# Sequence: smallest number greater than previous term with exactly n divisors

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

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

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

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

L(item) sequence(15)
print(item)```
Output:
```1
2
4
6
16
18
64
66
100
112
1024
1035
4096
4288
4624
```

## 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()
CARD a
BYTE i

a=1
FOR i=1 TO 15
DO
WHILE CountDivisors(a)#i
DO
a==+1
OD
IF i>1 THEN
Print(", ")
FI
PrintC(a)
OD
RETURN```
Output:
```1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624
```

Translation of: Go
```with Ada.Text_IO;

procedure Show_Sequence is

function Count_Divisors (N : in Natural) return Natural is
Count : Natural := 0;
I     : Natural;
begin
I := 1;
while I**2 <= N loop
if N mod I = 0 then
if I = N / I then
Count := Count + 1;
else
Count := Count + 2;
end if;
end if;
I := I + 1;
end loop;

return Count;
end Count_Divisors;

procedure Show (Max : in Natural) is
N : Natural := 1;
Begin
Put_Line ("The first" & Max'Image & "terms of the sequence are:");
for Divisors in 1 .. Max loop
while Count_Divisors (N) /= Divisors loop
N := N + 1;
end loop;
Put (N'Image);
end loop;
New_Line;
end Show;

begin
Show (15);
end Show_Sequence;
```
Output:
```The first 15terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624```

## ALGOL 68

Translation of: Go

with a small optimisation.

```BEGIN

PROC count divisors = ( INT n )INT:
BEGIN
INT i2, count := 0;
FOR i WHILE ( i2 := i * i ) < n DO
IF n MOD i = 0 THEN count +:= 2 FI
OD;
IF i2 = n THEN count + 1 ELSE count FI
END # count divisors # ;

INT max = 15;

print( ( "The first ", whole( max, 0 ), " terms of the sequence are:" ) );
INT next := 1;
FOR i WHILE next <= max DO
IF next = count divisors( i ) THEN
print( ( " ", whole( i, 0 ) ) );
next +:= 1
FI
OD

END```
Output:
```The first 15 terms of the sequence are: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## ALGOL W

Translation of: Go

...via Algol 68 and with a small optimisation.

```begin
integer max, next, i;

integer procedure countDivisors ( Integer value n ) ;
begin
integer count, i, i2;
count := 0;
i     := 1;
while  begin i2 := i * i;
i2 < n
end do begin
if n rem i = 0 then count := count + 2;
i := i + 1
end;
if i2 = n then count + 1 else count
end countDivisors ;

max := 15;
write( i_w := 1, s_w := 0, "The first ", max, " terms of the sequence are: " );
i := next := 1;
while next <= max do begin
if next = countDivisors( i ) then begin
writeon( i_w := 1, s_w := 0, " ", i );
next := next +  1
end;
i := i + 1
end;
write()
end.
```
Output:
```The first 15 terms of the sequence are:  1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## Arturo

```i: new 0
next: new 1
MAX: 15
while [next =< MAX][
if next = size factors i [
prints ~"|i| "
inc 'next
]
inc 'i
]
print ""
```
Output:
`1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624`

## AutoHotkey

Translation of: Go
```MAX := 15
next := 1, i := 1
while (next <= MAX)
if (next = countDivisors(A_Index))
Res.= A_Index ", ", next++
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, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624```

## AWK

```# syntax: GAWK -f SEQUENCE_SMALLEST_NUMBER_GREATER_THAN_PREVIOUS_TERM_WITH_EXACTLY_N_DIVISORS.AWK
# converted from Kotlin
BEGIN {
limit = 15
printf("first %d terms:",limit)
n = 1
while (n <= limit) {
if (n == count_divisors(++i)) {
printf(" %d",i)
n++
}
}
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 18 64 66 100 112 1024 1035 4096 4288 4624
```

## BASIC

### BASIC256

Translation of: FreeBASIC
```UPTO = 15
i = 2
nfound = 1

print 1; " ";    #special case

while nfound < UPTO
n = divisors(i)
if n = nfound + 1 then
nfound += 1
print i; " ";
end if
i += 1
end while
end

function divisors(n)
#find the number of divisors of an integer
r = 2
for i = 2 to n\2
if n mod i = 0 then r += 1
next i
return r
end function
```

### Gambas

Translation of: FreeBASIC
```Public Sub Main()

Dim UPTO As Integer = 15, i As Integer = 2
Dim n As Integer, nfound As Integer = 1

Print 1; " ";    'special case

While nfound < UPTO
n = divisors(i)
If n = nfound + 1 Then
nfound += 1
Print i; " ";
End If
i += 1
Wend

End Sub

Function divisors(n As Integer) As Integer
'find the number of divisors of an integer

Dim r As Integer = 2, i As Integer
For i = 2 To n \ 2
If n Mod i = 0 Then r += 1
Next
Return r

