Twoja funkcja przyjmuje liczbę naturalną i zwraca najmniejszą liczbę naturalną, która ma dokładnie taką liczbę dzielników, w tym siebie.

Przykłady:

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]
 ...

Funkcja nie musi zwracać listy dzielników, są one tylko dla przykładów.

APL, 25 24 23 znaków

f←{({+/⍵=⍵∧⍳⍵}¨⍳2*⍵)⍳⍵}

Definiuje funkcję, fktórej można następnie użyć do obliczenia liczb:

> f 13
4096

> f 14
192

Rozwiązanie wykorzystuje fakt, że LCM (n, x) == n iff x dzieli n . Zatem blok {+/⍵=⍵∧⍳⍵}po prostu oblicza liczbę dzielników. Ta funkcja jest stosowana do wszystkich liczb od 1 do 2 ^ d ¨⍳2*⍵ . Wynikowa lista jest następnie przeszukiwana pod kątem samej d ( ⍳⍵), która jest pożądaną funkcją f (d) .

GolfScript, 29 28 znaków

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

Edycja: Pojedynczy znak może zostać zapisany, jeśli ograniczymy wyszukiwanie do <2 ^ n, dzięki Peterowi Taylorowi za ten pomysł.

Poprzednia wersja:

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

Próba w GolfScript, uruchom online .

Przykłady:

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

Kod zawiera zasadniczo trzy bloki, które zostały szczegółowo wyjaśnione w poniższych wierszach.

# 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)

Python: 64

Przeglądając rozwiązanie Bakuriu i włączając sugestię grc, a także sztuczkę z rozwiązania R plannapusa, otrzymujemy:

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

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)

Powyższe podniesie wartość RuntimeError: maximum recursion depth exceededprzy niewielkich nakładach w CPython, a nawet ustawienie limitu na ogromną liczbę prawdopodobnie spowoduje pewne problemy. W implementacjach Pythona, które optymalizują rekurencję ogona, powinno działać dobrze.

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

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.

J, 33 chars

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

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

   f 19
262144

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]

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

APL 33

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

Example:

F 6
12           

APL (25)

{⍵{⍺=+/0=⍵|⍨⍳⍵:⍵⋄⍺∇⍵+1}1}

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

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 :)

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.

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.

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

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.

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.

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

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.

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;}

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)

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)