# Sequence: smallest number with exactly n divisors

Sequence: smallest number with exactly n divisors
You are encouraged to solve this task according to the task description, using any language you may know.

Calculate the sequence where each term   an   is the smallest natural number that has exactly   n   divisors.

Task

Show here, on this page, at least the first  15  terms of the sequence.

Related tasks

See also

## 11l

Translation of: Python
```F divisors(n)
V divs = [1]
L(ii) 2 .< Int(n ^ 0.5) + 3
I n % ii == 0
divs.append(ii)
divs.append(Int(n / ii))
divs.append(n)
R Array(Set(divs))

F sequence(max_n)
V n = 0
[Int] r
L
n++
V ii = 0
I n > max_n
L.break
L
ii++
I divisors(ii).len == n
r.append(ii)
L.break
R r

L(item) sequence(15)
print(item)```
Output:
```1
2
4
6
16
12
64
24
36
48
1024
60
4096
192
144
```

## Action!

Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU.

```CARD FUNC CountDivisors(CARD a)
CARD i,count

i=1 count=0
WHILE i*i<=a
DO
IF a MOD i=0 THEN
IF i=a/i THEN
count==+1
ELSE
count==+2
FI
FI
i==+1
OD
RETURN (count)

PROC Main()
DEFINE MAX="15"
CARD a,count
BYTE i
CARD ARRAY seq(MAX)

FOR i=0 TO MAX-1
DO
seq(i)=0
OD

i=0 a=1
WHILE i<MAX
DO
count=CountDivisors(a)
IF count<=MAX AND seq(count-1)=0 THEN
seq(count-1)=a
i==+1
FI
a==+1
OD

FOR i=0 TO MAX-1
DO
IF i>0 THEN
Print(", ")
FI
PrintC(seq(i))
OD
RETURN```
Output:
```1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144
```

## ALGOL 68

Translation of: C
```BEGIN

PROC count divisors = ( INT n )INT:
BEGIN
INT count := 0;
FOR i WHILE i*i <= n DO
IF n MOD i = 0 THEN
count +:= IF i = n OVER i THEN 1 ELSE 2 FI
FI
OD;
count
END # count divisors # ;

INT max = 15;
[ max ]INT seq;FOR i TO max DO seq[ i ] := 0 OD;
INT found := 0;
FOR i WHILE found < max DO
IF INT divisors = count divisors( i );
divisors <= max
THEN
# have an i with an appropriate number of divisors                 #
IF seq[ divisors ] = 0 THEN
# this is the first i with that many divisors                  #
seq[ divisors ] := i;
found          +:= 1
FI
FI
OD;
print( ( "The first ", whole( max, 0 ), " terms of the sequence are:", newline ) );
FOR i TO max DO
print( ( whole( seq( i ), 0 ), " " ) )
OD;
print( ( newline ) )

END```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```

## APL

```((0+.=⍳|⊢)¨∘(⍳2÷⍨2∘*)⍳⍳)15
```
Output:
`1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144`

## Arturo

```firstNumWithDivisors: function [n][
i: 0
while ø [
if n = size factors i -> return i
i: i+1
]
]

print map 1..15 => firstNumWithDivisors
```
Output:
`1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144`

## AutoHotkey

Translation of: Go
```max := 15
seq := [], n := 0
while (n < max)
if ((k := countDivisors(A_Index)) <= max) && !seq[k]
seq[k] := A_Index, n++
for i, v in seq
res .= v ", "
MsgBox % "The first " . max . " terms of the sequence are:`n" Trim(res, ", ")
return

countDivisors(n){
while (A_Index**2 <= n)
if !Mod(n, A_Index)
count += A_Index = n/A_Index ? 1 : 2
return count
}
```

Outputs:

```The first 15 terms of the sequence are:
1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144```

## AWK

```# syntax: GAWK -f SEQUENCE_SMALLEST_NUMBER_WITH_EXACTLY_N_DIVISORS.AWK
# converted from Kotlin
BEGIN {
limit = 15
printf("first %d terms:",limit)
i = 1
n = 0
while (n < limit) {
k = count_divisors(i)
if (k <= limit && seq[k-1]+0 == 0) {
seq[k-1] = i
n++
}
i++
}
for (i=0; i<limit; i++) {
printf(" %d",seq[i])
}
printf("\n")
exit(0)
}
function count_divisors(n,  count,i) {
for (i=1; i*i<=n; i++) {
if (n % i == 0) {
count += (i == n / i) ? 1 : 2
}
}
return(count)
}
```
Output:
```first 15 terms: 1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```

## BASIC

### BASIC256

Translation of: Ring
```print "the first 15 terms of the sequence are:"
for n = 1 to 15
for m = 1 to 4100
pnum = 0
for p = 1 to 4100
if m % p = 0 then pnum += 1
next p
if pnum = n then
print m; " ";
exit for
end if
next m
next n
```

### Chipmunk Basic

Translation of: Ring
Works with: Chipmunk Basic version 3.6.4
Works with: QBasic
Works with: MSX BASIC
Works with: Run BASIC
```100 print "the first 15 terms of the sequence are:"
110 for n = 1 to 15
120   for m = 1 to 4100
130     pnum = 0
140     for p = 1 to 4100
150       if m mod p = 0 then pnum = pnum+1
160     next p
170     if pnum = n then print m;" "; : goto 190
180   next m
190 next n
200 print "done..."
210 end
```

### GW-BASIC

Works with: PC-BASIC version any
Works with: BASICA

The Chipmunk Basic solution works without any changes.

### MSX Basic

Works with: MSX BASIC version any

The Chipmunk Basic solution works without any changes.

### QBasic

Translation of: Ring
Works with: QBasic version 1.1
Works with: QuickBasic version 4.5
```PRINT "the first 15 terms of the sequence are:"
FOR n = 1 to 15
FOR m = 1 to 4100
pnum = 0
FOR p = 1 to 4100
IF m MOD p = 0 then pnum = pnum + 1
NEXT p
IF pnum = n then
PRINT m;
EXIT FOR
END IF
NEXT m
NEXT n
```

### Run BASIC

The Chipmunk Basic solution works without any changes.

### True BASIC

Translation of: Ring
```PRINT "the first 15 terms of the sequence are:"
FOR n = 1 to 15
FOR m = 1 to 4100
LET pnum = 0
FOR p = 1 to 4100
IF remainder(m, p) = 0 then LET pnum = pnum+1
NEXT p
IF pnum = n then
PRINT m;
EXIT FOR
END IF
NEXT m
NEXT n
PRINT "done..."
END
```

### XBasic

Translation of: Ring
Works with: Windows XBasic
```PROGRAM  "program name"
VERSION  "0.0000"

DECLARE FUNCTION  Entry ()

FUNCTION  Entry ()
PRINT "the first 15 terms of the sequence are:"
FOR n = 1 TO 15
FOR m = 1 TO 4100
pnum = 0
FOR p = 1 TO 4100
IF m MOD p = 0 THEN INC pnum
NEXT p
IF pnum = n THEN
PRINT m;
EXIT FOR
END IF
NEXT m
NEXT n
END FUNCTION
END PROGRAM
```

### PureBasic

Translation of: FreeBASIC
```Procedure.i divisors(n)
;find the number of divisors of an integer
Define.i r, i
r = 2
For i = 2 To n/2
If n % i = 0: r + 1
EndIf
Next i
ProcedureReturn r
EndProcedure

OpenConsole()
Define.i UPTO, i, n, nfound

UPTO = 15
i = 2
nfound = 1
Dim sfound.i(UPTO)
sfound(1) = 1

While nfound < UPTO
n = divisors(i)
If n <= UPTO And sfound(n) = 0:
nfound + 1
sfound(n) = i
EndIf
i + 1
Wend

Print("1 ")    ;special case
For i = 2 To UPTO
Print(Str(sfound(i)) + " ")
Next i

