19
Jelly, 15 11 10 bytes
Hð+,_ðH²_½
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The following binary code works with this version of the Jelly interpreter.
0000000: 48 98 2b 2c 5f 98 48 8a 5f 90 H.+,_.H._.
Idea
This is based on the fact that
Code
Hð+,_ðH²_½ Left input: s -- Right input: p
ð ð This is a link fork. We define three links, call the left and right
...
19
05AB1E, 2 bytes
ÑO
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How?
Ñ Divisors
O Sum
16
PowerShell v3+, 450 bytes
param($n)function f{param($a)for($i=2;$a-gt1){if(!($a%$i)){$i;$a/=$i}else{$i++}}}
$y=($x=@((f $n)-split'(.)'-ne''|sort))|?{$_-eq(f $_)}
$a,$b=$x
$a=,$a
while($b){$z,$b=$b;$a=$a+($a+$y|%{$c="$_";0..$c.Length|%{-join($c[0..$_]+$z+$c[++$_..$c.Length])};"$z$c";"$c$z"})|select -u}
$x=-join($x|sort -des)
$l=@();$a|?{$_-eq(f $_)}|%{$j=$_;...
16
Brachylog, 4 bytes
fk+?
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The predicate succeeds for perfect inputs and fails for imperfect inputs, printing true. or false. if run as a complete program (except on the last test case which takes more than a minute on TIO).
The input's
f factors
k without the last element
+ sum to
? the input.
15
Pyth, 1 byte
P
I like Pyth's chances in this challenge.
14
Python 2, 55 bytes
f=lambda n,k=2:n/k*[0]and(f(n,k+1),[k]+f(n/k,k))[n%k<1]
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14
Husk, 35 31 30 29 26 25 24 22 20 19 15 bytes
-7 bytes thanks to @Zgarb!
Saved an extra 4 bytes, indirectly, thanks to Zgarb
ḋhΣhgφṁȯ`Jḋ2⁰ṗp
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Explaination
φ -- Define a recursive function which calls itself ⁰ and is applied to an Integer
ṁ p -- map then concatenate over its prime factors
ṗ -- ...
14
Jelly, 15 bytes
¹ÆFṛÇṭ¥Ị?@/€$Ẓ?
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A full program that takes the number and outputs the answer.
This isn't a link because I am using "last link" to refer to itself, so it would break as a link on its own. I fix this for my test suite by making the last link a single atom that calls the next link. For whatever reason, "this link&...
13
Unicorn, 4650 2982 1874 1546
[ ✨✨✨✨✨✨✨✨✨✨✨✨✨✨✨✨✨✨✨✨ 🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄 ( 🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈🌈 🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄🦄 🌈�...
12
SageMath, 31 Bytes
N=input()
print N,"=",factor(N)
Test case:
83891573479027823458394579234582347590825792034579235923475902312344444
Outputs:
83891573479027823458394579234582347590825792034579235923475902312344444 = 2^2 * 3^2 * 89395597 * 98966790508447596609239 * 263396636003096040031295425789508274613
12
Proof that every repunit has a novel prime factor
Using Zsigmondy's Theorem, the proof is simple. From Wikipedia:
In number theory, Zsigmondy's theorem, named after Karl Zsigmondy,
states that if a > b > 0 are coprime integers, then for any integer n
≥ 1, there is a prime number p (called a primitive prime divisor) that
divides an − bn and does not ...
11
Python 2, 53 bytes
f=lambda n,i=2:n/i*[f]and[f(n,i+1),[i]+f(n/i)][n%i<1]
Tries each potential divisor i in turn. If i is a divisor, prepends it and restarts with n/i. Else, tries the next-highest divisor. Because divisors are checked in increasing order, only the prime ones are found.
As a program, for 55 bytes:
n=input();i=2
while~-n:
if n%i:i+=1
...
10
Julia, 95 93 bytes
g(x)=reduce(vcat,map(p->map(sum,p),partitions([keys(factor(x))...])))
f(a,b)=g(a)∩g(b)!=[]
The primary function is f and it calls a helper function g.
Ungolfed:
function g(x::Integer)
# Find the sum of each combination of prime factors of the input
return reduce(vcat, map(p -> map(sum, p), partitions([keys(factor(x))...]))...
10
Jelly, 12 bytes
Æf*³<‘Ạ
1Ç#Ṫ
Takes n and k (one-indexed) as command-line arguments.
Try it online!
How it works
1Ç#Ṫ Main link. Left argument: n. Right argument: k
1 Set the return value to 1.
Ç# Execute the helper link above for r = 1, 2, 3, ... until k of them return
a truthy value. Yield the list of all k matches.
Ṫ ...
