Sequence: smallest number with exactly n divisors

From Rosetta Code
Sequence: smallest number with exactly n divisors is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

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.

See also
Related tasks

ALGOL 68[edit]

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

C[edit]

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++[edit]

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 

F#[edit]

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

Go[edit]

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]

Java[edit]

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]

Kotlin[edit]

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]

Perl[edit]

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

Perl 6[edit]

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

Phix[edit]

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.

REXX[edit]

/*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 /*step through a number of divisors. */
do j=1+(i\==1) by 1+(i\==1) /*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 /*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 do; #= 1; end /*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

Sidef[edit]

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]

zkl[edit]

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