Ultra useful primes
An ultra-useful prime is a member of the sequence where each a(n) is the smallest positive integer k such that 2(2n) - k is prime.
k must always be an odd number since 2 to any power is always even.
- Task
- Find and show here, on this page, the first 10 elements of the sequence.
- Stretch
- Find and show the next several elements. (The numbers get really big really fast. Only nineteen elements have been identified as of this writing.)
- See also
Factor
<lang factor>USING: io kernel lists lists.lazy math math.primes prettyprint ;
- useful ( -- list )
1 lfrom [ 2^ 2^ 1 lfrom [ - prime? ] with lfilter car ] lmap-lazy ;
10 useful ltake [ pprint bl ] leach nl</lang>
- Output:
1 3 5 15 5 59 159 189 569 105
Perl
<lang perl>use strict; use warnings; use feature 'say'; use bigint; use ntheory 'is_prime';
sub useful {
my @n = @_; my @u; for my $n (@n) { my $p = 2**(2**$n); LOOP: for (my $k = 1; $k < $p; $k += 2) { is_prime($p-$k) and push @u, $k and last LOOP; } } @u
}
say join ' ', useful 1..13;</lang>
- Output:
1 3 5 15 5 59 159 189 569 105 1557 2549 2439
Phix
with javascript_semantics atom t0 = time() include mpfr.e mpz p = mpz_init() function a(integer n) mpz_ui_pow_ui(p,2,power(2,n)) mpz_sub_si(p,p,1) integer k = 1 while not mpz_prime(p) do k += 2 mpz_sub_si(p,p,2) end while return k end function for i=1 to 10 do printf(1,"%d ",a(i)) end for if machine_bits()=64 then ?elapsed(time()-t0) for i=11 to 13 do printf(1,"%d ",a(i)) end for end if ?elapsed(time()-t0)
- Output:
1 3 5 15 5 59 159 189 569 105 "0.0s" 1557 2549 2439 "1 minute and 1s"
Raku
The first 10 take less than a quarter second. 11 through 13, a little under 30 seconds. Drops off a cliff after that.
<lang perl6>sub useful ($n) {
(|$n).map: { my $p = 1 +< ( 1 +< $_ ); ^$p .first: ($p - *).is-prime }
}
put useful 1..10;
put useful 11..13;</lang>
- Output:
1 3 5 15 5 59 159 189 569 105 1557 2549 2439
Wren
An embedded version as Wren-CLI (with BigInt) will be very slow for this task.
The following takes about 17.5 seconds to get to n = 13 but 7 minutes 15 seconds to reach n = 14. I didn't bother after that. <lang ecmascript>import "./gmp" for Mpz import "./fmt" for Fmt
var a = Fn.new { |n|
var p = Mpz.one.lsh(1 << n) var k = 1 while (true) { var q = p - k if (q.probPrime(15) > 0) return k k = k + 2 }
}
System.print(" n k") System.print("----------") for (n in 1..14) Fmt.print("$2d $d", n, a.call(n))</lang>
- Output:
n k ---------- 1 1 2 3 3 5 4 15 5 5 6 59 7 159 8 189 9 569 10 105 11 1557 12 2549 13 2439 14 13797