PrintN(#CRLF\$ + "Press ENTER to exit"): Input()
CloseConsole()
```

### Yabasic

Translation of: FreeBASIC
```UPTO = 15
i = 2
nfound = 1
dim sfound(UPTO)
sfound(1) = 1

while nfound < UPTO
n = divisors(i)
if n <= UPTO and sfound(n) = 0 then
nfound = nfound + 1
sfound(n) = i
fi
i = i + 1
end while

print 1, " ";    //special case
for i = 2 to UPTO
print sfound(i), " ";
next i
print
end

sub divisors(n)
local r, i

//find the number of divisors of an integer
r = 2
for i = 2 to n / 2
if mod(n, i) = 0  r = r + 1
next i
return r
end sub
```

## BCPL

```get "libhdr"

manifest \$( LENGTH = 15 \$)

let divisors(n) = valof
\$(  let count = 0 and i = 1
while i*i <= n
\$(  if n rem i = 0 then
count := count + (i = n/i -> 1, 2)
i := i + 1
\$)
resultis count
\$)

let sequence(n, seq) be
\$(  let found = 0 and i = 1
for i=1 to n do seq!i := 0
while found < n
\$(  let divs = divisors(i)
if divs <= n & seq!divs = 0
\$(  found := found + 1
seq!divs := i
\$)
i := i + 1
\$)
\$)

let start() be
\$(  let seq = vec LENGTH
sequence(LENGTH, seq)

for i=1 to LENGTH do writef("%N ", seq!i)
wrch('*N')
\$)```
Output:
`1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144`

## C

Translation of: Go
```#include <stdio.h>

#define MAX 15

int count_divisors(int n) {
int i, count = 0;
for (i = 1; i * i <= n; ++i) {
if (!(n % i)) {
if (i == n / i)
count++;
else
count += 2;
}
}
return count;
}

int main() {
int i, k, n, seq[MAX];
for (i = 0; i < MAX; ++i) seq[i] = 0;
printf("The first %d terms of the sequence are:\n", MAX);
for (i = 1, n = 0; n <  MAX; ++i) {
k = count_divisors(i);
if (k <= MAX && seq[k - 1] == 0) {
seq[k - 1] = i;
++n;
}
}
for (i = 0; i < MAX; ++i) printf("%d ", seq[i]);
printf("\n");
return 0;
}
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```

## C++

Translation of: C
```#include <iostream>

#define MAX 15

using namespace std;

int count_divisors(int n) {
int count = 0;
for (int i = 1; i * i <= n; ++i) {
if (!(n % i)) {
if (i == n / i)
count++;
else
count += 2;
}
}
return count;
}

int main() {
int i, k, n, seq[MAX];
for (i = 0; i < MAX; ++i) seq[i] = 0;
cout << "The first " << MAX << " terms of the sequence are:" << endl;
for (i = 1, n = 0; n <  MAX; ++i) {
k = count_divisors(i);
if (k <= MAX && seq[k - 1] == 0) {
seq[k - 1] = i;
++n;
}
}
for (i = 0; i < MAX; ++i) cout << seq[i] << " ";
cout << endl;
return 0;
}
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 12 64 24 36 48 1024 60 0 192 144
```

## Cowgol

```include "cowgol.coh";

const AMOUNT := 15;
typedef I is uint16;

sub divisors(n: I): (count: I) is
var i: I := 1;
count := 0;

while i*i <= n loop
if n%i == 0 then
if n/i == i then
count := count + 1;
else
count := count + 2;
end if;
end if;
i := i + 1;
end loop;
end sub;

var seq: I[AMOUNT+1];
MemZero(&seq as [uint8], @bytesof seq);

var found: I := 0;
var i: I := 1;

while found < AMOUNT loop
var divs := divisors(i) as @indexof seq;
if divs <= AMOUNT and seq[divs] == 0 then
found := found + 1;
seq[divs] := i;
end if;
i := i + 1;
end loop;

var j: @indexof seq := 1;
while j <= AMOUNT loop
print_i16(seq[j]);
print_char(' ');
j := j + 1;
end loop;
print_nl();```
Output:
`1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144`

## Delphi

Works with: Delphi version 6.0

```{This code would normally be in a separate library. It is provided for clarity}

function GetAllProperDivisors(N: Integer;var IA: TIntegerDynArray): integer;
{Make a list of all the "proper dividers" for N}
{Proper dividers are the of numbers the divide evenly into N}
var I: integer;
begin
SetLength(IA,0);
for I:=1 to N-1 do
if (N mod I)=0 then
begin
SetLength(IA,Length(IA)+1);
IA[High(IA)]:=I;
end;
Result:=Length(IA);
end;

function GetAllDivisors(N: Integer;var IA: TIntegerDynArray): integer;
{Make a list of all the "proper dividers" for N, Plus N itself}
begin
Result:=GetAllProperDivisors(N,IA)+1;
SetLength(IA,Length(IA)+1);
IA[High(IA)]:=N;
end;

{-------------------------------------------------------------------------------}

procedure SequenceWithNdivisors(Memo: TMemo);
var N,N2: integer;
var IA: TIntegerDynArray;
begin
for N:=1 to 15 do
begin
N2:=0;
repeat Inc(N2) until N=GetAllDivisors(N2,IA);
Memo.Lines.Add(IntToStr(N)+' - '+IntToStr(N2));
end;
end;
```
Output:
```1 - 1
2 - 2
3 - 4
4 - 6
5 - 16
6 - 12
7 - 64
8 - 24
9 - 36
10 - 48
11 - 1024
12 - 60
13 - 4096
14 - 192
15 - 144

Elapsed Time: 54.950 ms.

```

## EasyLang

```func cntdiv n .
i = 1
while i <= sqrt n
if n mod i = 0
cnt += 1
if i <> sqrt n
cnt += 1
.
.
i += 1
.
return cnt
.
len seq[] 15
i = 1
while n < 15
k = cntdiv i
if k <= 15 and seq[k] = 0
seq[k] = i
n += 1
.
i += 1
.
for v in seq[]
print v
.```

## F#

This task uses Extensible Prime Generator (F#)

```// Find Antı-Primes plus. Nigel Galloway: April 9th., 2019
// Increasing the value 14 will increase the number of anti-primes plus found
let fI=primes|>Seq.take 14|>Seq.map bigint|>List.ofSeq
let N=Seq.reduce(*) fI
let fG g=Seq.unfold(fun ((n,i,e) as z)->Some(z,(n+1,i+1,(e*g)))) (1,2,g)
let fE n i=n|>Seq.collect(fun(n,e,g)->Seq.map(fun(a,c,b)->(a,c*e,g*b)) (i|>Seq.takeWhile(fun(g,_,_)->g<=n))|> Seq.takeWhile(fun(_,_,n)->n<N))
let fL=let mutable g=0 in (fun n->g<-g+1; n=g)
let n=Seq.concat(Seq.scan(fun n g->fE n (fG g)) (seq[(2147483647,1,1I)]) fI)|>List.ofSeq|>List.groupBy(fun(_,n,_)->n)|>List.sortBy(fun(n,_)->n)|>List.takeWhile(fun(n,_)->fL n)
for n,g in n do printfn "%d->%A" n (g|>List.map(fun(_,_,n)->n)|>List.min)
```
Output:
```1->1
2->2
3->4
4->6
5->16
6->12
7->64
8->24
9->36
10->48
11->1024
12->60
13->4096
14->192
15->144
16->120
17->65536
18->180
19->262144
20->240
21->576
22->3072
23->4194304
24->360
25->1296
26->12288
27->900
28->960
29->268435456
30->720
31->1073741824
32->840
33->9216
34->196608
35->5184
36->1260
37->68719476736
38->786432
39->36864
40->1680
41->1099511627776
42->2880
43->4398046511104
44->15360
45->3600
46->12582912
47->70368744177664
48->2520
49->46656
50->6480
51->589824
52->61440
53->4503599627370496
54->6300
55->82944
56->6720
57->2359296
58->805306368
Real: 00:00:01.079, CPU: 00:00:01.080, GC gen0: 47, gen1: 0
```

## Factor

```USING: fry kernel lists lists.lazy math math.primes.factors
prettyprint sequences ;

: A005179 ( -- list )
1 lfrom [
1 swap '[ dup divisors length _ = ] [ 1 + ] until
] lmap-lazy ;

