Proper Divisor mash-up

Question

A proper divisor is a divisor of a number n, which is not n itself. For example, the proper divisors of 12 are 1, 2, 3, 4 and 6.

You will be given an integer x, x ≥ 2, x ≤ 1000. Your task is to sum all the highest proper divisors of the integers from 2 to x (inclusive) (OEIS A280050).

Example (with x = 6):

• Find all the integers between 2 and 6 (inclusive): 2,3,4,5,6.

• Get the proper divisors of all of them, and pick the highest ones from each number:

  • 2 -> 1
  • 3 -> 1
  • 4 -> 1, 2
  • 5 -> 1
  • 6 -> 1, 2, 3.

• Sum the highest proper divisors: 1 + 1 + 2 + 1 + 3 = 8.

• The final result is 8.

Test Cases

Input  |  Output
-------+---------
       |
 2     | 1
 4     | 4
 6     | 8
 8     | 13
 15    | 41
 37    | 229
 100   | 1690
 1000  | 165279

Rules

• Default Loopholes apply.

• You can take input and provide output by any standard method.

• This is code-golf, the shortest valid submission in each language wins! Have Fun!

Answer 1

Oasis, 4 bytes

Code:

nj+U

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Explanation:

Extended version:

nj+00

    0   = a(0)
   0    = a(1)

a(n) =

n       # Push n
 j      # Get the largest divisor under n
  +     # Add to a(n - 1)

Answer 2

Brachylog, 10 bytes

⟦bb{fkt}ᵐ+

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Answer 3

Pyth, 13 9 8 bytes

1 byte thanks to jacoblaw.

tsm*FtPh

Test suite.

How it works

The largest proper divisor is the product of the prime factors except the smallest one.

Answer 4

Husk, 7 bytes

ṁȯΠtptḣ

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Explanation

Husk has no built-in for computing the divisors directly (yet), so I'm using prime factorization instead. The largest proper divisor of a number is the product of its prime factors except the smallest one. I map this function over the range from 2 to the input, and sum the results.

ṁȯΠtptḣ  Define a function:
      ḣ  Range from 1 to input.
     t   Remove the first element (range from 2).
ṁ        Map over the list and take sum:
 ȯ        The composition of
    p     prime factorization,
   t      tail (remove smallest prime) and
  Π       product.

Answer 5

Python 2, 50 bytes

f=lambda n,k=2:n/k and(f(n,k+1),n/k+f(n-1))[n%k<1]

This is slow and can't even cope with input 15 on TIO.

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However, memoization (thanks @musicman523) can be used to verify all test cases.

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Alternate version, 52 bytes

At the cost of 2 bytes, we can choose whether to compute f(n,k+1) or n/k+f(n-1).

f=lambda n,k=2:n>1and(n%k and f(n,k+1)or n/k+f(n-1))

With some trickery, this works for all test cases, even on TIO.

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Answer 6

Jelly, 6 bytes

ÆḌ€Ṫ€S

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How it works

ÆḌ€Ṫ€S
ÆḌ€    map proper divisor (1 would become empty array)
           implicitly turns argument into 1-indexed range
   Ṫ€  map last element
     S sum

Answer 7

JavaScript (ES6), 40 bytes

f=(n,i=2)=>n<2?0:n%i?f(n,i+1):n/i+f(n-1)

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

A number equals the product of its highest proper divisor and its smallest prime factor.

Answer 8

05AB1E, 9 8 bytes

-1 Byte thanks to Leaky Nun's prime factor trick in his Pyth answer

L¦vyÒ¦PO

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Explanation

L¦vyÒ¦PO
L¦       # Range [2 .. input]
  vy     # For each...
    Ò¦    # All prime factors except the first one
      P   # Product
       O  # Sum with previous results
         # Implicit print

Alternative 8 Byte solution (That doesnt work on TIO)

L¦vyѨθO    

and ofc alternative 9 Byte solution (That works on TIO)

L¦vyѨ®èO    

Answer 9

Retina, 31 24 bytes

7 bytes thanks to Martin Ender.

.+
$*
M!&`(1+)(?=\1+$)
1

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How it works

The regex /^(1+)\1+$/ captures the largest proper divisor of a certain number represented in unary. In the code, the \1+ is turned to a lookahead syntax.

