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]

Try it online!

Explanation

Yes.

answered Dec 14, 2017 at 3:03

ბიმო

16.8k2 gold badges41 silver badges104 bronze badges

7

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

Add a comment |

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

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!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 15, 2017 at 3:25

ATaco

10.8k27 silver badges77 bronze badges

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

Try it online!

answered Dec 14, 2017 at 5:27

Mr. Xcoder

42.4k9 gold badges79 silver badges214 bronze badges

Add a comment |

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}}

Try it online!

Ruby, 28 bytes

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

Try it online!

edited Dec 20, 2017 at 0:48

answered Dec 14, 2017 at 1:48

Jordan

10.1k1 gold badge31 silver badges47 bronze badges

Add a comment |

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

5,3642 gold badges15 silver badges33 bronze badges

Add a comment |

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

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.

answered Dec 14, 2017 at 8:16

Sanchises

9,4961 gold badge35 silver badges61 bronze badges

Add a comment |

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

5

R, 20 bytes

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

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Dec 14, 2017 at 9:25

plannapus

8,91921 silver badges46 bronze badges

Add a comment |

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

4

Ohm v2, 4 bytes

D@%Σ

Try it online!

answered Dec 14, 2017 at 8:06

Cinaski

2,2581 gold badge7 silver badges17 bronze badges

Add a comment |

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

13.2k4 gold badges17 silver badges45 bronze badges

Add a comment |

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)))

Try it online!

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

edited Dec 19, 2017 at 19:27

answered Dec 19, 2017 at 8:06

Max00355

1512 bronze badges

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
Dec 19, 2017 at 8:27
\$\begingroup\$ Oh wow! I didn't even think of that. Thanks \$\endgroup\$
– Max00355
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
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
Dec 19, 2017 at 19:52

Add a comment |

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.

Try it online!

answered Oct 7, 2022 at 1:23

manatwork

20.2k5 gold badges47 silver badges78 bronze badges

Add a comment |

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

4

Julia 1.0, 16 bytes

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

Try it online!

answered Oct 9, 2022 at 19:38

Ashlin Harris

1,9912 silver badges7 bronze badges

Add a comment |

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

192k20 gold badges175 silver badges636 bronze badges

Add a comment |

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))

Try it online!

answered Dec 14, 2017 at 1:31

Neil

2,5259 silver badges26 bronze badges

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

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

Add a comment |

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

12.9k1 gold badge41 silver badges99 bronze badges

Add a comment |

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

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.

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;}

Try it online!

answered Dec 14, 2017 at 11:02

Olivier Grégoire

14.4k3 gold badges31 silver badges56 bronze badges

1

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

Add a comment |

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

3

Mathematica, 18 bytes

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

Try it online!

answered Dec 14, 2017 at 11:30

totallyhuman

17.2k3 gold badges31 silver badges88 bronze badges

2

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

Add a comment |

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

7,8722 gold badges17 silver badges37 bronze badges

answered Dec 14, 2017 at 2:00

ETHproductions

50k6 gold badges93 silver badges239 bronze badges

Add a comment |

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

3

4, 67 bytes

4 doesn't have any modulo built in.

3.79960101002029980200300023049903204040310499040989804102020195984

Try it online!

answered Dec 14, 2017 at 21:32

Uriel

13.2k4 gold badges17 silver badges45 bronze badges

Add a comment |

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%∑

Try it Online!

Vyxal `s`, 2 bytes

ɽ%

Try it Online!

answered Jun 19, 2021 at 22:38

Underslash

1,7291 gold badge6 silver badges14 bronze badges

Add a comment |

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

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

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

4722 silver badges10 bronze badges

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

Add a comment |

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

2

Ruby, 23 bytes

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

Try it online!

answered Dec 14, 2017 at 9:49

G B

21.9k1 gold badge20 silver badges53 bronze badges

Add a comment |

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

2

Julia 0.4, 15 bytes

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

Try it online!

answered Dec 14, 2017 at 16:20

Lescurel

2912 silver badges7 bronze badges

Add a comment |

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

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]

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

17.3k2 gold badges36 silver badges70 bronze badges

answered Dec 14, 2017 at 9:20

Neil

170k12 gold badges72 silver badges275 bronze badges

Add a comment |

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;

Try it online!

edited Dec 22, 2017 at 15:55

answered Dec 17, 2017 at 20:51

NK1406

7396 silver badges16 bronze badges

Add a comment |

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

2

Japt `-mx`, 3 bytes

N%U

Try it online!

answered Aug 20, 2018 at 18:06

Luis felipe De jesus Munoz

10.6k2 gold badges28 silver badges82 bronze badges

Add a comment |

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

Your Answer

Sign up or log in

Post as a guest

Not the answer you're looking for? Browse other questions tagged code-golfsequencedivision or ask your own question.

Linked

Related

Hot Network Questions

Vyxal `s`, 2 bytes

Japt `-mx`, 3 bytes

Not the answer you're looking for? Browse other questions tagged
code-golf
sequence
division
or ask your own question.