15 A005179 ltake list>array .
```
Output:
```{ 1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144 }
```

## FreeBASIC

```#define UPTO 15

function divisors(byval n as ulongint) as uinteger
'find the number of divisors of an integer
dim as integer r = 2, i
for i = 2 to n\2
if n mod i = 0 then r += 1
next i
return r
end function

dim as ulongint i = 2
dim as integer n, nfound = 1, sfound(UPTO) = {1}

while nfound < UPTO
n = divisors(i)
if n<=UPTO andalso sfound(n)=0 then
nfound += 1
sfound(n) = i
end if
i+=1
wend

print 1;" ";    'special case
for i = 2 to UPTO
print sfound(i);" ";
next i
print
end
```
Output:
` 1  2  4  6  16  12  64  24  36  48  1024  60  4096  192  144`

## Go

```package main

import "fmt"

func countDivisors(n int) int {
count := 0
for i := 1; i*i <= n; i++ {
if n%i == 0 {
if i == n/i {
count++
} else {
count += 2
}
}
}
return count
}

func main() {
const max = 15
seq := make([]int, max)
fmt.Println("The first", max, "terms of the sequence are:")
for i, n := 1, 0; n < max; i++ {
if k := countDivisors(i); k <= max && seq[k-1] == 0 {
seq[k-1] = i
n++
}
}
fmt.Println(seq)
}
```
Output:
```The first 15 terms of the sequence are:
[1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144]
```

## Haskell

Without any subtlety or optimisation, but still fast at this modest scale:

```import Data.List (find, group, sort)
import Data.Maybe (mapMaybe)
import Data.Numbers.Primes (primeFactors)

------------------------- A005179 ------------------------

a005179 :: [Int]
a005179 =
mapMaybe
( \n ->
find
((n ==) . succ . length . properDivisors)
[1 ..]
)
[1 ..]

--------------------------- TEST -------------------------
main :: IO ()
main = print \$ take 15 a005179

------------------------- GENERIC ------------------------

properDivisors :: Int -> [Int]
properDivisors =
init
. sort
. foldr
(flip ((<*>) . fmap (*)) . scanl (*) 1)
[1]
. group
. primeFactors
```
Output:
`[1,2,4,6,16,12,64,24,36,48,1024,60,4096,192,144]`

## JavaScript

Defining a generator in terms of re-usable generic functions, and taking the first 15 terms:

```(() => {
'use strict';

// a005179 :: () -> [Int]
const a005179 = () =>
fmapGen(
n => find(
compose(
eq(n),
succ,
length,
properDivisors
)
)(enumFrom(1)).Just
)(enumFrom(1));

// ------------------------TEST------------------------
// main :: IO ()
const main = () =>
console.log(
take(15)(
a005179()
)
);

// [1,2,4,6,16,12,64,24,36,48,1024,60,4096,192,144]

// -----------------GENERIC FUNCTIONS------------------

// Just :: a -> Maybe a
const Just = x => ({
type: 'Maybe',
Nothing: false,
Just: x
});

// Nothing :: Maybe a
const Nothing = () => ({
type: 'Maybe',
Nothing: true,
});

// Tuple (,) :: a -> b -> (a, b)
const Tuple = a =>
b => ({
type: 'Tuple',
'0': a,
'1': b,
length: 2
});

// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
fs.reduce(
(f, g) => x => f(g(x)),
x => x
);

// enumFrom :: Enum a => a -> [a]
function* enumFrom(x) {
// A non-finite succession of enumerable
// values, starting with the value x.
let v = x;
while (true) {
yield v;
v = succ(v);
}
};

// eq (==) :: Eq a => a -> a -> Bool
const eq = a =>
// True when a and b are equivalent in the terms
// defined below for their shared data type.
b => a === b;

// find :: (a -> Bool) -> Gen [a] -> Maybe a
const find = p => xs => {
const mb = until(tpl => {
const nxt = tpl[0];
return nxt.done || p(nxt.value);
})(
tpl => Tuple(tpl[1].next())(
tpl[1]
)
)(Tuple(xs.next())(xs))[0];
return mb.done ? (
Nothing()
) : Just(mb.value);
}

// fmapGen <\$> :: (a -> b) -> Gen [a] -> Gen [b]
const fmapGen = f =>
function*(gen) {
let v = take(1)(gen);
while (0 < v.length) {
yield(f(v[0]))
v = take(1)(gen)
}
};

// group :: [a] -> [[a]]
const group = xs => {
// A list of lists, each containing only equal elements,
// such that the concatenation of these lists is xs.
const go = xs =>
0 < xs.length ? (() => {
const
h = xs[0],
i = xs.findIndex(x => h !== x);
return i !== -1 ? (
[xs.slice(0, i)].concat(go(xs.slice(i)))
) : [xs];
})() : [];
return go(xs);
};

// length :: [a] -> Int
const length = xs =>
// Returns Infinity over objects without finite
// length. This enables zip and zipWith to choose
// the shorter argument when one is non-finite,
// like cycle, repeat etc
(Array.isArray(xs) || 'string' === typeof xs) ? (
xs.length
) : Infinity;

// liftA2List :: (a -> b -> c) -> [a] -> [b] -> [c]
const liftA2List = f => xs => ys =>
// The binary operator f lifted to a function over two
// lists. f applied to each pair of arguments in the
// cartesian product of xs and ys.
xs.flatMap(
x => ys.map(f(x))
);

// mul (*) :: Num a => a -> a -> a
const mul = a => b => a * b;

// properDivisors :: Int -> [Int]
const properDivisors = n =>
// The ordered divisors of n,
// excluding n itself.
1 < n ? (
sort(group(primeFactors(n)).reduce(
(a, g) => liftA2List(mul)(a)(
scanl(mul)([1])(g)
),
[1]
)).slice(0, -1)
) : [];

// primeFactors :: Int -> [Int]
const primeFactors = n => {
// A list of the prime factors of n.
const
go = x => {
const
root = Math.floor(Math.sqrt(x)),
m = until(
([q, _]) => (root < q) || (0 === (x % q))
)(
([_, r]) => [step(r), 1 + r]
)(
[0 === x % 2 ? 2 : 3, 1]
)[0];
return m > root ? (
[x]
) : ([m].concat(go(Math.floor(x / m))));
},
step = x => 1 + (x << 2) - ((x >> 1) << 1);
return go(n);
};

// scanl :: (b -> a -> b) -> b -> [a] -> [b]
const scanl = f => startValue => xs =>
xs.reduce((a, x) => {
const v = f(a[0])(x);
return Tuple(v)(a[1].concat(v));
}, Tuple(startValue)([startValue]))[1];

// sort :: Ord a => [a] -> [a]
const sort = xs => xs.slice()
.sort((a, b) => a < b ? -1 : (a > b ? 1 : 0));

// succ :: Enum a => a -> a
const succ = x =>
1 + x;

// take :: Int -> [a] -> [a]
// take :: Int -> String -> String
const take = n =>
// The first n elements of a list,
// string of characters, or stream.
xs => 'GeneratorFunction' !== xs
.constructor.constructor.name ? (
xs.slice(0, n)
) : [].concat.apply([], Array.from({
length: n
}, () => {
const x = xs.next();
return x.done ? [] : [x.value];
}));

// until :: (a -> Bool) -> (a -> a) -> a -> a
const until = p => f => x => {
let v = x;
while (!p(v)) v = f(v);
return v;
};

// MAIN ---
return main();
})();
```
Output:
`[1,2,4,6,16,12,64,24,36,48,1024,60,4096,192,144]`

## J

Rather than factoring by division, this algorithm uses a sieve to tally the factors. Of the whole numbers below ten thousand these are the smallest with n divisors. The list is fully populated through the first 15.

