Ruth-Aaron numbers: Difference between revisions

This uses a large amount of memory - too much for Algol 68G under Windows (and possibly under Linux).<br>

With max number set to 1 000 000, Algol 68G can find the first triple using factors in a few seconds (the loop to find the first divisors triple must be commented out or removed) - Real time: 0.941 s on TIO.RUN for the cutdown version.

# of the prime factors or divisors are equal #

INT max number = 99 000 000; # max number we will consider #

print( ( newline, "First Ruth-Aaron triple (factors): ", whole( fra3, 0 ) ) );

print( ( newline, "First Ruth-Aaron triple (divisors): ", whole( dra3, 0 ) ) )

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<pre>

=={{header|C++}}==

This takes about 2 minutes 24 seconds (3.2GHz Quad-Core Intel Core i5).

#include <iostream>

 

dsum2 = dsum3;

}

 

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=={{header|Factor}}==

{{works with|Factor|0.99 2022-04-03}}

math.primes.factors prettyprint ranges sequences ;

 

 

"First 30 Ruth-Aaron numbers (divisors):" print

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<pre>

{{libheader|Go-rcu}}

Takes about 4.5 minutes.

 

import (

fmt.Println("\nFirst Ruth-Aaron triple (divisors):")

fmt.Println(resT[0])

 

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=={{header|Haskell}}==

import Data.List.Split ( chunksOf )

 

putStrLn "First 30 Ruth-Aaron numbers( divisors ):"

mapM_ (\nlin -> putStrLn $ foldl1 ( ++ ) $ map (\st -> formatNumber (maxlen2 + 2) st )

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<pre>First 30 Ruth-Aaaron numbers ( factors ) :

 

Thus:

30{.1 2+/~I. 2 =/\ +/@q: 1+i.100000

5 6

26642 26643

35456 35457

 

=={{header|Julia}}==

using Primes

 

println("\nRuth Aaron triple starts at: ", findfirst(ruthaarontriple, 1:100000000))

println("\nRuth Aaron factor triple starts at: ", findfirst(ruthaaronfactorstriple, 1:10000000))

<pre>

30 Ruth Aaron numbers:

==={{header|Free Pascal}}===

all depends on fast prime decomposition.

// gets factors of consecutive integers fast

// limited to 1.2e11

writeln('First Ruth-Aaron triple (divisors):');

writeln(findfirstTripplesFactor(false):10);

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<pre>

 

=={{header|Perl}}==

 

use strict;

$n++;

}

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<pre>

{{libheader|Phix/online}}

You can run this online [http://phix.x10.mx/p2js/ruthaaron.htm here].

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

<span style="color: #008080;">procedure</span> <span style="color: #000000;">ruth_aaron</span><span style="color: #0000FF;">(</span><span style="color: #004080;">bool</span> <span style="color: #000000;">d</span><span style="color: #0000FF;">,</span> <span style="color: #004080;">integer</span> <span style="color: #000000;">n</span><span style="color: #0000FF;">=</span><span style="color: #000000;">30</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">l</span><span style="color: #0000FF;">=</span><span style="color: #000000;">2</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">i</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span><span style="color: #0000FF;">)</span>

<span style="color: #000000;">ruth_aaron</span><span style="color: #0000FF;">(</span><span style="color: #004600;">true</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">1</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">3</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">89460000</span><span style="color: #0000FF;">)</span> <span style="color: #000080;font-style:italic;">-- (0.1s)

--ruth_aaron(true, 1, 3) -- (24 minutes 30s)</span>

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<pre>

<code>primefactors</code> is defined at [[Prime decomposition#Quackery]].

 

witheach

[ tuck != if

raf echo

cr cr

 

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=={{header|Raku}}==

 

 

my @pf = lazy (^∞).hyper(:1000batch).map: *.&prime-factors.sum;

# Really, really, _really_ slow. 186(!) minutes... but with no cheating or "leg up".

put "\nFirst Ruth-Aaron triple (Divisors):\n" ~

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<pre>First 30 Ruth-Aaron numbers (Factors):

 

=={{header|Sidef}}==

say 30.by {|n| (sopfr(n) == sopfr(n+1)) && (n > 0) }.join(' ')

 

say "\nFirst 30 Ruth-Aaron numbers (divisors):"

 

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However, with nearly 90 million trios of numbers to slog through, it takes around 68 minutes to find the first triple based on divisors.

import "./seq" for Lst

import "./fmt" for Fmt

 

System.print("\nFirst Ruth-Aaron triple (divisors):")

 

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=={{header|XPL0}}==

int N, AllDiv; \all divisors vs. only prime divisors

int F, F0, S, Q;

CrLf(0);

Ruth(false);

 

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