfect", p);
count++;
}
k = kk + 1;
kk += k;
}
writefln("There are %d perfect numbers <= 35,000,000", count);
}
- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to 100 = 25
Count of primes up to 1000 = 168
Count of primes up to 10000 = 1229
Count of primes up to 100000 = 9592
Count of primes up to 1000000 = 78498
Count of primes up to 10000000 = 664579
6 is a number that is perfect
28 is a number that is perfect
496 is a number that is perfect
8128 is a number that is perfect
33550336 is a number that is perfect
There are 5 perfect numbers <= 35,000,000
Delphi
See #Pascal.
Dyalect
func chowla(n) {
var sum = 0
var i = 2
var j = 0
while i * i <= n {
if n % i == 0 {
j = n / i
var app = if i == j {
0
} else {
j
}
sum += i + app
}
i += 1
}
return sum
}
func sieve(limit) {
var c = Array.Empty(limit)
var i = 3
while i * 3 < limit {
if !c[i] && (chowla(i) == 0) {
var j = 3 * i
while j < limit {
c[j] = true
j += 2 * i
}
}
i += 2
}
return c
}
for i in 1..37 {
print("chowla(\(i)) = \(chowla(i))")
}
var count = 1
var limit = 10000000
var power = 100
var c = sieve(limit)
var i = 3
while i < limit {
if !c[i] {
count += 1
}
if i == power - 1 {
print("Count of primes up to \(power) = \(count)")
power *= 10
}
i += 2
}
count = 0
limit = 35000000
var k = 2
var kk = 3
var p
i = 2
while true {
p = k * kk
if p > limit {
break
}
if chowla(p) == p - 1 {
print("\(p) is a number that is perfect")
count += 1
}
k = kk + 1
kk += k
}
print("There are \(count) perfect numbers <= 35,000,000")- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1000 = 168 Count of primes up to 10000 = 1229 Count of primes up to 100000 = 9592 Count of primes up to 1000000 = 78498 Count of primes up to 10000000 = 664579 6 is a number that is perfect 28 is a number that is perfect 496 is a number that is perfect 8128 is a number that is perfect 33550336 is a number that is perfect There are 5 perfect numbers <= 35,000,000
EasyLang
fastfunc chowla n .
sum = 0
i = 2
while i * i <= n
if n mod i = 0
j = n div i
if i = j
sum += i
else
sum += i + j
.
.
i += 1
.
return sum
.
proc sieve . c[] .
i = 3
while i * 3 <= len c[]
if c[i] = 0
if chowla i = 0
j = 3 * i
while j <= len c[]
c[j] = 1
j += 2 * i
.
.
.
i += 2
.
.
proc commatize n . s$ .
s$[] = strchars n
s$ = ""
l = len s$[]
for i = 1 to len s$[]
if i > 1 and l mod 3 = 0
s$ &= ","
.
l -= 1
s$ &= s$[i]
.
.
print "chowla number from 1 to 37"
for i = 1 to 37
print " " & i & ": " & chowla i
.
proc main . .
print ""
len c[] 10000000
count = 1
sieve c[]
power = 100
i = 3
while i <= len c[]
if c[i] = 0
count += 1
.
if i = power - 1
commatize power p$
commatize count c$
print "There are " & c$ & " primes up to " & p$
power *= 10
.
i += 2
.
print ""
limit = 35000000
count = 0
i = 2
k = 2
kk = 3
repeat
p = k * kk
until p > limit
if chowla p = p - 1
commatize p s$
print s$ & " is a perfect number"
count += 1
.
k = kk + 1
kk += k
i += 1
.
commatize limit s$
print "There are " & count & " perfect mumbers up to " & s$
.
main
- Output:
chowla number from 1 to 37 1: 0 2: 0 3: 0 4: 2 5: 0 6: 5 7: 0 8: 6 9: 3 10: 7 11: 0 12: 15 13: 0 14: 9 15: 8 16: 14 17: 0 18: 20 19: 0 20: 21 21: 10 22: 13 23: 0 24: 35 25: 5 26: 15 27: 12 28: 27 29: 0 30: 41 31: 0 32: 30 33: 14 34: 19 35: 12 36: 54 37: 0 There are 25 primes up to 100 There are 168 primes up to 1,000 There are 1,229 primes up to 10,000 There are 9,592 primes up to 100,000 There are 78,498 primes up to 1,000,000 There are 664,579 primes up to 10,000,000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8,128 is a perfect number 33,550,336 is a perfect number There are 5 perfect mumbers up to 35,000,000
EMal
fun chowla = int by int n
int sum = 0
int j = 0
for int i = 2; i * i <= n; i++ do
if n % i == 0 do sum += i + when(i == (j = n / i), 0, j) end
end
return sum
end
fun sieve = List by int limit
List c = logic[].with(limit)
for int i = 3; i * 3 < limit; i += 2
if c[i] or chowla(i) != 0 do continue end
for int j = 3 * i; j < limit; j += 2 * i do c[j] = true end
end
return c
end
# find and display (1 per line) for the 1st 37 integers
for int i = 1; i <= 37; i++ do writeLine("chowla(" + i + ") = " + chowla(i)) end
int count = 1
int limit = 10000000
int power = 100
List c = sieve(limit)
for int i = 3; i < limit; i += 2
if not c[i] do count++ end
if i == power - 1
writeLine("Count of primes up to " + power + " = " + count)
power *= 10
end
end
count = 0
limit = 35000000
int k = 2
int kk = 3
int p
for int i = 2; ; i++
if (p = k * kk) > limit do break end
if chowla(p) == p - 1
writeLine(p + " is a number that is perfect")
count++
end
k = kk + 1
kk += k
end
writeLine("There are " + count + " perfect numbers <= 35,000,000")It takes about fifteen minutes to complete on my i7-8650U with 8.00GB of RAM.
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1000 = 168 Count of primes up to 10000 = 1229 Count of primes up to 100000 = 9592 Count of primes up to 1000000 = 78498 Count of primes up to 10000000 = 664579 6 is a number that is perfect 28 is a number that is perfect 496 is a number that is perfect 8128 is a number that is perfect 33550336 is a number that is perfect There are 5 perfect numbers <= 35,000,000
Factor
USING: formatting fry grouping.extras io kernel math
math.primes.factors math.ranges math.statistics sequences
tools.memory.private ;
IN: rosetta-code.chowla-numbers
: chowla ( n -- m )
dup 1 = [ 1 - ] [ [ divisors sum ] [ - 1 - ] bi ] if ;
: show-chowla ( n -- )
[1,b] [ dup chowla "chowla(%02d) = %d\n" printf ] each ;
: count-primes ( seq -- )
dup 0 prefix [ [ 1 + ] dip 2 <range> ] 2clump-map
[ [ chowla zero? ] count ] map cum-sum
[ [ commas ] bi@ "Primes up to %s: %s\n" printf ] 2each ;
: show-perfect ( n -- )
[ 2 3 ] dip '[ 2dup * dup _ > ] [
dup [ chowla ] [ 1 - = ] bi
[ commas "%s is perfect\n" printf ] [ drop ] if
[ nip 1 + ] [ nip dupd + ] 2bi
] until 3drop ;
: chowla-demo ( -- )
37 show-chowla nl { 100 1000 10000 100000 1000000 10000000 }
count-primes nl 35e7 show-perfect ;
MAIN: chowla-demo
- Output:
chowla(01) = 0 chowla(02) = 0 chowla(03) = 0 chowla(04) = 2 chowla(05) = 0 chowla(06) = 5 chowla(07) = 0 chowla(08) = 6 chowla(09) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Primes up to 100: 25 Primes up to 1,000: 168 Primes up to 10,000: 1,229 Primes up to 100,000: 9,592 Primes up to 1,000,000: 78,498 Primes up to 10,000,000: 664,579 6 is perfect 28 is perfect 496 is perfect 8,128 is perfect 33,550,336 is perfect
Fortran
This compiler implements the Fortran-77 standard. The VAX/VMS operating system runs on simulated hardware using the open source opensimh platform.
Run time on a Raspberry Pi 4 Model B Rev 1.1 (Raspbian GNU/Linux 10 buster) was 7h 21m
PROGRAM CHOWLA
CALL PUT_1ST_37
CALL PUT_PRIME
CALL PUT_PERFECT
END
INTEGER*4 FUNCTION CHOWLA1(N)
C The Chowla number of N is the sum of the divisors of N
C excluding unity and N where N is a positive integer
IMPLICIT INTEGER*4 (A-Z)
IF (N .LE. 0) STOP 'Argument to Chowla function must be > 0'
SUM = 0
I = 2
100 CONTINUE
IF (I * I .GT. N) GOTO 200
IF (MOD(N, I) .NE. 0) GOTO 110
J = N / I
SUM = SUM + I
IF ( I .NE. J) SUM = SUM + J
110 CONTINUE
I = I + 1
GOTO 100
200 CONTINUE
CHOWLA1 = SUM
RETURN
END
SUBROUTINE PUT_1ST_37
IMPLICIT INTEGER*4 (A-Z)
DO 100 I = 1, 37
PRINT 900, I, CHOWLA1(I)
100 CONTINUE
RETURN
900 FORMAT(1H , 'CHOWLA(', I2, ') = ', I2)
END
SUBROUTINE PUT_PRIME
IMPLICIT INTEGER*4 (A-Z)
PARAMETER LIMIT = 10000000
COUNT = 0
POWER = 100
DO 200 N = 2, LIMIT
IF (CHOWLA1(N) .EQ. 0) COUNT = COUNT + 1
IF (MOD(N, POWER) .NE. 0) GOTO 100
PRINT 900, COUNT, POWER
POWER = POWER * 10
100 CONTINUE
200 CONTINUE
RETURN
900 FORMAT(1H ,'There are ', I12, ' primes < ', I12)
END
SUBROUTINE PUT_PERFECT
IMPLICIT INTEGER*4 (A-Z)
PARAMETER LIMIT = 35000000
COUNT = 0
K = 2
KK = 3
100 CONTINUE
P = K * KK
IF (P .GT. LIMIT) GOTO 300
IF (CHOWLA1(P) .NE. P - 1) GOTO 200
PRINT 900, P
COUNT = COUNT + 1
200 CONTINUE
K = KK + 1
KK = KK + K
GOTO 100
300 CONTINUE
PRINT 910, COUNT, LIMIT
RETURN
900 FORMAT(1H , I10, ' is a perfect number')
910 FORMAT(1H , 'There are ', I10, ' perfect numbers < ', I10)
END
- Output:
CHOWLA( 1) = 0 CHOWLA( 2) = 0 CHOWLA( 3) = 0 CHOWLA( 4) = 2 CHOWLA( 5) = 0 CHOWLA( 6) = 5 CHOWLA( 7) = 0 CHOWLA( 8) = 6 CHOWLA( 9) = 3 CHOWLA(10) = 7 CHOWLA(11) = 0 CHOWLA(12) = 15 CHOWLA(13) = 0 CHOWLA(14) = 9 CHOWLA(15) = 8 CHOWLA(16) = 14 CHOWLA(17) = 0 CHOWLA(18) = 20 CHOWLA(19) = 0 CHOWLA(20) = 21 CHOWLA(21) = 10 CHOWLA(22) = 13 CHOWLA(23) = 0 CHOWLA(24) = 35 CHOWLA(25) = 5 CHOWLA(26) = 15 CHOWLA(27) = 12 CHOWLA(28) = 27 CHOWLA(29) = 0 CHOWLA(30) = 41 CHOWLA(31) = 0 CHOWLA(32) = 30 CHOWLA(33) = 14 CHOWLA(34) = 19 CHOWLA(35) = 12 CHOWLA(36) = 54 CHOWLA(37) = 0 There are 25 primes < 100 There are 168 primes < 1000 There are 1229 primes < 10000 There are 9592 primes < 100000 There are 78498 primes < 1000000 There are 664579 primes < 10000000
6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect numberThere are 5 perfect numbers < 35000000
FreeBASIC
' Chowla_numbers
#include "string.bi"
Dim Shared As Long limite
limite = 10000000
Dim Shared As Boolean c(limite)
Dim As Long count, topenumprimo, a
count = 1
topenumprimo = 100
Dim As Longint p, k, kk, limitenumperfect
limitenumperfect = 35000000
k = 2: kk = 3
Declare Function chowla(Byval n As Longint) As Longint
Declare Sub sieve(Byval limite As Long, c() As Boolean)
Function chowla(Byval n As Longint) As Longint
Dim As Long i, j, r
i = 2
Do While i * i <= n
j = n \ i
If n Mod i = 0 Then
r += i
If i <> j Then r += j
End If
i += 1
Loop
chowla = r
End Function
Sub sieve(Byval limite As Long, c() As Boolean)
Dim As Long i, j
Redim As Boolean c(limite - 1)
i = 3
Do While i * 3 < limite
If Not c(i) Then
If chowla(i) = false Then
j = 3 * i
Do While j < limite
c(j) = true
j += 2 * i
Loop
End If
End If
i += 2
Loop
End Sub
Print "Chowla numbers"
For a = 1 To 37
Print "chowla(" & Trim(Str(a)) & ") = " & Trim(Str(chowla(a)))
Next a
' Si chowla(n) = falso and n > 1 Entonces n es primo
Print: Print "Contando los numeros primos hasta: "
sieve(limite, c())
For a = 3 To limite - 1 Step 2
If Not c(a) Then count += 1
If a = topenumprimo - 1 Then
Print Using "########## hay"; topenumprimo;
Print count
topenumprimo *= 10
End If
Next a
' Si chowla(n) = n - 1 and n > 1 Entonces n es un número perfecto
Print: Print "Buscando numeros perfectos... "
count = 0
Do
p = k * kk : If p > limitenumperfect Then Exit Do
If chowla(p) = p - 1 Then
Print Using "##########,# es un numero perfecto"; p
count += 1
End If
k = kk + 1 : kk += k
Loop
Print: Print "Hay " & count & " numeros perfectos <= " & Format(limitenumperfect, "###############################,#")
Print: Print "Pulsa una tecla para salir"
Sleep
End
- Output:
Chowla numbers
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Contando los numeros primos hasta:
100 hay 25
1000 hay 168
10000 hay 1229
100000 hay 9592
1000000 hay 78498
10000000 hay 664579
Buscando numeros perfectos...