```sieve=: 3 :0
range=. <. + i.@:|@:-
tally=. y#0
for_i.#\tally do.
j=. }:^:(y<:{:)i * 1 range >: <. y % i
tally=. j >:@:{`[`]} tally
end.
/:~({./.~ {."1) tally,.i.#tally
)
```
```|: sieve 10000
0 1 2 3 4  5  6  7  8  9 10   11 12   13  14  15  16  18  20  21   22  24   25  27  28  30  32   33   35   36   40   42   45   48   50   54   56   60   64
0 1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144 120 180 240 576 3072 360 1296 900 960 720 840 9216 5184 1260 1680 2880 3600 2520 6480 6300 6720 5040 7560
```

## Java

Translation of: C
```import java.util.Arrays;

public class OEIS_A005179 {

static int count_divisors(int n) {
int count = 0;
for (int i = 1; i * i <= n; ++i) {
if (n % i == 0) {
if (i == n / i)
count++;
else
count += 2;
}
}
return count;
}

public static void main(String[] args) {
final int max = 15;
int[] seq = new int[max];
System.out.printf("The first %d terms of the sequence are:\n", max);
for (int i = 1, n = 0; n < max; ++i) {
int k = count_divisors(i);
if (k <= max && seq[k - 1] == 0) {
seq[k- 1] = i;
n++;
}
}
System.out.println(Arrays.toString(seq));
}
}
```
Output:
```The first 15 terms of the sequence are:
[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144]
```

## jq

This entry uses a streaming approach to avoid constructing unnecessary arrays.

```# divisors as an unsorted stream (without calling sqrt)
def divisors:
if . == 1 then 1
else . as \$n
| label \$out
| range(1; \$n) as \$i
| (\$i * \$i) as \$i2
| if \$i2 > \$n then break \$out
else if \$i2 == \$n
then \$i
elif (\$n % \$i) == 0
then \$i, (\$n/\$i)
else empty
end
end
end;

def count(s): reduce s as \$x (0; .+1);

# smallest number with exactly n divisors
def A005179:
. as \$n
| first( range(1; infinite) | select( count(divisors) == \$n ));

# The task:
"The first 15 terms of the sequence are:",
[range(1; 16) | A005179]```

Invocation: jq -ncr -f A005179.jq

Output:
```The first 15 terms of the sequence are:
[1,2,4,6,16,12,64,24,36,48,1024,60,4096,192,144]```

## Julia

Works with: Julia version 1.2
```using Primes

numfactors(n) = reduce(*, e+1 for (_,e) in factor(n); init=1)

A005179(n) = findfirst(k -> numfactors(k) == n, 1:typemax(Int))

println("The first 15 terms of the sequence are:")
println(map(A005179, 1:15))
```
Output:
```First 15 terms of OEIS sequence A005179:
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```

## Kotlin

Translation of: Go
```// Version 1.3.21

const val MAX = 15

fun countDivisors(n: Int): Int {
var count = 0
var i = 1
while (i * i <= n) {
if (n % i == 0) {
count += if (i == n / i) 1 else 2
}
i++
}
return count
}

fun main() {
var seq = IntArray(MAX)
println("The first \$MAX terms of the sequence are:")
var i = 1
var n = 0
while (n < MAX) {
var k = countDivisors(i)
if (k <= MAX && seq[k - 1] == 0) {
seq[k - 1] = i
n++
}
i++
}
println(seq.asList())
}
```
Output:
```The first 15 terms of the sequence are:
[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144]
```

## Maple

```with(NumberTheory):

countDivisors := proc(x::integer)
return numelems(Divisors(x));
end proc:

sequenceValue := proc(x::integer)
local count:
for count from 1 to infinity while not countDivisors(count) = x do end:
return count;
end proc:

seq(sequenceValue(number), number = 1..15);```
Output:
```
1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144

```

## Mathematica / Wolfram Language

```Take[SplitBy[SortBy[{DivisorSigma[0, #], #} & /@ Range[100000], First], First][[All, 1]], 15] // Grid
```
Output:
```1	1
2	2
3	4
4	6
5	16
6	12
7	64
8	24
9	36
10	48
11	1024
12	60
13	4096
14	192
15	144```

## MiniScript

```divisors = function(n)
divs = {1: 1}
divs[n] = 1
i = 2
while i * i <= n
if n % i == 0 then
divs[i] = 1
divs[n / i] = 1
end if
i += 1
end while
return divs.indexes
end function

counts = []
for i in range(1, 15)
j = 1
while divisors(j).len != i
j += 1
end while
counts.push(j)
end for

print "The first 15 terms in the sequence are:"
print counts.join(", ")
```
Output:
```The first 15 terms in the sequence are:
1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144```

## Miranda

```main :: [sys_message]
main = [Stdout (show (take 15 a005179) ++ "\n")]

a005179 :: [num]
a005179 = map smallest_n_divisors [1..]

smallest_n_divisors :: num->num
smallest_n_divisors n = hd [i | i<-[1..]; n = #divisors i]

divisors :: num->[num]
divisors n = [d | d<-[1..n]; n mod d=0]```
Output:
`[1,2,4,6,16,12,64,24,36,48,1024,60,4096,192,144]`

## Modula-2

```MODULE A005179;
FROM InOut IMPORT WriteCard, WriteLn;

CONST Amount = 15;
VAR found, i, ndivs: CARDINAL;
A: ARRAY [1..Amount] OF CARDINAL;

PROCEDURE divisors(n: CARDINAL): CARDINAL;
VAR count, i: CARDINAL;
BEGIN
count := 0;
i := 1;
WHILE i*i <= n DO
IF n MOD i = 0 THEN
INC(count);
IF n DIV i # i THEN
INC(count);
END;
END;
INC(i);
END;
RETURN count;
END divisors;

BEGIN
FOR i := 1 TO Amount DO A[i] := 0; END;

found := 0;
i := 1;
WHILE found < Amount DO
ndivs := divisors(i);
IF (ndivs <= Amount) AND (A[ndivs] = 0) THEN
INC(found);
A[ndivs] := i;
END;
INC(i);
END;

FOR i := 1 TO Amount DO
WriteCard(A[i], 4);
WriteLn;
END;
END A005179.
```
Output:
```   1
2
4
6
16
12
64
24
36
48
1024
60
4096
192
144```

## Nanoquery

Translation of: Java
```def count_divisors(n)
count = 0
for (i = 1) ((i * i) <= n) (i += 1)
if (n % i) = 0
if i = (n / i)
count += 1
else
count += 2
end
end
end
return count
end

max = 15
seq = {0} * max
print format("The first %d terms of the sequence are:\n", max)
i = 1
for (n = 0) (n < max) (i += 1)
k = count_divisors(i)
if (k <= max)
if seq[k - 1] = 0
seq[k - 1] = i
n += 1
end
end
end
println seq```
Output:
```The first 15 terms of the sequence are:
[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144]```

## Nim

Translation of: Kotlin
```import strformat

const MAX = 15

func countDivisors(n: int): int =
var count = 0
var i = 1
while i * i <= n:
if n mod i == 0:
if i == n div i:
inc count, 1
else:
inc count, 2
inc i
count

var sequence: array[MAX, int]
echo fmt"The first {MAX} terms of the sequence are:"
var i = 1
var n = 0
while n < MAX:
var k = countDivisors(i)
if k <= MAX and sequence[k - 1] == 0:
sequence[k - 1] = i
inc n
inc i
echo sequence
```
Output:
```The first 15 terms of the sequence are:
[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144]
```

## Perl

Library: ntheory
```use strict;
use warnings;
use ntheory 'divisors';

print "First 15 terms of OEIS: A005179\n";
for my \$n (1..15) {
my \$l = 0;
while (++\$l) {
print "\$l " and last if \$n == divisors(\$l);
}
}
```
Output:
```First 15 terms of OEIS: A005179
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144```

## Phix

### naive

```with javascript_semantics
constant limit = 15
sequence res = repeat(0,limit)
integer found = 0, n = 1
while found<limit do
integer k = length(factors(n,1))
if k<=limit and res[k]=0 then
res[k] = n
found += 1
end if
n += 1
end while
printf(1,"The first %d terms are: %V\n",{limit,res})
```
Output:
```The first 15 terms are: {1,2,4,6,16,12,64,24,36,48,1024,60,4096,192,144}
```

You would need something quite a bit smarter to venture over a limit of 28.

