Ramanujan primes/twins: Difference between revisions

=={{header|F_Sharp|F#}}==

This task uses [http://www.rosettacode.org/wiki/Ramanujan_primes#F.23 Ramanujan primes (F#)]

// Twin Ramanujan primes. Nigel Galloway: September 9th., 2021

printfn $"There are %d{rP 1000000|>Seq.pairwise|>Seq.filter(fun(n,g)->n=g-2)|>Seq.length} twins in the first million Ramanujan primes"

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<pre>

{{trans|Wren}}

{{libheader|Go-rcu}}

import (

fmt.Println()

}

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=={{header|Julia}}==

function rajpairs(N, verbose, interval = 2)

rajpairs(100000, false)

@time rajpairs(1000000, true)

<pre>

There are 74973 twins in the first 1000000 Ramanujan primes.

=={{header|Mathematica}}/{{header|Wolfram Language}}==

l = PrimePi[Range[35 10^6]] - PrimePi[Range[35 10^6]/2];

ramanujanprimes = GatherBy[Transpose[{Range[2, Length[l] + 1], l}], Last][[All, -1, 1]];

ramanujanprimes = Take[Sort@ramanujanprimes, 10^6];

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<pre>74973</pre>

{{trans|Go}}

We use Phix algorithm in its Go translation.

let t0 = now()

main()

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Can't do better than the <code>ntheory</code> module.

{{libheader|ntheory}}

use warnings;

use ntheory <ramanujan_primes nth_ramanujan_prime>;

printf "There are %s twins in the first %s Ramanujan primes.\n\n",

comma( scalar grep { $rp->[$_+1] == $rp->[$_]+2 } 0 .. $limit-2 ), comma $limit;

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<pre>The 100,000th Ramanujan prime is 2,916,539

Calculating pi(p) - pi(floor(pi/2) for all (normal) primes below that one millionth, and then

filtering them based on that list is significantly faster, and finally we can scan for twins.

<span style="color: #008080;">with</span> <span style="color: #008080;">javascript_semantics</span>

<span style="color: #004080;">sequence</span> <span style="color: #000000;">pi</span> <span style="color: #0000FF;">=</span> <span style="color: #0000FF;">{}</span>

<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"There are %,d twins in the first %,d Ramanujan primes\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">twins</span><span style="color: #0000FF;">,</span><span style="color: #7060A8;">length</span><span style="color: #0000FF;">(</span><span style="color: #000000;">r</span><span style="color: #0000FF;">)})</span>

<span style="color: #0000FF;">?</span><span style="color: #7060A8;">elapsed</span><span style="color: #0000FF;">(</span><span style="color: #7060A8;">time</span><span style="color: #0000FF;">()-</span><span style="color: #000000;">t0</span><span style="color: #0000FF;">)</span>

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using a smaller limit:

Timing is informational only. Will vary by system specs and load.

{{libheader|ntheory}}

use Lingua::EN::Numbers;

}

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<pre>The 100,000th Ramanujan prime is 2,916,539

<br>

As calculating the first 1 million Ramanujan primes individually is out of the question, we follow the Phix entry's clever strategy here which brings this task comfortably within our ambit.

import "/math" for Int, Math

import "/fmt" for Fmt

Fmt.print("There are $,d twins in the first $,d Ramanujan primes.", twins, limit)

System.print("Took %(Math.toPlaces(System.clock -start, 2, 0)) seconds.\n")

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