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Sequence: smallest number with exactly n divisors

From Rosetta Code
Task
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.

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

AWK[edit]

 
# 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

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

Factor[edit]

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 }

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]

Haskell[edit]

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

import Data.Numbers.Primes (primeFactors)
import Data.List (find, group, sort)
import Data.Maybe (catMaybes)
 
a005179 :: [Int]
a005179 =
catMaybes $
(\n -> find ((n ==) . succ . length . properDivisors) [1 ..]) <$> [1 ..]
 
properDivisors :: Int -> [Int]
properDivisors =
init .
sort .
foldr --
(flip ((<*>) . fmap (*)) . scanl (*) 1)
[1] .
group . primeFactors
 
main :: IO ()
main = print $ take 15 a005179
Output:
[1,2,4,6,16,12,64,24,36,48,1024,60,4096,192,144]

JavaScript[edit]

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[edit]

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=. <. + [email protected]:|@:-
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[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]

Julia[edit]

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[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]

Maple[edit]

 
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

Nanoquery[edit]

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[edit]

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[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

Phix[edit]

naive[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.

advanced[edit]

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.

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[edit]

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
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 = "" -- (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) then short &= "*" end if
short &= sprintf("%d",p)
if e!=1 then
short &= 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="2^0" then short = "1" end if
printf(1,"%4d : %25s %30s %s\n",{i,short,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

Python[edit]

Procedural[edit]

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[edit]

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]

Raku[edit]

(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[edit]

some optimization[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; 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[edit]

/*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.


Ruby[edit]

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]

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]

Tcl[edit]

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

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