End Function
```

### 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, found

UPTO = 15
i = 2
nfound = 1

Print("1 ")    ;special case

While nfound < UPTO
n = divisors(i)
If n = nfound + 1:
nfound + 1
Print(Str(i) + " ")
EndIf
i + 1
Wend
PrintN(#CRLF\$ + "Press ENTER to exit"): Input()
CloseConsole()```

### QBasic

Translation of: FreeBASIC
Works with: QBasic version 1.1
Works with: QuickBasic version 4.5
```FUNCTION divisors (n)
'find the number of divisors of an integer
r = 2
FOR i = 2 TO n \ 2
IF n MOD i = 0 THEN r = r + 1
NEXT i
divisors = r
END FUNCTION

UPTO = 15
i = 2
nfound = 1

PRINT 1;    'special case

WHILE nfound < UPTO
n = divisors(i)
IF n = nfound + 1 THEN
nfound = nfound + 1
PRINT i;
END IF
i = i + 1
WEND
```

### Run BASIC

Translation of: FreeBASIC
Works with: Just BASIC
Works with: Liberty BASIC
```UPTO = 15
i = 2
nfound = 1

print 1; " ";    'special case

while nfound < UPTO
n = divisors(i)
if n = nfound + 1 then
nfound = nfound + 1
print i; " ";
end if
i = i + 1
wend
print
end

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

### XBasic

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

DECLARE FUNCTION  Entry ()
DECLARE  FUNCTION divisors (n)

FUNCTION  Entry ()
UPTO = 15
i = 2
nfound = 1

PRINT 1;    'special case

DO WHILE nfound < UPTO
n = divisors(i)
IF n = nfound + 1 THEN
INC nfound
PRINT i;
END IF
INC i
LOOP
END FUNCTION

FUNCTION divisors (n)
'find the number of divisors of an integer
r = 2
FOR i = 2 TO n / 2
IF n MOD i = 0 THEN INC r
NEXT i
RETURN r
END FUNCTION
END PROGRAM
```

### Yabasic

Translation of: FreeBASIC
```UPTO = 15
i = 2
nfound = 1

print 1, " ";    //special case

while nfound < UPTO
n = divisors(i)
if n = nfound + 1 then
nfound = nfound + 1
print i, " ";
fi
i = i + 1
end while
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
```

## 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, next = 1;
printf("The first %d terms of the sequence are:\n", MAX);
for (i = 1; next <= MAX; ++i) {
if (next == count_divisors(i)) {
printf("%d ", i);
next++;
}
}
printf("\n");
return 0;
}
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## 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() {
cout << "The first " << MAX << " terms of the sequence are:" << endl;
for (int i = 1, next = 1; next <= MAX; ++i) {
if (next == count_divisors(i)) {
cout << i << " ";
next++;
}
}
cout << endl;
return 0;
}
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

### Alternative

Translation of: Pascal

More terms and quicker than the straightforward C version above. Try It Online!

```#include <cstdio>
#include <chrono>

using namespace std::chrono;

const int MAX = 32;

unsigned int getDividersCnt(unsigned int n) {
unsigned int d = 3, q, dRes, res = 1;
while (!(n & 1)) { n >>= 1; res++; }
while ((d * d) <= n) { q = n / d; if (n % d == 0) { dRes = 0;
do { dRes += res; n = q; q /= d; } while (n % d == 0);
res += dRes; } d += 2; } return n != 1 ? res << 1 : res; }

int main() { unsigned int i, nxt, DivCnt;
printf("The first %d anti-primes plus are: ", MAX);
auto st = steady_clock::now(); i = nxt = 1; do {
if ((DivCnt = getDividersCnt(i)) == nxt ) { printf("%d ", i);
if ((++nxt > 4) && (getDividersCnt(nxt) == 2))
i = (1 << (nxt - 1)) - 1; } i++; } while (nxt <= MAX);
printf("%d ms", (int)(duration<double>(steady_clock::now() - st).count() * 1000));
}
```
Output:
`The first 32 anti-primes plus are: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752 268435456 268437200 1073741824 1073741830 223 ms`

## Delphi

Works with: Delphi version 6.0

```{These routines would normally be in a library, but the shown here 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 SmallestWithNDivisors(Memo: TMemo);
var N,Count: integer;
var IA: TIntegerDynArray;
begin
Count:=1;
for N:=1 to high(Integer) do
if Count=GetAllDivisors(N,IA) then
begin
Inc(Count);
if Count>15 then break;
end;
end;
```
Output:
```1 - 1
2 - 2
3 - 4
4 - 6
5 - 16
6 - 18
7 - 64
8 - 66
9 - 100
10 - 112
11 - 1024
12 - 1035
13 - 4096
14 - 4288
15 - 4624

