# Erdős-Nicolas numbers

**Erdős-Nicolas numbers**

You are encouraged to solve this task according to the task description, using any language you may know.

- Definition

An **Erdős–Nicolas number** is a positive integer which is not perfect but is equal to the sum of its first **k** divisors (arranged in ascending order and including one) for some value of **k** greater than one.

- Examples

24 is an Erdős–Nicolas number because the sum of its first 6 divisors (1, 2, 3, 4, 6 and 8) is equal to 24 and it is not perfect because 12 is also a divisor.

6 is not an Erdős–Nicolas number because it is perfect (1 + 2 + 3 = 6).

48 is not an Erdős–Nicolas number because its divisors are: 1, 2, 3, 4, 6, 8, 12, 16, 24 and 48. The first seven of these add up to 36, but the first eight add up to 52 which is more than 48.

- Task

Find and show here the first 8 Erdős–Nicolas numbers and the number of divisors needed (i.e. the value of 'k') to satisfy the definition.

- Stretch

Do the same for any further Erdős–Nicolas numbers which you have the patience for.

- Note

As all known Erdős–Nicolas numbers are even you may assume this to be generally true in order to quicken up the search. However, it is not obvious (to me at least) why this should necessarily be the case.

- Reference

## ALGOL 68

Builds tables of proper divisor counts and sums and finds the numbers whilst doing it. This means that the numbers are not found in numerical order.

```
BEGIN # find some Erdos-Nicolas numbers: numbers equal to the sum of their #
# first k proper divisors but k is not the count of all their proper #
# divisors ( so the numbers aren't perfect ) #
INT max number = 2 000 000; # largest number we will consider #
# construct tables of the divisor counts and divisor sums and check for #
# the numbers as we do it - note they will not necessarily be found in #
# order #
[ 1 : max number ]INT dsum; FOR i TO UPB dsum DO dsum[ i ] := 1 OD;
[ 1 : max number ]INT dcount; FOR i TO UPB dcount DO dcount[ i ] := 1 OD;
FOR i FROM 2 TO UPB dsum
DO FOR j FROM i + i BY i TO UPB dsum DO
# have another proper divisor #
IF dsum[ j ] = j THEN
# the divisor sum is currently equal to the number but is #
# about to increase, so we have an Erdos-Nicolas number #
print( ( whole( j, -10 ), " equals the sum of its first "
, whole( dcount[ j ], 0 ), " divisors"
, newline
)
)
FI;
dsum[ j ] +:= i;
dcount[ j ] +:= 1
OD
OD
END
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors

With max number set to 200 000 000, another two numbers can be found. Note that that would probably be too large for ALGOL 68G under Windows, so another compiler would need to be used:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors 61900800 equals the sum of its first 280 divisors 91963648 equals the sum of its first 142 divisors

## Arturo

```
erdosNicolas: function [n][
facts: factors n
if facts > 2 [
loop 1..(size facts)-2 'k [
if n = sum first.n:k facts -> return @[n, k]
]
]
return ø
]
cnt: 0
i: 2
while [cnt < 8][
if enNum: <= erdosNicolas i [
print[enNum\0 "equals the sum of its first" enNum\1 "divisors"]
cnt: cnt + 1
]
i: i + 2
]
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 392448 equals the sum of its first 68 divisors 714240 equals the sum of its first 113 divisors 1571328 equals the sum of its first 115 divisors

## BASIC

Rosetta Code problem: https://rosettacode.org/wiki/Erdős-Nicolas_numbers

by Jjuanhdez, 02/2023

### BASIC256

```
limite = 400000
dim DSum(limite+1) fill 1
dim DCount(limite+1) fill 1
for i = 2 to limite
j = i + i
while j <= limite
if DSum[j] = j then
print rjust(j,8); " equals the sum of its first "; rjust(DCount[j],3); " divisors"
end if
DSum[j] += i
DCount[j] += 1
j += i
end while
next i
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors

### FreeBASIC

```
Dim As Uinteger limite = 2e6
Dim As Uinteger DSum(limite+1), DCount(limite+1)
Dim As Integer i, j
For i = 0 To limite
DSum(i) = 1
DCount(i) = 1
Next i
For i = 2 To limite
j = i + i
While j <= limite
If DSum(j) = j Then
Print Using "######## equals the sum of its first ### divisors"; j; DCount(j)
End If
DSum(j) += i
DCount(j) += 1
j += i
Wend
Next i
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors

### QBasic

```
limite = 50000
DIM DSum(limite + 1), DCount(limite + 1)
FOR i = 0 TO limite
DSum(i) = 1
DCount(i) = 1
NEXT i
FOR i = 2 TO limite
j = i + i
WHILE j <= limite
IF DSum(j) = j THEN
PRINT USING "######## equals the sum of its first ### divisors"; j; DCount(j)
END IF
DSum(j) = DSum(j) + i
DCount(j) = DCount(j) + 1
j = j + i
WEND
NEXT i
```

### Run BASIC

```
limite = 1000000-1
'Maximum array size is 1,000,001 elements
dim DSum(limite+1)
dim DCount(limite+1)
for i = 0 to limite
DSum(i) = 1
DCount(i) = 1
next i
for i = 2 to limite
j = i + i
while j <= limite
if DSum(j) = j then
print using("########", j); " equals the sum of its first"; using("###", DCount(j)); " divisors"
end if
DSum(j) = DSum(j) + i
DCount(j) = DCount(j) + 1
j = j + i
wend
next i
```

### PureBasic

```
OpenConsole()
limite.l = 2e6
Dim DSum.l(limite+1)
Dim DCount.l(limite+1)
For i.l = 0 To limite
DSum(i) = 1
DCount(i) = 1
Next i
For i = 2 To limite
j.l = i + i
While j <= limite
If DSum(j) = j
PrintN(RSet(Str(j), 8) + " equals the sum of its first " + RSet(Str(DCount(j)), 3) + " divisors")
EndIf
DSum(j) = DSum(j) + i
DCount(j) = DCount(j) + 1
j = j + i
Wend
Next i
PrintN(#CRLF$ + "--- terminado, pulsa RETURN---"): Input()
CloseConsole()
```

- Output:

Same as FreeBASIC entry.

### Yabasic

```
limite = 2e6
dim DSum(limite+1), DCount(limite+1)
for i = 0 to limite
DSum(i) = 1
DCount(i) = 1
next i
for i = 2 to limite
j = i + i
while j <= limite
if DSum(j) = j print j using ("########"), " equals the sum of its first ", DCount(j) using ("###"), " divisors"
DSum(j) = DSum(j) + i
DCount(j) = DCount(j) + 1
j = j + i
wend
next i
```

- Output:

Same as FreeBASIC entry.

## C

Run time about 46 seconds.

```
#include <stdio.h>
#include <stdlib.h>
int main() {
const int maxNumber = 100000000;
int *dsum = (int *)malloc((maxNumber + 1) * sizeof(int));
int *dcount = (int *)malloc((maxNumber + 1) * sizeof(int));
int i, j;
for (i = 0; i <= maxNumber; ++i) {
dsum[i] = 1;
dcount[i] = 1;
}
for (i = 2; i <= maxNumber; ++i) {
for (j = i + i; j <= maxNumber; j += i) {
if (dsum[j] == j) {
printf("%8d equals the sum of its first %d divisors\n", j, dcount[j]);
}
dsum[j] += i;
++dcount[j];
}
}
free(dsum);
free(dcount);
return 0;
}
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors 61900800 equals the sum of its first 280 divisors 91963648 equals the sum of its first 142 divisors

### slight improvement

dsum and dcount are in "far" distance -> cache miss. Using an struct "typedef struct { int divsum, divcnt;} divs;" improves runtime to 32 s. Calculating the count of divisors only at output saves 50% space and runtime too.

```
#include <stdio.h>
#include <stdlib.h>
void get_div_cnt(int n){
int lmt,f,divcnt,divsum;
divsum = 1;
divcnt = 1;
lmt = n/2;
f = 2;
for (;;) {
if (f > lmt ) break;
if (!(n % f)){
divsum +=f;
divcnt++;
}
if (divsum == n) break;
f++;
}
printf("%8d equals the sum of its first %d divisors\n", n, divcnt);
}
int main() {
const int maxNumber = 100*1000*1000;
int *dsum = (int *)malloc((maxNumber + 1) * sizeof(int));
int i, j;
for (i = 0; i <= maxNumber; ++i) {
dsum[i] = 1;
}
for (i = 2; i <= maxNumber; ++i) {
for (j = i + i; j <= maxNumber; j += i) {
if (dsum[j] == j) get_div_cnt(j);
dsum[j] += i;
}
}
free(dsum);
return 0;
}
```

- TIO.RUN:

the same. compiler flags -march=native -O2 Real time: 23.893 s User time: 23.475 s Sys. time: 0.203 s CPU share: 99.10 % //Sys. time: up to 8s for version before??? //with lmt 2,000,000 the results on TIO.RUN are more stable. version before User time: 0.304 s CPU share: 98.93 % slight improvement version User time: 0.139 s CPU share: 98.80 %

## C#