6 es un numero perfecto
28 es un numero perfecto
496 es un numero perfecto
8,128 es un numero perfecto
33,550,336 es un numero perfecto
Hay 5 numeros perfectos <= 35.000.000
Pulsa una tecla para salir
FutureBasic
local fn Chowla( n as NSUInteger ) as NSUInteger
NSUInteger i, j, r = 0
i = 2
while ( i * i <= n )
j = n / i
if ( n mod i == 0 )
r += i
if ( i != j )
r += j
end if
end if
i++
wend
end fn = r
local fn DoIt
NSUInteger n, count = 0, power = 100, limit, k, kk, p = 0
for n = 1 to 37
printf @"chowla(%u) = %u", n, fn Chowla( n )
next
for n = 2 to 10000000
if ( fn Chowla(n) == 0 ) then count ++
if ( n mod power == 0 ) then printf @"There are %u primes < %-7u", count, power : power *= 10
next
count = 0
limit = 350000000
k = 2 : kk = 3
do
p = k * kk
if ( fn Chowla( p ) == p - 1 )
printf @"%9u is a perfect number", p
count++
end if
k = kk + 1
kk = kk + k
until ( p > limit )
printf @"There are %u perfect numbers < %u", count, limit
end fn
fn DoIt
HandleEvents- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
There are 25 primes < 100
There are 168 primes < 1,000
There are 1,229 primes < 10,000
There are 9,592 primes < 100,000
There are 78,498 primes < 1,000,000
There are 664,579 primes < 10,000,000
6 is a perfect number
28 is a perfect number
496 is a perfect number
8,128 is a perfect number
33,550,336 is a perfect number
There are 5 perfect numbers < 35,000,000
Go
package main
import "fmt"
func chowla(n int) int {
if n < 1 {
panic("argument must be a positive integer")
}
sum := 0
for i := 2; i*i <= n; i++ {
if n%i == 0 {
j := n / i
if i == j {
sum += i
} else {
sum += i + j
}
}
}
return sum
}
func sieve(limit int) []bool {
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
c := make([]bool, limit)
for i := 3; i*3 < limit; i += 2 {
if !c[i] && chowla(i) == 0 {
for j := 3 * i; j < limit; j += 2 * i {
c[j] = true
}
}
}
return c
}
func commatize(n int) string {
s := fmt.Sprintf("%d", n)
le := len(s)
for i := le - 3; i >= 1; i -= 3 {
s = s[0:i] + "," + s[i:]
}
return s
}
func main() {
for i := 1; i <= 37; i++ {
fmt.Printf("chowla(%2d) = %d\n", i, chowla(i))
}
fmt.Println()
count := 1
limit := int(1e7)
c := sieve(limit)
power := 100
for i := 3; i < limit; i += 2 {
if !c[i] {
count++
}
if i == power-1 {
fmt.Printf("Count of primes up to %-10s = %s\n", commatize(power), commatize(count))
power *= 10
}
}
fmt.Println()
count = 0
limit = 35000000
for i := uint(2); ; i++ {
p := 1 << (i - 1) * (1<<i - 1) // perfect numbers must be of this form
if p > limit {
break
}
if chowla(p) == p-1 {
fmt.Printf("%s is a perfect number\n", commatize(p))
count++
}
}
fmt.Println("There are", count, "perfect numbers <= 35,000,000")
}
- Output:
chowla( 1) = 0 chowla( 2) = 0 chowla( 3) = 0 chowla( 4) = 2 chowla( 5) = 0 chowla( 6) = 5 chowla( 7) = 0 chowla( 8) = 6 chowla( 9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1,000 = 168 Count of primes up to 10,000 = 1,229 Count of primes up to 100,000 = 9,592 Count of primes up to 1,000,000 = 78,498 Count of primes up to 10,000,000 = 664,579 6 is a perfect number 28 is a perfect number 496 is a perfect number 8,128 is a perfect number 33,550,336 is a perfect number There are 5 perfect numbers <= 35,000,000
Groovy
class Chowla {
static int chowla(int n) {
if (n < 1) throw new RuntimeException("argument must be a positive integer")
int sum = 0
int i = 2
while (i * i <= n) {
if (n % i == 0) {
int j = (int) (n / i)
sum += (i == j) ? i : i + j
}
i++
}
return sum
}
static boolean[] sieve(int limit) {
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
boolean[] c = new boolean[limit]
for (int i = 3; i < limit / 3; i += 2) {
if (!c[i] && chowla(i) == 0) {
for (int j = 3 * i; j < limit; j += 2 * i) {
c[j] = true
}
}
}
return c
}
static void main(String[] args) {
for (int i = 1; i <= 37; i++) {
printf("chowla(%2d) = %d\n", i, chowla(i))
}
println()
int count = 1
int limit = 10_000_000
boolean[] c = sieve(limit)
int power = 100
for (int i = 3; i < limit; i += 2) {
if (!c[i]) {
count++
}
if (i == power - 1) {
printf("Count of primes up to %,10d = %,7d\n", power, count)
power *= 10
}
}
println()
count = 0
limit = 35_000_000
int i = 2
while (true) {
int p = (1 << (i - 1)) * ((1 << i) - 1) // perfect numbers must be of this form
if (p > limit) break
if (chowla(p) == p - 1) {
printf("%,d is a perfect number\n", p)
count++
}
i++
}
printf("There are %,d perfect numbers <= %,d\n", count, limit)
}
}
- Output:
chowla( 1) = 0 chowla( 2) = 0 chowla( 3) = 0 chowla( 4) = 2 chowla( 5) = 0 chowla( 6) = 5 chowla( 7) = 0 chowla( 8) = 6 chowla( 9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1,000 = 168 Count of primes up to 10,000 = 1,229 Count of primes up to 100,000 = 9,592 Count of primes up to 1,000,000 = 78,498 Count of primes up to 10,000,000 = 664,579 6 is a perfect number 28 is a perfect number 496 is a perfect number 8,128 is a perfect number 33,550,336 is a perfect number There are 5 perfect numbers <= 35,000,000
Haskell
Uses arithmoi Library: https://hackage.haskell.org/package/arithmoi-0.11.0.0
compiled with "-O2 -threaded -rtsopts"
import Control.Concurrent (setNumCapabilities)
import Control.Monad.Par (runPar, get, spawnP)
import Control.Monad (join, (>=>))
import Data.List.Split (chunksOf)
import Data.List (intercalate, mapAccumL, genericTake, genericDrop)
import Data.Bifunctor (bimap)
import GHC.Conc (getNumProcessors)
import Math.NumberTheory.Primes (factorise, unPrime)
import Text.Printf (printf)
chowla :: Word -> Word
chowla 1 = 0
chowla n = f n
where
f = (-) =<< pred . product . fmap sumFactor . factorise
sumFactor (n, e) = foldr (\p s -> s + unPrime n^p) 1 [1..e]
chowlas :: [Word] -> [(Word, Word)]
chowlas [] = []
chowlas xs = runPar $ join <$>
(mapM (spawnP . fmap ((,) <*> chowla)) >=> mapM get) (chunksOf (10^6) xs)
chowlaPrimes :: [(Word, Word)] -> (Word, Word) -> (Word, Word)
chowlaPrimes chowlas range = (count chowlas, snd range)
where
isPrime (1, n) = False
isPrime (_, n) = n == 0
count = fromIntegral . length . filter isPrime . between range
between (min, max) = genericTake (max - pred min) . genericDrop (pred min)
chowlaPerfects :: [(Word, Word)] -> [Word]
chowlaPerfects = fmap fst . filter isPerfect
where
isPerfect (1, _) = False
isPerfect (n, c) = c == pred n
commas :: (Show a, Integral a) => a -> String
commas = reverse . intercalate "," . chunksOf 3 . reverse . show
main :: IO ()
main = do
cores <- getNumProcessors
setNumCapabilities cores
printf "Using %d cores\n" cores
mapM_ (uncurry (printf "chowla(%2d) = %d\n")) $ take 37 allChowlas
mapM_ (uncurry (printf "There are %8s primes < %10s\n"))
(chowlaP
[ (1, 10^2)
, (succ $ 10^2, 10^3)
, (succ $ 10^3, 10^4)
, (succ $ 10^4, 10^5)
, (succ $ 10^5, 10^6)
, (succ $ 10^6, 10^7) ])
mapM_ (printf "%10s is a perfect number.\n" . commas) perfects
printf "There are %2d perfect numbers < 35,000,000\n" $ length perfects
where
chowlaP = fmap (bimap commas commas) . snd
. mapAccumL (\total (count, max) -> (total + count, (total + count, max))) 0
. fmap (chowlaPrimes $ take (10^7) allChowlas)
perfects = chowlaPerfects allChowlas
allChowlas = chowlas [1..35*10^6]
- Output:
Using 4 cores
chowla( 1) = 0
chowla( 2) = 0
chowla( 3) = 0
chowla( 4) = 2
chowla( 5) = 0
chowla( 6) = 5
chowla( 7) = 0
chowla( 8) = 6
chowla( 9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
There are 25 primes < 100
There are 168 primes < 1,000
There are 1,229 primes < 10,000
There are 9,592 primes < 100,000
There are 78,498 primes < 1,000,000
There are 664,579 primes < 10,000,000
6 is a perfect number.
28 is a perfect number.
496 is a perfect number.
8,128 is a perfect number.
33,550,336 is a perfect number.
There are 5 perfect numbers < 35,000,000
J
Solution:
chowla=: >: -~ >:@#.~/.~&.q: NB. sum of factors - (n + 1)
intsbelow=: (2 }. i.)"0
countPrimesbelow=: +/@(0 = chowla)@intsbelow
findPerfectsbelow=: (#~ <: = chowla)@intsbelow
Tasks:
(] ,. chowla) >: i. 37 NB. chowla numbers 1-37
1 0
2 0
3 0
4 2
5 0
6 5
7 0
8 6
9 3
10 7
11 0
12 15
13 0
14 9
15 8
16 14
17 0
18 20
19 0
20 21
21 10
22 13
23 0
24 35
25 5
26 15
27 12
28 27
29 0
30 41
31 0
32 30
33 14
34 19
35 12
36 54
37 0
countPrimesbelow 100 1000 10000 100000 1000000 10000000
25 168 1229 9592 78498 664579
findPerfectsbelow 35000000
6 28 496 8128 33550336
Java
public class Chowla {
public static void main(String[] args) {
int[] chowlaNumbers = findChowlaNumbers(37);
for (int i = 0; i < chowlaNumbers.length; i++) {
System.out.printf("chowla(%d) = %d%n", (i+1), chowlaNumbers[i]);
}
System.out.println();
int[][] primes = countPrimes(100, 10_000_000);
for (int i = 0; i < primes.length; i++) {
System.out.printf(Locale.US, "There is %,d primes up to %,d%n", primes[i][1], primes[i][0]);
}
System.out.println();
int[] perfectNumbers = findPerfectNumbers(35_000_000);
for (int i = 0; i < perfectNumbers.length; i++) {
System.out.printf("%d is a perfect number%n", perfectNumbers[i]);
}
System.out.printf(Locale.US, "There are %d perfect numbers < %,d%n", perfectNumbers.length, 35_000_000);
}
public static int chowla(int n) {
if (n < 0) throw new IllegalArgumentException("n is not positive");
int sum = 0;
for (int i = 2, j; i * i <= n; i++)
if (n % i == 0) sum += i + (i == (j = n / i) ? 0 : j);
return sum;
}
protected static int[][] countPrimes(int power, int limit) {
int count = 0;
int[][] num = new int[countMultiplicity(limit, power)][2];
for (int n = 2, i=0; n <= limit; n++) {
if (chowla(n) == 0) count++;
if (n % power == 0) {
num[i][0] = power;
num[i][1] = count;
i++;
power *= 10;
}
}
return num;
}
protected static int countMultiplicity(int limit, int start) {
int count = 0;
int cur = limit;
while(cur >= start) {
count++;
cur = cur/10;
}
return count;
}
protected static int[] findChowlaNumbers(int limit) {
int[] num = new int[limit];
for (int i = 0; i < limit; i++) {
num[i] = chowla(i+1);
}
return num;
}
protected static int[] findPerfectNumbers(int limit) {
int count = 0;
int[] num = new int[count];
int k = 2, kk = 3, p;
while ((p = k * kk) < limit) {
if (chowla(p) == p - 1) {
num = increaseArr(num);
num[count++] = p;
}
k = kk + 1;
kk += k;
}
return num;
}
private static int[] increaseArr(int[] arr) {
int[] tmp = new int[arr.length + 1];
System.arraycopy(arr, 0, tmp, 0, arr.length);
return tmp;
}
}
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There is 25 primes up to 100 There is 168 primes up to 1,000 There is 1,229 primes up to 10,000 There is 9,592 primes up to 100,000 There is 78,498 primes up to 1,000,000 There is 664,579 primes up to 10,000,000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers < 35,000,000
jq
Works with gojq, the Go implementation of jq
The "brute-force" computation of the perfect number beyond 8,128 took many hours.
