Sum the powers that be

A simple but hopefully not quite trivial challenge:

Write a program or function that adds up the kth powers dividing a number n. More specifically:

• Input: two positive integers n and k (or an ordered pair of integers, etc.)
• Output: the sum of all of the positive divisors of n that are kth powers of integers

For example, 11! = 39916800 has six divisors that are cubes, namely 1, 8, 27, 64, 216, and 1728. Therefore given inputs 39916800 and 3, the program should return their sum, 2044.

Other test cases:

{40320, 1} -> 159120
{40320, 2} -> 850
{40320, 3} -> 73
{40320, 4} -> 17
{40320, 5} -> 33
{40320, 6} -> 65
{40320, 7} -> 129
{40320, 8} -> 1
{46656, 1} -> 138811
{46656, 2} -> 69700
{46656, 3} -> 55261
{46656, 4} -> 1394
{46656, 5} -> 8052
{46656, 6} -> 47450
{46656, 7} -> 1
{1, [any positive integer]} -> 1

This is code golf, so the shorter your code, the better. I welcome golfed code in all kinds of different languages, even if some other language can get away with fewer bytes than yours.

• When I first saw your challenge, I had the weird feeling that it was a Metallica song title. – Arnauld Feb 10 '17 at 14:54
• What? There's no Mathematica built-in for this? – boboquack Feb 10 '17 at 22:47

05AB1E, 9 bytes

DLImDŠÖÏO

Try it online!

Explanation

Example input 46656, 3

D          # duplicate first input
# STACK: 46656, 46656
L         # range [1 ... first input]
# STACK: 46656, [1 ... 46656]
Im       # each to the power of second input
# STACK: 46656, [1, 8, 27 ...]
D      # duplicate
# STACK: 46656, [1, 8, 27 ...], [1, 8, 27 ...]
Š     # move down 2 spots on the stack
# STACK: [1, 8, 27 ...], 46656, [1, 8, 27 ...]
Ö    # a mod b == 0
# STACK: [1, 8, 27 ...], [1,1,1,1,0 ...]
Ï   # keep only items from first list which are true in second
# STACK: [1, 8, 27, 64, 216, 729, 1728, 5832, 46656]
O  # sum
# OUTPUT: 55261

Mathematica, 28 bytes

Tr[Divisors@#⋂Range@#^#2]&

Unnamed function taking n and k as inputs in that order.

• DivisorSum is frustratingly close to being useful here. – ngenisis Feb 10 '17 at 19:24

Haskell, 37 35 34 bytes

n!k=sum[x^k|x<-[1..n],nmodx^k<1]

Try it online! Usage:

Prelude> 40320 ! 1
159120

The code is quite inefficient because it always computes 1^k, 2^k, ..., n^k.

Edit: Saved one byte thanks to Zgarb.

Explanation:

n!k=             -- given n and k, the function ! returns
sum[x^k|        -- the sum of the list of all x^k
x<-[1..n],    -- where x is drawn from the range 1 to n
nmodx^k<1]  -- and n modulus x^k is less than 1, that is x^k divides n
• mod n(x^k) can be nmodx^k. – Zgarb Feb 10 '17 at 14:08

Python 2, 54 52 bytes

lambda x,n:sum(i**n*(x%i**n<1)for i in range(1,-~x))

Thanks @Rod for cutting off 2 bytes.

• You can replace x%i**n==0 with x%i**n<1, and move to the other side as i**n*(x%i**n<1) – Rod Feb 10 '17 at 12:05

Ruby, 45 bytes

->n,m{(1..n).reduce{|a,b|n%(c=b**m)<1?a+c:a}}

Would be shorter using "sum" in Ruby 2.4. Time to upgrade?

• Time to upgrade. – Yytsi Feb 10 '17 at 10:34

MATL, 10 bytes

t:i^\~5M*s

Try it online!