```
using System;
class ErdosNicolasNumbers
{
static void Main(string[] args)
{
const int limit = 100_000_000;
int[] divisorSum = new int[limit + 1];
int[] divisorCount = new int[limit + 1];
for (int i = 0; i <= limit; i++)
{
divisorSum[i] = 1;
divisorCount[i] = 1;
}
for (int index = 2; index <= limit / 2; index++)
{
for (int number = 2 * index; number <= limit; number += index)
{
if (divisorSum[number] == number)
{
Console.WriteLine($"{number,8} equals the sum of its first {divisorCount[number],3} divisors");
}
divisorSum[number] += index;
divisorCount[number]++;
}
}
}
}
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors 61900800 equals the sum of its first 280 divisors 91963648 equals the sum of its first 142 divisors

## C++

```
#include <iomanip>
#include <iostream>
#include <vector>
int main() {
const int max_number = 100000000;
std::vector<int> dsum(max_number + 1, 1);
std::vector<int> dcount(max_number + 1, 1);
for (int i = 2; i <= max_number; ++i) {
for (int j = i + i; j <= max_number; j += i) {
if (dsum[j] == j) {
std::cout << std::setw(8) << j
<< " equals the sum of its first " << dcount[j]
<< " divisors\n";
}
dsum[j] += i;
++dcount[j];
}
}
}
```

- Output:

## Delphi

Translation from the C version. Runs in 38 seconds

```
const MaxNumber = 100000000;
var DSum: array [0..MaxNumber-1] of integer;
var DCount: array [0..MaxNumber-1] of integer;
procedure ShowErdosNicolasNumbers(Memo: TMemo);
var I,J: integer;
begin
for I:=0 to MaxNumber-1 do
begin
DSum[I]:=1;
DCount[I]:=1;
end;
for I:=2 to MaxNumber-1 do
begin
J:=I*2;
while J<MaxNumber do
begin
if dsum[J] = j then
begin
Memo.Lines.Add(Format('%8d equals the sum of its first %d divisors', [j, dcount[j]]));
end;
Inc(dsum[J],I);
Inc(DCount[J]);
Inc(J,I);
end;
end;
end;
```

- Output:

## EasyLang

```
limit = 2000000
for i to limit
dsum[] &= 1
dcnt[] &= 1
.
for i = 2 to limit
j = i + i
while j <= limit
if dsum[j] = j
print j & " equals the sum of its first " & dcnt[j] & " divisors"
.
dsum[j] += i
dcnt[j] += 1
j += i
.
.
```

## FutureBasic

```
_limit = 500000
void local fn ErdosNicolasNumbers
long i, j, sum( _limit ), count( _limit )
for i = 0 to _limit
sum(i) = 1
count(i) = 1
next
for i = 2 to _limit
j = i + i
while ( j <= _limit )
if sum(j) == j then printf @"%8ld == sum of its first %3ld divisors", j, count(j)
sum(j) = sum(j) + i
count(j) = count(j) + 1
j = j + i
wend
next
end fn
fn ErdosNicolasNumbers
HandleEvents
```

- Output:

24 == sum of its first 6 divisors 2016 == sum of its first 31 divisors 8190 == sum of its first 43 divisors 42336 == sum of its first 66 divisors 45864 == sum of its first 66 divisors 392448 == sum of its first 68 divisors

## Go

This runs in about 30 seconds which, surprisingly, is faster than C++ (43 seconds) - usually Go is a good bit slower.

```
package main
import "fmt"
func main() {
const maxNumber = 100000000
dsum := make([]int, maxNumber+1)
dcount := make([]int, maxNumber+1)
for i := 0; i <= maxNumber; i++ {
dsum[i] = 1
dcount[i] = 1
}
for i := 2; i <= maxNumber; i++ {
for j := i + i; j <= maxNumber; j += i {
if dsum[j] == j {
fmt.Printf("%8d equals the sum of its first %d divisors\n", j, dcount[j])
}
dsum[j] += i
dcount[j]++
}
}
}
```

- Output:

## J

Implementation:

```
divisors=: {{ /:~ ,*/@> { (^ i.@>:)&.>/__ q: y}} ::_:
erdosnicolas=: {{ y e. +/\ _2}. divisors y }}"0
```