Answer 10

Mathematica, 30 bytes

Divisors[i][[-2]]~Sum~{i,2,#}&

Answer 11

Python 2 (PyPy), 73 71 70 bytes

n=input();r=[0]*n;d=1
while n:n-=1;r[d+d::d]=n/d*[d];d+=1
print sum(r)

Not the shortest Python answer, but this just breezes through the test cases. TIO handles inputs up to 30,000,000 without breaking a sweat; my desktop computer handles 300,000,000 in a minute.

At the cost of 2 bytes, the condition n>d could be used for a ~10% speed-up.

Thanks to @xnor for the r=[0]*n idea, which saved 3 bytes!

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Answer 12

Haskell, 48 46 43 bytes

f 2=1
f n=until((<1).mod n)pred(n-1)+f(n-1)

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Edit: @rogaos saved two bytes. Thanks!

Edit II: ... and @xnor another 3 bytes.

Answer 13

Japt, 8+2=10 8 6 bytes

òâ1 xo

Test it

• 1 byte saved thanks to ETHproductions.

---

Explanation

    :Implicit input of integer U.
ò   :Generate an array of integers from 1 to U, inclusive
â   :Get the divisors of each number,
1   :  excluding itself.
x   :Sum the main array
o   :by popping the last element from each sub-array.
    :Implicit output of result

Answer 14

MATL, 12 bytes

q:Q"@Z\l_)vs

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Explanation

        % Implicitly grab input (N)
q       % Subtract one
:       % Create an array [1...(N-1)]
Q       % Add one to create [2...N]
"       % For each element
  @Z\   % Compute the divisors of this element (including itself)
  l_)   % Grab the next to last element (the largest that isn't itself)
  v     % Vertically concatenate the entire stack so far
  s     % Sum the result

Answer 15

PHP, 56 bytes

for($i=1;$v||$argn>=$v=++$i;)$i%--$v?:$v=!$s+=$v;echo$s;

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Answer 16

Prolog (SWI), 72 bytes

f(A,B):-A=2,B=1;C is A-1,f(C,D),between(2,A,E),divmod(A,E,S,0),B is D+S.

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Answer 17

Cubix, 27 39 bytes

?%\(W!:.U0IU(;u;p+qu.@Op\;;

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Cubified

      ? % \
      ( W !
      : . U
0 I U ( ; u ; p + q u .
@ O p \ ; ; . . . . . .
. . . . . . . . . . . .
      . . .
      . . .
      . . .

Watch It Run

• 0IU Set up the stack with an accumulator, and the starting integer. U-turn into the outer loop
• :(? duplicate the current top of stack, decrement and test
• \pO@ if zero loop around the cube to a mirror, grab the bottom of stack, output and halt
• %\! if positive, mod, relect and test.
  • u;.W if truthy, u-turn, remove mod result and lane change back into inner loop
  • U;p+qu;;\( if falsey, u-turn, remove mod result, bring accumulator to top, add current integer (top) divisor push to bottom and u-turn. Clean up the stack to have just accumulator and current integer, decrement the integer and enter the outer loop again.

Answer 18

C# (.NET Core), 74 72 bytes

n=>{int r=0,j;for(;n>1;n--)for(j=n;--j>0;)if(n%j<1){r+=j;j=0;}return r;}

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• 2 bytes shaved thanks to Kevin Cruijssen.

Answer 19

Actually, 12 bytes

u2x⌠÷R1@E⌡MΣ

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Answer 20

Python 3, 78 75 73 71 bytes

Not even close to Leaky nun's python answer in byte count.

f=lambda z:sum(max(i for i in range(1,y)if 1>y%i)for y in range(2,z+1))

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Answer 21

Python 3, 69 63 59 bytes

4 bytes thanks to Dennis.

f=lambda n:n-1and max(j for j in range(1,n)if n%j<1)+f(n-1)

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I set the recursion limit to 2000 for this to work for 1000.

Answer 22

Charcoal, 37 bytes

Ａ⁰βＦ…·²Ｎ«Ａ⟦⟧δＦ⮌…¹ι«¿¬﹪ικ⊞δκ»Ａ⁺β⌈δβ»Ｉβ

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Link is to the verbose version. It took me almost all day to figure out how could I solve a non-ASCII-art-related question in Charcoal, but finally I got it and I am very proud of me. :-D

Yes, I am sure this can be golfed a lot. I just translated my C# answer and I am sure things can be done differently in Charcoal. At least it solves the 1000 case in a couple of seconds...

