Sequence: smallest number greater than previous term with exactly n divisors: Difference between revisions
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integer next = 1 |
integer next = 1 |
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atom n = 1 |
atom n = 1 |
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atom t0 = time() |
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while next<=limit do |
while next<=limit do |
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integer k = length(factors(n,1)) |
integer k = length(factors(n,1)) |
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Revision as of 21:27, 13 May 2019
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
<lang algol68>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</lang>
- 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
<lang c>#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;
}</lang>
- 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++
<lang cpp>#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;
}</lang>
- 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
<lang 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()
}</lang>
- 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
<lang java>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();
}
}</lang>
- Output:
The first 15 terms of the sequence are: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
Kotlin
<lang scala>// 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()
}</lang>
- 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!
<lang pascal>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.</lang>
- 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
<lang perl>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);
}
}</lang>
- Output:
First 15 terms of OEIS: A069654 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624
Perl 6
<lang perl6>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 }; </lang>
- 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. <lang Phix>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})</lang>
- 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}
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). <lang rexx>/*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".*/</lang>
- 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
<lang 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
</lang>
- 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...
Sidef
<lang ruby>func n_divisors(n, from=1) {
from..Inf -> first_by { .sigma0 == n }
}
with (1) { |from|
say 15.of { from = n_divisors(_+1, from) }
}</lang>
- Output:
[1, 2, 4, 6, 16, 18, 64, 66, 100, 112, 1024, 1035, 4096, 4288, 4624]
zkl
<lang zkl>fcn countDivisors(n)
{ [1..(n).toFloat().sqrt()] .reduce('wrap(s,i){ s + (if(0==n%i) 1 + (i!=n/i)) },0) }</lang>
<lang zkl>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();</lang>
- Output:
The first 15 anti-primes plus are: 1 2 4 6 16 18 64 66 100 112 1024 1035 4096 4288 4624