def add(stream): reduce stream as $x (0; . + $x);
# input should be an integer
def commatize:
def digits: tostring | explode | reverse;
if . == null then ""
elif . < 0 then "-" + ((- .) | commatize)
else [foreach digits[] as $d (-1; .+1;
# "," is 44
(select(. > 0 and . % 3 == 0)|44), $d)]
| reverse
| implode
end;
def count(stream): reduce stream as $i (0; . + 1);
def lpad($len): tostring | ($len - length) as $l | (" " * $l)[:$l] + .;
# To take advantage of gojq's arbitrary-precision integer arithmetic:
def power($b): . as $in | reduce range(0;$b) as $i (1; . * $in);
# unordered
def proper_divisors:
. as $n
| if $n > 1 then 1,
( range(2; 1 + (sqrt|floor)) as $i
| if ($n % $i) == 0 then $i,
(($n / $i) | if . == $i then empty else . end)
else empty
end)
else empty
end;
def chowla:
if . == 1 then 0
else add(proper_divisors) - 1
end;
# Input: a positive integer
def is_chowla_prime:
. > 1 and chowla == 0;
# In the interests of green(er) computing ...
def chowla_primes($n):
2, range(3; $n; 2) | select(is_chowla_prime);
def report_chowla_primes:
reduce range(2; 10000000) as $i (null;
if $i | is_chowla_prime
then if $i < 10000000 then .[7] += 1 else . end
| if $i < 1000000 then .[6] += 1 else . end
| if $i < 100000 then .[5] += 1 else . end
| if $i < 10000 then .[4] += 1 else . end
| if $i < 1000 then .[3] += 1 else . end
| if $i < 100 then .[2] += 1 else . end
else . end)
| (range(2;8) as $i
| "10 ^ \($i) \(.[$i]|commatize|lpad(16))") ;
def is_chowla_perfect:
(. > 1) and (chowla == . - 1);
def task:
" n\("chowla"|lpad(16))",
(range(1;38) | "\(lpad(3)): \(chowla|lpad(10))"),
"\n n \("Primes < n"|lpad(10))",
report_chowla_primes,
# "\nPerfect numbers up to 35e6",
# (range(1; 35e6) | select(is_chowla_perfect) | commatize)
""
;
task- Output:
n chowla 1: 0 2: 0 3: 0 4: 2 5: 0 6: 5 7: 0 8: 6 9: 3 10: 7 11: 0 12: 15 13: 0 14: 9 15: 8 16: 14 17: 0 18: 20 19: 0 20: 21 21: 10 22: 13 23: 0 24: 35 25: 5 26: 15 27: 12 28: 27 29: 0 30: 41 31: 0 32: 30 33: 14 34: 19 35: 12 36: 54 37: 0 n Primes < n 10 ^ 2 25 10 ^ 3 168 10 ^ 4 1,229 10 ^ 5 9,592 10 ^ 6 78,498 10 ^ 7 664,579 Perfect numbers up to 35e6 6 28 496 8,128 33,550,336
Julia
using Primes, Formatting
function chowla(n)
if n < 1
throw("Chowla function argument must be positive")
elseif n < 4
return zero(n)
else
f = [one(n)]
for (p,e) in factor(n)
f = reduce(vcat, [f*p^j for j in 1:e], init=f)
end
return sum(f) - one(n) - n
end
end
function countchowlas(n, asperfect=false, verbose=false)
count = 0
for i in 2:n # 1 is not prime or perfect so skip
chow = chowla(i)
if (asperfect && chow == i - 1) || (!asperfect && chow == 0)
count += 1
verbose && println("The number $(format(i, commas=true)) is ", asperfect ? "perfect." : "prime.")
end
end
count
end
function testchowla()
println("The first 37 chowla numbers are:")
for i in 1:37
println("Chowla($i) is ", chowla(i))
end
for i in [100, 1000, 10000, 100000, 1000000, 10000000]
println("The count of the primes up to $(format(i, commas=true)) is $(format(countchowlas(i), commas=true))")
end
println("The count of perfect numbers up to 35,000,000 is $(countchowlas(35000000, true, true)).")
end
testchowla()
- Output:
The first 37 chowla numbers are: Chowla(1) is 0 Chowla(2) is 0 Chowla(3) is 0 Chowla(4) is 2 Chowla(5) is 0 Chowla(6) is 5 Chowla(7) is 0 Chowla(8) is 6 Chowla(9) is 3 Chowla(10) is 7 Chowla(11) is 0 Chowla(12) is 15 Chowla(13) is 0 Chowla(14) is 9 Chowla(15) is 8 Chowla(16) is 14 Chowla(17) is 0 Chowla(18) is 20 Chowla(19) is 0 Chowla(20) is 21 Chowla(21) is 10 Chowla(22) is 13 Chowla(23) is 0 Chowla(24) is 35 Chowla(25) is 5 Chowla(26) is 15 Chowla(27) is 12 Chowla(28) is 27 Chowla(29) is 0 Chowla(30) is 41 Chowla(31) is 0 Chowla(32) is 30 Chowla(33) is 14 Chowla(34) is 19 Chowla(35) is 12 Chowla(36) is 54 Chowla(37) is 0 The count of the primes up to 100 is 25 The count of the primes up to 1,000 is 168 The count of the primes up to 10,000 is 1,229 The count of the primes up to 100,000 is 9,592 The count of the primes up to 1,000,000 is 78,498 The count of the primes up to 10,000,000 is 664,579 The number 6 is perfect. The number 28 is perfect. The number 496 is perfect. The number 8,128 is perfect. The number 33,550,336 is perfect. The count of perfect numbers up to 35,000,000 is 5.
Kotlin
// Version 1.3.21
fun chowla(n: Int): Int {
if (n < 1) throw RuntimeException("argument must be a positive integer")
var sum = 0
var i = 2
while (i * i <= n) {
if (n % i == 0) {
val j = n / i
sum += if (i == j) i else i + j
}
i++
}
return sum
}
fun sieve(limit: Int): BooleanArray {
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
val c = BooleanArray(limit)
for (i in 3 until limit / 3 step 2) {
if (!c[i] && chowla(i) == 0) {
for (j in 3 * i until limit step 2 * i) c[j] = true
}
}
return c
}
fun main() {
for (i in 1..37) {
System.out.printf("chowla(%2d) = %d\n", i, chowla(i))
}
println()
var count = 1
var limit = 10_000_000
val c = sieve(limit)
var power = 100
for (i in 3 until limit step 2) {
if (!c[i]) count++
if (i == power - 1) {
System.out.printf("Count of primes up to %,-10d = %,d\n", power, count)
power *= 10
}
}
println()
count = 0
limit = 35_000_000
var i = 2
while (true) {
val p = (1 shl (i - 1)) * ((1 shl i) - 1) // perfect numbers must be of this form
if (p > limit) break
if (chowla(p) == p - 1) {
System.out.printf("%,d is a perfect number\n", p)
count++
}
i++
}
println("There are $count perfect numbers <= 35,000,000")
}
- Output:
Same as Go example.
Lua
function chowla(n)
local sum = 0
local i = 2
local j = 0
while i * i <= n do
if n % i == 0 then
j = math.floor(n / i)
sum = sum + i
if i ~= j then
sum = sum + j
end
end
i = i + 1
end
return sum
end
function sieve(limit)
-- True denotes composite, false denotes prime.
-- Only interested in odd numbers >= 3
local c = {}
local i = 3
while i * 3 < limit do
if not c[i] and (chowla(i) == 0) then
local j = 3 * i
while j < limit do
c[j] = true
j = j + 2 * i
end
end
i = i + 2
end
return c
end
function main()
for i = 1, 37 do
print(string.format("chowla(%d) = %d", i, chowla(i)))
end
local count = 1
local limit = math.floor(1e7)
local power = 100
local c = sieve(limit)
local i = 3
while i < limit do
if not c[i] then
count = count + 1
end
if i == power - 1 then
print(string.format("Count of primes up to %10d = %d", power, count))
power = power * 10
end
i = i + 2
end
count = 0
limit = 350000000
local k = 2
local kk = 3
local p = 0
i = 2
while true do
p = k * kk
if p > limit then
break
end
if chowla(p) == p - 1 then
print(string.format("%10d is a number that is perfect", p))
count = count + 1
end
k = kk + 1
kk = kk + k
i = i + 1
end
print(string.format("There are %d perfect numbers <= 35,000,000", count))
end
main()
- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to 100 = 25
Count of primes up to 1000 = 168
Count of primes up to 10000 = 1229
Count of primes up to 100000 = 9592
Count of primes up to 1000000 = 78498
Count of primes up to 10000000 = 664579
6 is a number that is perfect
28 is a number that is perfect
496 is a number that is perfect
8128 is a number that is perfect
33550336 is a number that is perfect
There are 5 perfect numbers <= 35,000,000
MAD
NORMAL MODE IS INTEGER
INTERNAL FUNCTION(N)
ENTRY TO CHOWLA.
SUM = 0
THROUGH LOOP, FOR I=2, 1, I*I.G.N
J = N/I
WHENEVER J*I.E.N
SUM = SUM + I
WHENEVER I.NE.J, SUM = SUM + J
END OF CONDITIONAL
LOOP CONTINUE
FUNCTION RETURN SUM
END OF FUNCTION
VECTOR VALUES CHWFMT = $7HCHOWLA(,I2,4H) = ,I2*$
THROUGH CH37, FOR CH=1, 1, CH.G.37
CH37 PRINT FORMAT CHWFMT, CH, CHOWLA.(CH)
VECTOR VALUES PRIMES =
0 $10HTHERE ARE ,I6,S1,13HPRIMES BELOW ,I8*$
POWER = 100
COUNT = 0
THROUGH PRM, FOR CH=2, 1, CH.G.10000000
WHENEVER CHOWLA.(CH).E.0, COUNT = COUNT + 1
WHENEVER (CH/POWER)*POWER.E.CH
PRINT FORMAT PRIMES, COUNT, POWER
POWER = POWER * 10
PRM END OF CONDITIONAL
COUNT = 0
LIMIT = 35000000
VECTOR VALUES PERFCT = $I8,S1,20HIS A PERFECT NUMBER.*$
VECTOR VALUES PRFCNT =
0 $10HTHERE ARE ,I1,S1,22HPERFECT NUMBERS BELOW ,I8*$
K = 2
KK = 3
LOOP CH = K * KK
WHENEVER CH.G.LIMIT, TRANSFER TO DONE
WHENEVER CHOWLA.(CH).E.CH-1
PRINT FORMAT PERFCT, CH
COUNT = COUNT + 1
END OF CONDITIONAL
K = KK + 1
KK = KK + K
TRANSFER TO LOOP
DONE PRINT FORMAT PRFCNT, COUNT, LIMIT
END OF PROGRAM- Output:
