Biorąc pod uwagę, że masz nieskończoną sekwencję liczb zdefiniowaną następująco:

1: 1 = 1
2: 1 + 2 = 3
3: 1 + 3 = 4
4: 1 + 2 + 4 = 7
5: 1 + 5 = 6
6: 1 + 2 + 3 + 6 = 12
7: 1 + 7 = 8
...

Sekwencja jest sumą dzielników n, w tym 1 i n.

Biorąc pod uwagę dodatnią liczbę całkowitą xjako dane wejściowe, oblicz najniższą liczbę, nktóra da wynik większy niż x.

Przypadki testowe

f(100) = 48, ∑ = 124
f(25000) = 7200, ∑ = 25389
f(5000000) = 1164240, ∑ = 5088960

Oczekiwany wynik

Twój program powinien zwrócić zarówno n sumę dzielników, jak i tak:

$ ./challenge 100
48,124

Zasady

Jest to kod-golf, więc wygrywa najkrótszy kod w bajtach w każdym języku.

Brachylog , 9 bajtów

∧;S?hf+S>

Ten program pobiera dane wejściowe z „zmiennej wyjściowej” ., a dane wyjściowe z „zmiennej wejściowej” ?. Wypróbuj online!

Wyjaśnienie

∧;S?hf+S>
∧;S        There is a pair [N,S]
   ?       which equals the output
    h      such that its first element's
     f     factors'
      +    sum
       S   equals S,
        >  and is greater than the input.

Zmienna niejawna Njest wyliczana w kolejności rosnącej, więc dla wyniku używana jest jej najniższa dopuszczalna wartość.

Galaretka , 18 12 11 10 bajtów

1Æs>¥#ḢṄÆs

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-1 bajt dzięki Mr. Xcoder !

Jak to działa

1Æs>¥#ḢṄÆs - Main link. Argument: n (integer)
1   ¥#     - Find the first n integers where...
 Æs        -   the divisor sum
   >       -   is greater than the input
       Ṅ   - Print...
      Ḣ    -   the first element
        Æs - then print the divisor sum

Wolfram Language (Mathematica) , 53 bajty

{#,f@#}&@@Select[Range[x=#]+1,(f=Tr@*Divisors)@#>x&]&

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Próbuje wszystkich wartości od 2 do x + 1, gdzie x jest wartością wejściową.

( SelectZwraca listę wszystkich wartości, które działają, ale funkcja {#,f@#}&przyjmuje je wszystkie jako dane wejściowe, a następnie ignoruje wszystkie dane wejściowe oprócz pierwszej.)

R , 71 bajtów

function(x)for(n in 1:x){z=sum(which(n%%1:n==0));if(z>x)return(c(n,z))}

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Łuska , 12 11 bajtów

§eVḟ>⁰moΣḊN

-1 bajt, dzięki @Zgarb!

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R , 73 bajty

n=scan();while(1){d=(x=1:T)[!T%%x];if(sum(d)>n)break;T=T+1};cat(T,sum(d))

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Outgolfed by duckmayr .

Japt , 15 bajtów

[@<(V=Xâ x}a V]

Spróbuj

Wyjaśnienie

Implicit input of integer U. [] is our array wrapper. For the first element, @ }a is a function that run continuously until it returns a truthy value, passing itself an incrementing integer (starting at 0) each time, and outputting the final value of that integer. â gets the divisors of the current integer (X), x sums them and that result is assigned to variable V. < checks if U is less than V. The second element in the array is then just V.

Clojure, 127 bytes

(defn f[n](reduce +(filter #(zero?(rem n %))(range 1(inc n)))))
(defn e[n](loop[i 1 n n](if(>(f i)n){i,(f i)}(recur(inc i)n))))

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thanks to @steadybox for -4 bytes!

Ruby, 58 bytes

Full program because I'm not sure if lambdas are allowed. /shrug

gets
$.+=1until$_.to_i.<v=(1..$.).sum{|n|$.%n<1?n:0}
p$.,v

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Explanation

gets     # read line ($_ is used instead of v= because it cuts a space)
$.+=1    # $. is "lines read" variable which starts at 1 because we read 1 line
    until     # repeat as long as the next part is not true
$_.to_i  # input, as numeric
  .<v=   # is <, but invoked as function to lower operator prescedence
  (1..$.)        # Range of 1 to n
  .sum{|n|       # .sum maps values into new ones and adds them together
     $.%n<1?n:0  # Factor -> add to sum, non-factor -> 0
  }
p$.,v    # output n and sum

JavaScript (ES6), 61 58 bytes

f=(n,i=1,s=j=0)=>j++<i?f(n,i,i%j?s:s+j):s>n?[i,s]:f(n,++i)

<input type=number min=0 oninput=o.textContent=f(this.value)><pre id=o>

Edit: Saved 3 bytes thanks to @Arnauld.

05AB1E, 11 bytes

>LʒÑO‹}нDÑO

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Leaves the output on the stack, as allowed per meta consensus. I added ) for the sake of visualization, but the program also implicitly prints the top of the stack.