Elapsed Time: 58.590 ms.

```

## Dyalect

Translation of: Go
```func countDivisors(n) {
var count = 0
var i = 1
while i * i <= n {
if n % i == 0 {
if i == n / i {
count += 1
} else {
count += 2
}
}
i += 1
}
return count
}

let max = 15
print("The first \(max) terms of the sequence are:")
var (i, next) = (1, 1)
while next <= max {
if next == countDivisors(i) {
print("\(i) ", terminator: "")
next += 1
}
i += 1
}

print()```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624```

## EasyLang

Translation of: AWK
```func ndivs n .
i = 1
while i <= sqrt n
if n mod i = 0
cnt += 2
if i = n div i
cnt -= 1
.
.
i += 1
.
return cnt
.
n = 1
while n <= 15
i += 1
if n = ndivs i
write i & " "
n += 1
.
.```

## F#

### First 28 are easy with a Naive implementation

```// Nigel Galloway: November 19th., 2017
let fN g=[1..(float>>sqrt>>int)g]|>List.fold(fun Σ n->if g%n>0 then Σ else if g/n=n then Σ+1 else Σ+2) 0
let A069654=let rec fG n g=seq{match g-fN n with 0->yield n; yield! fG(n+1)(g+1) |_->yield! fG(n+1)g} in fG 1 1

A069654 |> Seq.take 28|>Seq.iter(printf "%d "); printfn ""
```
Output:
```1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752
```

## Factor

```USING: io kernel math math.primes.factors prettyprint sequences ;

: next ( n num -- n' num' )
[ 2dup divisors length = ] [ 1 + ] do until [ 1 + ] dip ;

: A069654 ( n -- seq )
[ 2 1 ] dip [ [ next ] keep ] replicate 2nip ;

"The first 15 terms of the sequence are:" print 15 A069654 .
```
Output:
```The first 15 terms of the sequence are:
{ 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 }
```

## 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

print 1;" ";    'special case

while nfound < UPTO
n = divisors(i)
if n = nfound + 1 then
nfound += 1
print i;" ";
end if
i+=1
wend
print
end```
Output:
` 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624`

## 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
fmt.Println("The first", max, "terms of the sequence are:")
for i, next := 1, 1; next <= max; i++ {
if next == countDivisors(i) {
fmt.Printf("%d ", i)
next++
}
}
fmt.Println()
}
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

```import Text.Printf (printf)

sequence_A069654 :: [(Int,Int)]
sequence_A069654 = go 1 \$ (,) <*> countDivisors <\$> [1..]
where go t ((n,c):xs) | c == t    = (t,n):go (succ t) xs
| otherwise = go t xs
countDivisors n = foldr f 0 [1..floor \$ sqrt \$ realToFrac n]
where f x r | n `mod` x == 0 = if n `div` x == x then r+1 else r+2
| otherwise      = r

main :: IO ()
main = mapM_ (uncurry \$ printf "a(%2d)=%5d\n") \$ take 15 sequence_A069654
```
Output:
```a( 1)=    1
a( 2)=    2
a( 3)=    4
a( 4)=    6
a( 5)=   16
a( 6)=   18
a( 7)=   64
a( 8)=   66
a( 9)=  100
a(10)=  112
a(11)= 1024
a(12)= 1035
a(13)= 4096
a(14)= 4288
a(15)= 4624
```

## J

```sieve=: 3 :0
NB. sieve y  returns a vector of y boxes.
NB. In each box is an array of 2 columns.
NB. The first column is the factor tally
NB. and the second column is a number with
NB. that many factors.
NB. Rather than factoring, the algorithm
NB. counts prime seive cell hits.
NB. The boxes are not ordered by factor tally.
range=. <. + i.@:|@:-
tally=. y#0
for_i.#\tally do.
j=. }:^:(y<:{:)i * 1 range >: <. y % i
tally=. j >:@:{`[`]} tally
end.
(</.~ {."1) (,. i.@:#)tally
)
```
```   A=: sieve 100000
B=: /:~ A
missing=: [: (-.~i.@:#) (<0 0)&{&>
C=: ({.~ {.@:missing) B
D=:{:"1 L:_1 C
(] , [ {~ (I. {:))&.>/@:|. D
+-----------------------------------------------------------------------+
|0 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572|
+-----------------------------------------------------------------------+
```

Intermediate steps and explanation for a smaller sieve:

```   ] A=: sieve 60
+---+---+----+----+----+----+----+----+----+-----+
|0 0|1 1|2  2|3  4|4  6|6 12|5 16|8 24|9 36|10 48|
|   |   |2  3|3  9|4  8|6 18|    |8 30|    |     |
|   |   |2  5|3 25|4 10|6 20|    |8 40|    |     |
|   |   |2  7|3 49|4 14|6 28|    |8 42|    |     |
|   |   |2 11|    |4 15|6 32|    |8 54|    |     |
|   |   |2 13|    |4 21|6 44|    |8 56|    |     |
|   |   |2 17|    |4 22|6 45|    |    |    |     |
|   |   |2 19|    |4 26|6 50|    |    |    |     |
|   |   |2 23|    |4 27|6 52|    |    |    |     |
|   |   |2 29|    |4 33|    |    |    |    |     |
|   |   |2 31|    |4 34|    |    |    |    |     |
|   |   |2 37|    |4 35|    |    |    |    |     |
|   |   |2 41|    |4 38|    |    |    |    |     |
|   |   |2 43|    |4 39|    |    |    |    |     |
|   |   |2 47|    |4 46|    |    |    |    |     |
|   |   |2 53|    |4 51|    |    |    |    |     |
|   |   |2 59|    |4 55|    |    |    |    |     |
|   |   |    |    |4 57|    |    |    |    |     |
|   |   |    |    |4 58|    |    |    |    |     |
+---+---+----+----+----+----+----+----+----+-----+