CHOWLA( 1) = 0
CHOWLA( 2) = 0
CHOWLA( 3) = 0
CHOWLA( 4) = 2
CHOWLA( 5) = 0
CHOWLA( 6) = 5
CHOWLA( 7) = 0
CHOWLA( 8) = 6
CHOWLA( 9) = 3
CHOWLA(10) = 7
CHOWLA(11) = 0
CHOWLA(12) = 15
CHOWLA(13) = 0
CHOWLA(14) = 9
CHOWLA(15) = 8
CHOWLA(16) = 14
CHOWLA(17) = 0
CHOWLA(18) = 20
CHOWLA(19) = 0
CHOWLA(20) = 21
CHOWLA(21) = 10
CHOWLA(22) = 13
CHOWLA(23) = 0
CHOWLA(24) = 35
CHOWLA(25) = 5
CHOWLA(26) = 15
CHOWLA(27) = 12
CHOWLA(28) = 27
CHOWLA(29) = 0
CHOWLA(30) = 41
CHOWLA(31) = 0
CHOWLA(32) = 30
CHOWLA(33) = 14
CHOWLA(34) = 19
CHOWLA(35) = 12
CHOWLA(36) = 54
CHOWLA(37) = 0
THERE ARE 25 PRIMES BELOW 100
THERE ARE 168 PRIMES BELOW 1000
THERE ARE 1229 PRIMES BELOW 10000
THERE ARE 9592 PRIMES BELOW 100000
THERE ARE 78498 PRIMES BELOW 1000000
THERE ARE 664579 PRIMES BELOW 10000000
6 IS A PERFECT NUMBER.
28 IS A PERFECT NUMBER.
496 IS A PERFECT NUMBER.
8128 IS A PERFECT NUMBER.
33550336 IS A PERFECT NUMBER.
THERE ARE 5 PERFECT NUMBERS BELOW 35000000
Maple
ChowlaFunction := n -> NumberTheory:-SumOfDivisors(n) - n - 1;
PrintChowla := proc(n::posint) local i;
printf("Integer : Chowla Number\n");
for i to n do
printf("%d : %d\n", i, ChowlaFunction(i));
end do;
end proc:
countPrimes := n -> nops([ListTools[SearchAll](0, map(ChowlaFunction, [seq(1 .. n)]))]);
findPerfect := proc(n::posint) local to_check, found, k;
to_check := map(ChowlaFunction, [seq(1 .. n)]);
found := [];
for k to n do
if to_check(k) = k - 1 then
found := [found, k];
end if;
end do;
end proc:
PrintChowla(37);
countPrimes(100);
countPrimes(1000);
countPrimes(10000);
countPrimes(100000);
countPrimes(1000000);
countPrimes(10000000);
findPerfect(35000000)- Output:
Integer : Chowla Number 1 : -1 2 : 0 3 : 0 4 : 2 5 : 0 6 : 5 7 : 0 8 : 6 9 : 3 10 : 7 11 : 0 12 : 15 13 : 0 14 : 9 15 : 8 16 : 14 17 : 0 18 : 20 19 : 0 20 : 21 21 : 10 22 : 13 23 : 0 24 : 35 25 : 5 26 : 15 27 : 12 28 : 27 29 : 0 30 : 41 31 : 0 32 : 30 33 : 14 34 : 19 35 : 12 36 : 54 37 : 0 25 168 1229 9592 78498 664579 [6, 28, 496, 8128, 33550336]
Mathematica / Wolfram Language
ClearAll[Chowla]
Chowla[0 | 1] := 0
Chowla[n_] := DivisorSigma[1, n] - 1 - n
Table[{i, Chowla[i]}, {i, 37}] // Grid
PrintTemporary[Dynamic[n]];
i = 1; Do[If[Chowla[n] == 0, i++], {n, 3, 100, 2}]; i
i = 1; Do[If[Chowla[n] == 0, i++], {n, 3, 1000, 2}]; i
i = 1; Do[If[Chowla[n] == 0, i++], {n, 3, 10000, 2}]; i
i = 1; Do[If[Chowla[n] == 0, i++], {n, 3, 100000, 2}]; i
i = 1; Do[If[Chowla[n] == 0, i++], {n, 3, 1000000, 2}]; i
i = 1; Do[If[Chowla[n] == 0, i++], {n, 3, 10000000, 2}]; i
Reap[Do[If[Chowla[n] == n - 1, Sow[n]], {n, 1, 35 10^6}]][[2, 1]]
- Output:
25
168
1229
9592
78498
664579
{1, 6, 28, 496, 8128, 33550336}
Nim
import strformat
import strutils
func chowla(n: uint64): uint64 =
var sum = 0u64
var i = 2u64
var j: uint64
while i * i <= n:
if n mod i == 0:
j = n div i
sum += i
if i != j:
sum += j
inc i
sum
for n in 1u64..37:
echo &"chowla({n}) = {chowla(n)}"
var count = 0
var power = 100u64
for n in 2u64..10_000_000:
if chowla(n) == 0:
inc count
if n mod power == 0:
echo &"There are {insertSep($count, ','):>7} primes < {insertSep($power, ','):>10}"
power *= 10
count = 0
const limit = 350_000_000u64
var k = 2u64
var kk = 3u64
var p: uint64
while true:
p = k * kk
if p > limit:
break
if chowla(p) == p - 1:
echo &"{insertSep($p, ','):>10} is a perfect number"
inc count
k = kk + 1
kk += k
echo &"There are {count} perfect numbers < {insertSep($limit, ',')}"
- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
There are 25 primes < 100
There are 168 primes < 1,000
There are 1,229 primes < 10,000
There are 9,592 primes < 100,000
There are 78,498 primes < 1,000,000
There are 664,579 primes < 10,000,000
6 is a perfect number
28 is a perfect number
496 is a perfect number
8,128 is a perfect number
33,550,336 is a perfect number
There are 5 perfect numbers < 350,000,000
PARI/GP
chowla(n) = {
if (n < 1, error("Chowla function argument must be positive"));
if (n < 4, return(0));
my(divs = divisors(n));
sum(i=1, #divs, divs[i]) - n - 1;
}
\\ Function to count Chowla numbers
countchowlas(n, asperfect = 1, verbose = 1) = {
my(count = 0, chow, i);
for (i = 2, n,
chow = chowla(i);
if ( (asperfect && (chow == i - 1)) || ((!asperfect) && (chow == 0)),
count++;
if (verbose, print("The number " i " is " if (asperfect, "perfect.", "prime.")));
);
);
count;
}
\\ Main execution block
{
print("The first 37 chowla numbers are:");
for (i = 1, 37, printf("Chowla(%s) is %s\n", Str(i), Str(chowla(i)) ) );
m=100;
while(m<=10000000, print("The count of the primes up to " m " is " countchowlas(m, 0, 0)); m=m*10);
print("The count of perfect numbers up to 35,000,000 is " countchowlas(35000000, 1, 1));
}- Output:
The first 37 chowla numbers are: Chowla(1) is 0 Chowla(2) is 0 Chowla(3) is 0 Chowla(4) is 2 Chowla(5) is 0 Chowla(6) is 5 Chowla(7) is 0 Chowla(8) is 6 Chowla(9) is 3 Chowla(10) is 7 Chowla(11) is 0 Chowla(12) is 15 Chowla(13) is 0 Chowla(14) is 9 Chowla(15) is 8 Chowla(16) is 14 Chowla(17) is 0 Chowla(18) is 20 Chowla(19) is 0 Chowla(20) is 21 Chowla(21) is 10 Chowla(22) is 13 Chowla(23) is 0 Chowla(24) is 35 Chowla(25) is 5 Chowla(26) is 15 Chowla(27) is 12 Chowla(28) is 27 Chowla(29) is 0 Chowla(30) is 41 Chowla(31) is 0 Chowla(32) is 30 Chowla(33) is 14 Chowla(34) is 19 Chowla(35) is 12 Chowla(36) is 54 Chowla(37) is 0 The count of the primes up to 100 is 25 The count of the primes up to 1000 is 168 The count of the primes up to 10000 is 1229 The count of the primes up to 100000 is 9592 The count of the primes up to 1000000 is 78498 The count of the primes up to 10000000 is 664579 The number 6 is perfect. The number 28 is perfect. The number 496 is perfect. The number 8128 is perfect. The number 33550336 is perfect. The count of perfect numbers up to 35000000 is 5.
Pascal
but not using a sieve, cause a sieve doesn't need precalculated small primes.
So runtime is as bad as trial division.
program Chowla_numbers;
{$IFDEF FPC}
{$MODE Delphi}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
SysUtils
{$IFDEF FPC}
,StrUtils{for Numb2USA}
{$ENDIF}
;
{$IFNDEF FPC}
function Numb2USA(const S: string): string;
var
I, NA: Integer;
begin
I := Length(S);
Result := S;
NA := 0;
while (I > 0) do
begin
if ((Length(Result) - I + 1 - NA) mod 3 = 0) and (I <> 1) then
begin
Insert(',', Result, I);
Inc(NA);
end;
Dec(I);
end;
end;
{$ENDIF}
function Chowla(n: NativeUint): NativeUint;
var
Divisor, Quotient: NativeUint;
begin
result := 0;
Divisor := 2;
while sqr(Divisor) < n do
begin
Quotient := n div Divisor;
if Quotient * Divisor = n then
inc(result, Divisor + Quotient);
inc(Divisor);
end;
if sqr(Divisor) = n then
inc(result, Divisor);
end;
procedure Count10Primes(Limit: NativeUInt);
var
n, i, cnt: integer;
begin
writeln;
writeln(' primes til | count');
i := 100;
n := 2;
cnt := 0;
repeat
repeat
// Ord (true) = 1 ,Ord (false) = 0
inc(cnt, ORD(chowla(n) = 0));
inc(n);
until n > i;
writeln(Numb2USA(IntToStr(i)): 12, '|', Numb2USA(IntToStr(cnt)): 10);
i := i * 10;
until i > Limit;
end;
procedure CheckPerf;
var
k, kk, p, cnt, limit: NativeInt;
begin
writeln;
writeln(' number that is perfect');
cnt := 0;
limit := 35000000;
k := 2;
kk := 3;
repeat
p := k * kk;
if p > limit then
BREAK;
if chowla(p) = (p - 1) then
begin
writeln(Numb2USA(IntToStr(p)): 12);
inc(cnt);
end;
k := kk + 1;
inc(kk, k);
until false;
end;
var
I: integer;
begin
for I := 2 to 37 do
writeln('chowla(', I: 2, ') =', chowla(I): 3);
Count10Primes(10 * 1000 * 1000);
CheckPerf;
end.
- Output:
chowla( 2) = 0
chowla( 3) = 0
chowla( 4) = 2
chowla( 5) = 0
chowla( 6) = 5
chowla( 7) = 0
chowla( 8) = 6
chowla( 9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
primes til | count
100| 25
1,000| 168
10,000| 1,229
100,000| 9,592
1,000,000| 78,498
10,000,000| 664,579
number that is perfect
6
28
496
8,128
33,550,336
real 1m54,534s
PascalABC.NET
function chowla(n: Integer): Integer;
begin
result := 0;
var i := 2;
while i * i <= n do
begin
if n mod i = 0 then
result += i + (if i = n div i then 0 else n div i);
i += 1
end;
end;
function sieve(limit: Integer): array of Boolean;
begin
result := new Boolean[limit];
for var i := 3 to limit div 3 step 2 do
if not result[i] and (chowla(i) = 0) Then
for var j := 3 * i to limit step 2 * i do
result[j] := true;
end;
begin
for var i := 1 To 37 do
WriteLn('chowla(', i, ') = ', chowla(i));
var count := 1;
var limit := 10_000_000;
var power := 100;
var c := sieve(limit);
for var i := 3 To limit Step 2 do
begin
if not c[i] Then count += 1;
if i = power - 1 Then begin
WriteLn('Count of primes up to ', power:8, ' = ', count);
power *= 10;
end;
end;
count := 0;
limit := 35_000_000;
var k := 2; var kk := 3;
while True do
begin
var p := k * kk;
if p > limit Then break;
if chowla(p) = p - 1 Then begin
WriteLn(p:8, ' is a number that is perfect');
count += 1;
end;
k := kk + 1;
kk += k;
end;
WriteLn('There are ', count, ' perfect numbers <= 35,000,000')
end.
Perl
use strict;
use warnings;
use ntheory 'divisor_sum';
sub comma { reverse ((reverse shift) =~ s/(.{3})/$1,/gr) =~ s/^,//r }
sub chowla {
my($n) = @_;
$n < 2 ? 0 : divisor_sum($n) - ($n + 1);
}
sub prime_cnt {
my($n) = @_;
my $cnt = 1;
for (3..$n) {
$cnt++ if $_%2 and chowla($_) == 0
}
$cnt;
}
sub perfect {
my($n) = @_;
my @p;
for my $i (1..$n) {
push @p, $i if $i > 1 and chowla($i) == $i-1;
}
# map { push @p, $_ if $_ > 1 and chowla($_) == $_-1 } 1..$n; # speed penalty
@p;
}
printf "chowla(%2d) = %2d\n", $_, chowla($_) for 1..37;
print "\nCount of primes up to:\n";
printf "%10s %s\n", comma(10**$_), comma(prime_cnt(10**$_)) for 2..7;
my @perfect = perfect(my $limit = 35_000_000);
printf "\nThere are %d perfect numbers up to %s: %s\n",
1+$#perfect, comma($limit), join(' ', map { comma($_) } @perfect);
- Output:
chowla( 1) = 0
chowla( 2) = 0
chowla( 3) = 0
chowla( 4) = 2
chowla( 5) = 0
chowla( 6) = 5
chowla( 7) = 0
chowla( 8) = 6
chowla( 9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to:
100 25
1,000 168
10,000 1,229
100,000 9,592
1,000,000 78,498
10,000,000 664,579
There are 5 perfect numbers up to 35,000,000: 6 28 496 8,128 33,550,336
Phix
function chowla(atom n) return sum(factors(n)) end function function sieve(integer limit) -- True denotes composite, false denotes prime. -- Only interested in odd numbers >= 3 sequence c = repeat(false,limit) for i=3 to floor(limit/3) by 2 do -- if not c[i] and chowla(i)==0 then if not c[i] then -- (see note below) for j=3*i to limit by 2*i do c[j] = true end for end if end for return c end function atom limit = 1e7, count = 1, pow10 = 100, t0 = time() sequence s = {} for i=1 to 37 do s &= chowla(i) end for printf(1,"chowla[1..37]: %V\n",{s}) s = sieve(limit) for i=3 to limit by 2 do if not s[i] then count += 1 end if if i==pow10-1 then printf(1,"Count of primes up to %,d = %,d\n", {pow10, count}) pow10 *= 10 end if end for count = 0 limit = iff(machine_bits()=32?1.4e11:2.4e18) --limit = power(2,iff(machine_bits()=32?53:64)) -- (see note below) integer i=2 while true do atom p = power(2,i-1)*(power(2,i)-1) -- perfect numbers must be of this form if p>limit then exit end if if chowla(p)==p-1 then printf(1,"%,d is a perfect number\n", p) count += 1 end if i += 1 end while printf(1,"There are %d perfect numbers <= %,d\n",{count,limit}) ?elapsed(time()-t0)
The use of chowla() in sieve() does not actually achieve anything other than slow it down, so I took it out.
