Pierpont primes: Difference between revisions

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<lang perl6>use ntheory:from<Perl5> <is_prime>;

sub pierpont ($~~type~~ is copy = 1) {

sub pierpont ($kind is copy = 1) {

fail "Unknown type: $kind. Must be one of 1 (default) or 2" if $kind !== 1|2;

gather {

my $po3 = 0;

fail "Unknown type: $~~type~~. Must be one of 1 (default) or 2" if $~~type~~ !== 1|2;

~~take 2 if~~ $~~type~~ == 1;

my $add-one = 3;

my @iterators = [2,4,8 … *].iterator, [3,9,27 … *].iterator;

$~~type~~ = -1 if $~~type~~ == 2;

~~state~~ $po3 = 0;

~~state~~ $~~add-one~~ = 3;

~~state~~ @iterators = [2,4,8 … *].iterator, [3,9,27 … *].iterator;

gather {

take 2 if $kind == 1;

$kind = -1 if $kind == 2;

my @head = @iterators».pull-one;

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@head[$key] = @iterators[$key].pull-one;

take $min + $~~type~~ if "{$min + $~~type~~}".&is_prime;

take $min + $kind if "{$min + $kind}".&is_prime;

if $min >= $add-one {

++$po3;

@iterators.push: ([|((2,4,8).map: * * 3 ** $po3) … *]).iterator;

@head[+@iterators - 1] = @iterators[+@iterators - 1].pull-one;

$add-one = @head[+@iterators - 1] = @iterators[+@iterators - 1].pull-one;

$add-one *= 3;

}

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say "\n250th Pierpont prime of the second kind: " ~ pierpont(2)[249];

# And the question absolutely ~~noone~~ was asking:

# And the question absolutely nobody was asking:

say "\n1000th Pierpont prime of the first kind:\n" ~ pierpont[999];