19
05AB1E, 2 bytes
ÑO
Try it online!
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]
Try it online!
13
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
Try it online!
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 -- ...
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
SageMath, 31 Bytes
N=input()
print N,"=",factor(N)
Test case:
83891573479027823458394579234582347590825792034579235923475902312344444
Outputs:
83891573479027823458394579234582347590825792034579235923475902312344444 = 2^2 * 3^2 * 89395597 * 98966790508447596609239 * 263396636003096040031295425789508274613
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
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
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 ...
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
Neim, 3 bytes
𝐕𝐬𝔼
Try it online!
(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 ...
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
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
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 ...
8
05AB1E, 2 bytes
ÓZ
Try it online!
How?
Ó exponents of prime factors
Z maximum
8
05AB1E, 13 bytes
Encoder, 8 bytes
0ì¥ĀηOØP
Try it online!
Explanation
0ì # prepend 0 to input
¥ # calculate deltas
Ā # truthify each
η # calculate prefixes
O # sum each
Ø # get the prime at that index
P # product
Decoder, 5 bytes
Ò.ØÉJ
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Explanation
Ò # get ...
8
MATL, 7 bytes
Z\tn2/)
Try it online!
For this explanation, we will use '12' as a sample input. Explanation:
Z\ % Divisors.
% Stack:
% [1 2 3 4 6 12]
t % Duplicate.
% Stack:
% [1 2 3 4 6 12]
% [1 2 3 4 6 12]
n % Number of elements.
% Stack:
% 6
% [1 2 3 4 6 12]
...
8
R, 33 29 bytes
!2*(n=scan())-(x=1:n)%*%!n%%x
Try it online!
Returns TRUE for perfect numbers and FALSE for imperfect ones.
8
JavaScript (ES6), 91 83 79 76 bytes
f=(b,n=b,k=n)=>(g=d=>k<2?d<n:k%d?g(d+1):b*~-b*~b%d&&g(d,k/=d))(2)?n:f(b,n+1)
Try it online!
How?
Given the base \$b\$, we look for the smallest integer \$n\ge b\$ with a prime divisor \$d\$ which is neither \$n\$ itself nor a divisor of \$b-1\$, \$b\$ or \$b+1\$.
Commented
f = ( // ...
8
Haskell, 53 bytes
f b=[x|x<-[b..],k<-[2..x-1],mod x k+gcd(b^3-b)x<2]!!0
Try it online!
7
Total 36,757,269,913 cycles
830B assembled
Number Time (s) Cycles
8831269065180497 0.1 1148
2843901546547359024 55.0 9535194
6111061272747645669 351.4 60559378
11554045868611683619 0.8 135135
6764921230558061729 1.0 155407
16870180535862877896 43067.5 7126449414
...
7
Pyth - 17 12 11 bytes
Thanks to @FryAmTheEggman for fixing my answer and saving a byte.
@FmsMty{PdQ
Test Suite.
7
Pyth, 11 bytes
t@FmsMy{PdQ
Input in the form 30,7.
t@FmsMy{PdQ implicit: Q=input tuple
y powerset of
{ unique elements of
Pd prime factorizations of d
sM Map sum over each element of the powerset
sMy{Pd lambda d: all sums of unique prime factors of d
m Q Map over Q. Produces a ...
Only top voted, non community-wiki answers of a minimum length are eligible