- Output:
chowla[1..37]: {0,0,0,2,0,5,0,6,3,7,0,15,0,9,8,14,0,20,0,21,10,13,0,35,5,15,12,27,0,41,0,30,14,19,12,54,0}
Count of primes up to 100 = 25
Count of primes up to 1,000 = 168
Count of primes up to 10,000 = 1,229
Count of primes up to 100,000 = 9,592
Count of primes up to 1,000,000 = 78,498
Count of primes up to 10,000,000 = 664,579
6 is a perfect number
28 is a perfect number
496 is a perfect number
8,128 is a perfect number
33,550,336 is a perfect number
8,589,869,056 is a perfect number
137,438,691,328 is a perfect number
2,305,843,008,139,952,128 is a perfect number
There are 8 perfect numbers <= 9,223,372,036,854,775,808
Note that 32-bit only finds the first 7 perfect numbers, but does so in 0.4s, whereas 64-bit takes just under 45s to find the 8th one. Using the theoretical (power 2) limits, those times become 4s and 90s respectively, without finding anything else. Obviously 1.4e11 and 2.4e18 were picked to minimise the run times.
Picat
table
chowla(1) = 0.
chowla(2) = 0.
chowla(3) = 0.
chowla(N) = C, N>3 =>
Max = floor(sqrt(N)),
Sum = 0,
foreach (X in 2..Max, N mod X == 0)
Y := N div X,
Sum := Sum + X + Y
end,
if (N == Max * Max) then
Sum := Sum - Max
end,
C = Sum.
main =>
foreach (I in 1..37)
printf("chowla(%d) = %d\n", I, chowla(I))
end,
Ranges = {100, 1000, 10000, 100000, 1000000, 10000000},
foreach (Range in Ranges)
Count = 0,
foreach (I in 2..Range)
if (chowla(I) == 0) then
Count := Count + 1
end
end,
printf("There are %d primes less than %d.\n", Count, Range)
end,
Limit = 35000000,
Count = 0,
foreach (I in 2..Limit)
if (chowla(I) == I-1) then
printf("%d is a perfect number\n", I),
Count := Count + 1
end
end,
printf("There are %d perfect numbers less than %d.\n", Count, Limit).- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There are 25 primes less than 100. There are 168 primes less than 1000. There are 1229 primes less than 10000. There are 9592 primes less than 100000. There are 78498 primes less than 1000000. There are 664579 primes less than 10000000. 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers less than 35000000.
PicoLisp
(de accu1 (Var Key)
(if (assoc Key (val Var))
(con @ (inc (cdr @)))
(push Var (cons Key 1)) )
Key )
(de factor (N)
(let
(R NIL
D 2
L (1 2 2 . (4 2 4 2 4 6 2 6 .))
M (sqrt N) )
(while (>= M D)
(if (=0 (% N D))
(setq M
(sqrt (setq N (/ N (accu1 'R D)))) )
(inc 'D (pop 'L)) ) )
(accu1 'R N)
(mapcar
'((L)
(make
(for N (cdr L)
(link (** (car L) N)) ) ) )
R ) ) )
(de chowla (N)
(let F (factor N)
(-
(sum
prog
(make
(link 1)
(mapc
'((A)
(chain
(mapcan
'((B)
(mapcar '((C) (* C B)) (made)) )
A ) ) )
F ) ) )
N
1 ) ) )
(de prime (N)
(and (> N 1) (=0 (chowla N))) )
(de perfect (N)
(and
(> N 1)
(= (chowla N) (dec N))) )
(de countP (N)
(let C 0
(for I N
(and (prime I) (inc 'C)) )
C ) )
(de listP (N)
(make
(for I N
(and (perfect I) (link I)) ) ) )
(for I 37
(prinl "chowla(" I ") = " (chowla I)) )
(prinl "Count of primes up to 100 = " (countP 100))
(prinl "Count of primes up to 1000 = " (countP 1000))
(prinl "Count of primes up to 10000 = " (countP 10000))
(prinl "Count of primes up to 100000 = " (countP 100000))
(prinl "Count of primes up to 1000000 = " (countP 1000000))
(prinl "Count of primes up to 10000000 = " (countP 10000000))
(println (listP 35000000))- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1000 = 168 Count of primes up to 10000 = 1229 Count of primes up to 100000 = 9592 Count of primes up to 1000000 = 78498 Count of primes up to 10000000 = 664579 (6 28 496 8128 33550336)
PowerBASIC
#COMPILE EXE
#DIM ALL
#COMPILER PBCC 6
FUNCTION chowla(BYVAL n AS LONG) AS LONG
REGISTER i AS LONG, j AS LONG
LOCAL r AS LONG
i = 2
DO WHILE i * i <= n
j = n \ i
IF n MOD i = 0 THEN
r += i
IF i <> j THEN
r += j
END IF
END IF
INCR i
LOOP
FUNCTION = r
END FUNCTION
FUNCTION chowla1(BYVAL n AS QUAD) AS QUAD
LOCAL i, j, r AS QUAD
i = 2
DO WHILE i * i <= n
j = n \ i
IF n MOD i = 0 THEN
r += i
IF i <> j THEN
r += j
END IF
END IF
INCR i
LOOP
FUNCTION = r
END FUNCTION
SUB sieve(BYVAL limit AS LONG, BYREF c() AS INTEGER)
LOCAL i, j AS LONG
REDIM c(limit - 1)
i = 3
DO WHILE i * 3 < limit
IF NOT c(i) THEN
IF chowla(i) = 0 THEN
j = 3 * i
DO WHILE j < limit
c(j) = -1
j += 2 * i
LOOP
END IF
END IF
i += 2
LOOP
END SUB
FUNCTION PBMAIN () AS LONG
LOCAL i, count, limit, power AS LONG
LOCAL c() AS INTEGER
LOCAL s AS STRING
LOCAL s30 AS STRING * 30
LOCAL p, k, kk, r, ql AS QUAD
FOR i = 1 TO 37
s = "chowla(" & TRIM$(STR$(i)) & ") = " & TRIM$(STR$(chowla(i)))
CON.PRINT s
NEXT i
count = 1
limit = 10000000
power = 100
CALL sieve(limit, c())
FOR i = 3 TO limit - 1 STEP 2
IF ISFALSE c(i) THEN count += 1
IF i = power - 1 THEN
RSET s30 = FORMAT$(power, "#,##0")
s = "Count of primes up to " & s30 & " =" & STR$(count)
CON.PRINT s
power *= 10
END IF
NEXT i
ql = 2 ^ 61
k = 2: kk = 3
RESET count
DO
p = k * kk : IF p > ql THEN EXIT DO
IF chowla1(p) = p - 1 THEN
RSET s30 = FORMAT$(p, "#,##0")
s = s30 & " is a number that is perfect"
CON.PRINT s
count += 1
END IF
k = kk + 1 : kk += k
LOOP
s = "There are" & STR$(count) & " perfect numbers <= " & FORMAT$(ql, "#,##0")
CON.PRINT s
CON.PRINT "press any key to exit program"
CON.WAITKEY$
END FUNCTION- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to 100 = 25
Count of primes up to 1,000 = 168
Count of primes up to 10,000 = 1229
Count of primes up to 100,000 = 9592
Count of primes up to 1,000,000 = 78498
Count of primes up to 10,000,000 = 664579
6 is a number that is perfect
28 is a number that is perfect
496 is a number that is perfect
8,128 is a number that is perfect
33,550,336 is a number that is perfect
8,589,869,056 is a number that is perfect
137,438,691,328 is a number that is perfect
2,305,843,008,139,952,130 is a number that is perfect
There are 8 perfect numbers <= 2,305,843,009,213,693,950
press any key to exit program
Prolog
chowla(1, 0).
chowla(2, 0).
chowla(N, C) :-
N > 2,
Max is floor(sqrt(N)),
findall(X, (between(2, Max, X), N mod X =:= 0), Xs),
findall(Y, (member(X1, Xs), Y is N div X1, Y \= Max), Ys),
!,
sum_list(Xs, S1),
sum_list(Ys, S2),
C is S1 + S2.
prime_count(Upper, Upper, Count, Count) :-
!.
prime_count(Lower, Upper, Add, Count) :-
chowla(Lower, 0),
!,
Lower1 is Lower + 1,
Add1 is Add + 1,
prime_count(Lower1, Upper, Add1, Count).
prime_count(Lower, Upper, Add, Count) :-
Lower1 is Lower + 1,
prime_count(Lower1, Upper, Add, Count).
perfect_numbers(Upper, Upper, AccNums, Nums) :-
!,
reverse(AccNums, Nums).
perfect_numbers(Lower, Upper, AccNums, Nums) :-
Perfect is Lower - 1,
chowla(Lower, Perfect),
!,
Lower1 is Lower + 1,
AccNums1 = [Lower|AccNums],
perfect_numbers(Lower1, Upper, AccNums1, Nums).
perfect_numbers(Lower, Upper, AccNums, Nums) :-
Lower1 is Lower + 1,
perfect_numbers(Lower1, Upper, AccNums, Nums).
main :-
% Chowla numbers
forall(between(1, 37, N), (
chowla(N, C),
format('chowla(~D) = ~D\n', [N, C])
)),
% Prime numbers
Ranges = [100, 1000, 10000, 100000, 1000000, 10000000],
forall(member(Range, Ranges), (
prime_count(2, Range, 0, PrimeCount),
format('There are ~D primes less than ~D.\n', [PrimeCount, Range])
)),
% Perfect numbers
Limit = 35000000,
perfect_numbers(2, Limit, [], Nums),
forall(member(Perfect, Nums), (
format('~D is a perfect number.\n', [Perfect])
)),
length(Nums, PerfectCount),
format('There are ~D perfect numbers < ~D.\n', [PerfectCount, Limit]).
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There are 25 primes less than 100. There are 168 primes less than 1,000. There are 1,229 primes less than 10,000. There are 9,592 primes less than 100,000. There are 78,498 primes less than 1,000,000. There are 664,579 primes less than 10,000,000. 6 is a perfect number. 28 is a perfect number. 496 is a perfect number. 8,128 is a perfect number. 33,550,336 is a perfect number. There are 5 perfect numbers < 35,000,000.
Python
Uses underscores to separate digits in numbers, and th sympy library to aid calculations.
