17
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Your function takes a natural number and returns the smallest natural number that has exactly that amount of divisors, including itself.

Examples:

f(1) =  1 [1]
f(2) =  2 [1, 2]
f(3) =  4 [1, 2, 4]
f(4) =  6 [1, 2, 3, 6]
f(5) = 16 [1, 2, 4, 8, 16]
f(6) = 12 [1, 2, 3, 4, 6, 12]
 ...

The function doesn't have to return the list of divisors, they are only here for the examples.

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  • 2
    \$\begingroup\$ Is this code-golf or code-challenge? \$\endgroup\$ – marinus Apr 3 '13 at 12:49
  • \$\begingroup\$ Oops, forgot about that tag, code-golf! \$\endgroup\$ – SteeveDroz Apr 3 '13 at 13:06
  • 11
    \$\begingroup\$ A005179 \$\endgroup\$ – Peter Taylor Apr 3 '13 at 15:28

22 Answers 22

6
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APL, 25 24 23 characters

f←{({+/⍵=⍵∧⍳⍵}¨⍳2*⍵)⍳⍵}

Defines a function f which can then be used to calculate the numbers:

> f 13
4096

> f 14
192

The solution utilizes the fact that LCM(n,x)==n iff x divides n. Thus, the block {+/⍵=⍵∧⍳⍵} simply calculates the number of divisors. This function is applied to all numbers from 1 to 2^d ¨⍳2*⍵. The resulting list is then searched for d itself (⍳⍵) which is the desired function f(d).

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  • \$\begingroup\$ Congrats! 23 characters... wow! \$\endgroup\$ – SteeveDroz Apr 8 '13 at 7:54
  • \$\begingroup\$ 19: {⍵⍳⍨(+/⊢=⊢∧⍳)¨⍳2*⍵} \$\endgroup\$ – Adám Jun 28 '16 at 19:40
  • \$\begingroup\$ I don't think you need to define f. \$\endgroup\$ – Zacharý Jul 9 '17 at 14:50
5
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GolfScript, 29 28 characters

{.{\.,{)1$\%},,-=}+2@?,?}:f;

Edit: A single char can be saved if we restrict the search to <2^n, thanks to Peter Taylor for this idea.

Previous Version:

{.{\)..,{)1$\%},,-@=!}+do}:f;

An attempt in GolfScript, run online.

Examples:

13 f p  # => 4096
14 f p  # => 192
15 f p  # => 144

The code contains essentially three blocks which are explained in detail in the following lines.

# Calculate numbers of divisors
#         .,{)1$\%},,-    
# Input stack: n
# After application: D(n)

.,          # push array [0 .. n-1] to stack
{           # filter array by function
  )         #   take array element and increase by one
  1$\%      #   test division of n ($1) by this value
},          # -> List of numbers x where n is NOT divisible by x+1
,           # count these numbers. Stack now is n xd(n)
-           # subtracting from n yields the result



# Test if number of divisors D(n) is equal to d
#         {\D=}+   , for D see above
# Input stack: n d
# After application: D(n)==d

{
  \         # swap stack -> d n
  D         # calculate D(n) -> d D(n)
  =         # compare
}+          # consumes d from stack and prepends it to code block         



# Search for the first number which D(n) is equal to d
#         .T2@?,?    , for T see above
# Input stack: d
# After application: f(d)

.           # duplicate -> d d
T           # push code block (!) for T(n,d) -> d T(n,d)
2@?         # swap and calculate 2^d -> T(n,d) 2^d
,           # make array -> T(n,d) [0 .. 2^d-1]
?           # search first element in array where T(n,d) is true -> f(d)
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  • \$\begingroup\$ Seems to go into an infinite loop for input 1. \$\endgroup\$ – Peter Taylor Apr 3 '13 at 16:05
  • \$\begingroup\$ My best solution so far borrows quite heavily from yours, to the extent that I think it deserves to be a comment rather than a separate answer. \$\endgroup\$ – Peter Taylor Apr 3 '13 at 16:31
4
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Python: 64

Revising Bakuriu's solution and incorporating grc's suggestion as well as the trick from plannapus's R solution, we get:

f=lambda n,k=1:n-sum(k%i<1for i in range(1,k+1))and f(n,k+1)or k
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4
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Python: 66

f=lambda n,k=1:n==sum(k%i<1for i in range(1,k+1))and k or f(n,k+1)

The above will raise a RuntimeError: maximum recursion depth exceeded with small inputs in CPython, and even setting the limit to a huge number it will probably give some problems. On python implementations that optimize tail recursion it should work fine.

