Sum of the first n elements of the sequence of 9's complement

Question

Let's consider the following sequence:

$$9,8,7,6,5,4,3,2,1,0,89,88,87,86,85,84,83,82,81,80,79,78,77,76,75,74,73,72,71...$$ This is the sequence of \$9\$'s complement of a number: that is, \$ a(x) = 10^d - 1 - x \$ where \$ d \$ is the number of digits in \$ x \$. (A061601 in the OEIS).Your task is to add the first \$n\$ elements.

Input

A number \$n∈[0,10000]\$.

Output

The sum of the first \$n\$ elements of the sequence.

Test cases

0 -> 0
1 -> 9
10 -> 45
100 -> 4050
1000 -> 408600
10000 -> 40904100

Standard code-golf rules apply. The shortest code in bytes wins.

Do the test cases make sense? (I guess that's why the 0 confusion in the answers.) 0 -> 0 works if you take 0 as a 0-digit number, and the formula is 10^d - 1 - x as stated. But 1 -> 9 only works if the formula is 10^d - x. I think the 0 case should just be removed, and the formula changed to be without the - 1. — Sundar R, Feb 23 at 10:26
@SundarR I presume the confusion may result from indexing the sequence from 0 and then taking sum of first elements. So taking sum of first 0 elements results in 0 and taking first element (of index 0) results in 9. — pajonk, Feb 23 at 10:50
@pajonk Ahh thank you. That was a reading fail on my part, I assumed it was "sum up to n's complement" and didn't read where it clearly says "sum of first n elements". — Sundar R, Feb 23 at 11:05

Merijn van Tilborg · Accepted Answer · 2022-02-23 07:58:22Z

5

R, 37 bytes

f=function(n)sum(10^nchar(1:n-1)-1:n)

Try it online!

note I left out the check for 0 as I do not think it is part of the sequence as refered to A061601 in the OEIS reference. Adding the check would result in the exact same answer of Giuseppe, which would surely be preferable including the correct answer including 0.

edited Feb 23 at 7:58

answered Feb 22 at 14:15

Merijn van Tilborg

1815 bronze badges

\$\begingroup\$ Welcome to Code Golf, and nice answer! \$\endgroup\$
– NoHaxJustRadvylf
Feb 22 at 14:26
1

\$\begingroup\$ This is solid. It doesn't appear that this works for n=0 so you do need to adjust for that. Also, anonymous functions are perfectly fine (as long as they're not recursive), so you can drop the f= from your answer as well :-) \$\endgroup\$
– Giuseppe
Feb 22 at 14:27
\$\begingroup\$ We cross posted the same answer, I learned your *!!n part which is very elegant to include the n=0. I do think though if an OP asks for a oeis defined sequence, the offset value or in other words the value not part of that defined sequence should not be addressed in the answer. \$\endgroup\$
– Merijn van Tilborg
Feb 22 at 14:38
2

\$\begingroup\$ That's up to the discretion of the question asker, I believe. You can ask in a comment on the main post if you want, though. Also, I recommend looking at the tips for golfing in R; they can be helpful sometimes! \$\endgroup\$
– Giuseppe
Feb 22 at 16:38

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ZaMoC · Accepted Answer · 2022-02-22 18:01:47Z

4

Wolfram Language (Mathematica), 42 bytes

FromDigits[9-IntegerDigits@--n]~Sum~{n,#}&

Try it online!

-2 bytes from att

edited Feb 22 at 18:01

answered Feb 21 at 18:55

ZaMoC

23.1k2 gold badges27 silver badges75 bronze badges

\$\begingroup\$ 42 bytes \$\endgroup\$
– att
Feb 22 at 17:45

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Giuseppe · Accepted Answer · 2022-02-22 14:14:09Z

3

R, 39 bytes

function(n)sum(10^nchar(1:n-1)-1:n)*!!n

Try it online!

answered Feb 22 at 14:14

Giuseppe

26.7k3 gold badges28 silver badges96 bronze badges

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Kevin Cruijssen · Accepted Answer · 2022-02-22 14:48:47Z

05AB1E, 8 bytes

FNg°<N-O

Try it online or verify all test cases.

Explanation:

F         # Loop `N` in the range [0, (implicit) input):
 Ng       #  Push the length of `N`
   °      #  Take 10 to the power this length
    <     #  Decrease it by 1
     N-   #  Decrease it by `N`
       O  #  Take the sum of the values on the stack
          # (after which the last sum is output implicitly as result - 
          # or the implicit input if it was 0 and we never entered the loop)

mathcat · Accepted Answer · 2022-02-22 15:44:48Z

Vyxal `r`, 8 bytes

No, I'm not Jo King. I found another use for the r flag!!!

