27

我将此序列称为“耶稣序列”，因为它是mod的总和。</ pun>

对于这个序列，你把所有的正整数中号小于输入ñ，走的总和ñ模每米。换一种说法：

a_{n} = \sum_{m = 1}^{n - 1} n mod m

$a_n = \sum_{m=1}^{n-1}{n\bmod m}$

For example, take the term 14:

14 % 1 = 0
14 % 2 = 0
14 % 3 = 2
14 % 4 = 2
14 % 5 = 4
14 % 6 = 2
14 % 7 = 0
14 % 8 = 6
14 % 9 = 5
14 % 10 = 4
14 % 11 = 3
14 % 12 = 2
14 % 13 = 1
0+0+2+2+4+2+0+6+5+4+3+2+1=31

Your goal here is to write a function that implements this sequence. You should take the sequence term (this will be a positive integer from 1 to 2³¹) as the only input, and output the value of that term. This is OEIS A004125.

As always, standard loopholes apply and the shortest answer in bytes wins!

code-golf sequence division

— Nissa
source

7

05AB1E, 3 bytes

L%O

Try it online!

— Mr. Xcoder
source

8

Haskell, 22 bytes

f x=sum$mod x<$>[1..x]

Try it online!

Explanation

Yes.

— ბიმო
source

5

Great explanation

— Esolanging Fruit

7

Ruby, 28 27 23 bytes

-4 bytes thanks to @daniero.

->n{(1..n).sum{|i|n%i}}

Try it online!

Ruby, 28 bytes

f=->n{n>($.+=1)?n%$.+f[n]:0}

Try it online!

— Jordan
source

6

Funky, 25 bytes

n=>fors=~-i=1i<n)s+=n%i++

Just the Naïve answer, seems to work.

Try it online!

Desmos, 25 bytes.

f(x)=\sum_{n=1}^xmod(x,n)

Paste into Desmos, then run it by calling f.

When pasted into Desmos, the latex looks like this

The graph however looks like

Although it looks random and all over the place, that's the result of only supporting integers.

RProgN 2, 9 bytes

x=x³x\%S+

Explained

x=x³x\%S+
x=          # Store the input in "x"
  x         # Push the input to the stack.
   ³x\%     # Define a function which gets n%x
       S    # Create a stack from "x" with the previous function. Thus this gets the range from (1,x), and runs (i%x) on each element.
        +   # Get the sum of this stack.

Try it online!

ReRegex, 71 bytes

#import math
(_*)_a$/d<$1_>b$1a/(\d+)b/((?#input)%$1)+/\+a//u<#input
>a

Try it online!

ARBLE, 19 bytes

sum(range(1,a)|a%i)

Try it online!

MaybeLater, 56 bytes

whenf is*{n=0whenx is*{ifx>0{n=n+f%x x--}elseprintn}x=f}

Try it online!

— ATaco
source

向这一挑战提交的作品会永远结束吗？到目前为止，我每40分钟就会得到一个新的：P

— 妮莎（Nissa）

@StephenLeppik哦，我还有更多来，别担心。

— ATaco

@StephenLeppik I'd rather not, because they're of various quality across various languages.

— ATaco

@StephenLeppik I've combined them for you, begrudgingly.

— ATaco

4

Please don't do this. Separate languages -- even separate approaches -- should go in separate answers.

— Dennis

5

果冻，3个字节

%RS

说明

%RS
 R   Range(input)  [1...n]
%    Input (implicit) modulo [1...n]->[n%1,n%2...n%n]
  S  Sum of the above

在线尝试！

— dylnan
source

5

MATL, 4 bytes

t:\s

Try it online!

Explanation:

t      % Duplicate input. Stack: {n, n}
 :     % Range 1...n. Stack: {n, [1...n]}
  \    % Modulo. Stack: {[0,0,2,2,4,...]}
   s   % Sum. Implicitly display result.

— Sanchises
source

4

Ohm v2, 4 bytes

D@%Σ

Try it online!

— Cinaski
source

4

R，20个字节

sum((n=scan())%%1:n)

在线尝试！

— 浮游生物
source

4

Python 2, 44 bytes

lambda n:sum(map(lambda x:x%(n-x),range(n)))

Try it online!

EDIT: Changed range(0,n) to range(n)

— Max00355
source

2

Hello, and welcome to the site! range implicitly takes a first argument of 0, so you could shorten this by two bytes by doing range(n) instead.

— DJMcMayhem

哇！我什至没有想到这一点。谢谢

— Max00355

1

Welcome to PPCG! You can use a list comprehension instead of map for 38 bytes: Try it online!