Task example:

```
I.erdosnicolas i.1e7
24 2016 8190 42336 45864 392448 714240 1571328
(,. 1++/\@divisors i. ])@>24 2016 8190 42336 45864 392448 714240 1571328
24 6
2016 31
8190 43
42336 66
45864 66
392448 68
714240 113
1571328 115
```

## Java

```
import java.util.Arrays;
public final class ErdosNicolasNumbers {
public static void main(String[] aArgs) {
final int limit = 100_000_000;
int[] divisorSum = new int[limit + 1];
int[] divisorCount = new int[limit + 1];
Arrays.fill(divisorSum, 1);
Arrays.fill(divisorCount, 1);
for ( int index = 2; index <= limit / 2; index++ ) {
for ( int number = 2 * index; number <= limit; number += index ) {
if ( divisorSum[number] == number ) {
System.out.println(String.format("%8d", number) + " equals the sum of its first "
+ String.format("%3d", divisorCount[number]) + " divisors");
}
divisorSum[number] += index;
divisorCount[number]++;
}
}
}
}
```

- Output:

## jq

**Adapted from #Wren**

**Works with jq and gojq, the C and Go implementations of jq**

The following program will also work using jaq provided `sqrt` is defined appropriately and other minor adjustments are made.

```
# Output a stream of the (unsorted) proper divisors of . including 1
def proper_divisors:
. as $n
| if $n > 1 then 1,
( range(2; 1 + sqrt) as $i
| if ($n % $i) == 0 then $i,
(($n / $i) | if . == $i then empty else . end)
else empty
end)
else empty
end;
# Emit k if . is an Erdos-Nicolas number, otherwise emit 0
def erdosNicolas:
. as $n
| ([proper_divisors] | sort) as $divisors
| ($divisors|length) as $dc
| if $dc < 3 then 0
else {sum: ($divisors[0] + $divisors[1])}
# An Erdos-Nicolas is not perfect, and hence $dc-1 in the following line:
| first(
foreach range(2; $dc-1) as $i (.;
.sum += $divisors[$i]
| if .sum == $n then .emit = $i + 1
elif .sum > $n then .emit = 0
else .
end )
| select(.emit).emit ) // 0
end ;
limit(8;
range(2; infinite)
| . as $n
| erdosNicolas as $k
| select($k > 0)
| "\($n) from \($k)" )
```

- Output:

24 from 6 2016 from 31 8190 from 43 42336 from 66 45864 from 66 392448 from 68 714240 from 113 1571328 from 115

## Julia

```
using Primes
function isErdősNicolas_with_k(n)
@assert n > 2
d = [one(n)]
for (p, e) in eachfactor(n)
d = reduce(vcat, [d * p^j for j in 1:e], init=d)
end
sort!(d)
pop!(d)
len = length(d)
(len < 2 || sum(d) <= n) && return false, 0
for k in 2:len
sum(@view d[1:k]) == n && return true, k
end
return false, 0
end
for n in 3:2_000_000
isEN, k = isErdősNicolas_with_k(n)
isEN && println(lpad(n, 8), " equals the sum of its first $k divisors.")
end
```

- Output:

24 equals the sum of its first 6 divisors. 2016 equals the sum of its first 31 divisors. 8190 equals the sum of its first 43 divisors. 42336 equals the sum of its first 66 divisors. 45864 equals the sum of its first 66 divisors. 392448 equals the sum of its first 68 divisors. 714240 equals the sum of its first 113 divisors. 1571328 equals the sum of its first 115 divisors.

## Nim

To get better performances, we use 32 bits integers rather than 64 bits integers. We got the best (and excellent) execution time with compilation command: `nim c -d:release -d:lto --gc:boehm erdos_nicolas_numbers.nim`

```
import std/[sequtils, strformat]
proc main() =
const MaxNumber = 100_000_000i32
var dsum, dcount = repeat(1'i32, MaxNumber + 1)
for i in 2i32..MaxNumber:
for j in countup(i + i, MaxNumber, i):
if dsum[j] == j:
echo &"{j:>8} equals the sum of its first {dcount[j]} divisors"
inc dsum[j], i
inc dcount[j]
main()
```

- Output:

## Pascal

### Free Pascal

using Factors_of_an_integer#using_Prime_decomposition , using a sort of sieve.

```
program ErdoesNumb;
// gets factors of consecutive integers fast
// limited to 1.2e11
{$IFDEF FPC}
{$MODE DELPHI} {$OPTIMIZATION ON,ALL} {$COPERATORS ON}
{$ELSE}
{$APPTYPE CONSOLE}
{$ENDIF}
uses
sysutils
{$IFDEF WINDOWS},Windows{$ENDIF}
;
//######################################################################
//prime decomposition
const
//HCN(86) > 1.2E11 = 128,501,493,120 count of divs = 4096 7 3 1 1 1 1 1 1 1
HCN_DivCnt = 4096;
type
tItem = Uint64;
tDivisors = array [0..HCN_DivCnt] of tItem;
tpDivisor = pUint64;
const
SizePrDeFe = 32768;//*56 <= 64kb level I or 2 Mb ~ level 2 cache or more
type
tdigits = array [0..31] of Uint32;
//the first number with 11 different prime factors =
//2*3*5*7*11*13*17*19*23*29*31 = 2E11
//56 byte
tprimeFac = packed record
pfSumOfDivs,
pfRemain : Uint64;
pfDivCnt : Uint32;
pfMaxIdx : Uint32;
pfpotMax : array[0..11] of byte;
pfpotPrimIdx : array[0..9] of word;
end;
tpPrimeFac = ^tprimeFac;
tPrimeDecompField = array[0..SizePrDeFe-1] of tprimeFac;
tPrimes = array[0..65535] of Uint32;
var
{$ALIGN 8}
SmallPrimes: tPrimes;
{$ALIGN 32}
PrimeDecompField :tPrimeDecompField;
pdfIDX,pdfOfs: NativeInt;
procedure InitSmallPrimes;
//get primes. #0..65535.Sieving only odd numbers
const
MAXLIMIT = (821641-1) shr 1;
var
pr : array[0..MAXLIMIT] of byte;
p,j,d,flipflop :NativeUInt;
Begin
SmallPrimes[0] := 2;
fillchar(pr[0],SizeOf(pr),#0);
p := 0;
repeat
repeat
p +=1
until pr[p]= 0;
j := (p+1)*p*2;
if j>MAXLIMIT then
BREAK;
d := 2*p+1;
repeat
pr[j] := 1;
j += d;
until j>MAXLIMIT;
until false;
SmallPrimes[1] := 3;
SmallPrimes[2] := 5;
j := 3;
d := 7;
flipflop := (2+1)-1;//7+2*2,11+2*1,13,17,19,23
p := 3;
repeat
if pr[p] = 0 then
begin
SmallPrimes[j] := d;
inc(j);
end;
d += 2*flipflop;
p+=flipflop;
flipflop := 3-flipflop;
until (p > MAXLIMIT) OR (j>High(SmallPrimes));
end;
function CnvtoBASE(var dgt:tDigits;n:Uint64;base:NativeUint):NativeInt;
//n must be multiple of base aka n mod base must be 0
var
q,r: Uint64;
i : NativeInt;
Begin
fillchar(dgt,SizeOf(dgt),#0);
i := 0;
n := n div base;
result := 0;
repeat
r := n;
q := n div base;
r -= q*base;
n := q;
dgt[i] := r;
inc(i);
until (q = 0);
//searching lowest pot in base
result := 0;
while (result<i) AND (dgt[result] = 0) do
inc(result);
inc(result);
end;
function IncByBaseInBase(var dgt:tDigits;base:NativeInt):NativeInt;
var
q :NativeInt;
Begin
result := 0;
q := dgt[result]+1;
if q = base then
repeat
dgt[result] := 0;
inc(result);
q := dgt[result]+1;
until q <> base;
dgt[result] := q;
result +=1;
end;
function SieveOneSieve(var pdf:tPrimeDecompField):boolean;
var
dgt:tDigits;
i,j,k,pr,fac,n,MaxP : Uint64;
begin
n := pdfOfs;
if n+SizePrDeFe >= sqr(SmallPrimes[High(SmallPrimes)]) then
EXIT(FALSE);
//init
for i := 0 to SizePrDeFe-1 do
begin
with pdf[i] do
Begin
pfDivCnt := 1;
pfSumOfDivs := 1;
pfRemain := n+i;
pfMaxIdx := 0;
pfpotPrimIdx[0] := 0;
pfpotMax[0] := 0;
end;
end;
//first factor 2. Make n+i even
i := (pdfIdx+n) AND 1;
IF (n = 0) AND (pdfIdx<2) then
i := 2;
repeat
with pdf[i] do
begin
j := BsfQWord(n+i);
pfMaxIdx := 1;
pfpotPrimIdx[0] := 0;
pfpotMax[0] := j;
pfRemain := (n+i) shr j;
pfSumOfDivs := (Uint64(1) shl (j+1))-1;