] B=: /:~ A            NB. ascending sort by tally
+---+---+----+----+----+----+----+----+----+-----+
|0 0|1 1|2  2|3  4|4  6|5 16|6 12|8 24|9 36|10 48|
|   |   |2  3|3  9|4  8|    |6 18|8 30|    |     |
|   |   |2  5|3 25|4 10|    |6 20|8 40|    |     |
|   |   |2  7|3 49|4 14|    |6 28|8 42|    |     |
|   |   |2 11|    |4 15|    |6 32|8 54|    |     |
|   |   |2 13|    |4 21|    |6 44|8 56|    |     |
|   |   |2 17|    |4 22|    |6 45|    |    |     |
|   |   |2 19|    |4 26|    |6 50|    |    |     |
|   |   |2 23|    |4 27|    |6 52|    |    |     |
|   |   |2 29|    |4 33|    |    |    |    |     |
|   |   |2 31|    |4 34|    |    |    |    |     |
|   |   |2 37|    |4 35|    |    |    |    |     |
|   |   |2 41|    |4 38|    |    |    |    |     |
|   |   |2 43|    |4 39|    |    |    |    |     |
|   |   |2 47|    |4 46|    |    |    |    |     |
|   |   |2 53|    |4 51|    |    |    |    |     |
|   |   |2 59|    |4 55|    |    |    |    |     |
|   |   |    |    |4 57|    |    |    |    |     |
|   |   |    |    |4 58|    |    |    |    |     |
+---+---+----+----+----+----+----+----+----+-----+

NB. truncate to the fully populated length

NB. missing lists the unavailable factor tallies
missing=: [: (-.~i.@:#) (<0 0)&{&>
] C=: ({.~ {.@:missing) B   NB. retain tally counts with no holes
+---+---+----+----+----+----+----+
|0 0|1 1|2  2|3  4|4  6|5 16|6 12|
|   |   |2  3|3  9|4  8|    |6 18|
|   |   |2  5|3 25|4 10|    |6 20|
|   |   |2  7|3 49|4 14|    |6 28|
|   |   |2 11|    |4 15|    |6 32|
|   |   |2 13|    |4 21|    |6 44|
|   |   |2 17|    |4 22|    |6 45|
|   |   |2 19|    |4 26|    |6 50|
|   |   |2 23|    |4 27|    |6 52|
|   |   |2 29|    |4 33|    |    |
|   |   |2 31|    |4 34|    |    |
|   |   |2 37|    |4 35|    |    |
|   |   |2 41|    |4 38|    |    |
|   |   |2 43|    |4 39|    |    |
|   |   |2 47|    |4 46|    |    |
|   |   |2 53|    |4 51|    |    |
|   |   |2 59|    |4 55|    |    |
|   |   |    |    |4 57|    |    |
|   |   |    |    |4 58|    |    |
+---+---+----+----+----+----+----+

NB. Remove the tally column from each box (by retaining the values)
,.L:_1 D=:{:"1 L:_1 C
+-+-+--+--+--+--+--+
|0|1| 2| 4| 6|16|12|
| | | 3| 9| 8|  |18|
| | | 5|25|10|  |20|
| | | 7|49|14|  |28|
| | |11|  |15|  |32|
| | |13|  |21|  |44|
| | |17|  |22|  |45|
| | |19|  |26|  |50|
| | |23|  |27|  |52|
| | |29|  |33|  |  |
| | |31|  |34|  |  |
| | |37|  |35|  |  |
| | |41|  |38|  |  |
| | |43|  |39|  |  |
| | |47|  |46|  |  |
| | |53|  |51|  |  |
| | |59|  |55|  |  |
| | |  |  |57|  |  |
| | |  |  |58|  |  |
+-+-+--+--+--+--+--+

NB. finally enlist successively larger values
(] , [ {~ (I. {:))&.>/@:|. D
+---------------+
|0 1 2 4 6 16 18|
+---------------+
```

## Java

Translation of: C
```public class AntiPrimesPlus {