# https://docs.sympy.org/latest/modules/ntheory.html#sympy.ntheory.factor_.divisors
from sympy import divisors
def chowla(n):
return 0 if n < 2 else sum(divisors(n, generator=True)) - 1 -n
def is_prime(n):
return chowla(n) == 0
def primes_to(n):
return sum(chowla(i) == 0 for i in range(2, n))
def perfect_between(n, m):
c = 0
print(f"\nPerfect numbers between [{n:_}, {m:_})")
for i in range(n, m):
if i > 1 and chowla(i) == i - 1:
print(f" {i:_}")
c += 1
print(f"Found {c} Perfect numbers between [{n:_}, {m:_})")
if __name__ == '__main__':
for i in range(1, 38):
print(f"chowla({i:2}) == {chowla(i)}")
for i in range(2, 6):
print(f"primes_to({10**i:_}) == {primes_to(10**i):_}")
perfect_between(1, 1_000_000)
print()
for i in range(6, 8):
print(f"primes_to({10**i:_}) == {primes_to(10**i):_}")
perfect_between(1_000_000, 35_000_000)
- Output:
chowla( 1) == 0 chowla( 2) == 0 chowla( 3) == 0 chowla( 4) == 2 chowla( 5) == 0 chowla( 6) == 5 chowla( 7) == 0 chowla( 8) == 6 chowla( 9) == 3 chowla(10) == 7 chowla(11) == 0 chowla(12) == 15 chowla(13) == 0 chowla(14) == 9 chowla(15) == 8 chowla(16) == 14 chowla(17) == 0 chowla(18) == 20 chowla(19) == 0 chowla(20) == 21 chowla(21) == 10 chowla(22) == 13 chowla(23) == 0 chowla(24) == 35 chowla(25) == 5 chowla(26) == 15 chowla(27) == 12 chowla(28) == 27 chowla(29) == 0 chowla(30) == 41 chowla(31) == 0 chowla(32) == 30 chowla(33) == 14 chowla(34) == 19 chowla(35) == 12 chowla(36) == 54 chowla(37) == 0 primes_to(100) == 25 primes_to(1_000) == 168 primes_to(10_000) == 1_229 primes_to(100_000) == 9_592 Perfect numbers between [1, 1_000_000) 6 28 496 8_128 Found 4 Perfect numbers between [1, 1_000_000) primes_to(1_000_000) == 78_498 primes_to(10_000_000) == 664_579 Perfect numbers between [1_000_000, 35_000_000) 33_550_336 Found 1 Perfect numbers between [1_000_000, 35_000_000)
Python: Numba
(Elementary) use of the numba library needs
- library install and import
- use of `@jit` decorator on some functions
- Rewrite to remove use of `sum()`
- Splitting one function for the jit compiler to digest.
from numba import jit
# https://docs.sympy.org/latest/modules/ntheory.html#sympy.ntheory.factor_.divisors
from sympy import divisors
@jit
def chowla(n):
return 0 if n < 2 else sum(divisors(n, generator=True)) - 1 -n
@jit
def is_prime(n):
return chowla(n) == 0
@jit
def primes_to(n):
acc = 0
for i in range(2, n):
if chowla(i) == 0:
acc += 1
return acc
@jit
def _perfect_between(n, m):
for i in range(n, m):
if i > 1 and chowla(i) == i - 1:
yield i
def perfect_between(n, m):
c = 0
print(f"\nPerfect numbers between [{n:_}, {m:_})")
for i in _perfect_between(n, m):
print(f" {i:_}")
c += 1
print(f"Found {c} Perfect numbers between [{n:_}, {m:_})")
- Output:
Same as above for use of same __main__ block.
Speedup - not much, subjectively...
Racket
#lang racket
(require racket/fixnum)
(define cache-size 35000000)
(define chowla-cache (make-fxvector cache-size -1))
(define (chowla/uncached n)
(for/sum ((i (sequence-filter (λ (x) (zero? (modulo n x))) (in-range 2 (add1 (quotient n 2)))))) i))
(define (chowla n)
(if (> n cache-size)
(chowla/uncached n)
(let ((idx (sub1 n)))
(if (negative? (fxvector-ref chowla-cache idx))
(let ((c (chowla/uncached n))) (fxvector-set! chowla-cache idx c) c)
(fxvector-ref chowla-cache idx)))))
(define (prime?/chowla n)
(and (> n 1)
(zero? (chowla n))))
(define (perfect?/chowla n)
(and (> n 1)
(= n (add1 (chowla n)))))
(define (make-chowla-sieve n)
(let ((v (make-vector n 0)))
(for* ((i (in-range 2 n)) (j (in-range (* 2 i) n i))) (vector-set! v j (+ i (vector-ref v j))))
(for ((i (in-range 1 n))) (fxvector-set! chowla-cache (sub1 i) (vector-ref v i)))))
(module+
main
(define (count-and-report-primes limit)
(printf "Primes < ~a: ~a~%" limit (for/sum ((i (sequence-filter prime?/chowla (in-range 2 (add1 limit))))) 1)))
(for ((i (in-range 1 (add1 37)))) (printf "(chowla ~a) = ~a~%" i (chowla i)))
(make-chowla-sieve cache-size)
(for-each count-and-report-primes '(1000 10000 100000 1000000 10000000))
(let ((ns (for/list ((n (sequence-filter perfect?/chowla (in-range 2 35000000)))) n)))
(printf "There are ~a perfect numbers <= 35000000: ~a~%" (length ns) ns)))
- Output:
(chowla 1) = 0 (chowla 2) = 0 (chowla 3) = 0 (chowla 4) = 2 (chowla 5) = 0 (chowla 6) = 5 (chowla 7) = 0 (chowla 8) = 6 (chowla 9) = 3 (chowla 10) = 7 (chowla 11) = 0 (chowla 12) = 15 (chowla 13) = 0 (chowla 14) = 9 (chowla 15) = 8 (chowla 16) = 14 (chowla 17) = 0 (chowla 18) = 20 (chowla 19) = 0 (chowla 20) = 21 (chowla 21) = 10 (chowla 22) = 13 (chowla 23) = 0 (chowla 24) = 35 (chowla 25) = 5 (chowla 26) = 15 (chowla 27) = 12 (chowla 28) = 27 (chowla 29) = 0 (chowla 30) = 41 (chowla 31) = 0 (chowla 32) = 30 (chowla 33) = 14 (chowla 34) = 19 (chowla 35) = 12 (chowla 36) = 54 (chowla 37) = 0 cpu time: 23937 real time: 23711 gc time: 151 Primes < 1000: 168 Primes < 10000: 1229 Primes < 100000: 9592 Primes < 1000000: 78498 Primes < 10000000: 664579 There are 5 perfect numbers <= 35000000: (6 28 496 8128 33550336)
Raku
(formerly Perl 6) Much like in the Totient function task, we are using a thing poorly suited to finding prime numbers, to find large quantities of prime numbers.
- (From the task's author): the object is not in the finding of prime numbers, but in verifying that the Chowla function operates correctly (and can be used for such a purpose, whatever the efficacy). These types of comments belong in the discussion page. Whether or not this function is poorly suited for finding prime numbers (or anything else) is not part of this task's purpose or objective.
(For a more reasonable test, reduce the orders-of-magnitude range in the "Primes count" line from 2..7 to 2..5)
sub comma { $^i.flip.comb(3).join(',').flip }
sub schnitzel (\Radda, \radDA = 0) {
Radda.is-prime ?? !Radda !! ?radDA ?? Radda
!! sum flat (2 .. Radda.sqrt.floor).map: -> \RAdda {
my \RADDA = Radda div RAdda;
next if RADDA * RAdda !== Radda;
RAdda !== RADDA ?? (RAdda, RADDA) !! RADDA
}
}
my \chowder = cache (1..Inf).hyper(:8degree).grep( !*.&schnitzel: 'panini' );
my \mung-daal = lazy gather for chowder -> \panini {
my \gazpacho = 2**panini - 1;
take gazpacho * 2**(panini - 1) unless schnitzel gazpacho, panini;
}
printf "chowla(%2d) = %2d\n", $_, .&schnitzel for 1..37;
say '';
printf "Count of primes up to %10s: %s\n", comma(10**$_),
comma chowder.first( * > 10**$_, :k) for 2..7;
say "\nPerfect numbers less than 35,000,000";
.&comma.say for mung-daal[^5];
- Output:
chowla( 1) = 0 chowla( 2) = 0 chowla( 3) = 0 chowla( 4) = 2 chowla( 5) = 0 chowla( 6) = 5 chowla( 7) = 0 chowla( 8) = 6 chowla( 9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100: 25 Count of primes up to 1,000: 168 Count of primes up to 10,000: 1,229 Count of primes up to 100,000: 9,592 Count of primes up to 1,000,000: 78,498 Count of primes up to 10,000,000: 664,579 Perfect numbers less than 35,000,000 6 28 496 8,128 33,550,336
REXX
Version 1
/*REXX program computes/displays chowla numbers (and may count primes & perfect numbers.*/
parse arg LO HI . /*obtain optional arguments from the CL*/
if LO=='' | LO=="," then LO= 1 /*Not specified? Then use the default.*/
perf= LO<0; LO= abs(LO) /*Negative? Then determine if perfect.*/
if HI=='' | HI=="," then HI= LO /*Not specified? Then use the default.*/
prim= HI<0; HI= abs(HI) /*Negative? Then determine if a prime.*/
numeric digits max(9, length(HI) + 1 ) /*use enough decimal digits for // */
w= length( commas(HI) ) /*W: used in aligning output numbers.*/
tell= \(prim | perf) /*set boolean value for showing chowlas*/
p= 0 /*the number of primes found (so far).*/
do j=LO to HI; #= chowla(j) /*compute the cholwa number for J. */
if tell then say right('chowla('commas(j)")", w+9) ' = ' right( commas(#), w)
else if #==0 then if j>1 then p= p+1
if perf then if j-1==# & j>1 then say right(commas(j), w) ' is a perfect number.'
end /*j*/
if prim & \perf then say 'number of primes found for the range ' commas(LO) " to " ,
commas(HI) " (inclusive) is: " commas(p)
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
chowla: procedure; parse arg x; if x<2 then return 0; odd= x // 2
s=0 /* [↓] use EVEN or ODD integers. ___*/
do k=2+odd by 1+odd while k*k<x /*divide by all the integers up to √ X */
if x//k==0 then s=s + k + x%k /*add the two divisors to the sum. */
end /*k*/ /* [↓] adkust for square. ___*/
if k*k==x then s=s + k /*Was X a square? If so, add √ X */
return s /*return " " " " " */
/*──────────────────────────────────────────────────────────────────────────────────────*/
commas: parse arg _; do k=length(_)-3 to 1 by -3; _= insert(',', _, k); end; return _
- output when using the input of: 1 37
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0
- output when using the input of: 1 -100
number of primes found for the range 1 to 100 (inclusive) is: 25
- output when using the input of: 1 -1000
number of primes found for the range 1 to 1,000 (inclusive) is: 168
- output when using the input of: 1 -10000
number of primes found for the range 1 to 10,000 (inclusive) is: 1,229
- output when using the input of: 1 -100000
number of primes found for the range 1 to 100,000 (inclusive) is: 9,592
- output when using the input of: 1 -1000000
number of primes found for the range 1 to 1,000,000 (inclusive) is: 78.498
- output when using the input of: 1 -10000000
number of primes found for the range 1 to 10,000,000 (inclusive) is: 664,579
- output when using the input of: 1 -100000000
number of primes found for the range 1 to 100,000,000 (inclusive) is: 5,761,455
- output when using the input of: -1 35000000
6 is a perfect number.
28 is a perfect number.
496 is a perfect number.
8,128 is a perfect number.
33,550,336 is a perfect number.
Version 2
Libraries: How to use
Library: Numbers
Library: Functions
Library: Settings
Library: Abend
Optimization 1: Use a sieve in procedure PrimeCount, excluding all multiples of primes in the rest of the process. A factor 5 faster, still very slow for n = 100e6.
Optimization 2: Use a well-known and proven property of perfect numbers in procedure PerfectNumbers (as in other entries). Still the 8th perfect number remains out of reach.
include Settings
say version; say 'Chowla numbers'; say
numeric digits 12
call ChowlaNumbers
call PrimeCount
call PerfectNumbers
exit
ChowlaNumbers:
procedure expose glob.
call Time('r'); sep = Copies('-',12)
say sep; say ' n Chowla(n)'; say sep
m = 37
do i = 1 to m
say Right(i,2) Right(Chowla(i),9)
end
say sep
say Format(Time('e'),,3) 'seconds'
say
return
PrimeCount:
procedure expose glob. work.
call Time('r'); sep = Copies('-',17)
say sep; say Right('n',8) Right('Primes<n',8); say sep
d = 1; work. = 1; m = 1e7; n = 0
do i = 1 by 2 to m+1
e = Xpon(i)
if e > d then do
say Right(10**e,8) Right(n,8)
d = e
end
if work.i = 0 then
iterate i
if Chowla(i) = 0 then do
n = n+1
if i > 1 then do
j = i+i
do while j < m
work.j = 0; j = j+i
end
end
end
end
say sep
say Format(Time('e'),,3) 'seconds'
say
return
PerfectNumbers:
procedure expose divi.
call Time('r'); sep = Copies('-',12)
say sep; say Right('Perfect',12); say sep
k = 2; kk = 3; m = 140e9; n = 0
do forever
p = k * kk
if p > 140e9 then
leave
if Chowla(p) = p-1 then do
n = n+1
say Right(p,12)
end
k = kk+1; kk = kk+k
end
say sep
say n 'perfect numbers found below 140e9'
say Format(Time('e'),,3) 'seconds'
say
return
Chowla:
/* Chowla numbers */
procedure expose divi.