— Mr. Xcoder

You are right, but that was used in Neil's answer, so I wasn't sure if copying it would have been the best thing. Unless I am missing something here of course. I wanted to post the alternative, even if it was a bit longer.

— Max00355

3

JavaScript (ES6), 26 bytes

f=(n,k=n)=>n&&k%n+f(n-1,k)

Demo

Show code snippet

f=(n,k=n)=>n&&k%n+f(n-1,k)

for(n = 1; n < 30; n++) {
  console.log('a(' + n + ') = ' + f(n))
}

Expand snippet

— Arnauld
source

3

Python 3, 37 bytes

lambda a:sum(a%k for k in range(1,a))

Try it online!

— Neil
source

4 answers in 10 minutes? Wow.

— Nissa

1

Must be some very gifted programmers. ;)

— Neil

3

Charcoal, 9 bytes

ＩΣＥＮ﹪Ｉθ⊕ι

Try it online!

Link is to the verbose version of the code:

Print(Cast(Sum(Map(InputNumber(),Modulo(Cast(q),++(i))))));

— Charlie
source

3

Standard ML (MLton), 53 51 bytes

fn& =>let fun f 1a=a|f%a=f(% -1)(a+ &mod%)in f&0end

Try it online!

Ungolfed:

fn n =>
   let fun f 1 a = a
         | f x a = f (x-1) (a + n mod x)
   in  
       f n 0
   end

Previous 53 byte version:

fn n=>foldl op+0(List.tabulate(n-1,fn i=>n mod(i+1)))

Try it online!

Explanation:

List.tabulate takes an integer x and a function f and generates the list [f 0, f 1, ..., f(x-1)]. Given some number n, we call List.tabulate with n-1 and the function fn i=>n mod(i+1) to avoid dividing by zero. The resulting list is summed with foldl op+0.

— Laikoni
source

3

Java (OpenJDK 8), 45 bytes

n->{int m=n,s=0;for(;m-->1;)s+=n%m;return s;}

Try it online!

— Olivier Grégoire
source

1

+1 For using the goes to (-->) operator.

— raznagul

3

Mathematica，18个字节

#~Mod~i~Sum~{i,#}&

在线尝试！

— 完全
source

2

same bytes Tr[#~Mod~Range@#]&

— J42161217

3

APL (Dyalog), 5 bytes

+/⍳|⊢

Try it online!

How?

Monadic train -

+/ - sum

⊢ - n

| - vectorized modulo

⍳ - the range of n

— Uriel
source

3

Japt, 6 5 bytes

Saved 1 byte thanks to @Shaggy

Æ%XÃx

Test it online!

How it works

         Implicit: U = input number
Æ        Create the range [0, U),
 %XÃ       mapping each item X to U%X. U%0 gives NaN.
    x    Sum, ignoring non-numbers.
         Implicit: output result of last expression

— ETHproductions
source

2

05AB1E, 6 bytes

ÎGIN%+

Try it online!

My first 05AB1E program ;)

~~Btw I got two 39s, 1 for JS6 and 1 for python, but I was too late~~

Explanation:

ÎGIN%+
Î                      # Push 0, then input, stack = [(accumulator = 0), I]
 G                     # For N in range(1, I), stack = [(accumulator)]
  IN                   # Push input, then N, stack = [(accumulator), I, N]
    %                  # Calculate I % N, stack = [(accumulator), I % N]
     +                 # Add the result of modulus to accumulator

— Shieru Asakoto
source

2

Ruby, 23 bytes

->n{(1..n).sum{|i|n%i}}

Try it online!

— G B
source

2

Julia 0.4, 15 bytes

x->sum(x%[1:x])

Try it online!

— Lescurel
source

2

Add++, 14 bytes

L,RAdx$p@BcB%s

Try it online!

How it works

L,   - Create a lambda function.
     - Example argument:     [7]
  R  - Range;        STACK = [[1 2 3 4 5 6 7]]
  A  - Argument;     STACK = [[1 2 3 4 5 6 7] 7]
  d  - Duplicate;    STACK = [[1 2 3 4 5 6 7] 7 7]
  x  - Repeat;       STACK = [[1 2 3 4 5 6 7] 7 [7 7 7 7 7 7 7]]
  $p - Swap and pop; STACK = [[1 2 3 4 5 6 7] [7 7 7 7 7 7 7]]
  @  - Reverse;      STACK = [[7 7 7 7 7 7 7] [1 2 3 4 5 6 7]]
  Bc - Zip;          STACK = [[7 1] [7 2] [7 3] [7 4] [7 5] [7 6] [7 7]]
  B% - Modulo each;  STACK = [0, 1, 1, 3, 2, 1, 0]
  s  - Sum;          STACK = [8]

— caird coinheringaahing
source

2

4, 67 bytes

4 doesn't have any modulo built in.