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;
System.out.printf("The first %d terms of the sequence are:\n", max);
for (int i = 1, next = 1; next <= max; ++i) {
if (next == count_divisors(i)) {
System.out.printf("%d ", i);
next++;
}
}
System.out.println();
}
}
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

```# The number of prime factors (as in prime factorization)
def numfactors:
. as \$num
| reduce range(1; 1 + sqrt|floor) as \$i (null;
if (\$num % \$i) == 0
then (\$num / \$i) as \$r
| if \$i == \$r then .+1 else .+2 end
else .
end );

# Output: a stream
def A06954:
foreach range(1; infinite) as \$i ({k: 0};
.j = .k + 1
| until( \$i == (.j | numfactors); .j += 1)
| .k = .j ;
.j ) ;

"First 15 terms of OEIS sequence A069654: ",
[limit(15; A06954)]```
Output:
```First 15 terms of OEIS sequence A069654:
[1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624]
```

## Julia

Translation of: Perl
```using Primes

function numfactors(n)
f = [one(n)]
for (p,e) in factor(n)
f = reduce(vcat, [f*p^j for j in 1:e], init=f)
end
length(f)
end

function A06954(N)
println("First \$N terms of OEIS sequence A069654: ")
k = 0
for i in 1:N
j = k
while (j += 1) > 0
if i == numfactors(j)
print("\$j ")
k = j
break
end
end
end
end

A06954(15)
```
Output:
```First 15 terms of OEIS sequence A069654:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## 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() {
println("The first \$MAX terms of the sequence are:")
var i = 1
var next = 1
while (next <= MAX) {
if (next == countDivisors(i)) {
print("\$i ")
next++
}
i++
}
println()
}
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## Mathematica / Wolfram Language

```res = {#, DivisorSigma[0, #]} & /@ Range[100000];
highest = 0;
filter = {};
Do[
If[r[[2]] == highest + 1,
AppendTo[filter, r[[1]]];
highest = r[[2]]
]
,
{r, res}
]
Take[filter, 15]
```
Output:
`{1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624}`

## Maxima

```sngptend(n):=block([i:1,count:1,result:[]],
while count<=n do (if length(listify(divisors(i)))=count then (result:endcons(i,result),count:count+1),i:i+1),
result)\$

/* Test case */
sngptend(15);
```
Output:
```[1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624]
```

## MiniScript

This GUI implementation is for use with Mini Micro.

```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 = []
j = 1
for i in range(1, 15)
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, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624```

## Nim

```import strformat

const MAX = 15

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

var i, next = 1
echo fmt"The first {MAX} terms of the sequence are:"
while next <= MAX:
if next == countDivisors(i):
write(stdout, fmt"{i} ")
inc next
inc i
write(stdout, "\n")
```
Output:
```The first 15 terms of the sequence are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## Pascal

Counting divisors by prime factorisation.
If divCnt= Count of divisors is prime then the only candidate ist n = prime^(divCnt-1). There will be more rules. If divCnt is odd then the divisors of divCnt are a^(even_factor*i)*..*k^(even_factor*j). I think of next = 33 aka 11*3 with the solution 1031^2 * 2^10=1,088,472,064 with a big distance to next= 32 => 1073741830.
Try it online!

```program AntiPrimesPlus;
{\$IFDEF FPC}
{\$MODE Delphi}
{\$ELSE}
{\$APPTYPE CONSOLE} // delphi
{\$ENDIF}
uses
sysutils,math;
const
MAX =32;

function getDividersCnt(n:Uint32):Uint32;
// getDividersCnt by dividing n into its prime factors
// aka n = 2250 = 2^1*3^2*5^3 has (1+1)*(2+1)*(3+1)= 24 dividers
var
divi,quot,deltaRes,rest : Uint32;
begin
result := 1;

//divi  := 2; //separat without division
while Not(Odd(n)) do
Begin
n := n SHR 1;
inc(result);
end;

//from now on only odd numbers
divi  := 3;
while (sqr(divi)<=n) do
Begin
DivMod(n,divi,quot,rest);
if rest = 0 then
Begin
deltaRes := 0;
repeat
inc(deltaRes,result);
n := quot;
DivMod(n,divi,quot,rest);
until rest <> 0;
inc(result,deltaRes);
end;
inc(divi,2);
end;
//if last factor of n is prime
IF n <> 1 then
result := result*2;
end;

var
T0 : Int64;
i,next,DivCnt: Uint32;
begin
writeln('The first ',MAX,' anti-primes plus are:');
T0:= GetTickCount64;
i := 1;
next := 1;
repeat
DivCnt := getDividersCnt(i);
IF DivCnt= next then
Begin
write(i,' ');
inc(next);
//if next is prime then only prime( => mostly 2 )^(next-1) is solution
IF (next > 4) AND (getDividersCnt(next) = 2) then
i := 1 shl (next-1) -1;// i is incremented afterwards
end;
inc(i);
until Next > MAX;
writeln;
writeln(GetTickCount64-T0,' ms');
end.
```
Output:
```The first 32 anti-primes plus are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752 268435456 268437200 1073741824 1073741830
525 ms```

## Perl

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

print "First 15 terms of OEIS: A069654\n";
my \$m = 0;
for my \$n (1..15) {
my \$l = \$m;
while (++\$l) {
print("\$l "), \$m = \$l, last if \$n == divisors(\$l);
}
}
```
Output:
```First 15 terms of OEIS: A069654
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624```

## Phix

Uses the optimisation trick from pascal, of n:=power(2,next-1) when next is a prime>4.