arg x
/* Formula */
return Aliquot(x)-1
include Numbers
include Functions
include Abend
- Output:
REXX-Regina_3.9.6(MT) 5.00 29 Apr 2024
Chowla numbers
------------
n Chowla(n)
------------
1 0
2 0
3 0
4 2
5 0
6 5
7 0
8 6
9 3
10 7
11 0
12 15
13 0
14 9
15 8
16 14
17 0
18 20
19 0
20 21
21 10
22 13
23 0
24 35
25 5
26 15
27 12
28 27
29 0
30 41
31 0
32 30
33 14
34 19
35 12
36 54
37 0
------------
0.001 seconds
-----------------
n Primes<n
-----------------
100 25
1000 168
10000 1229
100000 9592
1000000 78498
10000000 664579
-----------------
511.718 seconds
------------
Perfect
------------
6
28
496
8128
33550336
8589869056
137438691328
------------
7 perfect numbers found below 140e9
1.234 seconds
Ruby
def chowla(n)
sum = 0
i = 2
while i * i <= n do
if n % i == 0 then
sum = sum + i
j = n / i
if i != j then
sum = sum + j
end
end
i = i + 1
end
return sum
end
def main
for n in 1 .. 37 do
puts "chowla(%d) = %d" % [n, chowla(n)]
end
count = 0
power = 100
for n in 2 .. 10000000 do
if chowla(n) == 0 then
count = count + 1
end
if n % power == 0 then
puts "There are %d primes < %d" % [count, power]
power = power * 10
end
end
count = 0
limit = 350000000
k = 2
kk = 3
loop do
p = k * kk
if p > limit then
break
end
if chowla(p) == p - 1 then
puts "%d is a perfect number" % [p]
count = count + 1
end
k = kk + 1
kk = kk + k
end
puts "There are %d perfect numbers < %d" % [count, limit]
end
main()
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 There are 25 primes < 100 There are 168 primes < 1000 There are 1229 primes < 10000 There are 9592 primes < 100000 There are 78498 primes < 1000000 There are 664579 primes < 10000000 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers < 350000000
Rust
fn chowla(n: usize) -> usize {
let mut sum = 0;
let mut i = 2;
while i * i <= n {
if n % i == 0 {
sum += i;
let j = n / i;
if i != j {
sum += j;
}
}
i += 1;
}
sum
}
fn sieve(limit: usize) -> Vec<bool> {
let mut c = vec![false; limit];
let mut i = 3;
while i * i < limit {
if !c[i] && chowla(i) == 0 {
let mut j = 3 * i;
while j < limit {
c[j] = true;
j += 2 * i;
}
}
i += 2;
}
c
}
fn main() {
for i in 1..=37 {
println!("chowla({}) = {}", i, chowla(i));
}
let mut count = 1;
let limit = 1e7 as usize;
let mut power = 100;
let c = sieve(limit);
for i in (3..limit).step_by(2) {
if !c[i] {
count += 1;
}
if i == power - 1 {
println!("Count of primes up to {} = {}", power, count);
power *= 10;
}
}
count = 0;
let limit = 35000000;
let mut k = 2;
let mut kk = 3;
loop {
let p = k * kk;
if p > limit {
break;
}
if chowla(p) == p - 1 {
println!("{} is a number that is perfect", p);
count += 1;
}
k = kk + 1;
kk += k;
}
println!("There are {} perfect numbers <= 35,000,000", count);
}
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1000 = 168 Count of primes up to 10000 = 1229 Count of primes up to 100000 = 9592 Count of primes up to 1000000 = 78498 Count of primes up to 10000000 = 664579 6 is a number that is perfect 28 is a number that is perfect 496 is a number that is perfect 8128 is a number that is perfect 33550336 is a number that is perfect There are 5 perfect numbers <= 35,000,000
Scala
This solution uses a lazily-evaluated iterator to find and sum the divisors of a number, and speeds up the large searches using parallel vectors.
object ChowlaNumbers {
def main(args: Array[String]): Unit = {
println("Chowla Numbers...")
for(n <- 1 to 37){println(s"$n: ${chowlaNum(n)}")}
println("\nPrime Counts...")
for(i <- (2 to 7).map(math.pow(10, _).toInt)){println(f"$i%,d: ${primesPar(i).size}%,d")}
println("\nPerfect Numbers...")
print(perfectsPar(35000000).toVector.sorted.zipWithIndex.map{case (n, i) => f"${i + 1}%,d: $n%,d"}.mkString("\n"))
}
def primesPar(num: Int): ParVector[Int] = ParVector.range(2, num + 1).filter(n => chowlaNum(n) == 0)
def perfectsPar(num: Int): ParVector[Int] = ParVector.range(6, num + 1).filter(n => chowlaNum(n) + 1 == n)
def chowlaNum(num: Int): Int = Iterator.range(2, math.sqrt(num).toInt + 1).filter(n => num%n == 0).foldLeft(0){case (s, n) => if(n*n == num) s + n else s + n + (num/n)}
}
- Output:
Chowla Numbers... 1: 0 2: 0 3: 0 4: 2 5: 0 6: 5 7: 0 8: 6 9: 3 10: 7 11: 0 12: 15 13: 0 14: 9 15: 8 16: 14 17: 0 18: 20 19: 0 20: 21 21: 10 22: 13 23: 0 24: 35 25: 5 26: 15 27: 12 28: 27 29: 0 30: 41 31: 0 32: 30 33: 14 34: 19 35: 12 36: 54 37: 0 Prime Counts... 100: 25 1,000: 168 10,000: 1,229 100,000: 9,592 1,000,000: 78,498 10,000,000: 664,579 Perfect Numbers... 1: 6 2: 28 3: 496 4: 8,128 5: 33,550,336
Swift
Uses Grand Central Dispatch to perform concurrent prime counting and perfect number searching
import Foundation
@inlinable
public func chowla<T: BinaryInteger>(n: T) -> T {
stride(from: 2, to: T(Double(n).squareRoot()+1), by: 1)
.lazy
.filter({ n % $0 == 0 })
.reduce(0, {(s: T, m: T) in
m*m == n ? s + m : s + m + (n / m)
})
}
extension Dictionary where Key == ClosedRange<Int> {
subscript(n: Int) -> Value {
get {
guard let key = keys.first(where: { $0.contains(n) }) else {
fatalError("dict does not contain range for \(n)")
}
return self[key]!
}
set {
guard let key = keys.first(where: { $0.contains(n) }) else {
fatalError("dict does not contain range for \(n)")
}
self[key] = newValue
}
}
}
let lock = DispatchSemaphore(value: 1)
var perfect = [Int]()
var primeCounts = [
1...100: 0,
101...1_000: 0,
1_001...10_000: 0,
10_001...100_000: 0,
100_001...1_000_000: 0,
1_000_001...10_000_000: 0
]
for i in 1...37 {
print("chowla(\(i)) = \(chowla(n: i))")
}
DispatchQueue.concurrentPerform(iterations: 35_000_000) {i in
let chowled = chowla(n: i)
if chowled == 0 && i > 1 && i < 10_000_000 {
lock.wait()
primeCounts[i] += 1
lock.signal()
}
if chowled == i - 1 && i > 1 {
lock.wait()
perfect.append(i)
lock.signal()
}
}
let numPrimes = primeCounts
.sorted(by: { $0.key.lowerBound < $1.key.lowerBound })
.reduce(into: [(Int, Int)](), {counts, oneCount in
guard !counts.isEmpty else {
counts.append((oneCount.key.upperBound, oneCount.value))
return
}
counts.append((oneCount.key.upperBound, counts.last!.1 + oneCount.value))
})
for (upper, count) in numPrimes {
print("Number of primes < \(upper) = \(count)")
}
for p in perfect {
print("\(p) is a perfect number")
}
- Output:
chowla(1) = 0 chowla(2) = 0 chowla(3) = 0 chowla(4) = 2 chowla(5) = 0 chowla(6) = 5 chowla(7) = 0 chowla(8) = 6 chowla(9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Number of primes < 100 = 25 Number of primes < 1000 = 168 Number of primes < 10000 = 1229 Number of primes < 100000 = 9592 Number of primes < 1000000 = 78498 Number of primes < 10000000 = 664579 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number
Visual Basic
Option Explicit
Private Declare Function AllocConsole Lib "kernel32.dll" () As Long
Private Declare Function FreeConsole Lib "kernel32.dll" () As Long
Dim mStdOut As Scripting.TextStream
Function chowla(ByVal n As Long) As Long
Dim j As Long, i As Long
i = 2
Do While i * i <= n
j = n \ i
If n Mod i = 0 Then
chowla = chowla + i
If i <> j Then
chowla = chowla + j
End If
End If
i = i + 1
Loop
End Function
Function sieve(ByVal limit As Long) As Boolean()
Dim c() As Boolean
Dim i As Long
Dim j As Long
i = 3
ReDim c(limit - 1)
Do While i * 3 < limit
If Not c(i) Then
If (chowla(i) = 0) Then
j = 3 * i
Do While j < limit
c(j) = True
j = j + 2 * i
Loop
End If
End If
i = i + 2
Loop
sieve = c()
End Function
Sub Display(ByVal s As String)
Debug.Print s
mStdOut.Write s & vbNewLine
End Sub
Sub Main()
Dim i As Long
Dim count As Long
Dim limit As Long
Dim power As Long
Dim c() As Boolean
Dim p As Long
Dim k As Long
Dim kk As Long
Dim s As String * 30
Dim mFSO As Scripting.FileSystemObject
Dim mStdIn As Scripting.TextStream
AllocConsole
Set mFSO = New Scripting.FileSystemObject
Set mStdIn = mFSO.GetStandardStream(StdIn)
Set mStdOut = mFSO.GetStandardStream(StdOut)
For i = 1 To 37
Display "chowla(" & i & ")=" & chowla(i)
Next i
count = 1
limit = 10000000
power = 100
c = sieve(limit)
For i = 3 To limit - 1 Step 2
If Not c(i) Then
count = count + 1
End If
If i = power - 1 Then
RSet s = FormatNumber(power, 0, vbUseDefault, vbUseDefault, True)
Display "Count of primes up to " & s & " = " & FormatNumber(count, 0, vbUseDefault, vbUseDefault, True)
power = power * 10
End If
Next i
count = 0: limit = 35000000
k = 2: kk = 3
Do
p = k * kk
If p > limit Then
Exit Do
End If
If chowla(p) = p - 1 Then
RSet s = FormatNumber(p, 0, vbUseDefault, vbUseDefault, True)
Display s & " is a number that is perfect"
count = count + 1
End If
k = kk + 1
kk = kk + k
Loop
Display "There are " & CStr(count) & " perfect numbers <= 35.000.000"
mStdOut.Write "press enter to quit program."
mStdIn.Read 1
FreeConsole
End Sub
- Output:
chowla(1)=0
chowla(2)=0
chowla(3)=0
chowla(4)=2
chowla(5)=0
chowla(6)=5
chowla(7)=0
chowla(8)=6
chowla(9)=3
chowla(10)=7
chowla(11)=0
chowla(12)=15
chowla(13)=0
chowla(14)=9
chowla(15)=8
chowla(16)=14
chowla(17)=0
chowla(18)=20
chowla(19)=0
chowla(20)=21
chowla(21)=10
chowla(22)=13
chowla(23)=0
chowla(24)=35
chowla(25)=5
chowla(26)=15
chowla(27)=12
chowla(28)=27
chowla(29)=0
chowla(30)=41
chowla(31)=0
chowla(32)=30
chowla(33)=14
chowla(34)=19
chowla(35)=12
chowla(36)=54
chowla(37)=0
Count of primes up to 100 = 25
Count of primes up to 1.000 = 168
Count of primes up to 10.000 = 1.229
Count of primes up to 100.000 = 9.592
Count of primes up to 1.000.000 = 78.498
Count of primes up to 10.000.000 = 664.579
6 is a number that is perfect
28 is a number that is perfect
496 is a number that is perfect
8.128 is a number that is perfect
33.550.336 is a number that is perfect
There are 5 perfect numbers <= 35.000.000
press enter to quit program.
Visual Basic .NET
Imports System
Module Program
Function chowla(ByVal n As Integer) As Integer
chowla = 0 : Dim j As Integer, i As Integer = 2
While i * i <= n
j = n / i : If n Mod i = 0 Then chowla += i + (If(i = j, 0, j))
i += 1
End While
End Function
Function sieve(ByVal limit As Integer) As Boolean()
Dim c As Boolean() = New Boolean(limit - 1) {}, i As Integer = 3
While i * 3 < limit
If Not c(i) AndAlso (chowla(i) = 0) Then
Dim j As Integer = 3 * i
While j < limit : c(j) = True : j += 2 * i : End While
End If : i += 2
End While
Return c
End Function
Sub Main(args As String())
For i As Integer = 1 To 37
Console.WriteLine("chowla({0}) = {1}", i, chowla(i))
Next
Dim count As Integer = 1, limit As Integer = CInt((10000000.0)), power As Integer = 100,
c As Boolean() = sieve(limit)
For i As Integer = 3 To limit - 1 Step 2
If Not c(i) Then count += 1
If i = power - 1 Then
Console.WriteLine("Count of primes up to {0,10:n0} = {1:n0}", power, count)
power = power * 10
End If
Next
count = 0 : limit = 35000000
Dim p As Integer, k As Integer = 2, kk As Integer = 3
While True
p = k * kk : If p > limit Then Exit While
If chowla(p) = p - 1 Then
Console.WriteLine("{0,10:n0} is a number that is perfect", p)
count += 1
End If
k = kk + 1 : kk += k
End While
Console.WriteLine("There are {0} perfect numbers <= 35,000,000", count)
If System.Diagnostics.Debugger.IsAttached Then Console.ReadKey()
End Sub
End Module
- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to 100 = 25
Count of primes up to 1,000 = 168
Count of primes up to 10,000 = 1,229
Count of primes up to 100,000 = 9,592
Count of primes up to 1,000,000 = 78,498
Count of primes up to 10,000,000 = 664,579
6 is a number that is perfect
28 is a number that is perfect
496 is a number that is perfect
8,128 is a number that is perfect
33,550,336 is a number that is perfect
There are 5 perfect numbers <= 35,000,000
More Cowbell
One can get a little further, but that 8th perfect number takes nearly a minute to verify. The 9th takes longer than I have patience. If you care to see the 9th and 10th perfect numbers, change the 31 to 61 or 89 where indicated by the comment.