### advanced

Using the various formula from the OEIS:A005179 link above.
get_primes() and product() have recently been added as new builtins, if necessary see Extensible_prime_generator and Deconvolution/2D+#Phix.

```with javascript_semantics
constant limit = iff(machine_bits()=32?58:66)
sequence found = repeat(0,limit)
integer n = 1

procedure populate_found(integer i)
while found[i]=0 do
integer k = length(factors(n,1))
if k<=limit and found[k]=0 then
found[k] = n
end if
n += 1
end while
end procedure

for i=1 to limit do
sequence f = factors(i,1)
integer lf = length(f)
atom ri
if lf<=2 then                       ri = power(2,i-1)                               -- prime (or 1)
elsif lf=3 then                     ri = power(6,f[2]-1)                            -- p^2 (eg f={1,5,25})
elsif f[2]>2 -- (see note)
and f[\$] = power(f[2],lf-1) then  ri = power(product(get_primes(-(lf-1))),f[2]-1) -- p^k (eg f={1,3,9,27})
elsif lf=4 then                     ri = power(2,f[3]-1)*power(3,f[2]-1)            -- p*q (eg f={1,2,3,6})
else populate_found(i)              ri = found[i]                                   -- do the rest manually
end if
printf(1,"%d->%d\n",{i,ri})
end for
```

Note: the f[2]>2 test should really be something more like >log(get_primes(-(lf-1))[\$])/log(2), apparently, but everything seems ok within the IEEE 754 53/64 bit limits this imposes. It takes longer, afaict, to print the answers than it did to calculate them, tee hee!

Output:

64-bit (as shown) manages 8 more answers than 32-bit, which as per limit halts on 58: on 32 bit the accuracy limit is 2^53, hence the result for 59, which is 2^58, would get printed wrong since the first /10 needed to print it rounds to the nearest 16 or so. It is quite probably perfectly accurate internally up to much higher limits, but proving/showing that is a bit of a problem, which would in turn probably be easiest to solve by simply rewriting this to use gmp/mpir.

```1->1
2->2
3->4
4->6
5->16
6->12
7->64
8->24
9->36
10->48
11->1024
12->60
13->4096
14->192
15->144
16->120
17->65536
18->180
19->262144
20->240
21->576
22->3072
23->4194304
24->360
25->1296
26->12288
27->900
28->960
29->268435456
30->720
31->1073741824
32->840
33->9216
34->196608
35->5184
36->1260
37->68719476736
38->786432
39->36864
40->1680
41->1099511627776
42->2880
43->4398046511104
44->15360
45->3600
46->12582912
47->70368744177664
48->2520
49->46656
50->6480
51->589824
52->61440
53->4503599627370496
54->6300
55->82944
56->6720
57->2359296
58->805306368
59->288230376151711744
60->5040
61->1152921504606846976
62->3221225472
63->14400
64->7560
65->331776
66->46080
```

### insane

A rather silly (but successful) attempt to reverse engineer all the rules up to 2000.
I got it down to just 11 of them, with only 1 being a complete fudge. Obviously, the fewer cases each covers, the less sound it is, and those mini-tables for np/p2/p3/p5 and adj are not exactly, um, scientific. Completes in about 0.1s

Library: Phix/mpfr
```with javascript_semantics
include mpfr.e
mpz r = mpz_init(),
pn = mpz_init()
sequence rule_names = {},
rule_counts = {}
for i=1 to 2000 do
sequence pf = prime_factors(i,true), ri, adj
integer lf = length(pf), np, p2, p3, p5, p, e
string what
if lf>10 then ?9/0 end if
if lf<=1 then                                   what = "prime (proper rule)"
np = 1
adj = {i}
elsif pf[\$]=2 then                              what = "2^k (made up rule)"
np = lf-1
p2 = {2,4,4,4,4,4,4,8,8}[np]
p3 = {2,2,2,2,4,4,4,4,4}[np]
np = {2,2,3,4,4,5,6,6,7}[np]
adj = {p2,p3}
elsif pf[\$]=3
and pf[\$-1]=2 then                            what = "2^k*3 (made up rule)"
np = lf-1
p2 = {3,3,4,4,4,6,6,6,6}[np]
p3 = {2,2,3,3,3,4,4,4,4}[np]
np = {2,3,3,4,5,5,6,7,8}[np]
adj = {p2,p3}
elsif lf>4
and pf[\$-1]=2 then                            what="2^k*p (made up rule)"
np = lf-1
adj = {0,4}
elsif lf>4
and pf[\$]=3
and pf[\$-1]=3
and pf[\$-2]=2 then                            what="2^k*3^2*p (made up rule)"
np = lf-4
p3 = {3,3,3,4,4}[np]
p5 = {2,2,2,3,3}[np]
np = {4,5,6,6,7}[np]
adj = {6,p3,p5}
elsif lf>4
and pf[\$]=3
and pf[\$-2]=3
and pf[\$-4]=2 then                            what="2^k*3^3*p (made up rule)"
np = lf-1
adj = {6}
elsif lf>5
and pf[\$]>3
and pf[\$-1]=3
and pf[\$-4]=3
and pf[2]=3
and (pf[1]=2 or pf[\$]>5) then                 what="2^k*3^4*p (made up rule)"
np = lf
adj = {}
elsif lf>4
and pf[\$-1]=3
and pf[\$-4]=3
and (lf>5 or pf[\$]=3) then                    what="[2^k]*3^(>=4)*p (made up rule)"
np = lf-1
adj = {9,pf[\$]}&reverse(pf[1..\$-3]) -- <bsg>
elsif lf>=7
and pf[\$]>3
and pf[\$-1]=3
and pf[\$-2]=2 then                            what="2^k*3*p (made up rule)"
np = lf-1
adj = {0,4,3}
elsif i=1440
and pf={2,2,2,2,2,3,3,5} then                 what="1440 (complete fudge)"
-- nothing quite like this, nothing to build any pattern from...
np = 7
adj = {6,5,3,2,2,2,2}
else                                            what="general (proper rule)"
-- (note this incorporates the p^2, (p>2)^k, p*q, and p*m*q rules)
np = lf
adj = {}
end if
ri = get_primes(-np)
for j=1 to length(adj) do
integer aj = adj[j]
if aj!=0 then pf[-j] = aj end if
end for
for j=1 to np do
ri[j] = {ri[j],pf[-j]-1}
end for

string short_form = ""  -- (eg "2^2*3^3" form)
mpz_set_si(r,1)     -- (above as big int)
for j=1 to length(ri) do
{p, e} = ri[j]
if length(short_form) then short_form &= "*" end if
short_form &= sprintf("%d",p)
if e!=1 then
short_form &= sprintf("^%d",{e})
end if
mpz_ui_pow_ui(pn,p,e)
mpz_mul(r,r,pn)
end for
if i<=15 or remainder(i-1,250)>=248 or i=1440 then
string rs = mpz_get_str(r)
if length(rs)>20 then
rs[6..-6] = sprintf("<-- %d digits -->",length(rs)-10)
end if
if short_form="2^0" then short_form = "1" end if
printf(1,"%4d : %25s %30s %s\n",{i,short_form,rs,what})
end if
integer k = find(what,rule_names)
if k=0 then
rule_names = append(rule_names,what)
rule_counts = append(rule_counts,1)
else
rule_counts[k] += 1
end if
end for
integer lr = length(rule_names)
printf(1,"\nrules(%d):\n",lr)
sequence tags = custom_sort(rule_counts, tagset(lr))
for i=1 to lr do
integer ti = tags[-i]
printf(1,"  %30s:%d\n",{rule_names[ti],rule_counts[ti]})
end for
{r,pn} = mpz_free({r,pn})
```
Output:
```   1 :                         1                              1 prime (proper rule)
2 :                         2                              2 prime (proper rule)
3 :                       2^2                              4 prime (proper rule)
4 :                       2*3                              6 2^k (made up rule)
5 :                       2^4                             16 prime (proper rule)
6 :                     2^2*3                             12 2^k*3 (made up rule)
7 :                       2^6                             64 prime (proper rule)
8 :                     2^3*3                             24 2^k (made up rule)
9 :                   2^2*3^2                             36 general (proper rule)
10 :                     2^4*3                             48 general (proper rule)
11 :                      2^10                           1024 prime (proper rule)
12 :                   2^2*3*5                             60 2^k*3 (made up rule)
13 :                      2^12                           4096 prime (proper rule)
14 :                     2^6*3                            192 general (proper rule)
15 :                   2^4*3^2                            144 general (proper rule)
249 :                  2^82*3^2    43521<-- 16 digits -->22336 general (proper rule)
250 :             2^4*3^4*5^4*7                        5670000 general (proper rule)
499 :                     2^498   81834<-- 140 digits -->97344 prime (proper rule)
500 :          2^4*3^4*5^4*7*11                       62370000 general (proper rule)
749 :                 2^106*3^6    59143<-- 25 digits -->22656 general (proper rule)
750 :        2^4*3^4*5^4*7^2*11                      436590000 general (proper rule)
999 :          2^36*3^2*5^2*7^2                757632231014400 general (proper rule)
1000 :       2^4*3^4*5^4*7*11*13                      810810000 general (proper rule)
1249 :                    2^1248   48465<-- 366 digits -->22656 prime (proper rule)
1250 :        2^4*3^4*5^4*7^4*11                    21392910000 general (proper rule)
1440 :    2^5*3^4*5^2*7*11*13*17                     1102701600 1440 (complete fudge)
1499 :                    2^1498   87686<-- 441 digits -->37344 prime (proper rule)
1500 :     2^4*3^4*5^4*7^2*11*13                     5675670000 general (proper rule)
1749 :             2^52*3^10*5^2    66483<-- 12 digits -->57600 general (proper rule)
1750 :        2^6*3^4*5^4*7^4*11                    85571640000 general (proper rule)
1999 :                    2^1998   28703<-- 592 digits -->57344 prime (proper rule)
2000 :    2^4*3^4*5^4*7*11*13*17                    13783770000 general (proper rule)

rules(11):
general (proper rule):1583
prime (proper rule):304
2^k*p (made up rule):59
2^k*3*p (made up rule):9
2^k*3^3*p (made up rule):9
2^k (made up rule):9
2^k*3 (made up rule):9
[2^k]*3^(>=4)*p (made up rule):8
2^k*3^2*p (made up rule):5
2^k*3^4*p (made up rule):4
1440 (complete fudge):1
```