```with javascript_semantics
constant limit = 15
sequence res = repeat(0,limit)
integer next = 1
atom n = 1
while next<=limit do
integer k = length(factors(n,1))
if k=next then
res[k] = n
next += 1
if next>4 and is_prime(next) then
n := power(2,next-1)-1 -- (-1 for +=1 next)
end if
end if
n += 1
end while
printf(1,"The first %d terms are: %v\n",{limit,res})
```
Output:
```The first 15 terms are: {1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624}
```

You can raise the limit to 32, the 25th takes about 4s but the rest are all near-instant, however I lost patience waiting for the 33rd.
(The last two trailing .0 just mean "not a 31-bit integer" and don't appear when run on 64 bit.)

```The first 32 terms are: {1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624,
4632,65536,65572,262144,262192,263169,269312,4194304,
4194306,4477456,4493312,4498641,4498752,268435456,
268437200,1073741824.0,1073741830.0}
```

## PL/M

Translation of: Go

via Algol 68

```100H: /* FIND THE SMALLEST NUMBER > THE PREVIOUS ONE WITH EXACTLY N DIVISORS */

/* CP/M BDOS SYSTEM CALL */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
/* CONSOLE OUTPUT ROUTINES */
PR\$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C );  END;
PR\$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S );  END;
PR\$NL:     PROCEDURE; CALL PR\$STRING( .( 0DH, 0AH, '\$' ) );       END;
PR\$NUMBER: PROCEDURE( N );
DECLARE V ADDRESS, N\$STR( 6 ) BYTE INITIAL( '.....\$' ), W BYTE;
N\$STR( W := LAST( N\$STR ) - 1 ) = '0' + ( ( V := N ) MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR\$STRING( .N\$STR( W ) );
END PR\$NUMBER;

/* RETURNS THE DIVISOR COUNT OF N */
DECLARE ( I, I2, COUNT ) ADDRESS;
COUNT = 0;
I     = 1;
DO WHILE( ( I2 := I * I ) < N );
IF N MOD I = 0 THEN COUNT = COUNT + 2;
I = I + 1;
END;
IF I2 = N THEN RETURN ( COUNT + 1 ); ELSE RETURN ( COUNT );
END COUNT\$DIVISORS ;

DECLARE MAX LITERALLY '15';
DECLARE ( I, NEXT ) ADDRESS;

CALL PR\$STRING( .'THE FIRST \$' );
CALL PR\$NUMBER( MAX );
CALL PR\$STRING( .' TERMS OF THE SEQUENCE ARE:\$' );
NEXT = 1;
I    = 1;
DO WHILE( NEXT <= MAX );
IF NEXT = COUNT\$DIVISORS( I ) THEN DO;
CALL PR\$CHAR( ' ' );
CALL PR\$NUMBER( I );
NEXT = NEXT + 1;
END;
I = I + 1;
END;

EOF```
Output:
```THE FIRST 15 TERMS OF THE SEQUENCE ARE: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## Polyglot:PL/I and PL/M

Works with: 8080 PL/M Compiler

... under CP/M (or an emulator)

Should work with many PL/I implementations.
The PL/I include file "pg.inc" can be found on the Polyglot:PL/I and PL/M page. Note the use of text in column 81 onwards to hide the PL/I specifics from the PL/M compiler.

Translation of: PL/M
``` /* FIND THE SMALLEST NUMBER > THE PREVIOUS ONE WITH EXACTLY N DIVISORS */

sequence_100H: procedure options                                                (main);

/* PROGRAM-SPECIFIC %REPLACE STATEMENTS MUST APPEAR BEFORE THE %INCLUDE AS */
/* E.G. THE CP/M PL/I COMPILER DOESN'T LIKE THEM TO FOLLOW PROCEDURES      */
/* PL/I                                                                      */
%replace maxnumber by         15;
/* PL/M */                                                                   /*
DECLARE  MAXNUMBER LITERALLY '15';
/* */