A more verbose version, which shouldn't have such limitations, is the following 79 bytes solution:

def f(n,k=1):
    while 1:
        if sum(k%i<1for i in range(1,k+1))==n:return k
        k+=1
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  • \$\begingroup\$ I'm hitting the recursion limit on 11, 13, 17, 19, and others. \$\endgroup\$ – Steven Rumbalski Apr 3 '13 at 17:31
  • \$\begingroup\$ @StevenRumbalski Nobody mentioned that the program should work with arbitrary integers. Unfortunately the numbers grow up pretty fast even with small inputs. \$\endgroup\$ – Bakuriu Apr 3 '13 at 19:00
  • \$\begingroup\$ You can save some chars by using replacing if else with and or and ==1 with <1: f=lambda n,k=1:n==sum(k%i<1for i in range(1,k+1))and k or f(n,k+1) \$\endgroup\$ – grc Apr 4 '13 at 1:18
  • \$\begingroup\$ Because I find 66 a little too evil, you can save 2 characters if you use sum(k%-~i<1for i in range(k)) \$\endgroup\$ – Volatility Apr 25 '13 at 21:51
  • \$\begingroup\$ f=lambda n,k=1:n==sum(k%-~i<1for i in range(k))or-~f(n,k+1) saves 7 bytes. \$\endgroup\$ – Dennis Oct 15 '16 at 5:32
4
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Mathematica 38 36

(For[i=1,DivisorSum[++i,1&]!=#,];i)&

Usage:

(For[i=1,DivisorSum[++i,1&]!=#,];i)&@200

Result:

498960

Edit

Some explanation:

DivisorSum[n,form] represents the sum of form[i] for all i that divide n.

As form[i] I am using the function 1 &, that returns always 1, so effectively computing the sum of the divisors in a terse way.

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  • \$\begingroup\$ There was no code-golf tag so I gave a long answer! oops \$\endgroup\$ – DavidC Apr 3 '13 at 13:23
  • \$\begingroup\$ @DavidCarraher I just guessed :) \$\endgroup\$ – Dr. belisarius Apr 3 '13 at 13:28
  • \$\begingroup\$ I thought I knew what DivisorSum returns (the sum of the divisors) but I don't see how that is instrumental for answering the question posed. Would you explain how it works. BTW, I think you should include timing data for n=200; the function is remarkably fast, given all the numbers it had to check. \$\endgroup\$ – DavidC Apr 3 '13 at 13:31
  • \$\begingroup\$ @DavidCarraher See edit. Re: timings - My machine is way toooo slow :( \$\endgroup\$ – Dr. belisarius Apr 3 '13 at 13:35
  • \$\begingroup\$ Does Mathematica not have enough built-ins for the more sophisticated approach around factoring to be shorter? If that's the case, I'm disappointed. \$\endgroup\$ – Peter Taylor Apr 3 '13 at 16:41
3
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J, 33 chars

Fairly quick, goes through all smaller numbers and computes number of divisors based on factorization.

   f=.>:@]^:([~:[:*/[:>:_&q:@])^:_&1

   f 19
262144
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3
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Haskell 54

Quick and dirty (so readable and non-tricky) solution:

f k=head[x|x<-[k..],length[y|y<-[1..x],mod x y==0]==k]
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  • \$\begingroup\$ The edit did not make the answer any shorter, but it is maybe more haskell-like. Also I have always included the trailing newline to my code length, is this wrong? \$\endgroup\$ – shiona Apr 5 '13 at 17:04
  • \$\begingroup\$ I thought you've miscounted; the main purpose for the edit was to update the count. The change in code itself was minor. I think the other entries here don't count the trailing newline either, like e.g. the entry for J (33 chars). \$\endgroup\$ – Will Ness Apr 13 '13 at 19:52
2
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K, 42

Inefficient recursive solution that blows up the stack quite easily

{{$[x=+/a=_a:y%!1+y;y;.z.s[x;1+y]]}[x;0]} 

.

k){{$[x=+/a=_a:y%!1+y;y;.z.s[x;1+y]]}[x;0]}14
192
k){{$[x=+/a=_a:y%!1+y;y;.z.s[x;1+y]]}[x;0]}13
'stack
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2
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APL 33

F n            
i←0              
l:i←i+1          
→(n≠+/0=(⍳i)|i)/l
i 

Example:

F 6
12           
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2
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APL (25)