10
Python3, 49 47 bytes
def f(x):
l=x**.5//1
while x%l:l-=1
return l
Explanation
l=x**.5//1 → Assign l the largest integer less than equal to the square root of x
while x%l:l-=1 → While l does not evenly divide x, decrement l.
Edits
Mention Python3, not Python2
Use ...//1 to save two bytes. (Decimals are okay! Thanks @Rod)
10
JavaScript (ES6), 45 44 bytes
Takes input as (n)(p1), where \$n\$ is 0-indexed.
n=>g=(p,d=2)=>n?~p%d?g(p,d+1):--n?g(p*d):d:p
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Commented
n => // n = 0-based index of the requested term
g = ( // g is a recursive function taking:
p, // p = current prime product
d = 2 // d ...
10
Scala 3, 49 44 bytes
a=>b=>(-b*b to b*b)map(p=>a-p->p)find(_*_==b)
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Takes (a)(b) and returns an Option[(Int, Int)]. It's now a little more inefficient since it goes from \$-b^2\$ to \$b^2\$ instead of \$-|b|\$ to \$|b|\$, including values that \$p\$ and \$q\$ could never be, but it saves 4 bytes.
a => b =>
(-b*b to b*b) ...
9
Husk, 11 10 bytes
Saved one byte thanks to Zgarb!
Ωεo?oṗ←¬Ep
Returns 1 for unique, 0 otherwise
Try it online! Or returning the first 50
Explanation:
Ωε Until the result is small (either 1 or 0), we repeat the following
p Get the prime factors
o? If ...
E they are all equal:
ȯṗ← Get the index ...
9
R, 33 29 bytes
!2*(n=scan())-(x=1:n)%*%!n%%x
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Returns TRUE for perfect numbers and FALSE for imperfect ones.
9
Neim, 3 bytes
𝐕𝐬𝔼
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(I don't actually know how to run all of the test cases at once, since I started learning Neim about fifteen minutes ago, but I did check them individually.)
Prints 0 for imperfect, 1 for perfect.
𝐕 Pop an int from the stack and push its proper divisors,
implicitly reading the int from a line of input as ...
9
05AB1E, 6 bytes
This produces and infinite output stream.
λλP>fW
Try it online! (link includes a slightly modified version, λ£λP>fW, which instead outputs the first \$n\$ terms)
Explanation
Very straightforward. Given \$p_1\$ and \$n\$, the program does the following:
Starts with \$p_1\$ as an initial parameter for the infinite stream (which is ...
9
Haskell, 53 bytes
f b=[x|x<-[b..],k<-[2..x-1],mod x k+gcd(b^3-b)x<2]!!0
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9
JavaScript (ES7), 37 36 34 31 bytes
Saved 3 bytes thanks to @tsh
Expects (a)(b).
a=>b=>[b=a/2+(a*a/4-b)**.5,a-b]
Try it online!
9
Wolfram Language (Mathematica), 23 21 18 bytes
Solve[x x+#2==x#]&
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8
Java - 124
String f(int t){int s=2,h=3,p=s,i;String o="";for(;p!=t&&s*s<t;p=p<t?p*h++:p/s++);if(p==t)for(i=s;i<h;o+++=i+" ");return o;}
Starting at 2, this loops until the start number is > the square root of the target (or target is reached exactly). If the product is low, it multiplies by the high number and ...
8
Python - 104 97 95 92 try it
n=input()
s=i=2
c=1
while s<n:
s*=i+c;c+=1
if s==n:print range(i,i+c)
if s/n:i+=1;s,c=i,1
If n is, e.g., set to 120 beforehand, the program outputs the two solutions:
[2, 3, 4, 5]
[4, 5, 6]
8
JavaScript ES6, 45 39 37* bytes
(q,p)=>[x=p/2+Math.sqrt(p*p/4-q),p-x]
* Thanks to Dennis!
8
CJam, 26 23 bytes
{_mfs$:XW%i){mfs$X=},^}
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Explanation
Concatenating two numbers always gives a bigger result than multiplying them. So the largest number we possibly need to consider is the largest number we can form from the digits of the input's prime factorisation, which is just all digits sorted in descending order. For the given ...
8
Mathematica, 38 30 bytes
Thanks @MartinEnder for 8 bytes!
Join@@Table@@@FactorInteger@#&
8
Jelly, 22 20 19 bytes
-1 thanks to Erik the Outgolfer (tail zeros from both sides, t, rather than from the right œr)
ÆfÆC$ÐLŒṘO%3ḟ2Ḋt0ṖḄ
A monadic link taking an integer greater than 2 and returning an integer greater than 0 (2 would return 0 as per the original spec too).
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How?
This almost exactly replicates the description ...
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