/* PL/I DEFINITIONS                                                             */
%include 'pg.inc';
/* PL/M DEFINITIONS: CP/M BDOS SYSTEM CALL AND CONSOLE I/O ROUTINES, ETC. */    /*
DECLARE BINARY LITERALLY 'ADDRESS', CHARACTER LITERALLY 'BYTE';
DECLARE FIXED  LITERALLY ' ',       BIT       LITERALLY 'BYTE';
DECLARE STATIC LITERALLY ' ',       RETURNS   LITERALLY ' ';
DECLARE FALSE  LITERALLY '0',       TRUE      LITERALLY '1';
DECLARE HBOUND LITERALLY 'LAST',    SADDR     LITERALLY '.';
BDOSF: PROCEDURE( FN, ARG )BYTE;
DECLARE FN BYTE, ARG ADDRESS; GOTO 5;   END;
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5;   END;
PRCHAR:   PROCEDURE( C );   DECLARE C BYTE;      CALL BDOS( 2, C ); END;
PRSTRING: PROCEDURE( S );   DECLARE S ADDRESS;   CALL BDOS( 9, S ); END;
PRNL:     PROCEDURE;        CALL PRCHAR( 0DH ); CALL PRCHAR( 0AH ); END;
PRNUMBER: PROCEDURE( N );
DECLARE V ADDRESS, N\$STR( 6 ) BYTE, W BYTE;
N\$STR( W := LAST( N\$STR ) ) = '\$';
N\$STR( W := W - 1 ) = '0' + ( ( V := N ) MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL BDOS( 9, .N\$STR( W ) );
END PRNUMBER;
RETURN( A MOD B );
END MODF;
MIN: PROCEDURE( A, B ) ADDRESS;
DECLARE  ( A, B ) ADDRESS;
IF A < B THEN RETURN( A ); ELSE RETURN( B );
END MIN;
/* END LANGUAGE DEFINITIONS */

COUNTDIVISORS: PROCEDURE( N )RETURNS /* THE DIVISOR COUNT OF N */            (
FIXED BINARY                                    )
;
DECLARE N                 FIXED BINARY;
DECLARE ( I, I2, COUNT )  FIXED BINARY;
COUNT = 0;
I     = 1;
I2    = 1;
DO WHILE( I2 < N );
IF MODF( N, I ) = 0 THEN COUNT = COUNT + 2;
I  = I + 1;
I2 = I * I;
END;
IF I2 = N THEN RETURN ( COUNT + 1 ); ELSE RETURN ( COUNT );
END COUNTDIVISORS ;

DECLARE ( I, NEXT ) FIXED BINARY;

CALL PRSTRING( SADDR( 'THE FIRST \$' ) );
CALL PRNUMBER( MAXNUMBER );
CALL PRSTRING( SADDR( ' TERMS OF THE SEQUENCE ARE:\$' ) );
NEXT = 1;
I    = 1;
DO WHILE( NEXT <= MAXNUMBER );
IF NEXT = COUNTDIVISORS( I ) THEN DO;
CALL PRCHAR( ' ' );
CALL PRNUMBER( I );
NEXT = NEXT + 1;
END;
I = I + 1;
END;

EOF: end sequence_100H;```
Output:
```THE FIRST 15 TERMS OF THE SEQUENCE ARE: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## Python

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

if __name__ == '__main__':
for item in sequence(15):
print(item)
```

Output:

```1
2
4
6
16
18
64
66
100
112
1024
1035
4096
4288
4624
```

## Quackery

`factors` is defined at Factors of an integer#Quackery.

```  [ stack ]                is terms (   --> s )

[ temp put
[] terms put
0 1
[ dip 1+
over factors size
over = if
[ over
terms take
swap join
terms put
1+ ]
terms share size
temp share = until ]
terms take
temp release
dip 2drop ]            is task  ( n --> [ )

Output:
`[ 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 ]`

## R

```#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
A06954 <- function(terms)
{
out <- 1
while((resultCount <- length(out)) != terms)
{
n <- resultCount + 1
out[n] <- out[resultCount]
while(divisorCount(out[n]) != n) out[n] <- out[n] + 1
}
out
}
print(A06954(15))```
Output:
`[1]    1    2    4    6   16   18   64   66  100  112 1024 1035 4096 4288 4624`

## 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;

my \$m = 1;
put "First \$limit terms of OEIS:A069654";
put (1..\$limit).map: -> \$n { my \$ = \$m = first { \$n == .&div-count }, \$m..Inf };
```
Output:
```First 15 terms of OEIS:A069654
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## REXX

Programming note:   this Rosetta Code task (for 15 sequence numbers) doesn't require any optimization,   but the code was optimized for listing higher numbers.

The method used is to find the number of proper divisors   (up to the integer square root of X),   and add one.