## PROMAL

Builds a table of divisor counts wihout division.

```;;; Find the smallest number with n divisors, n in 1..15

PROGRAM SmallestN
INCLUDE LIBRARY

CON WORD maxDivisors =   15 ; number of sequence elements to find
CON WORD maxNUmber   = 6000 ; maximum number we will consider

WORD dc [ maxNumber + 1 ]
WORD firstWithDivisors [ maxDivisors + 1 ]
WORD found
WORD divisors
WORD d
WORD n
WORD j

BEGIN
; compute a table of divisor counts
FOR n = 1 TO maxNumber
dc[ n ] = 0
FOR n = 1 TO maxNumber
j = n
WHILE j <= maxNumber
dc[ j ] = dc[ j ] + 1
j = j + n
; find the first number wih the required divisor counts
FOR n = 0 TO maxDivisors
firstWithDivisors[ n ] = 0
found = 0
n = 0
WHILE found < maxDivisors
n = n + 1
divisors = dc[ n ]
IF divisors <= maxDivisors
IF firstWithDivisors[ divisors ] = 0
; first number with this number of divisors
found = found + 1
firstWithDivisors[ divisors ] = n
FOR d = 1 TO maxDivisors
OUTPUT " #W", firstWithDivisors[ d ]
END```
Output:
``` 1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```

## Python

### Procedural

```def divisors(n):
divs = [1]
for ii in range(2, int(n ** 0.5) + 3):
if n % ii == 0:
divs.append(ii)
divs.append(int(n / ii))
divs.append(n)
return list(set(divs))

def sequence(max_n=None):
n = 0
while True:
n += 1
ii = 0
if max_n is not None:
if n > max_n:
break
while True:
ii += 1
if len(divisors(ii)) == n:
yield ii
break

if __name__ == '__main__':
for item in sequence(15):
print(item)
```
Output:
```1
2
4
6
16
12
64
24
36
48
1024
60
4096
192
144```

### Functional

Aiming for functional transparency, speed of writing, high levels of code reuse, and ease of maintenance.

Not optimized, but still fast at this modest scale.

```'''Smallest number with exactly n divisors'''

from itertools import accumulate, chain, count, groupby, islice, product
from functools import reduce
from math import sqrt, floor
from operator import mul

# a005179 :: () -> [Int]
def a005179():
'''Integer sequence: smallest number with exactly n divisors.'''
return (
next(
x for x in count(1)
if n == 1 + len(properDivisors(x))
) for n in count(1)
)

# --------------------------TEST---------------------------
# main :: IO ()
def main():
'''First 15 terms of a005179'''
print(main.__doc__ + ':\n')
print(
take(15)(
a005179()
)
)

# -------------------------GENERIC-------------------------

# properDivisors :: Int -> [Int]
def properDivisors(n):
'''The ordered divisors of n, excluding n itself.
'''
def go(a, x):
return [a * b for a, b in product(
a,
accumulate(chain([1], x), mul)
)]
return sorted(
reduce(go, [
list(g) for _, g in groupby(primeFactors(n))
], [1])
)[:-1] if 1 < n else []

# primeFactors :: Int -> [Int]
def primeFactors(n):
'''A list of the prime factors of n.
'''
def f(qr):
r = qr[1]
return step(r), 1 + r

def step(x):
return 1 + (x << 2) - ((x >> 1) << 1)

def go(x):
root = floor(sqrt(x))

def p(qr):
q = qr[0]
return root < q or 0 == (x % q)

q = until(p)(f)(
(2 if 0 == x % 2 else 3, 1)
)[0]
return [x] if q > root else [q] + go(x // q)

return go(n)

# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
'''The prefix of xs of length n,
or xs itself if n > length xs.
'''
return lambda xs: (
xs[0:n]
if isinstance(xs, (list, tuple))
else list(islice(xs, n))
)

# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
'''The result of repeatedly applying f until p holds.
The initial seed value is x.
'''
def go(f, x):
v = x
while not p(v):
v = f(v)
return v
return lambda f: lambda x: go(f, x)

# MAIN ---
if __name__ == '__main__':
main()
```
Output:
`[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144]`

## Quackery

`factors` is defined at Factors of an integer.

``` [ 0
[ 1+ 2dup
factors size =
until ]
nip ]            is nfactors ( n --> n )

15 times [ i^ 1+ nfactors echo sp ]```
Output:
`1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144`

## R

Could probably be speeded up by caching the results of divisorCount, but it's quick enough already. Not to mention, delightfully short.

```#Need to add 1 to account for skipping n. Not the most efficient way to count divisors, but quite clear.
divisorCount <- function(n) length(Filter(function(x) n %% x == 0, seq_len(n %/% 2))) + 1
smallestWithNDivisors <- function(n)
{
i <- 1
while(divisorCount(i) != n) i <- i + 1
i
}
print(sapply(1:15, smallestWithNDivisors))
```
Output:
`[1]    1    2    4    6   16   12   64   24   36   48 1024   60 4096  192  144`

## Raku

(formerly Perl 6)

Works with: Rakudo version 2019.03
```sub div-count (\x) {
return 2 if x.is-prime;
+flat (1 .. x.sqrt.floor).map: -> \d {
unless x % d { my \y = x div d; y == d ?? y !! (y, d) }
}
}

my \$limit = 15;

put "First \$limit terms of OEIS:A005179";
put (1..\$limit).map: -> \$n { first { \$n == .&div-count }, 1..Inf };
```
Output:
```First 15 terms of OEIS:A005179
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144```

## REXX

### some optimization

