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

From Rosetta Code
Task
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.

Calculate the sequence where each term an is the smallest natural number greater than the previous term, 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: Go
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;
 
print( ( "The first ", whole( max, 0 ), " terms of the sequence are:", newline ) );
INT next := 1;
FOR i WHILE next <= max DO
IF next = count divisors( i ) THEN
print( ( whole( i, 0 ), " " ) );
next +:= 1
FI
OD;
print( ( newline, newline ) )
 
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

AWK[edit]

 
# 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

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, 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++[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() {
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 

Dyalect[edit]

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
}
 
const 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

Factor[edit]

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 }

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
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 

Java[edit]

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 




Julia[edit]

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[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() {
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 

Pascal[edit]

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[edit]

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

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;
 
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

Phix[edit]

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

constant limit = 32
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 length(factors(next,1))=2 then
n := power(2,next-1)-1 -- n is incremented afterwards
end if
end if
n += 1
end while
printf(1,"The first %d terms are:\n",limit)
pp(res,{pp_Pause,0,pp_StrFmt,1})
Output:
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,
 4498752,268435456,268437200,1073741824,1073741830}

Python[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):
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
 

REXX[edit]

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 /*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:
──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[edit]

 
# 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...

Ruby[edit]

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[edit]

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]

zkl[edit]

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