How it works

Example with 46656, 6.

t      % Implicitly input n. Duplicate
% STACK: 46656, 46656
:      % Range
% STACK: 46656, [1 2 ... 46656]
i      % Input k
% STACK: 46656, [1 2 ... 46656], 6
^      % Power, element-wise
% STACK: 46656, [1 64 ... 46656^6]
\      % Modulo
% STACK: [0 0 0 1600 ...]
~      % Logically negate
% STACK: [true true true false ...]
5M     % Push second input to function \ again
% STACK: [true true true false ...], [1^6 2^6 ... 46656^6]
*      % Multiply, element-wise
% STACK: [1 64 729 0 ...]
s      % Sum of array: 47450
% Implicitly display

Jelly, 7 6 bytes

-1 byte thanks to Dennis (traverse an implicit range)
A clever efficiency save also by Dennis at 0-byte cost
(Previously ÆDf*€S would filter keep those divisors that are a power of k of any natural number up to n. But note that n can only ever have a divisor of ik if it has a divisor of i anyway!)

ÆDf*¥S

Try it online!

How?

ÆDf*¥S - Main link: n, k
ÆD     - divisors of n  -> divisors = [1, d1, d2, ..., n]
¥  - last two links as a dyadic chain
f    -     filter divisors keeping those that appear in:
*   -     exponentiate k with base divisors (vectorises)
- i.e. [v for v in [1, d1, d2, ..., n] if v in [1^k, d1^k, ..., n^k]]
S - sum

JavaScript (ES7), 56 53 bytes

Takes n and k in currying syntax (n)(k).

n=>k=>[...Array(n)].reduce(p=>n%(a=++i**k)?p:p+a,i=0)

Test cases

let f =

n=>k=>[...Array(n)].reduce(p=>n%(a=++i**k)?p:p+a,i=0)

console.log(f(40320)(1)) // -> 159120
console.log(f(40320)(2)) // -> 850
console.log(f(40320)(3)) // -> 73
console.log(f(40320)(4)) // -> 17
console.log(f(40320)(5)) // -> 33
console.log(f(40320)(6)) // -> 65
console.log(f(40320)(7)) // -> 129
console.log(f(40320)(8)) // -> 1
console.log(f(46656)(1)) // -> 138811
console.log(f(46656)(2)) // -> 69700
console.log(f(46656)(3)) // -> 55261
console.log(f(46656)(4)) // -> 1394
console.log(f(46656)(5)) // -> 8052
console.log(f(46656)(6)) // -> 47450
console.log(f(46656)(7)) // -> 1

Perl 6, 39 bytes

->\n,\k{sum grep n%%*,({++$**k}...*>n)} How it works ->\n,\k{ } # A lambda taking two arguments. ++$              # Increment an anonymous counter
**k           # and raise it to the power k,
{      }...       # generate a list by repeatedly doing that,
*>n    # until we reach a value greater than n.
grep n%%*,(              )   # Filter factors of n from the list.
sum                              # Return their sum.

Try it

Japt, 10 bytes

Saved lots of bytes thanks to @ETHproductions

òpV f!vU x

Explanation

òpV f!vU x
ò           // Creates a range from 0 to U
pV         // Raises each item to the power of V (Second input)
f       // Selects all items Z where
!vU    //   U is divisible by Z
//   (fvU would mean Z is divisible by U; ! swaps the arguments)
x  // Returns the sum of all remaining items

Test it online!

• Does vU detect numbers divisible by U, or numbers that divide U? – Greg Martin Feb 10 '17 at 18:43
• @GregMartin fvU filters to items that are divisible by U; f!vU filters to items that U is divisible by. ! swaps the arguments. – Oliver Feb 10 '17 at 18:47
• Cool, so the code looks right, but the explanation might need to be tweaked. – Greg Martin Feb 10 '17 at 18:53
• @GregMartin Should be clearer now. – ETHproductions Feb 10 '17 at 18:55

Scala 63 bytes

(n:Int,k:Int)=>1 to n map{Math.pow(_,k).toInt}filter{n%_==0}sum

Python 2, 50 bytes

f=lambda n,k,i=1:n/i and(n%i**k<1)*i**k+f(n,k,i+1)

Try it online! Large inputs may exceed the recursion depth depending on your system and implementation.