```/*REXX program finds and displays  the   smallest number   with  exactly   N   divisors.*/
parse arg N .                                    /*obtain optional argument from the CL.*/
if N=='' | N==","  then N= 15                    /*Not specified?  Then use the default.*/
say '──divisors──  ──smallest number with N divisors──' /*display title for the numbers.*/
@.=                                              /*the  @  array is used for memoization*/
do i=1  for N;     z= 1  +  (i\==1)          /*step through a number of divisors.   */
do j=z  by z                              /*now, search for a number that ≡ #divs*/
if @.j==.  then iterate                   /*has this number already been found?  */
d= #divs(j);    if d\==i  then iterate    /*get # divisors;  Is not equal?  Skip.*/
say center(i, 12)  right(j, 19)           /*display the #divs and the smallest #.*/
@.j= .                                    /*mark as having found #divs for this J*/
leave                                     /*found a number, so now get the next I*/
end   /*j*/
end      /*i*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
#divs: procedure; parse arg x 1 y                /*X and Y:  both set from 1st argument.*/
if x<7  then do; if x<3   then return x   /*handle special cases for one and two.*/
if x<5   then return x-1 /*   "      "      "    "  three & four*/
if x==5  then return 2   /*   "      "      "    "  five.       */
if x==6  then return 4   /*   "      "      "    "  six.        */
end
odd= x // 2                               /*check if   X   is  odd  or not.      */
if odd  then      #= 1                    /*Odd?   Assume  Pdivisors  count of 1.*/
else do;  #= 3;   y= x%2;  end    /*Even?     "        "        "    " 3.*/
/* [↑]   start with known num of Pdivs.*/
do k=3  for x%2-3  by 1+odd  while k<y  /*for odd numbers, skip evens.*/
if x//k==0  then do            /*if no remainder, then found a divisor*/
#=#+2;  y=x%k /*bump  #  Pdivs,  calculate limit  Y. */
if k>=y  then do;  #= #-1;  leave;  end      /*limit?*/
end                                          /*  ___ */
else if k*k>x  then leave        /*only divide up to √ x  */
end   /*k*/                    /* [↑]  this form of DO loop is faster.*/
return #+1                                /*bump "proper divisors" to "divisors".*/
```
output   when using the default input:
```──divisors──  ──smallest number with N divisors──
1                         1
2                         2
3                         4
4                         6
5                        16
6                        12
7                        64
8                        24
9                        36
10                       48
11                     1024
12                       60
13                     4096
14                      192
15                      144
```

### with memorization

```/*REXX program finds and displays  the   smallest number   with  exactly   N   divisors.*/
parse arg N lim .                                /*obtain optional argument from the CL.*/
if   N=='' |   N==","  then   N=      15         /*Not specified?  Then use the default.*/
if lim=='' | lim==","  then lim= 1000000         /* "      "         "   "   "     "    */
say '──divisors──  ──smallest number with N divisors──' /*display title for the numbers.*/
@.=.                                             /*the  @  array is used for memoization*/
do i=1  for N;      z= 1  +  (i\==1)         /*step through a number of divisors.   */
do j=z  by z                              /*now, search for a number that ≡ #divs*/
if @.j\==.  then if @.j\==i  then iterate /*if already been found, is it THE one?*/
d= #divs(j);       if j<lim  then @.j= d  /*get # divisors; if not too high, save*/
if d\==i    then iterate                  /*Is d ¬== i?     Then skip this number*/
say center(i, 12)  right(j, 19)           /*display the #divs and the smallest #.*/
@.j= d                                    /*mark as having found #divs for this J*/
leave                                     /*found a number, so now get the next I*/
end   /*j*/
end      /*i*/
exit                                             /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
#divs: procedure; parse arg x 1 y                /*X and Y:  both set from 1st argument.*/
if x<7  then do; if x<3   then return x   /*handle special cases for one and two.*/
if x<5   then return x-1 /*   "      "      "    "  three & four*/
if x==5  then return 2   /*   "      "      "    "  five.       */
if x==6  then return 4   /*   "      "      "    "  six.        */
end
odd= x // 2                               /*check if   X   is  odd  or not.      */
if odd  then      #= 1                    /*Odd?   Assume  Pdivisors  count of 1.*/
else do;  #= 3;   y= x%2;  end    /*Even?     "        "        "    " 3.*/
/* [↑]   start with known num of Pdivs.*/
do k=3  for x%2-3  by 1+odd  while k<y  /*for odd numbers, skip evens.*/
if x//k==0  then do            /*if no remainder, then found a divisor*/
#=#+2;  y=x%k /*bump  #  Pdivs,  calculate limit  Y. */
if k>=y  then do;  #= #-1;  leave;  end      /*limit?*/
end                                          /*  ___ */
else if k*k>x  then leave        /*only divide up to √ x  */
end   /*k*/                    /* [↑]  this form of DO loop is faster.*/
return #+1                                /*bump "proper divisors" to "divisors".*/
```
output   is identical to the 1st REXX version.

## Ring

```load "stdlib.ring"

see "working..." + nl
see "the first 15 terms of the sequence are:" + nl

for n = 1 to 15
for m = 1 to 4100
pnum = 0
for p = 1 to 4100
if (m % p = 0)
pnum = pnum + 1
ok
next
if pnum = n
see "" + m + " "
exit
ok
next
next

see nl + "done..." + nl```
Output:
```working...
the first 15 terms of the sequence are:
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
done...
```

## RPL

Works with: HP version 49g
```≪ {1}
2 ROT FOR j
1
DO 1 + UNTIL DUP DIVIS SIZE j == END
+
NEXT
≫ 'TASK' STO
```
```15 TASK
```
Output:
```1: {1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144}
```

Runs in 13 minutes on an HP-50g.

#### Faster implementation

With the hint given in the OEIS page that a(p)=2^(p-1) when p is prime:

```≪ {1}
2 ROT FOR j
IF j ISPRIME? THEN 2 j 1 - ^
ELSE
1 DO 1 + UNTIL DUP DIVIS SIZE j == END
END
+
NEXT
≫ 'TASK' STO
```

the same result is obtained in less than a minute.

## Ruby

```require 'prime'

def num_divisors(n)
n.prime_division.inject(1){|prod, (_p,n)| prod *= (n + 1) }
end

def first_with_num_divs(n)
(1..).detect{|i| num_divisors(i) == n }
end

p (1..15).map{|n| first_with_num_divs(n) }
```
Output:
```[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144]
```