{⍵{⍺=+/0=⍵|⍨⍳⍵:⍵⋄⍺∇⍵+1}1}
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  • \$\begingroup\$ Cheater! echo -n '{⍵{⍺=+/0=⍵|⍨⍳⍵:⍵⋄⍺∇⍵+1}1}' | wc -c gives me 47! But really, could you please give me a link to some easy tutorial for APL? I tried to google it and have read a few articles, but still in the end I always want to ask "Why are they doing this :( ?". I have never worked with any non-ASCII syntax language and want to find out if it has any real advantage. \$\endgroup\$ – XzKto Apr 4 '13 at 12:55
  • \$\begingroup\$ This is for Dyalog APL, which is what I use, you can download the Windows version for free at the same site. dyalog.com/MasteringDyalogAPL/MasteringDyalogAPL.pdf \$\endgroup\$ – marinus Apr 5 '13 at 11:19
  • \$\begingroup\$ Wow, looks like I can really understand this one. Thank you for the link! The only downside is that they have some very strange licensing policy but maybe I just need to improve my english ) \$\endgroup\$ – XzKto Apr 5 '13 at 12:21
2
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R - 47 characters

f=function(N){n=1;while(N-sum(!n%%1:n))n=n+1;n}

!n%%1:n gives a vector of booleans: TRUE when an integer from 1 to n is a divisor of n and FALSE if not. sum(!n%%1:n) coerces booleans to 0 if FALSE and 1 if TRUE and sums them, so that N-sum(...) is 0 when number of divisors is N. 0 is then interpreted as FALSE by while which then stops.

Usage:

f(6)
[1] 12
f(13)
[1] 4096
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2
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Javascript 70

function f(N){for(j=i=m=1;m-N||j-i;j>i?i+=m=j=1:m+=!(i%++j));return i}

Really there are only 46 meaningful characters:

for(j=i=m=1;m-N||j-i;j>i?i+=m=j=1:m+=!(i%++j))

I probably should learn a language with shorter syntax :)

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  • \$\begingroup\$ N=>eval("for(j=i=m=1;m-N||j-i;j>i?i+=m=j=1:m+=!(i%++j));i") \$\endgroup\$ – TuxCrafting Oct 15 '16 at 10:06
2
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Haskell: 49 characters

It could be seen as an improvement of the earlier Haskell solution, but it was conceived in its own right (warning: it's very slow):

f n=until(\i->n==sum[1|j<-[1..i],rem i j<1])(+1)1

It's quite an interesting function, for example note that f(p) = 2^(p-1), where p is a prime number.

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  • \$\begingroup\$ The efficient, as opposed to short, way to calculate it would be to factor n into primes (with repetition), sort descending, decrement each one, zip with an infinite sequence of primes, and then fold the product of p^(factor-1) \$\endgroup\$ – Peter Taylor Apr 3 '13 at 18:46
  • 2
    \$\begingroup\$ @PeterTaylor Not necessary. For n=16=2*2*2*2 solution is 2^3*3^1*5^1=120, not 2^1*3^1*5^1*7^1=210. \$\endgroup\$ – randomra Apr 3 '13 at 19:15
2
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C: 66 64 characters

An almost short solution:

i;f(n){while(n-g(++i));return i;}g(j){return j?!(i%j)+g(j-1):0;}

And my previous solution that doesn't recurse:

i;j;k;f(n){while(k-n&&++i)for(k=0,j=1;j<=i;k+=!(i%j++));return i;}

Much shorter solutions must exist.

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2
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Haskell (120C), a very efficient method

1<>p=[]
x<>p|mod x p>0=x<>(p+1)|1<2=(div x p<>p)++[p]
f k=product[p^(c-1)|(p,c)<-zip[r|r<-[2..k],2>length(r<>2)](k<>2)]

Test code:

main=do putStrLn$show$ f (100000::Integer)

This method is very fast. The idea is first to find the prime factors of k=p1*p2*...*pm, where p1 <= p2 <= ... <= pm. Then the answer is n = 2^(pm-1) * 3^(p(m-1)-1) * 5^(p(m-2)-1) ....

For example, factorizing k=18, we get 18 = 2 * 3 * 3. The first 3 primes is 2, 3, 5. So the answer n = 2^(3-1) * 3^(3-1) * 5^(2-1) = 4 * 9 * 5 = 180

You can test it under ghci:

*Main> f 18
180
*Main> f 10000000
1740652905587144828469399739530000
*Main> f 1000000000
1302303070391975081724526582139502123033432810000
*Main> f 100000000000
25958180173643524088357042948368704203923121762667635047013610000
*Main> f 10000000000000
6558313786906640112489895663139340360110815128467528032775795115280724604138270000
*Main> f 1000000000000000
7348810968806203597063900192838925279090695601493714327649576583670128003853133061160889908724790000
*Main> f 100000000000000000
71188706857499485011467278407770542735616855123676504522039680180114830719677927305683781590828722891087523475746870000
*Main> f 10000000000000000000
2798178979166951451842528148175504903754628434958803670791683781551387366333345375422961774196997331643554372758635346791935929536819490000
*Main> f 10000000000000000000000
6628041919424064609742258499702994184911680129293140595567200404379028498804621325505764043845346230598649786731543414049417584746693323667614171464476224652223383190000
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  • \$\begingroup\$ That's a poor golf score, but +1 for the path you've taken! \$\endgroup\$ – SteeveDroz Apr 17 '13 at 6:05
  • \$\begingroup\$ For 8=2*2*2 this algorithm give number 2*3*5=30. But best solution is 2^3*3=24 (for 8=2*4) \$\endgroup\$ – AMK Apr 17 '13 at 15:59
  • \$\begingroup\$ The solution is incorrect if the specified number of divisors contain a high power of small prime. So most likely listed solutions for powers of 10 are wrong. \$\endgroup\$ – AMK Apr 17 '13 at 16:09
  • \$\begingroup\$ @AMK Yes, you're right. Thanks for pointing that out. \$\endgroup\$ – Ray Apr 17 '13 at 16:27
2
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Brachylog, 2 bytes

fl

Try it online!

