Modulus Summation

Question

I call this sequence "the Jesus sequence", because it is the sum of mod.</pun>

For this sequence, you take all the positive integers m less than the input n, and take the sum of n modulo each m. In other words:

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

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!

ბიმო · Accepted Answer · 2017-12-14 03:03:08Z

10

Haskell, 22 bytes

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

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Explanation

Yes.

answered Dec 14, 2017 at 3:03

ბიმო

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7

\$\begingroup\$ Great explanation \$\endgroup\$
– Esolanging Fruit
Commented Dec 15, 2017 at 3:32

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Community · Accepted Answer · 2020-06-17 09:04:33Z

8

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

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ARBLE, 19 bytes

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

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MaybeLater, 56 bytes

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

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 15, 2017 at 3:25

ATaco

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\$\begingroup\$ Will the submissions to this challenge ever end? So far I've been getting a new one every 40 minutes :P \$\endgroup\$
– Nissa
Commented Dec 15, 2017 at 3:26
\$\begingroup\$ @StephenLeppik Oh I've still got more coming, don't worry. \$\endgroup\$
– ATaco
Commented Dec 15, 2017 at 3:33
\$\begingroup\$ @StephenLeppik I'd rather not, because they're of various quality across various languages. \$\endgroup\$
– ATaco
Commented Dec 15, 2017 at 3:42
\$\begingroup\$ @StephenLeppik I've combined them for you, begrudgingly. \$\endgroup\$
– ATaco
Commented Dec 15, 2017 at 3:52
5

\$\begingroup\$ Please don't do this. Separate languages -- even separate approaches -- should go in separate answers. \$\endgroup\$
– Dennis
Commented Dec 18, 2017 at 4:48

| Show 2 more comments

Mr. Xcoder · Accepted Answer · 2017-12-14 05:27:01Z

7

05AB1E, 3 bytes

L%O

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answered Dec 14, 2017 at 5:27

Mr. Xcoder

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Jordan · Accepted Answer · 2017-12-20 00:48:19Z

7

Ruby, 28 27 23 bytes

-4 bytes thanks to @daniero.

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

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Ruby, 28 bytes

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

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edited Dec 20, 2017 at 0:48

answered Dec 14, 2017 at 1:48

Jordan

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dylnan · Accepted Answer · 2017-12-14 01:28:04Z

5

Jelly, 3 bytes

%RS

Explanation

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

Try it online!

edited Dec 14, 2017 at 1:28

answered Dec 14, 2017 at 1:25

dylnan

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Sanchises · Accepted Answer · 2017-12-14 08:16:46Z

5

MATL, 4 bytes

t:\s

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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.

answered Dec 14, 2017 at 8:16

Sanchises

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Community · Accepted Answer · 2020-06-17 09:04:33Z

5

R, 20 bytes

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

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edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 14, 2017 at 9:25

plannapus

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Cinaski · Accepted Answer · 2017-12-14 08:06:33Z

4

Ohm v2, 4 bytes

D@%Σ

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answered Dec 14, 2017 at 8:06

Cinaski

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Uriel · Accepted Answer · 2017-12-14 12:39:40Z

4

APL (Dyalog), 5 bytes

+/⍳|⊢

Try it online!

How?

Monadic train -

+/ - sum

⊢ - n

| - vectorized modulo

⍳ - the range of n

answered Dec 14, 2017 at 12:39

Uriel

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Max00355 · Accepted Answer · 2017-12-19 19:27:37Z

4

Python 2, 44 bytes

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

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

edited Dec 19, 2017 at 19:27

answered Dec 19, 2017 at 8:06

Max00355

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2

\$\begingroup\$ 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. \$\endgroup\$
– DJMcMayhem
Commented Dec 19, 2017 at 8:27
\$\begingroup\$ Oh wow! I didn't even think of that. Thanks \$\endgroup\$
– Max00355
Commented Dec 19, 2017 at 19:26
1

\$\begingroup\$ Welcome to PPCG! You can use a list comprehension instead of map for 38 bytes: Try it online! \$\endgroup\$
– Mr. Xcoder
Commented Dec 19, 2017 at 19:40
\$\begingroup\$ 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. \$\endgroup\$
– Max00355
Commented Dec 19, 2017 at 19:52

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manatwork · Accepted Answer · 2022-10-07 01:23:37Z

4

Rust, 27 characters

|n|(1..n).map(|i|n%i).sum()

Based on Should rust anonymous functions fully specify parameter types? I assume that return type inference is also fine.

