TL;DR

Here we are with TASK #1 from The Weekly Challenge #141. Enjoy!

# The challenge

Write a script to find lowest 10 positive integers having exactly 8 divisors.

Example

24 is the first such number having exactly 8 divisors.
1, 2, 3, 4, 6, 8, 12 and 24.


# The questions

I canâ€™t think of any.

# The solution

Oh, the joys of has so much space for being maximally overkill!!!

Thereâ€™s a slightly boring demonstration why positive integers having exactly 8 divisors MUST be only in exactly one of three possible forms:

$p_1^7 \\ p_1^3 p_2 \\ p_1 p_2 p_3$

The first form is the easier to generate: just take primes in increasing value, and get their 7th power.

The second and third forms are trickier to generate in ascending order. As an exampleâ€¦ $2^3 7$ and $2 3^3$ are quite close to one another, so we have to be careful to generate them in order. But hey! We can always borrow the BasicPriorityQueue from the last weekâ€™s challenge, and feed it with multiple candidates, so that we can extract the best, right?

For the second case, we will keep two parallel tracks, one for generating values where $p_1 < p_2$, the other one - you guess? - for $p_1 > p_2$. Then, we will always proceed like this when we â€śuseâ€ť the best value from the lot:

• always produce one â€śnextâ€ť item with the higher number increasing to the following prime;
• also produce another item with the lower number increasing to the following prime if this the previous prime of the higher one.

In practice:

• if we start from $(2, 3)$ we just generate $(2, 5)$
• from $(2, 5)$ we generate $(2, 7)$ and $(3, 5)$ (because 3 comes after 2, and is immediately before 5 as a prime)
• from $(2, 7)$ we generate just $(2, 11)$

You can easily convince yourself that this is a good way to go.

Something similar can be done with the third form too. This time, though, itâ€™s easier to always try to increase any of the three numbers, filtering out the cases where values might get on each otherâ€™s way. In practice, our candidates from $(A, B, C)$ will be the following three:

$(A, B, succ(C)) \\ succ(B) < C \Rightarrow (A, succ(B), C) \\ succ(A) < B \Rightarrow (succ(A), B, C)$

Again, a BasicPriorityQueeu will help us figure out the best candidate as we move on.

Last, we consider these three sources of positive integers, and get the best (i.e. lower one) at each round, â€śadvancingâ€ť only the source where we take the item from.

Soâ€¦ Raku time!

#!/usr/bin/env raku
use v6;

sub next-prime-after ($p) { #$p is prime
state %nxt = 2 => 3, 3 => 5, 5 => 7, 7 => 11;
state $max= 7; while ($p > $max) {$max= %nxt{$max}; %nxt{$max} = $max+ 2; %nxt{$max} += 2 until %nxt{$max}.is-prime; } return %nxt{$p};
}

class BasicPriorityQueue {
has @!items;
has &!before;

submethod BUILD (:&!before = {$^a <$^b}, :@items) {
@!items = '-';
self.enqueue($_) for @items; } #method dequeue ($obj) <-- see below
method elems { @!items.end }
# method enqueue ($obj) <-- see below method is-empty { @!items.elems == 1 } method size { @!items.end } method top { @!items.end ?? @!items[1] !! Any } method dequeue () { # includes "sink" return unless @!items.end; my$r = @!items.pop;
($r, @!items[1]) = (@!items[1],$r) if @!items.end >= 1;
my $k = 1; while (my$j = $k * 2) <= @!items.end { ++$j if $j < @!items.end && &!before(@!items[$j + 1], @!items[$j]); last if &!before(@!items[$k], @!items[$j]); (@!items[$j, $k],$k) = (|@!items[$k,$j], $j); } return$r;
}

method enqueue ($obj) { # includes "swim" @!items.push:$obj;
my $k = @!items.end; (@!items[$k/2, $k],$k) = (|@!items[$k,$k/2], ($k/2).Int) while$k > 1 && &!before(@!items[$k], @!items[$k/2]);
return self;
}
}

class A7 {
has Int $!A = 2; method current () {$!A ** 7 }
method move-on () { $!A = next-prime-after($!A) }
}

class A3B1 {
has $!queue = BasicPriorityQueue.new( before => {$^a[2] < $^b[2]}, items => [[2, 3, 24, 0], [2, 3, 54, 1], ], ); method current () { return$!queue.top[2] }
method move-on () {
my ($A,$B, $value,$twist) = $!queue.dequeue.Slip; my$next-B = next-prime-after($B); my @new = [$A, $next-B],; # this always appears my$next-A = next-prime-after($A); @new.push: [$next-A, $B] if$next-A < $B && next-prime-after($next-A) == $B; # fork! for @new ->$item {
($A,$B) = $item.Slip;$item.push: $twist ?? ($A * $B ** 3) !! ($A ** 3 * $B);$item.push: $twist;$!queue.enqueue($item); } } } class A1B1C1 { has$!queue = BasicPriorityQueue.new(
before => {$^a[3] <$^b[3]},
items  => [[2, 3, 5, 30], ]
);
method current() { return $!queue.top[3] } method move-on() { my ($A, $B,$C, $value) =$!queue.dequeue.Slip;
my ($n-A,$n-B, $n-C) = ($A, $B,$C).map: {next-prime-after($^a)}; my @new = [$A, $B,$n-C], ;
@new.push: [$A,$n-B, $C] if$n-B < $C; @new.push: [$n-A, $B,$C] if $n-A <$B;
for @new -> $item { ($A, $B,$C) = $item.Slip;$item.push: $A *$B * $C;$!queue.enqueue($item); } } } class EnumerateEighters { has$!a7     = A7.new();
has $!a3b1 = A3B1.new(); has$!a1b1c1 = A1B1C1.new();
method get() {
my $A =$!a7.current;
my $B =$!a3b1.current;
my $C =$!a1b1c1.current;
my $retval = ($A, $B,$C).min;
if ($retval ==$A) { $!a7.move-on } elsif ($retval == $B) {$!a3b1.move-on }
else { $!a1b1c1.move-on } return$retval;
}
}

sub MAIN (Int $n = 10) { my$x = EnumerateEighters.new;
$x.get.put for 1 ..$n;
}


This was fun, but it took me a lot and I wasnâ€™t even too sure it was working. So, for the Perl implementation, I opted for the easy route of going brute force:

#!/usr/bin/env perl
use v5.24;
use warnings;
use experimental 'signatures';
no warnings 'experimental::signatures';

sub count_divisors ($n) { my$c = 2; # 1, $n for my$d (2 .. $n / 2) { ++$c unless $n %$d;
}
return $c; } sub number_divisors ($n) {
my $i = 1; my @retval; while ($n > 0) {
if (count_divisors($i) == 8) { push @retval,$i;
--$n; } ++$i;
}
return @retval;
}

say for number_divisors(shift // 10);


It sticks to the definition: we iterate over integers, checking for compliance to our requirement about the number of divisors, until we have enough of them.

Incrediblyâ€¦ the two programs print out the same list of numbers! Isnâ€™t this awesome?!?

Stay safe people!

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