It happens once in a blue moon on average

^{Okay, yes, I said that}

ʁƛL↵-‹;∑

Try it Online! Flagless 9 bytes

Explanation

So the r flag is basically a reverser, which reverses the order of the arguments. That makes it hard to use for long programs, but for short ones like this, it's useful sometimes.

ʁƛL↵-‹;∑
ʁ          create list with range(0, n)
 ƛ         open mapping lambda (with loop item n)
  L↵       push 10 to the length of n
    -      n - the latter (because it's reversed)
     ‹;    decrement by 1 and close lambda
       ∑   sum list

Jonathan Allan · Accepted Answer · 2022-02-22 16:31:01Z

2

Jelly, 7 bytes

’æċ⁵_)S

A monadic Link accepting a non-negative integer that yields a non-negative integer.

Try it online!

How?

’æċ⁵_)S - Link: non-negative integer, n
     )  - for each i in [1..n] (if n is 0 this will yield []):
’       -   decrement -> i-1
   ⁵    -   10
 æċ     -   ceil i-1 to the nearest (positive integer) power of 10
    _   -   subtract i
      S - sum

edited Feb 22 at 16:31

answered Feb 22 at 16:23

Jonathan Allan

92k7 gold badges60 silver badges251 bronze badges

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friddo · Accepted Answer · 2022-02-22 17:46:20Z

2

Python 3.8 (pre-release), 50 44 bytes

Gave a fun recursive solution a go :)

edit: -6 bytes thanks to solid.py

f=lambda n:n and f(m:=n-1)-n+10**len(str(m))

Try it online!

edited Feb 22 at 17:46

answered Feb 22 at 15:44

friddo

3331 silver badge6 bronze badges

1

\$\begingroup\$ Welcome to Code Golf, and nice answer! \$\endgroup\$
– NoHaxJustRadvylf
Feb 22 at 15:45
\$\begingroup\$ Actually, 44 bytes \$\endgroup\$
– solid.py
Feb 22 at 17:38
\$\begingroup\$ @solid.py yes! haha, that one was painfully obvious \$\endgroup\$
– friddo
Feb 22 at 17:46
\$\begingroup\$ Surely repeating n-1 saves a byte, as the assignment takes three bytes. \$\endgroup\$
– Neil
Feb 22 at 19:55

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att · Accepted Answer · 2022-02-22 22:30:04Z

2

C (gcc), 79 bytes

g(n,t,s){t=--n/10;s=n%10+1;n=t*45+s*(19-s)/2+10*(t?s*g(t+1)+(10-s)*g(t)-90:0);}

Try it online!

answered Feb 22 at 22:30

att

15.8k2 gold badges10 silver badges43 bronze badges

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amelies · Accepted Answer · 2022-02-23 19:25:52Z

2

Julia 1.0, 43 38 30 bytes

!n=n>0&&10^ndigits(~-n)-n+!~-n

Try it online!

-5 dingledooper, -8 MarcMush

edited Feb 23 at 19:25

answered Feb 22 at 15:12

amelies

3411 silver badge4 bronze badges

1

\$\begingroup\$ 38 bytes: !n=n>0&&sum(n->~n+10^ndigits(n),0:n-1) \$\endgroup\$
– dingledooper
Feb 22 at 16:40
1

\$\begingroup\$ 30 bytes with recursion: !n=n>0&&10^ndigits(~-n)-n+!~-n \$\endgroup\$
– MarcMush
Feb 23 at 19:03

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Shaggy · Accepted Answer · 2022-02-22 15:32:05Z

1

Japt `-mx`, 8 bytes

+1 byte to work around a bug in Japt.

aÓApUs l

Try it

answered Feb 22 at 15:32

Shaggy

35.5k4 gold badges29 silver badges88 bronze badges

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solid.py · Accepted Answer · 2022-02-22 17:28:26Z

1

Python 3, 49 bytes

-1 byte thanks to user friddo

lambda n:sum(~i+10**len(str(i))for i in range(n))

Try it online!

edited Feb 22 at 17:28

answered Feb 22 at 14:12

solid.py

6651 gold badge1 silver badge10 bronze badges

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rabbitgrowth · Accepted Answer · 2022-02-22 18:37:47Z

1

Dyalog APL, 17 bytes

+/⍳-⍨1-⍨10*≢∘⍕¨∘⍳

⎕IO←0.