Takes input through its output variable and outputs through its input variable.

f     The list of factors of
      the input variable
 l    has length equal to
      the output variable.

This exact same predicate, taking input through its input variable and outputting through its output variable, solves this challenge instead.

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  • \$\begingroup\$ Nice, but not eligible for that puzzle as the language is more recent than the question. \$\endgroup\$ – SteeveDroz Apr 25 at 7:07
  • \$\begingroup\$ When I was new here one of the first things I was told was that languages newer than questions aren't noncompeting anymore, and this is backed up by meta: codegolf.meta.stackexchange.com/questions/12877/… \$\endgroup\$ – Unrelated String Apr 25 at 7:16
  • \$\begingroup\$ Oh well, nevermind then. Apparently, rules are made to evolve and we must keep in mind that this site main purpose is to improve ourselves and have fun. Answer accepted! \$\endgroup\$ – SteeveDroz Apr 25 at 20:58
1
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C, 69 chars

Not the shortest, but the first C answer:

f(n,s){return--s?f(n,s)+!(n%s):1;}
x;
g(d){return++x,f(x,x)-d&&g(d),x;}

f(n,s) counts divisors of n in the range 1..s. So f(n,n) counts divisors of n.
g(d) loops (by recursion) until f(x,x)==d, then returns x.

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1
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Mathematica 38 36

(For[k=1,DivisorSigma[0, k]!= #,k++]; k)&

Usage

   (For[k = 1, DivisorSigma[0, k] != #, k++]; k) &[7]

(* 64 *)

First entry (before the code-golf tag was added to the question.)

A straightforward problem, given that Divisors[n] returns the divisors of n (including n) and Length[Divisors[n]] returns the number of such divisors.**

smallestNumber[nDivisors_] :=
   Module[{k = 1},
   While[Length[Divisors[k]] != nDivisors, k++];k]

Examples

Table[{i, nDivisors[i]}, {i, 1, 20}] // Grid

Mathematica graphics

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  • \$\begingroup\$ David, shorter and faster than Length@Divisors@n is DivisorSigma[0,n]. \$\endgroup\$ – Mr.Wizard May 13 '13 at 11:06
  • \$\begingroup\$ Thanks. I hadn't known about that use of DivisorSigma. \$\endgroup\$ – DavidC May 14 '13 at 2:46
  • \$\begingroup\$ math.stackexchange.com/questions/1551959/… \$\endgroup\$ – Sparr Nov 29 '15 at 19:49
1
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Jelly, 6 bytes (non-competing)

2*RÆdi

Try it online! or verify all test cases.

How it works

2*RÆdi  Main link. Argument: n (integer)

2*      Compute 2**n.
  R     Range; yield [1, ..., 2**n]. Note that 2**(n-1) has n divisors, so this
        range contains the number we are searching for.
   Æd   Divisor count; compute the number of divisors of each integer in the range.
     i  Index; return the first (1-based) index of n.
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  • \$\begingroup\$ Why do you do 2*? Is it that every number after that has more divisors than n? \$\endgroup\$ – Erik the Outgolfer Oct 15 '16 at 5:48
  • 2
    \$\begingroup\$ No; e.g., all primes have exactly two divisors. However, we are searching for the smallest positive integer with n divisors. Since 2**(n-1) belongs to that range, the smallest one does as well. \$\endgroup\$ – Dennis Oct 15 '16 at 5:50
0
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C++, 87 characters

int a(int d){int k=0,r,i;for(;r!=d;k++)for(i=2,r=1;i<=k;i++)if(!(k%i))r++;return k-1;}
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0
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Python2, 95 characters, Non-recursive

A bit more verbose than the other python solutions but it's non-recursive so it doesn't hit cpython's recursion limit:

from itertools import*
f=lambda n:next(i for i in count()if sum(1>i%(j+1)for j in range(i))==n)
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0
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Perl 6, 39 chars

{my \a=$=0;a++while $_-[+] a X%%1..a;a}

Example usage:

say (0..10).map: {my \a=$=0;a++while $_-[+] a X%%1..a;a}
(0 1 2 4 6 16 12 64 24 36 48)
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