pfDivCnt := j+1;
end;
i += 2;
until i >=SizePrDeFe;
//i now index in SmallPrimes
i := 0;
maxP := trunc(sqrt(n+SizePrDeFe))+1;
repeat
//search next prime that is in bounds of sieve
if n = 0 then
begin
repeat
inc(i);
pr := SmallPrimes[i];
k := pr-n MOD pr;
if k < SizePrDeFe then
break;
until pr > MaxP;
end
else
begin
repeat
inc(i);
pr := SmallPrimes[i];
k := pr-n MOD pr;
if (k = pr) AND (n>0) then
k:= 0;
if k < SizePrDeFe then
break;
until pr > MaxP;
end;
//no need to use higher primes
if pr*pr > n+SizePrDeFe then
BREAK;
//j is power of prime
j := CnvtoBASE(dgt,n+k,pr);
repeat
with pdf[k] do
Begin
pfpotPrimIdx[pfMaxIdx] := i;
pfpotMax[pfMaxIdx] := j;
pfDivCnt *= j+1;
fac := pr;
repeat
pfRemain := pfRemain DIV pr;
dec(j);
fac *= pr;
until j<= 0;
pfSumOfDivs *= (fac-1)DIV(pr-1);
inc(pfMaxIdx);
k += pr;
j := IncByBaseInBase(dgt,pr);
end;
until k >= SizePrDeFe;
until false;
//correct sum of & count of divisors
for i := 0 to High(pdf) do
Begin
with pdf[i] do
begin
j := pfRemain;
if j <> 1 then
begin
pfSumOFDivs *= (j+1);
pfDivCnt *=2;
end;
end;
end;
result := true;
end;
function NextSieve:boolean;
begin
dec(pdfIDX,SizePrDeFe);
inc(pdfOfs,SizePrDeFe);
result := SieveOneSieve(PrimeDecompField);
end;
function GetNextPrimeDecomp:tpPrimeFac;
begin
if pdfIDX >= SizePrDeFe then
if Not(NextSieve) then
EXIT(NIL);
result := @PrimeDecompField[pdfIDX];
inc(pdfIDX);
end;
function Init_Sieve(n:NativeUint):boolean;
//Init Sieve pdfIdx,pdfOfs are Global
begin
pdfIdx := n MOD SizePrDeFe;
pdfOfs := n-pdfIdx;
result := SieveOneSieve(PrimeDecompField);
end;
procedure InsertSort(pDiv:tpDivisor; Left, Right : NativeInt );
var
I, J: NativeInt;
Pivot : tItem;
begin
for i:= 1 + Left to Right do
begin
Pivot:= pDiv[i];
j:= i - 1;
while (j >= Left) and (pDiv[j] > Pivot) do
begin
pDiv[j+1]:=pDiv[j];
Dec(j);
end;
pDiv[j+1]:= pivot;
end;
end;
procedure GetDivisors(pD:tpPrimeFac;var Divs:tDivisors);
var
pDivs : tpDivisor;
pPot : UInt64;
i,len,j,l,p,k: Int32;
Begin
pDivs := @Divs[0];
pDivs[0] := 1;
len := 1;
l := 1;
with pD^ do
Begin
For i := 0 to pfMaxIdx-1 do
begin
//Multiply every divisor before with the new primefactors
//and append them to the list
k := pfpotMax[i];
p := SmallPrimes[pfpotPrimIdx[i]];
pPot :=1;
repeat
pPot *= p;
For j := 0 to len-1 do
Begin
pDivs[l]:= pPot*pDivs[j];
inc(l);
end;
dec(k);
until k<=0;
len := l;
end;
p := pfRemain;
If p >1 then
begin
For j := 0 to len-1 do
Begin
pDivs[l]:= p*pDivs[j];
inc(l);
end;
len := l;
end;
end;
//Sort. Insertsort much faster than QuickSort in this special case
InsertSort(pDivs,0,len-1);
//end marker
pDivs[len] :=0;
end;
var
pPrimeDecomp :tpPrimeFac;
Divs:tDivisors;
T0:Int64;
n,s : NativeUInt;
i : Int32;
Begin
T0 := GetTickCount64;
InitSmallPrimes;
Init_Sieve(0);
//jump over 0
pPrimeDecomp:= GetNextPrimeDecomp;
n := 1;
repeat
pPrimeDecomp:= GetNextPrimeDecomp;
s := pPrimeDecomp^.pfSumOfDivs;
if (s > 2*n) then
begin
s -= n;
// 75% of runtime
GetDivisors(pPrimeDecomp,Divs);
//calculate downwards. Not really an impact
For i := pPrimeDecomp^.pfDivCnt-2 downto 0 do
Begin
s -= Divs[i];
if s<n then
break;
if s = n then
writeln(Format('%8d equals the sum of its first %4d divisors',
[n,i]));
end;
end;
n += 1;
until n > 100*1000*1000+1;
T0 := GetTickCount64-T0;
writeln('runtime ',T0/1000:0:3,' s');
end.
```