APL (Dyalog), 32 bytes

{⍺<o←+/o/⍨0=⍵|⍨o←⍳⍵:⍵,o⋄⍺∇⍵+1}∘0

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⍵o⍺⍵.

SOGL V0.12, 14 bytes

1[:Λ∑:A.>?ao←I

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Explanation:

1               push 1
 [              while ToS != 0
  :Λ              get the divisors
    ∑             sum
     :A           save on variable A without popping
       .>?  ←     if greater than the input
          ao        output the variable A
            ←       and stop the program, implicitly outputting ToS - the counter
             I    increment the counter

C,  79  78 bytes

i,n,s;f(x){for(i=n=s=0;x>s;s+=n%++i?0:i)i-n||(++n,i=s=0);printf("%d %d",n,s);}

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MATL, 12 bytes

`@Z\sG>~}@6M

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Explanation

`      % Do...while
  @    %   Push iteration index (1-based)
  Z\   %   Array of divisors
  s    %   Sum of array
  G    %   Push input
  >~   %   Greater than; logical negate. This is the loop condition
}      % Finally (execute on loop exit)
  @    %   Push latest iteration index
  6M   %   Push latest sum of divisors again
       % End (implicit). Run new iteration if top of the stack is true
       % Display stack (implicit)

Gaia, 11 bytes

dΣ@>
↑#(:dΣ

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Leaves the output on the stack, as allowed per meta consensus. I added €. for the sake of visualization, but the program also implicitly prints the top of the stack.

Haskell, 59 bytes

f x=[(i,s)|i<-[1..],s<-[sum[d|d<-[1..i],i`mod`d<1]],s>x]!!0

-1 byte, thanks to @nimi!

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Ohm v2, 11 bytes

£^DVΣD³>‽;«

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Factor, 88

USE: math.primes.factors [ 0 0 [ drop 1 + dup divisors sum pick over > ] loop rot drop ]

Brute-force search. It's a quotation (lambda), call it with x on the stack, leaves n and f(n) on the stack.

As a word:

: f(n)>x ( x -- n f(n) )
  0 0 [ drop 1 + dup divisors sum pick over > ] loop rot drop ;

Python 3, 163 bytes

def f(x):
    def d(x):return[i for i in range(1,x+1) if x%i==0]
    return min(i for i in range(x) if sum(d(i)) >x),sum(d(min(i for i in range(x) if sum(d(i)) >x)))

Python 3, 100 bytes

d=lambda y:sum(i+1for i in range(y)if y%-~i<1)
f=lambda x:min((j,d(j))for j in range(x+1)if x<=d(j))

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Thanks to Jonathan Frech's comment on the previous python 3 attempt, I have just greatly expanded my knowledge of python syntax. I'd never have thought of the -~i for i+1 trick, which saves two characters.

However, that answer is 1) not minimal and 2) doesn't work for x=1 (due to an off-by-one error which is easy to make while going for brevity; I suggest everyone else check their answers for this edge case!).

Quick explanation: sum(i+1for i in range(y)if y%-~i<1) is equivalent to sum(i for i in range(1,y+1)if y%i<1) but saves two characters. Thanks again to Mr. Frech.

d=lambda y:sum(i+1for i in range(y)if y%-~i<1) therefore returns the divisors of y.

f=lambda x:min((j,d(j))for j in range(x+1)if x<=d(j)) is where I really did work. Since comparing a tuple works in dictionary order, we can compare j,d(j) as easily as we can compare j, and this lets us not have to find the minimal j, store it in a variable, and /then/ compute the tuple in a separate operation. Also, we have to have the <=, not <, in x<=d(j), because d(1) is 1 so if x is 1 you get nothing. This is also why we need range(x+1) and not range(x).

I'd previously had d return the tuple, but then I have to subscript it in f, so that takes three more characters.

Python 2, 81 bytes

def f(n):
 a=b=0
 while b<n:
	a+=1;i=b=0
	while i<a:i+=1;b+=i*(a%i<1)
 return a,b

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Java (OpenJDK 8), 91 bytes

i->{for(int j=0;j++<i;)for(int k=0,l=0;k++<j;)if((l+=j%k<1?k:0)>i)return k+","+l;return"";}

Try it online! (timeout on third test case)

Perl 5, 60 + 1 (-p) = 61 bytes

$"=$_;until($_>$"){$_=$/=0;$\--;$_+=$\%$/?0:$/until++$/>-$\}

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Output format:

sum-n

Clojure, 102 bytes

#(loop[i 1](let[s(apply +(for[j(range 1(inc i)):when(=(mod i j)0)]j))](if(> s %)[i s](recur(inc i)))))

PHP, 69 bytes

for(;$argv[1]>=$t;)for($t=$j=++$i;--$j;)$t+=$i%$j?0:$j;echo$i,',',$t;

Perl 6, 48 bytes

{first :kv,*>$_,map {sum grep $_%%*,1..$_},0..*}

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