Optimization was included when examining   even   or   odd   index numbers   (determine how much to increment the   do   loop).

```/*REXX program finds and displays   N   numbers of the   "anti─primes plus"   sequence. */
parse arg N .                                    /*obtain optional argument from the CL.*/
if N=='' | N==","  then N= 15                    /*Not specified?  Then use the default.*/
idx= 1                                           /*the maximum number of divisors so far*/
say '──index──  ──anti─prime plus──'             /*display a title for the numbers shown*/
#= 0                                             /*the count of anti─primes found  "  " */
do i=1  until #==N                       /*step through possible numbers by twos*/
d= #divs(i);  if d\==idx  then iterate   /*get # divisors;  Is too small?  Skip.*/
#= # + 1;     idx= idx + 1               /*found an anti─prime #;  set new minD.*/
say center(#, 8)  right(i, 15)           /*display the index and the anti─prime.*/
end   /*i*/
exit 0                                            /*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                           /*handle special cases for numbers < 7.*/
if x<3   then return x       /*   "      "      "    "  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       /*Even?     "        "        "    " 3.*/
end                          /* [↑]  Even,  so add 2 known divisors.*/
do k=3  for x%2-3  by 1+odd  while k<y /*for odd numbers, skip over the evens.*/
if x//k==0  then do                    /*if no remainder, then found a divisor*/
#= # + 2;   y= x % k  /*bump the # Pdivs;  calculate limit Y.*/
if k>=y  then do;   #= # - 1;   leave
end     /* [↑]  has the limit been reached?    */
end                   /*                         ___         */
else if k*k>x  then leave  /*only divide up to the   √ x          */
end   /*k*/                            /* [↑]  this form of DO loop is faster.*/
return # + 1                              /*bump "proper divisors" to "divisors".*/
```
output   when using the default input:
```──index──  ──anti─prime plus──
1                   1
2                   2
3                   4
4                   6
5                  16
6                  18
7                  64
8                  66
9                 100
10                112
11               1024
12               1035
13               4096
14               4288
15               4624
```

## Ring

```# Project : ANti-primes

see "working..." + nl
see "wait for done..." + nl + nl
see "the first 15 Anti-primes Plus are:" + nl + nl
num = 1
n = 0
result = list(15)
while num < 16
n = n + 1
div = factors(n)
if div = num
result[num] = n
num = num + 1
ok
end
see "["
for n = 1 to len(result)
if n < len(result)
see string(result[n]) + ","
else
see string(result[n]) + "]" + nl + nl
ok
next
see "done..." + nl

func factors(an)
ansum = 2
if an < 2
return(1)
ok
for nr = 2 to an/2
if an%nr = 0
ansum = ansum+1
ok
next
return ansum```
Output:
```working...
wait for done...

the first 15 Anti-primes Plus are:

[1,2,4,6,16,18,64,66,100,112,1024,1035,4096,4288,4624]

done...
```

## RPL

Works with: HP version 49g
```≪ {1}
2 ROT FOR j
DUP DUP SIZE GET
DO 1 + UNTIL DUP DIVIS SIZE j == END
+
NEXT
```
```15 TASK
```
Output:
```1: {1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624}
```

Runs in 10 minutes 55 on a HP-50g.

## Ruby

```require 'prime'

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

seq = Enumerator.new do |y|
cur = 0
(1..).each do |i|
if num_divisors(i) == cur + 1 then
y << i
cur += 1
end
end
end

p seq.take(15)
```
Output:
```[1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624]
```

## Sidef

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

with (1) { |from|
say 15.of { from = n_divisors(_+1, from) }
}
```
Output:
```[1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624]
```

## Swift

Includes an optimization borrowed from the Pascal example above.

```// 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
}
}

let limit = 32
var n = 1
var next = 1
while next <= limit {
if next == divisorCount(number: n) {
print(n, terminator: " ")
next += 1
if next > 4 && divisorCount(number: next) == 2 {
n = 1 << (next - 1) - 1;
}
}
n += 1
}
print()
```
Output:
```1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624 4632 65536 65572 262144 262192 263169 269312 4194304 4194306 4477456 4493312 4498641 4498752 268435456 268437200 1073741824 1073741830
```

## Wren

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

var limit = 24
var res = List.filled(limit, 0)
var next = 1
var n = 1
while (next <= limit) {
var k = Int.divisors(n).count
if (k == next) {
res[k-1] = n
next = next + 1
if (next > 4 && Int.isPrime(next)) n = 2.pow(next-1) - 1
}
n = n + 1
}
System.print("The first %(limit) terms are:")
System.print(res)
```
Output:
```The first 24 terms are:
[1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624, 4632, 65536, 65572, 262144, 262192, 263169, 269312, 4194304, 4194306]
```

## XPL0

```func Divs(N);   \Count divisors of N
int  N, D, C;
[C:= 0;
if N > 1 then
[D:= 1;
repeat  if rem(N/D) = 0 then C:= C+1;
D:= D+1;
until   D >= N;
];
return C;
];

int An, N;
[An:= 1;  N:= 0;
loop [if Divs(An) = N then
[IntOut(0, An);  ChOut(0, ^ );
N:= N+1;
if N >= 15 then quit;
];
An:= An+1;
];
]```
Output:
```1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```

## zkl

```fcn countDivisors(n)
{ [1..(n).toFloat().sqrt()] .reduce('wrap(s,i){ s + (if(0==n%i) 1 + (i!=n/i)) },0) }```
```n:=15;
println("The first %d anti-primes plus are:".fmt(n));
(1).walker(*).tweak(
fcn(n,rn){ if(rn.value==countDivisors(n)){ rn.inc(); n } else Void.Skip }.fp1(Ref(1)))
.walk(n).concat(" ").println();```
Output:
```The first 15 anti-primes plus are:
1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
```