- @TIO.RUN:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 392448 equals the sum of its first 68 divisors 714240 equals the sum of its first 113 divisors 1571328 equals the sum of its first 115 divisors 61900800 equals the sum of its first 280 divisors 91963648 equals the sum of its first 142 divisors runtime 15.760 s Real time: 15.909 s User time: 15.721 s CPU share: 99.09 %

## PascalABC.NET

```
const
max_number = 100_000_000;
begin
var dsum := arrfill(max_number + 1, 1);
var dcount := arrfill(max_number + 1, 1);
for var i := 2 to max_number do
for var j := i + i to max_number step i do
begin
if dsum[j] = j then
writeln(j:8, ' equals the sum of its first ', dcount[j], ' divisors');
dsum[j] += i;
dcount[j] += 1;
end;
end.
```

- Output:

2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors 61900800 equals the sum of its first 280 divisors 91963648 equals the sum of its first 142 divisors

## Perl

```
use v5.36;
use enum qw(False True);
use ntheory qw(divisors divisor_sum);
sub is_Erdos_Nicolas ($n) {
divisor_sum($n) > 2*$n or return False;
my $sum = 0;
my $count = 0;
foreach my $d (divisors($n)) {
++$count; $sum += $d;
return $count if ($sum == $n);
return False if ($sum > $n);
}
}
my ($n, $count) = (2, 0);
until ($count == 8) {
next unless 0 == ++$n % 2;
if (my $key = is_Erdos_Nicolas $n) {
printf "%8d == sum of its first %3d divisors\n", $n, $key;
$count++;
}
}
```

- Output:

24 == sum of its first 6 divisors 2016 == sum of its first 31 divisors 8190 == sum of its first 43 divisors 42336 == sum of its first 66 divisors 45864 == sum of its first 66 divisors 392448 == sum of its first 68 divisors 714240 == sum of its first 113 divisors 1571328 == sum of its first 115 divisors

## Phix

with javascript_semantics function erdos_nicolas(integer n) sequence divisors = factors(n) integer tot = 1 for i=1 to length(divisors)-1 do tot += divisors[i] if tot=n then return i+1 end if if tot>n then exit end if end for return 0 end function constant limit = 8 integer n = 2, count = 0 while count<limit do integer k = erdos_nicolas(n) if k>0 then printf(1,"%8d equals the sum of its first %d divisors.\n",{n,k}) count += 1 end if n += 2 end while

Aside: The default for factors() is to return neither 1 nor n, though you can change that if you want, ie ",1" -> 1 and n; ",-1" -> 1 but not n.

Output same as Julia

## Python

This is a translation of the "slight improvement" C code. I ran this script using PyPy7.3.13 (Python3.10.13) and it completes in ~10.5s on a AMD Ryzen 7 7800X3D.

```
# erdos-nicolas.py by Xing216
from time import perf_counter
start = perf_counter()
def get_div_cnt(n: int) -> None:
divcnt,divsum = 1,1
lmt = n/2
f = 2
while True:
if f > lmt: break
if not (n % f):
divsum += f
divcnt += 1
if divsum == n: break
f+=1
print(f"{n:>8} equals the sum of its first {divcnt} divisors")
max_number = 91963649
dsum = [1 for _ in range(max_number+1)]
for i in range(2, max_number + 1):
for j in range(i + i, max_number + 1, i):
if (dsum[j] == j): get_div_cnt(j)
dsum[j] += i
done = perf_counter() - start
print(f"Done in: {done:.3f}s")
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors 61900800 equals the sum of its first 280 divisors 91963648 equals the sum of its first 142 divisors Done in: 10.478s

## Raku