3.79960101002029980200300023049903204040310499040989804102020195984

Try it online!

— Uriel
source

2

Windows Batch (CMD), 63 bytes

@set s=0
@for /l %%i in (1,1,%1)do @set/as+=%1%%%%i
@echo %s%

Previous 64-byte version:

@set/ai=%2+1,s=%3+%1%%i
@if %i% neq %1 %0 %1 %i% %s%
@echo %s%

— Neil
source

2

T-SQL, 80 79 bytes

-1 byte thanks to @MickyT

WITH c AS(SELECT @ i UNION ALL SELECT i-1FROM c WHERE i>1)SELECT SUM(@%i)FROM c

Receives input from an integer parameter named @, something like this:

DECLARE @ int = 14;

Uses a Common Table Expression to generate numbers from 1 to n. Then uses that cte to sum up the moduluses.

Note: a cte needs a ; between the previous statement and the cte. Most code I've seen puts the ; right before the declaration, but in this case I can save a byte by having it in the input statement (since technically my code by itself is the only statement).

Try it (SEDE)

The less "SQL-y" way is only 76 bytes. This time the input variable is @i instead of @ (saves one byte). This one just does a while loop.

DECLARE @ int=2,@o int=0WHILE @<@i BEGIN SELECT @o+=@i%@,@+=1 END PRINT @o

— Brian J
source

2

PHP，61字节

-2个字节，用于删除结束标记

<?php $z=fgets(STDIN);for($y=1;$y<$z;$y++){$x+=$z%$y;}echo$x;

在线尝试！

— NK1406
source

2

Japt `-mx`, 3 bytes

N%U

Try it online!

— Luis felipe De jesus Munoz
source

2

Brachylog，9个字节

⟨gz⟦₆⟩%ᵐ+

在线尝试！

— Kroppeb
source

1

Husk, 5 bytes

ΣṠM%ḣ

Try it online!

Explanation

ΣṠM%ḣ  -- implicit input x, example: 5
 ṠM%   -- map (mod x) over the following..
    ḣ  -- ..the range [1..x]: [5%1,5%2,5%3,5%4,5%5] == [0,1,2,1,0]
Σ      -- sum: 0+1+2+1+0 == 4

— ბიმო
source

1

Perl 5, 23 + 1 (`-p`) = 24 bytes

//;map$\+=$'%$_,1..$_}{

Try it online!

— Xcali
source

1

Pyth，5个字节

s%LQS

s%LQS - Full program, inputs N from stdin and prints sum to stdout
s     - output the sum of
 %LQ  - the function (elem % N) mapped over 
    S - the inclusive range from 1..N

在线尝试！

— 戴夫·
source

Oh actually I found a different 5 byter than you, didn't read yours correctly

— Dave

模总和

05AB1E, 3 bytes

Haskell, 22 bytes

Explanation

Ruby, 28 27 23 bytes

Ruby, 28 bytes

Funky, 25 bytes

Desmos, 25 bytes.

RProgN 2, 9 bytes

Explained

ReRegex, 71 bytes

ARBLE, 19 bytes

MaybeLater, 56 bytes

果冻，3个字节

MATL, 4 bytes

Ohm v2, 4 bytes

R，20个字节

Python 2, 44 bytes

JavaScript (ES6), 26 bytes

Demo

Python 3, 37 bytes

Charcoal, 9 bytes

Standard ML (MLton), 53 51 bytes

Previous 53 byte version:

Java (OpenJDK 8), 45 bytes

Mathematica，18个字节

APL (Dyalog), 5 bytes

Japt, 6 5 bytes

How it works

05AB1E, 6 bytes

Ruby, 23 bytes

Julia 0.4, 15 bytes

Add++, 14 bytes

How it works

4, 67 bytes

Windows Batch (CMD), 63 bytes

T-SQL, 80 79 bytes

PHP，61字节

Japt -mx, 3 bytes

Brachylog，9个字节

Husk, 5 bytes

Explanation

Perl 5, 23 + 1 (-p) = 24 bytes

Pyth，5个字节

Japt `-mx`, 3 bytes

Perl 5, 23 + 1 (`-p`) = 24 bytes