Imports System.Numerics
Module Program
Function chowla(n As Integer) As Integer
chowla = 0 : Dim j As Integer, i As Integer = 2
While i * i <= n
If n Mod i = 0 Then j = n / i : chowla += i : If i <> j Then chowla += j
i += 1
End While
End Function
Function chowla1(ByRef n As BigInteger, x As Integer) As BigInteger
chowla1 = 1 : Dim j As BigInteger, lim As BigInteger = BigInteger.Pow(2, x - 1)
For i As BigInteger = 2 To lim
If n Mod i = 0 Then j = n / i : chowla1 += i : If i <> j Then chowla1 += j
Next
End Function
Function sieve(ByVal limit As Integer) As Boolean()
Dim c As Boolean() = New Boolean(limit - 1) {}, i As Integer = 3
While i * 3 < limit
If Not c(i) AndAlso (chowla(i) = 0) Then
Dim j As Integer = 3 * i
While j < limit : c(j) = True : j += 2 * i : End While
End If : i += 2
End While
Return c
End Function
Sub Main(args As String())
For i As Integer = 1 To 37
Console.WriteLine("chowla({0}) = {1}", i, chowla(i))
Next
Dim count As Integer = 1, limit As Integer = CInt((10000000.0)), power As Integer = 100,
c As Boolean() = sieve(limit)
For i As Integer = 3 To limit - 1 Step 2
If Not c(i) Then count += 1
If i = power - 1 Then
Console.WriteLine("Count of primes up to {0,10:n0} = {1:n0}", power, count)
power = power * 10
End If
Next
count = 0
Dim p As BigInteger, k As BigInteger = 2, kk As BigInteger = 3
For i As Integer = 2 To 31 ' if you dare, change the 31 to 61 or 89
If {2, 3, 5, 7, 13, 17, 19, 31, 61, 89}.Contains(i) Then
p = k * kk
If chowla1(p, i) = p Then
Console.WriteLine("{0,25:n0} is a number that is perfect", p)
st = DateTime.Now
count += 1
End If
End If
k = kk + 1 : kk += k
Next
Console.WriteLine("There are {0} perfect numbers <= {1:n0}", count, 25 * BigInteger.Pow(10, 18))
If System.Diagnostics.Debugger.IsAttached Then Console.ReadKey()
End Sub
End Module
- Output:
chowla(1) = 0
chowla(2) = 0
chowla(3) = 0
chowla(4) = 2
chowla(5) = 0
chowla(6) = 5
chowla(7) = 0
chowla(8) = 6
chowla(9) = 3
chowla(10) = 7
chowla(11) = 0
chowla(12) = 15
chowla(13) = 0
chowla(14) = 9
chowla(15) = 8
chowla(16) = 14
chowla(17) = 0
chowla(18) = 20
chowla(19) = 0
chowla(20) = 21
chowla(21) = 10
chowla(22) = 13
chowla(23) = 0
chowla(24) = 35
chowla(25) = 5
chowla(26) = 15
chowla(27) = 12
chowla(28) = 27
chowla(29) = 0
chowla(30) = 41
chowla(31) = 0
chowla(32) = 30
chowla(33) = 14
chowla(34) = 19
chowla(35) = 12
chowla(36) = 54
chowla(37) = 0
Count of primes up to 100 = 25
Count of primes up to 1,000 = 168
Count of primes up to 10,000 = 1,229
Count of primes up to 100,000 = 9,592
Count of primes up to 1,000,000 = 78,498
Count of primes up to 10,000,000 = 664,579
6 is a number that is perfect
28 is a number that is perfect
496 is a number that is perfect
8,128 is a number that is perfect
33,550,336 is a number that is perfect
8,589,869,056 is a number that is perfect
137,438,691,328 is a number that is perfect
2,305,843,008,139,952,128 is a number that is perfect
There are 8 perfect numbers <= 25,000,000,000,000,000,000
V (Vlang)
fn chowla(n int) int {
if n < 1 {
panic("argument must be a positive integer")
}
mut sum := 0
for i := 2; i*i <= n; i++ {
if n%i == 0 {
j := n / i
if i == j {
sum += i
} else {
sum += i + j
}
}
}
return sum
}
fn sieve(limit int) []bool {
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
mut c := []bool{len: limit}
for i := 3; i*3 < limit; i += 2 {
if !c[i] && chowla(i) == 0 {
for j := 3 * i; j < limit; j += 2 * i {
c[j] = true
}
}
}
return c
}
fn main() {
for i := 1; i <= 37; i++ {
println("chowla(${i:2}) = ${chowla(i)}")
}
println('')
mut count := 1
mut limit := int(1e7)
c := sieve(limit)
mut power := 100
for i := 3; i < limit; i += 2 {
if !c[i] {
count++
}
if i == power-1 {
println("Count of primes up to ${power:-10} = $count")
power *= 10
}
}
println('')
count = 0
limit = 35000000
for i := 2; ; i++ {
p := (1 << (i -1)) * ((1<<i) - 1)
if p > limit {
break
}
if chowla(p) == p-1 {
println("$p is a perfect number")
count++
}
}
println("There are $count perfect numbers <= 35,000,000")
}- Output:
chowla( 1) = 0 chowla( 2) = 0 chowla( 3) = 0 chowla( 4) = 2 chowla( 5) = 0 chowla( 6) = 5 chowla( 7) = 0 chowla( 8) = 6 chowla( 9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1,000 = 168 Count of primes up to 10,000 = 1,229 Count of primes up to 100,000 = 9,592 Count of primes up to 1,000,000 = 78,498 Count of primes up to 10,000,000 = 664,579 6 is a perfect number 28 is a perfect number 496 is a perfect number 8128 is a perfect number 33550336 is a perfect number There are 5 perfect numbers <= 35,000,000
Wren
import "./fmt" for Fmt
import "./math" for Int, Nums
var chowla = Fn.new { |n| (n > 1) ? Nums.sum(Int.properDivisors(n)) - 1 : 0 }
for (i in 1..37) Fmt.print("chowla($2d) = $d", i, chowla.call(i))
System.print()
var count = 1
var limit = 1e7
var c = Int.primeSieve(limit, false)
var power = 100
var i = 3
while (i < limit) {
if (!c[i]) count = count + 1
if (i == power - 1) {
Fmt.print("Count of primes up to $,-10d = $,d", power, count)
power = power * 10
}
i = i + 2
}
System.print()
count = 0
limit = 35 * 1e6
i = 2
while (true) {
var p = (1 << (i -1)) * ((1<<i) - 1) // perfect numbers must be of this form
if (p > limit) break
if (chowla.call(p) == p-1) {
Fmt.print("$,d is a perfect number", p)
count = count + 1
}
i = i + 1
}
System.print("There are %(count) perfect numbers <= 35,000,000")
- Output:
chowla( 1) = 0 chowla( 2) = 0 chowla( 3) = 0 chowla( 4) = 2 chowla( 5) = 0 chowla( 6) = 5 chowla( 7) = 0 chowla( 8) = 6 chowla( 9) = 3 chowla(10) = 7 chowla(11) = 0 chowla(12) = 15 chowla(13) = 0 chowla(14) = 9 chowla(15) = 8 chowla(16) = 14 chowla(17) = 0 chowla(18) = 20 chowla(19) = 0 chowla(20) = 21 chowla(21) = 10 chowla(22) = 13 chowla(23) = 0 chowla(24) = 35 chowla(25) = 5 chowla(26) = 15 chowla(27) = 12 chowla(28) = 27 chowla(29) = 0 chowla(30) = 41 chowla(31) = 0 chowla(32) = 30 chowla(33) = 14 chowla(34) = 19 chowla(35) = 12 chowla(36) = 54 chowla(37) = 0 Count of primes up to 100 = 25 Count of primes up to 1,000 = 168 Count of primes up to 10,000 = 1,229 Count of primes up to 100,000 = 9,592 Count of primes up to 1,000,000 = 78,498 Count of primes up to 10,000,000 = 664,579 6 is a perfect number 28 is a perfect number 496 is a perfect number 8,128 is a perfect number 33,550,336 is a perfect number There are 5 perfect numbers <= 35,000,000
XPL0
func Chowla(N); \Return sum of divisors
int N, Div, Sum, Quot;
[Div:= 2; Sum:= 0;
loop [Quot:= N/Div;
if Quot < Div then quit;
if Quot = Div and rem(0) = 0 then \N is a square
[Sum:= Sum+Quot; quit];
if rem(0) = 0 then
Sum:= Sum + Div + Quot;
Div:= Div+1;
];
return Sum;
];
int N, C, P;
[for N:= 1 to 37 do
[IntOut(0, N); Text(0, ": ");
IntOut(0, Chowla(N)); CrLf(0);
];
C:= 1; \count 2 as prime
N:= 3; \only check odd numbers
repeat if Chowla(N) = 0 then \N is prime
C:= C+1;
case N+1 of 100, 1000, 10_000, 100_000, 1_000_000, 10_000_000:
[Text(0, "There are "); IntOut(0, C); Text(0, " primes < ");
IntOut(0, N+1); CrLf(0)]
other [];
N:= N+2;
until N >= 10_000_000;
P:= 1; \perfect numbers are of form: 2^(P-1) * (2^P - 1)
loop [P:= P*2;
N:= P*(P*2-1);
if N > 35_000_000 then quit;
if Chowla(N) = N-1 then \N is perfect
[IntOut(0, N); CrLf(0)];
];
]- Output:
1: 0 2: 0 3: 0 4: 2 5: 0 6: 5 7: 0 8: 6 9: 3 10: 7 11: 0 12: 15 13: 0 14: 9 15: 8 16: 14 17: 0 18: 20 19: 0 20: 21 21: 10 22: 13 23: 0 24: 35 25: 5 26: 15 27: 12 28: 27 29: 0 30: 41 31: 0 32: 30 33: 14 34: 19 35: 12 36: 54 37: 0 There are 25 primes < 100 There are 168 primes < 1000 There are 1229 primes < 10000 There are 9592 primes < 100000 There are 78498 primes < 1000000 There are 664579 primes < 10000000 6 28 496 8128 33550336
zkl
fcn chowla(n){
if(n<1)
throw(Exception.ValueError("Chowla function argument must be positive"));
sum:=0;
foreach i in ([2..n.toFloat().sqrt()]){
if(n%i == 0){
j:=n/i;
if(i==j) sum+=i;
else sum+=i+j;
}
}
sum
}
fcn chowlaSieve(limit){
// True denotes composite, false denotes prime.
// Only interested in odd numbers >= 3
c:=Data(limit+100).fill(0); # slop at the end (for reverse wrap around)
foreach i in ([3..limit/3,2]){
if(not c[i] and chowla(i)==0)
{ foreach j in ([3*i..limit,2*i]){ c[j]=True } }
}
c
}fcn testChowla{
println("The first 37 Chowla numbers:\n",
[1..37].apply(chowla).concat(" ","[","]"), "\n");
count,limit,power := 1, (1e7).toInt(), 100;
c:=chowlaSieve(limit);
foreach i in ([3..limit-1,2]){
if(not c[i]) count+=1;
if(i == power - 1){
println("The count of the primes up to %10,d is %8,d".fmt(power,count));
power*=10;
}
}
println();
count, limit = 0, 35_000_000;
foreach i in ([2..]){
p:=(1).shiftLeft(i - 1) * ((1).shiftLeft(i)-1); // perfect numbers must be of this form
if(p>limit) break;
if(p-1 == chowla(p)){
println("%,d is a perfect number".fmt(p));
count+=1;
}
}
println("There are %,d perfect numbers <= %,d".fmt(count,limit));
}();- Output:
The first 37 Chowla numbers: [0 0 0 2 0 5 0 6 3 7 0 15 0 9 8 14 0 20 0 21 10 13 0 35 5 15 12 27 0 41 0 30 14 19 12 54 0] The count of the primes up to 100 is 25 The count of the primes up to 1,000 is 168 The count of the primes up to 10,000 is 1,229 The count of the primes up to 100,000 is 9,592 The count of the primes up to 1,000,000 is 78,498 The count of the primes up to 10,000,000 is 664,579 6 is a perfect number 28 is a perfect number 496 is a perfect number 8,128 is a perfect number 33,550,336 is a perfect number There are 5 perfect numbers <= 35,000,000
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