# Erdős-Nicolas numbers

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**Erdős-Nicolas numbers****draft**programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

- 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[edit]

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

## C++[edit]

```
#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:

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

## J[edit]

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

```
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
```

## Julia[edit]

```
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.

## Perl[edit]

```
use v5.36;
use ntheory 'divisors';
use enum qw(False True);
use List::AllUtils <firstidx sum>;
sub proper_divisors ($n) {
return 1 if $n == 0;
my @d = divisors($n);
pop @d;
@d;
}
sub is_Erdos_Nicolas ($n) {
my @divisors = proper_divisors($n);
return False unless sum(@divisors) > $n;
my $sum;
my $key = firstidx { $_ == $n } map { $sum += $_ } @divisors;
$key ? 1 + $key : False;
}
my($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[edit]

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

## Raku[edit]

```
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

## Wren[edit]

```
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