## Rust

```use itertools::Itertools;

const PRIMES: [u64; 15] = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47];
const MAX_DIVISOR: usize = 64;

struct DivisorSeq {
max_number_of_divisors: u64,
index: u64,
}

impl DivisorSeq {
fn new(max_number_of_divisors: u64) -> DivisorSeq {
DivisorSeq {
max_number_of_divisors,
index: 1,
}
}
}

impl Iterator for DivisorSeq {
type Item = u64;

fn next(&mut self) -> Option<u64> {
if self.max_number_of_divisors < self.index {
return None;
}
#[allow(unused_mut)]
let mut result: u64;
let divisors_of_divisor = get_divisors(self.index);
match divisors_of_divisor.len() {
1 | 2 => {
// when # divisors is a prime
result = 2_u64.pow(self.index as u32 - 1);
self.index += 1;
}
3 => {
// when # divisors is a prime square
result = 6_u64.pow(divisors_of_divisor[1] as u32 - 1);
self.index += 1;
}
4 => {
// when # divisors is a prime * non-prime
result = 2_u64.pow(divisors_of_divisor[2] as u32 - 1)
* 3_u64.pow(divisors_of_divisor[1] as u32 - 1);
self.index += 1;
}
8 if divisors_of_divisor
.iter()
.filter(|x| PRIMES.contains(x))
.count()
== 3 =>
{
// sphenic numbers, aka p*m*q, where, p, m and q are prime
let first_primes = divisors_of_divisor
.iter()
.filter(|x| PRIMES.contains(x))
.collect::<Vec<_>>();
result = 2_u64.pow(*first_primes[2] as u32 - 1)
* 3_u64.pow(*first_primes[1] as u32 - 1)
* 5_u64.pow(*first_primes[0] as u32 - 1);
self.index += 1;
}
_ => {
// brute force and slow: iterates over the numbers to find
// one with the appropriate number of divisors
let mut x: u64 = 1;
loop {
if get_divisors(x).len() as u64 == self.index {
break;
}
x += 1;
}

result = x;
self.index += 1;
}
}
Some(result)
}
}
/// Gets all divisors of a number
fn get_divisors(n: u64) -> Vec<u64> {
let mut results = Vec::new();

for i in 1..(n / 2 + 1) {
if n % i == 0 {
results.push(i);
}
}
results.push(n);
results
}

fn main() {
// simple version using factorizing numbers
// with rules applied from A005179 so speed up
// but as such it is slow in some cases, e.g for 52
let seq_iter = DivisorSeq::new(64);
println!("Simple method with rules");
println!("# divisors     Smallest number");
for (i, x) in seq_iter.enumerate() {
println!("{:>10}{:20}", i + 1, x);
}

// more advanced version using calculations based on number of
// prime factors and their exponent

// load initial result table with an initial value of 2**n for each item
let mut min_numbers = vec![0_u64; MAX_DIVISOR];
(0_usize..MAX_DIVISOR).for_each(|n| min_numbers[n] = 2_u64.pow(n as u32));

let prime_list = (1..15).map(|i| PRIMES[0..=i].to_vec()).collect::<Vec<_>>();

for pl in prime_list.iter() {
// calculate the max exponent a prime can get in a given prime-list
// to be able to produce the desired number of divisors
let max_exponent = 1 + MAX_DIVISOR as u32 / 2_u32.pow(pl.len() as u32 - 1);

// create a combination of exponents using cartesian product
let exponents = (1_usize..=pl.len())
.map(|_| 1_u32..=max_exponent)
.multi_cartesian_product()
.filter(|elt| {
let mut prev = None::<&u32>;
elt.iter().all(|x| match prev {
Some(n) if x > n => false,
_ => {
prev = Some(x);
true
}
})
});

// iterate throught he exponent combinations
for exp in exponents {
// calculate the number of divisors using the formula
// given primes of p, q, r
// and exponents of a1, a2, a3
// the # divisors is (a1+1)* (a2+1) *(a3+1)
let num_of_divisors = exp.iter().map(|x| x + 1).product::<u32>() as usize;

// and calculate the number with those primes and the given exponent set
let num = pl.iter().zip(exp.iter()).fold(1_u64, |mut acc, (p, e)| {
// check for overflow if numbers won't fit into u64
acc = match acc.checked_mul(p.pow(*e)) {
Some(z) => z,
_ => 0,
};
acc
});

// finally, if the number is less than what we have so far in the result table
// replace the result table with the smaller number
if num > 0
&& min_numbers.len() >= num_of_divisors
&& min_numbers[num_of_divisors - 1] > num
{
min_numbers[num_of_divisors - 1] = num;
}
}
}

println!("Advanced method");
println!("# divisors     Smallest number");
for (i, x) in min_numbers.iter().enumerate() {
println!("{:>10}{:20}", i + 1, x);
}
}
```
Output:
```Simple method with rules
# divisors     Smallest number
1                   1
2                   2
3                   4
4                   6
5                  16
6                  12
7                  64
8                  24
9                  36
10                  48
11                1024
12                  60
13                4096
14                 192
15                 144
16                 120
17               65536
18                 180
19              262144
20                 240
21                 576
22                3072
23             4194304
24                 360
25                1296
26               12288
27                2304
28                 960
29           268435456
30                 720
31          1073741824
32                 840
33                9216
34              196608
35                5184
36                1260
37         68719476736
38              786432
39               36864
40                1680
41       1099511627776
42                2880
43       4398046511104
44               15360
45                3600
46            12582912
47      70368744177664
48                2520
49               46656
50                6480
51              589824
52               61440
53    4503599627370496
54                6300
55               82944
56                6720
57             2359296
58           805306368
59  288230376151711744
60                5040
61 1152921504606846976
62          3221225472
63               14400
64                7560
```

## SETL

```program smallest_number_with_exactly_n_divisors;
print([a(n) : n in [1..15]]);

proc a(n);
return [i +:= 1 : until n = #[d : d in [1..i] | i mod d=0]](i);
end proc;
end program;```
Output:
`[1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144]`

## Sidef

```func n_divisors(n) {
1..Inf -> first_by { .sigma0 == n }
}

say 15.of { n_divisors(_+1) }
```
Output:
```[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144]
```

More efficient solution:

```func n_divisors(threshold, least_solution = Inf, k = 1, max_a = Inf, solutions = 1, n = 1) {

if (solutions == threshold) {
return n
}

if (solutions > threshold) {
return least_solution
}

var p = k.prime

for a in (1 .. max_a) {
n *= p
break if (n > least_solution)
least_solution = __FUNC__(threshold, least_solution, k+1, a, solutions * (a + 1), n)
}

return least_solution
}

say n_divisors(60)     #=> 5040
say n_divisors(1000)   #=> 810810000
```

## Swift

```// See https://en.wikipedia.org/wiki/Divisor_function
func divisorCount(number: Int) -> Int {
var n = number
var total = 1
// Deal with powers of 2 first
while n % 2 == 0 {
total += 1
n /= 2
}
// Odd prime factors up to the square root
var p = 3
while p * p <= n {
var count = 1
while n % p == 0 {
count += 1
n /= p
}
total *= count
p += 2
}
// If n > 1 then it's prime
if n > 1 {
total *= 2
}
return total
}

let limit = 15
var sequence = Array(repeating: 0, count: limit)
var count = 0
var n = 1
while count < limit {
let divisors = divisorCount(number: n)
if divisors <= limit && sequence[divisors - 1] == 0 {
sequence[divisors - 1] = n
count += 1
}
n += 1
}
for n in sequence {
print(n, terminator: " ")
}
print()
```
Output:
```1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```

## Tcl

```proc divCount {n} {
set cnt 0
for {set d 1} {(\$d * \$d) <= \$n} {incr d} {
if {0 == (\$n % \$d)} {
incr cnt
if {\$d < (\$n / \$d)} {
incr cnt
}
}
}
return \$cnt
}

proc A005179 {n} {
if {\$n >= 1} {
for {set k 1} {1} {incr k} {
if {\$n == [divCount \$k]} {
return \$k
}
}
}
return 0
}

proc show {func lo hi} {
puts "\${func}(\$lo..\$hi) ="
for {set n \$lo} {\$n <= \$hi} {incr n} {
puts -nonewline " [\$func \$n]"
}
puts ""
}

show A005179 1 15
```
Output:
```A005179(1..15) =
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144```

## Wren

Library: Wren-math
```import "./math" for Int

var limit = 22
var numbers = List.filled(limit, 0)
var i = 1
while (true) {
var nd = Int.divisors(i).count
if (nd <= limit && numbers[nd-1] == 0) {
numbers[nd-1] = i
if (numbers.all { |n| n > 0 }) break
}
i = i + 1
}
System.print("The first %(limit) terms are:")
System.print(numbers)
```
Output:
```The first 22 terms are:
[1, 2, 4, 6, 16, 12, 64, 24, 36, 48, 1024, 60, 4096, 192, 144, 120, 65536, 180, 262144, 240, 576, 3072]
```

## XPL0

```func Divisors(N); \Return number of divisors of N
int  N, Count, D;
[Count:= 0;
for D:= 1 to N do
if rem(N/D) = 0 then Count:= Count+1;
return Count;
];

int N, AN;
[for N:= 1 to 15 do
[AN:= 0;
repeat AN:= AN+1 until Divisors(AN) = N;
IntOut(0, AN);  ChOut(0, ^ );
];
]```
Output:
```1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```

## zkl

```fcn countDivisors(n)
{ [1.. n.toFloat().sqrt()].reduce('wrap(s,i){ s + (if(0==n%i) 1 + (i!=n/i)) },0) }
A005179w:=(1).walker(*).tweak(fcn(n){
var N=0,cache=Dictionary();
if(cache.find(n)) return(cache.pop(n));	// prune
while(1){
if(n == (d:=countDivisors(N+=1))) return(N);
if(n<d and not cache.find(d)) cache[d]=N;
}
});```
```N:=15;
println("First %d terms of OEIS:A005179".fmt(N));
A005179w.walk(N).concat(" ").println();```
Output:
```First 15 terms of OEIS:A005179
1 2 4 6 16 12 64 24 36 48 1024 60 4096 192 144
```