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answered Oct 7, 2022 at 1:23

manatwork

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Ashlin Harris · Accepted Answer · 2022-10-09 19:38:32Z

4

Julia 1.0, 16 bytes

~x=sum(x.%(1:x))

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answered Oct 9, 2022 at 19:38

Ashlin Harris

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Arnauld · Accepted Answer · 2017-12-14 01:30:26Z

3

JavaScript (ES6), 26 bytes

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

Demo

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

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

answered Dec 14, 2017 at 1:30

Arnauld

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Neil · Accepted Answer · 2017-12-14 01:31:11Z

3

Python 3, 37 bytes

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

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answered Dec 14, 2017 at 1:31

Neil

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\$\begingroup\$ 4 answers in 10 minutes? Wow. \$\endgroup\$
– Nissa
Commented Dec 14, 2017 at 1:31
2

\$\begingroup\$ Must be some very gifted programmers. ;) \$\endgroup\$
– Neil
Commented Dec 14, 2017 at 1:32

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Charlie · Accepted Answer · 2017-12-14 07:45:40Z

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))))));

answered Dec 14, 2017 at 7:45

Charlie

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Standard ML (MLton), 53 51 bytes

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

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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)))

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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.

Olivier Grégoire · Accepted Answer · 2017-12-14 11:02:46Z

3

Java (OpenJDK 8), 45 bytes

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

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answered Dec 14, 2017 at 11:02

Olivier Grégoire

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1

\$\begingroup\$ +1 For using the goes to (-->) operator. \$\endgroup\$
– raznagul
Commented Dec 15, 2017 at 12:07

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totallyhuman · Accepted Answer · 2017-12-14 11:30:31Z

3

Mathematica, 18 bytes

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

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answered Dec 14, 2017 at 11:30

totallyhuman

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2

\$\begingroup\$ same bytes Tr[#~Mod~Range@#]& \$\endgroup\$
– ZaMoC
Commented Dec 14, 2017 at 11:53

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Oliver · Accepted Answer · 2017-12-14 20:41:57Z

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

edited Dec 14, 2017 at 20:41

Oliver

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answered Dec 14, 2017 at 2:00

ETHproductions

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Uriel · Accepted Answer · 2017-12-14 21:32:17Z

3

4, 67 bytes

4 doesn't have any modulo built in.

3.79960101002029980200300023049903204040310499040989804102020195984

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answered Dec 14, 2017 at 21:32

Uriel

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Brian J · Accepted Answer · 2017-12-19 14:08:20Z

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

Underslash · Accepted Answer · 2021-06-19 22:38:25Z

3

Vyxal, 3 bytes

ɽ%∑

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Vyxal `s`, 2 bytes

ɽ%

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answered Jun 19, 2021 at 22:38

Underslash

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Shieru Asakoto · Accepted Answer · 2017-12-14 02:14:52Z

05AB1E, 6 bytes

ÎGIN%+

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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

Dave · Accepted Answer · 2017-12-14 03:06:50Z

2

Pyth, 5 bytes

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

Try it online!

edited Dec 14, 2017 at 3:06

answered Dec 14, 2017 at 2:55

Dave

4822 silver badges10 bronze badges

\$\begingroup\$ Oh actually I found a different 5 byter than you, didn't read yours correctly \$\endgroup\$
– Dave
Commented Dec 14, 2017 at 2:59

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G B · Accepted Answer · 2017-12-14 09:49:12Z

2

Ruby, 23 bytes

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

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answered Dec 14, 2017 at 9:49

G B

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Lescurel · Accepted Answer · 2017-12-14 16:20:39Z

2

Julia 0.4, 15 bytes

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

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answered Dec 14, 2017 at 16:20

Lescurel

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caird coinheringaahin g · Accepted Answer · 2017-12-14 17:33:23Z

Add++, 14 bytes

L,RAdx$p@BcB%s

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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]

DELETE_ME · Accepted Answer · 2017-12-16 03:36:55Z

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%

edited Dec 16, 2017 at 3:36

DELETE_ME

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answered Dec 14, 2017 at 9:20

Neil

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NK1406 · Accepted Answer · 2017-12-22 15:55:31Z

2

PHP, 61 bytes

-2 bytes for removing the closing tag

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

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edited Dec 22, 2017 at 15:55

answered Dec 17, 2017 at 20:51

NK1406

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Luis felipe De jesus Munoz · Accepted Answer · 2018-08-20 18:06:33Z

2

Japt `-mx`, 3 bytes

N%U

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answered Aug 20, 2018 at 18:06

Luis felipe De jesus Munoz

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Modulus Summation

75 Answers 75

Haskell, 22 bytes

Explanation

Funky, 25 bytes

Desmos, 25 bytes.

RProgN 2, 9 bytes

Explained

ReRegex, 71 bytes

ARBLE, 19 bytes

MaybeLater, 56 bytes

05AB1E, 3 bytes

Ruby, 28 27 23 bytes

Ruby, 28 bytes

Jelly, 3 bytes

MATL, 4 bytes

R, 20 bytes

Ohm v2, 4 bytes

APL (Dyalog), 5 bytes

Python 2, 44 bytes

Rust, 27 characters

Julia 1.0, 16 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 bytes

Japt, 6 5 bytes

How it works

4, 67 bytes

T-SQL, 80 79 bytes

Vyxal, 3 bytes

Vyxal s, 2 bytes

05AB1E, 6 bytes

Pyth, 5 bytes

Ruby, 23 bytes

Julia 0.4, 15 bytes

Add++, 14 bytes

How it works

Windows Batch (CMD), 63 bytes

PHP, 61 bytes

Japt -mx, 3 bytes

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