Answer 23

Pari/GP, 36 30 bytes

n->sum(i=2,n,i/divisors(i)[2])

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Answer 24

Python 2 (PyPy), 145 bytes

Because turning code-golf competitions into fastest-code competitions is fun, here is an O(n) algorithm that, on TIO, solves n = 5,000,000,000 in 30 seconds. (Dennis’s sieve is O(n log n).)

import sympy
n=input()
def g(i,p,k,s):
 while p*max(p,k)<=n:l=k*p;i+=1;p=sympy.sieve[i];s-=g(i,p,l,n/l*(n/l*k+k-2)/2)
 return s
print~g(1,2,1,-n)

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How it works

We count the size of the set

S = {(a, b) | 2 ≤ a ≤ n, 2 ≤ b ≤ largest-proper-divisor(a)},

by rewriting it as the union, over all primes p ≤ √n, of

S_p = {(p⋅d, b) | 2 ≤ d ≤ n/p, 2 ≤ b ≤ d},

and using the inclusion–exclusion principle:

|S| = ∑ (−1)^{m − 1} |S_p1 ∩ ⋯ ∩ S_pm| over m ≥ 1 and primes p₁ < ⋯ < p_m ≤ √n,

where

S_p1 ∩ ⋯ ∩ S_pm = {(p₁⋯p_m⋅e, b) | 1 ≤ e ≤ n/(p₁⋯p_m), 2 ≤ b ≤ p₁⋯p_{m − 1}e},
|S_p1 ∩ ⋯ ∩ S_pm| = ⌊n/(p₁⋯p_m)⌋⋅(p₁⋯p_{m − 1}⋅(⌊n/(p₁⋯p_m)⌋ + 1) − 2)/2.

The sum has C⋅n nonzero terms, where C converges to some constant that’s probably 6⋅(1 − ln 2)/π² ≈ 0.186544. The final result is then |S| + n − 1.

Answer 25

NewStack, 5 bytes

Luckily, there's actually a built in.

Nᵢ;qΣ

The breakdown:

Nᵢ       Add the first (user's input) natural numbers to the stack.
  ;      Perform the highest factor operator on whole stack.
   q     Pop bottom of stack.
    Σ    Sum stack.

In actual English:

Let's run an example for an input of 8.

Nᵢ: Make list of natural numbers from 1 though 8: 1, 2, 3, 4, 5, 6, 7, 8

;: Compute the greatest factors: 1, 1, 1, 2, 1, 3, 1, 4

q. Remove the first element: 1, 1, 2, 1, 3, 1, 4

Σ And take the sum: 1+1+2+1+3+1+4 = 13

Answer 26

Java 8, 78 74 72 bytes

n->{int r=0,j;for(;n>1;n--)for(j=n;j-->1;)if(n%j<1){r+=j;j=0;}return r;}

Port of @CarlosAlejo's C# answer.

Try it here.

Old answer (78 bytes):

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

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Explanation (of old answer):

n->{                    // Method with integer parameter and integer return-type
  int r=0,              //  Result-integers
      i=1,j,k;          //  Some temp integers
  for(;++i<=n;          //  Loop (1) from 2 to `n` (inclusive)
      r+=k)             //    And add `k` to the result after every iteration
    for(j=1,k=1;++j<i;  //   Inner loop (2) from `2` to `i` (exclusive)
      k=i%j<1?j:k       //    If `i` is dividable by `j`, replace `k` with `j`
    );                  //   End of inner loop (2)
                        //  End of loop (2) (implicit / single-line body)
  return r;             //  Return result-integer
}                       // End of method

Answer 27

Thunno 2 -S, 5 bytes

ı⁺FṫG

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Explanation

ı⁺FṫG  # Implicit input
       # Decrement the input
ı      # Map over [1..input-1]:
 ⁺     #  Increment the number
  Fṫ   #  Push the proper divisors
    G  #  Maximum of this list
       # Sum the list
       # Implicit output

Answer 28

Arturo, 32 bytes

$=>[∑map..2&=>[x:do factors&]]

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Answer 29

Lua, 74 bytes

c=0 for i=2,...do for j=1,i-1 do t=i%j<1 and j or t end c=c+t end print(c)

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Answer 30

J, 18 bytes

[:+/1}.&.q:@+}.@i.

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