JavaScript (ES7), 49 46 bytes

n=>g=(k,t=i=0,p=++i**k)=>p>n?t:g(k,t+p*!(n%p))
• Since you aren't recursing, why not n=>k=>? +1. – Yytsi Feb 12 '17 at 9:51
• @TuukkaX I came up with something better. (I actually had this earlier with i as a local, which costs 4 extra bytes, and forgot that I could abuse i in the same way that I did with my other formulation.) – Neil Feb 12 '17 at 10:17

PHP, 86 bytes

$n=$argv;$k=$argv;for($i=1;$i<=$n**(1/$k);$i++)if($n%$i**$k<1)$s+=$i**$k;echo$s;

Try it here !

Breakdown :

$n=$argv;$k=$argv;       # Assign variables from input
for($i=1;$i<=$n**(1/$k);$i++) # While i is between 1 AND kth root of n if($n%$i**$k<1)            #     if i^k is a divisor of n
$s+=$i**$k; # then add to s echo$s;                        # echo s (duh!)
• golfed, but not tested: for(;$x<$n=$argv;)$n%($x=++$i**$argv)?:$s+=$x;echo$s; 59 bytes; requires PHP 5.6 or later. – Titus Feb 12 '17 at 15:05

CJam, 20 bytes

Probably not optimally golfed, but I don't see any obvious changes to make...

ri:N,:)rif#{N\%!},:+

Try it online!

R*³%$ÐḟS Try it online! Bash + Unix utilities, 44 bytes bc<<<seq "-fx=%.f^$2;s+=($1%%x==0)*x;"$1s

Try it online!

Test runs:

for x in '40320 1' '40320 2' '40320 3' '40320 4' '40320 5' '40320 6' '40320 7' '40320 8' '46656 1' '46656 2' '46656 3' '46656 4' '46656 5' '46656 6' '46656 7' '1 1' '1 2' '1 3' '1 12' ; do echo -n "$x "; ./sumpowerdivisors$x; done

40320 1 159120
40320 2 850
40320 3 73
40320 4 17
40320 5 33
40320 6 65
40320 7 129
40320 8 1
46656 1 138811
46656 2 69700
46656 3 55261
46656 4 1394
46656 5 8052
46656 6 47450
46656 7 1
1 1 1
1 2 1
1 3 1
1 12 1

Python, 56 bytes

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

Try it online!

Fairly straightforward. The only noteworthy thing is that j**k**-1%1 always returns a float in [0,1) while n%j always returns a non-negative integer, so they can only be equal if both are 0.

Batch, 138 bytes

@set s=n
@for /l %%i in (2,1,%2)do @call set s=%%s%%*n
@set/at=n=0
:l
@set/an+=1,p=%s%,t+=p*!(%1%%p)
@if %p% lss %1 goto l
@echo %t%

Since Batch doesn't have a power operator, I'm abusing set/a as a form of eval. Very slow when k=1. 32-bit integer arithmetic limits the supported values of n and k:

n   k
(too slow)   1
<1366041600   2
<1833767424   3
<2019963136   4
<2073071593   5
<1838265625   6
<1801088541   7
<1475789056   8
<1000000000   9
<1073741824  10
<1977326743  11
<244140625  12
<1220703125  13
<268435456  14
<1073741824  15
<43046721  16
<129140163  17
<387420489  18
<1162261467  19
<1048576  20
...
<1073741824  30

R, 28 bytes direct, 43 bytes for function

if n,k in memory:

sum((n%%(1:n)^k==0)*(1:n)^k)

for a function:

r=function(n,k)sum((n%%(1:n)^k==0)*(1:n)^k)