```
use Prime::Factor;
sub is-Erdős-Nicolas ($n) {
my @divisors = $n.&proper-divisors: :s;
((@divisors.sum > $n) && (my $key = ([\+] @divisors).first: $n, :k)) ?? 1 + $key !! False
}
my $count;
(1..*).hyper(:2000batch).map( * × 2 ).map: {
if my $key = .&is-Erdős-Nicolas {
printf "%8d == sum of its first %3d divisors\n", $_, $key;
exit if ++$count >= 8;
}
}
```

- Output:

24 == sum of its first 6 divisors 2016 == sum of its first 31 divisors 8190 == sum of its first 43 divisors 42336 == sum of its first 66 divisors 45864 == sum of its first 66 divisors 392448 == sum of its first 68 divisors 714240 == sum of its first 113 divisors 1571328 == sum of its first 115 divisors

## Ring

```
see "works..." + nl
erdos = []
limit = 1600000
for n = 1 to limit
num = 0
sum = 0
erdos = []
for m = 1 to n/2
if n%m = 0
add(erdos,m)
ok
next
lenErdos = len(erdos)
for p = 1 to lenErdos
sum = sum + erdos[p]
if sum = n and p < lenErdos
num++
see "" + n + " equals the sum of its first " + p + " divisors" + nl
exit
ok
next
if num = 8
exit
ok
next
see "done..." + nl
```

- Output:

24 equals the sum of its first 6 divisors 2016 equals the sum of its first 31 divisors 8190 equals the sum of its first 43 divisors 42336 equals the sum of its first 66 divisors 45864 equals the sum of its first 66 divisors 714240 equals the sum of its first 113 divisors 392448 equals the sum of its first 68 divisors 1571328 equals the sum of its first 115 divisors

## Sidef

```
func is_Erdős_Nicolas(n) {
n.is_abundant || return false
var sum = 0
var count = 0
n.divisors.each {|d|
++count; sum += d
return count if (sum == n)
return false if (sum > n)
}
}
var count = 8 # how many terms to compute
^Inf -> by(2).each {|n|
if (is_Erdős_Nicolas(n)) { |v|
say "#{'%8s'%n} is the sum of its first #{'%3s'%v} divisors"
--count || break
}
}
```

- Output:

24 is the sum of its first 6 divisors 2016 is the sum of its first 31 divisors 8190 is the sum of its first 43 divisors 42336 is the sum of its first 66 divisors 45864 is the sum of its first 66 divisors 392448 is the sum of its first 68 divisors 714240 is the sum of its first 113 divisors 1571328 is the sum of its first 115 divisors

## Wren

### Version 1

```
import "./math" for Int
var erdosNicolas = Fn.new { |n|
var divisors = Int.properDivisors(n) // excludes n itself
var dc = divisors.count
if (dc < 3) return 0
var sum = divisors[0] + divisors[1]
for (i in 2...dc-1) {
sum = sum + divisors[i]
if (sum == n) return i + 1
if (sum > n) break
}
return 0
}
var limit = 8
var n = 2
var count = 0
while (true) {
var k = erdosNicolas.call(n)
if (k > 0) {
System.print("%(n) from %(k)")
count = count + 1
if (count == limit) return
}
n = n + 2
}
```

- Output:

24 from 6 2016 from 31 8190 from 43 42336 from 66 45864 from 66 392448 from 68 714240 from 113 1571328 from 115

### Version 2

This takes 7 minutes to run (about 10 times slower than C++) but finds 2 more numbers, albeit not in strictly increasing order.

If `maxNum` is set to 2 million, then it finds the first 8 in about 5.2 seconds which is more than 10 times faster than Version 1's 58 seconds.

```
import "./fmt" for Fmt
var maxNum = 1e8
var dsum = List.filled(maxNum+1,1)
var dcount = List.filled(maxNum+1, 1)
for (i in 2..maxNum) {
var j = i + i
while (j <= maxNum) {
if (dsum[j] == j) {
Fmt.print("$8d equals the sum of its first $d divisors", j, dcount[j])
}
dsum[j] = dsum[j] + i
dcount[j] = dcount[j] + 1
j = j + i
}
}
```

- Output:

## XPL0

Translation of Algol 68 and C++.

```
def Max = 2_000_000;
int DSum(Max+1), DCount(Max+1);
int I, J;
[for I:= 0 to Max do
[DSum(I):= 1;
DCount(I):= 1;
];
Format(8, 0);
for I:= 2 to Max do
[J:= I+I;
while J <= Max do
[if DSum(J) = J then
[RlOut(0, float(J));
Text(0, " equals the sum of its first ");
IntOut(0, DCount(J));
Text(0, " divisors^j^m");
];
DSum(J):= DSum(J)+I;
DCount(J):= DCount(J)+1;
J:= J+I;
]
]
]
```

- Output: