Output the nth Even Perfect Number

Question

Challenge

Given an integer, n, as input where 0 <= n <= 2^10, output the nth even perfect number.

Perfect Numbers

A perfect number is a number, x where the sum of its factors (excluding itself) equals x. For example, 6:

6: 1, 2, 3, 6

And, of course, 1 + 2 + 3 = 6, so 6 is perfect.

If a perfect number, x, is even, x mod 2 = 0.

Examples

The following are the first 10 even perfect numbers:

6
28
496
8128
33550336
8589869056
137438691328
2305843008139952128
2658455991569831744654692615953842176
191561942608236107294793378084303638130997321548169216

Note that you may index this however you wish: 6 may be the 1st or the 0th even perfect number.

Winning

Shortest code in bytes wins.

• Rohan Jhunjhunwala: @LeakyNun I think, that is an open question. If this question was output the nth odd perfect number... You would need a billion rep bounty to get it solved. blogs.ams.org/mathgradblog/2013/07/25/odd-perfect-numbers-exist (none exist below 10^300)
• Leaky Nun: What is the smallest odd perfect number?
• Dennis: An even number n is perfect iff there is a Mersenne prime p such that n = p(p+1)/2. There is no such formula for odd perfect numbers; moreover, it is unknown if odd perfect numbers even exist.
• Dennis: Not quite. There are only 49 known Mersenne primes.
• Ross Millikan: @BetaDecay: it is greater than 49ドル,ドル so the 60th perfect number is not known.

Answer 1

Jelly, 7 bytes

6Æṣ=$#Ṫ

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How it works

6Æṣ=$#Ṫ Main link. Argument: n
6 Set the return value to 6.
 # Execute the link to the left with argument k = 6, 7, 8, ... until n
 values of k result in a truthy value. Yield the array of matches.
 $ Combine the two links to the left into a monadic chain.
 Æṣ Compute the sum of k's proper divisors.
 = Compare the result with k.
 Ṫ Tail; extract the last match.

• Erik the Outgolfer: So many builtins regarding divisors...

Answer 2

Mathematica, 13 bytes

Not surprisingly, there is a built-in.

PerfectNumber

Example:

In[1]:= PerfectNumber[18] 
Out[1]= 33570832131986724437010877211080384841138028499879725454996241573482158\
> 45044404288204877880943769038844953577426084988557369475990617384115743842\
> 47301308070476236559422361748505091085378276585906423254824947614731965790\
> 74656099918600764404702181660294469121778737965822199901663478093006075022\
> 35922320184998563614417718592540207818507301504509772708485946474363553778\
> 15002849158802448863064617859829560720600134749556178514816801859885571366\
> 09224841817877083608951191123174885226416130683197710667392351007374503755\
> 40335253147622794359007165170269759424103195552989897121800121464177467313\
> 49444715625609571796578815564191221029354502997518133405151709561679510954\
> 53649485576150660101689160658011770193274226308280507786835049549112576654\
> 51011967045674593989019420525517538448448990932896764698816315598247156499\
> 81962616327512831278795091980742531934095804545624886643834653798850027355\
> 06153988851506645137759275553988219425439764732399824712438125054117523837\
> 43825674443705501944105100648997234160911797840456379499200487305751845574\
> 87014449512383771396204942879824895298272331406370148374088561561995154576\
> 69607964052126908149265601786094447595560440059050091763547114092255371397\
> 42580786755435211254219478481549478427620117084594927467463298521042107553\
> 17849183589266903954636497214522654057134843880439116344854323586388066453\
> 13826206591131266232422007835577345584225720310518698143376736219283021119\
> 28761789614688558486006504887631570108879621959364082631162227332803560330\
> 94756423908044994601567978553610182466961012539222545672409083153854682409\
> 31846166962495983407607141601251889544407008815874744654769507268678051757\
> 74695689121248545626112138666740771113961907153092335582317866270537439303\
> 50490226038824797423347994071302801487692985977437781930503487497407869280\
> 96033906295910199238181338557856978191860647256209708168229116156300978059\
> 19702685572687764976707268496046345276316038409383829227754491185785965832\
> 8888332628525056

• Paŭlo Ebermann: I think there is a standard loophole for that?
• Tim: @PaŭloEbermann correct, with 19 downvotes and a comment with 94 upvotes approving of it: codegolf.meta.stackexchange.com/a/1078/32933

Answer 3

MATL, 15 bytes

`@Z\s@E=vtsG<}n

Very slow. It keeps trying increasing numbers one by one until the n-th perfect number is found.

Try it online!

Explanation

` % Do...while
 @ % Push iteration index, k (starting at 1)
 Z\ % Array of divisors
 s % Sum
 @E % Push k. Multiply by 2
 = % Equal? If so, k is a perfect number
 v % Concatenate vertically. This gradually builds an array which at the k-th
 % iteration contains k zero/one values, where ones indicate perfect numbers
 ts % Duplicate. Sum of array
 G< % Push input. Less than? This is the loop condition: if true, proceed with
 % next iteration
} % Finally (execute right before exiting loop)
 n % Number of elements of the array
 % End (implicit). Display (implicit)

Answer 4

Pyth, 13 bytes

e.fqsf!%ZTStZ

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Please do not try any higher number. It just tests the even numbers one by one.

Answer 5

05AB1E, 8 bytes

μNNÑ ̈OQ1⁄2

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Explanation

μ # loop over increasing N until counter equals input
 N # push N
 NÑ # push factors of N
 ̈ # remove last factor (itself)
 O # sum factors
 Q # compare the sum to N for equality
 1⁄2 # if true, increase counter

Answer 6

Python 2, (削除) 198 153 83 78 77 75 (削除ここまで) 74 bytes

i=input()
j=0
while i:j+=1;i-=sum(x*(j%x<1)for x in range(1,j))==j
print j

Try it online!

Now it just reads like psuedocode.

• Saved (削除) 45 (削除ここまで) Countless Bytes because @Leaky Nun taught me about the sum function and list comprehension.

• Saved 2 bytes thanks to @shooqie's suggestion to remove the uncessary brackets.

We just iterate through every even number until we have found n perfect numbers.

• Leaky Nun: notice that your g is actually just sum.
• Rohan Jhunjhunwala: @LeakyNun serves me right, for not knowing the python libraries. I really should learn more than just java and SILOS.
• Leaky Nun: Also, I've replaced your f with a list comprehension
• Leaky Nun: more golfings
• Leaky Nun: no need for k, just decrement i.

Answer 7

PHP, 111 Bytes

0-Indexing

Works with the concept that a perfect number is a number where n=x*y x=2^i and y=2^(i+1)-1 and y must be prime

for(;!$r[$argn];$u?:$r[]=$z)for($z=2**++$n*($y=2**($n+1)-1),$u=0,$j=1;$j++<sqrt($y);)$y%$j?:$u++;echo$r[$argn];

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Answer 8

Python 3, 69 bytes

f=lambda n,k=1:n and-~f(n-(sum(j>>k%j*j for j in range(1,k))==k),k+1)

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Answer 9

Scala, 103 bytes

n=>Stream.from(1).filter(_%2==0).filter(x=>Stream.from(1).take(x-1).filter(x%_==0).sum==x).drop(n).head

Answer 10

Haskell, 61 Bytes

(!!)(filter(\x->x==sum[n|n<-[1..x-1],x`mod`n==0]||x==1)[1..])

• faubi: Since the index can start at 0, you don't need the ||x==1. You can also save bytes by moving the !! just before the closing parenthesis to make an operator section, and by replacing the filter with another list comprehension.

Answer 11

Haskell, 52 bytes

([n|n<-[1..],n==sum[d|d<-[1..div n 2],mod n d<1]]!!)

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Starts at n = 0. Not particularly fast

• naffetS: By our consensus, it's valid to omit the f= from the byte count.

Answer 12

JavaScript (ES6), 68 bytes

n=>eval(`for(x=5;n;s||n--)for(f=s=++x;f--;)(x/f-(x/f|0))||(s-=f);x`)

F=n=>eval('for(x=5;n;s||n--)for(f=s=++x;f--;)(x/f-(x/f|0))||(s-=f);x')
console.log(
 F(1),
 F(2),
 F(3),
 //F(4),
 //F(5),
)

Answer 13

Perl 6, 42 bytes

{(grep {$_==[+] grep $_%%*,^$_},^∞)[$_]}

The input index is 1-based.

Answer 14

Clojure, 79 bytes

#(nth(for[i(range):when(=(apply +(for[j(range 1 i):when(=(mod i j)0)]j))i)]i)%)

Following the spec, heavy usage of for's :when condition.

Answer 15

PowerShell, 152 bytes

$ErrorActionPreference='SilentlyContinue'
function p{param($x)(1..($x-1)|?{!($x%$_)})|%{$s+=$_};$s}
1..$args[0]|%{if(((p -x $_)-eq$_)-and(!($_%2))){$_}}

Try it online!