Try it online!

answered Feb 22 at 18:37

rabbitgrowth

3013 bronze badges

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tsh · Accepted Answer · 2022-02-23 07:53:25Z

1

JavaScript, 47 bytes

f=n=>(t=10**`${n}`.length)*n+n*~n/2+~-t*~t/11+9

Try it online!

There isn't any non-recursive implementation here. So maybe it is worth to include one. But it is longer than current answers...

If your language trunk divide result to int automatically, +~-t*~t/11 could be changed into -t*t/11.

$$ t = 10^{1+\lfloor\log_{10}n\rfloor} $$ $$ t\cdot n-\frac{n\cdot (n+1)}2-\frac{t^2-100}{11} $$

with special case where \$n=0\$

edited Feb 23 at 7:53

answered Feb 23 at 7:42

tsh

29.5k2 gold badges31 silver badges109 bronze badges

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Sundar R · Accepted Answer · 2022-02-23 11:11:54Z

1

MATL, 14 bytes

:q"10@Vn^q@-vs

Try it online!

answered Feb 23 at 11:11

Sundar R

6,43216 silver badges32 bronze badges

1

\$\begingroup\$ You could replace vs by ]vs :-P \$\endgroup\$
– Luis Mendo
Feb 24 at 16:00

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pxeger · Accepted Answer · 2022-02-23 15:30:02Z

1

Zsh, 35 bytes

for ((x=$1;x--;s+=1e$#x+~x)):
<<<$s

Attempt This Online!

edited Feb 23 at 15:30

answered Feb 22 at 20:58

pxeger

21.5k3 gold badges45 silver badges120 bronze badges

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Zaelin Goodman · Accepted Answer · 2022-02-24 12:54:36Z

1

PowerShell, 57 49 bytes

-8 bytes thanks to @mazzy!

0.."$args"-gt0|%{$s-=$_---("1e"+"$_".length)}
+$s

Try it online!

edited Feb 24 at 12:54

answered Feb 22 at 17:19

Zaelin Goodman

2,0292 silver badges11 bronze badges

\$\begingroup\$ 1e2 = 100, 1e3 = 1000, 1e4 = 10000... \$\endgroup\$
– mazzy
Feb 22 at 19:59
\$\begingroup\$ Try it online! ? \$\endgroup\$
– mazzy
Feb 22 at 20:13
1

\$\begingroup\$ @mazzy Ahhhhh, clever, thank you! I totally forgot about scientific notation in powershell!! And nice way to improve on the zero case \$\endgroup\$
– Zaelin Goodman
Feb 24 at 12:56

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Lince Assassino · Accepted Answer · 2022-02-24 16:03:13Z

1

Behaviour, 36 29 bytes

f=&#(a*&10^#(""+a)-1-a;>&a+b)

My own recently made esolang, so bear with me.

ungolfed and commented

f=&
    #(                  // deal with 0 case, convert nil into 0
      a * &             // generate array for n nine complements
        10^#(""+a)-1-a; // nine complements logic
      >&a+b             // reduce array by adding its items
    )

Test with:

f:0
f:1
f:10

edited Feb 24 at 16:03

answered Feb 23 at 20:18

Lince Assassino

1614 bronze badges

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m90 · Accepted Answer · 2022-02-22 14:10:12Z

0

JavaScript, 35 bytes

f=n=>n&&f(--n)+10**`${n}`.length+~n

Try it online!

answered Feb 22 at 14:10

m90

6,4461 gold badge7 silver badges28 bronze badges

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sinvec · Accepted Answer · 2022-02-22 14:31:28Z

0

J, 24 bytes

+/@((10&^@#@":->:)"0@i.)

Try it online!

answered Feb 22 at 14:31

sinvec

1,6452 silver badges26 bronze badges

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Wasif · Accepted Answer · 2022-02-22 16:40:52Z

0

Python 2, 42 bytes

f=lambda n:n and-~~n+10**len(`n-1`)+f(n-1)

Try it online!

answered Feb 22 at 16:40

Wasif

11.9k1 gold badge15 silver badges45 bronze badges

\$\begingroup\$ Remove ~~ to save two bytes. n is integer, so those two symbols are not needed \$\endgroup\$
– ophact
Feb 22 at 17:57

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G B · Accepted Answer · 2022-02-22 17:02:46Z

0

Ruby, 39 bytes

->n{n.times.sum{|x|10**x.to_s.size+~x}}

Try it online!

answered Feb 22 at 17:02

G B

18k1 gold badge18 silver badges46 bronze badges

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Neil · Accepted Answer · 2022-02-22 19:59:28Z

Charcoal, 12 bytes

Ｉ↨¹ＥＮ⁻Ｉ×Ｌι9ι

Try it online! Link is to verbose version of code. Explanation:

    Ｎ           Input as a number
   Ｅ            Map over implicit range
          9     Literal string `9`
       ×        Repeated by
         ι      Current value
        Ｌ       Length (of string representation)
      Ｉ         Cast to integer
     ⁻          Subtract
           ι    Current value
 ↨¹             Take the sum
Ｉ               Cast to string
                Implicitly print

(Sum() won't work for an input of 0.)

DeathIncarnate · Accepted Answer · 2022-02-22 21:19:53Z

0

Burlesque, 19 bytes

-.{Jln'9j.*j|-}GZ++

Try it online!

-.   # Decrement (Burlesque runs 0..N inclusive, need to exclude)
{
 J   # Duplicate
 ln  # Number of digits
 '9  # Character 9
 j   # Reorder stack
 .*  # Repeat 9 n_digits times as string
 j   # Reorder stack
 |-  # String subtract (parse and subtract)
}GZ  # Generate for range
++   # Sum

answered Feb 22 at 21:19

DeathIncarnate

2,2604 silver badges8 bronze badges

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alephalpha · Accepted Answer · 2022-02-23 01:43:18Z

0

Pari/GP, 30 bytes

n->sum(i=0,n-1,10^#Str(i)-1-i)

Try it online!

answered Feb 23 at 1:43

alephalpha

35k5 gold badges46 silver badges130 bronze badges

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Noodle9 · Accepted Answer · 2022-02-23 10:03:14Z

0

C (gcc), ^{62 50} 49 bytes

t;f(n){n=n?t=log10(--n?:1),exp10(t+1)+~n+f(n):0;}

Try it online!

Saved a whopping 12 bytes thanks to att!!!
Saved a byte thanks to ceilingcat!!!

edited Feb 23 at 10:03

answered Feb 22 at 19:33

Noodle9

18.4k3 gold badges16 silver badges42 bronze badges

\$\begingroup\$ 50 bytes \$\endgroup\$
– att
Feb 22 at 22:28
\$\begingroup\$ @att Very nice and corrected my use of log10 - thanks! :D \$\endgroup\$
– Noodle9
Feb 22 at 22:37
\$\begingroup\$ @ceilingcat Nice one - thanks! :D \$\endgroup\$
– Noodle9
Feb 23 at 10:04

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Aiden Chow · Accepted Answer · 2022-03-01 01:03:05Z

0

Desmos, 49 bytes

f(k)=∑_{n=2}^k(10^{1+floor(log(n-1))}-n)+9-0^k9

Would be 44 bytes if we didn't have to deal with 0...

Try It On Desmos!

Try It On Desmos! - Prettified

48 bytes, port of @tsh's answer

t=10^{log(n+0^n)}10
f(n)=tn-(nn+n)/2-(tt-100)/11

Try It On Desmos!

edited Mar 1 at 1:03

answered Feb 23 at 6:11

Aiden Chow

4,7271 gold badge4 silver badges36 bronze badges

\$\begingroup\$ how the heck do you use desmos? \$\endgroup\$
– DialFrost
Feb 23 at 6:49
\$\begingroup\$ @DialFrost it's hard to explain fully in a comment, but the gist is that Desmos uses mathematical formulae (like functions or expressions) in order to compute the output. these formulae are expressed through \$\LaTeX\$ (albeit with certain optimizations for byte count), which is the code you see. \$\endgroup\$
– Aiden Chow
Feb 23 at 6:57

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Sum of the first n elements of the sequence of 9's complement

Input

Output

Test cases

26 Answers 26

R, 37 bytes

Wolfram Language (Mathematica), 42 bytes

R, 39 bytes

05AB1E, 8 bytes

Vyxal r, 8 bytes

Explanation

Jelly, 7 bytes

How?

Python 3.8 (pre-release), 50 44 bytes

C (gcc), 79 bytes

Julia 1.0, 43 38 30 bytes

Japt -mx, 8 bytes

Python 3, 49 bytes

Dyalog APL, 17 bytes

JavaScript, 47 bytes

MATL, 14 bytes

Zsh, 35 bytes

PowerShell, 57 49 bytes

Behaviour, 36 29 bytes

JavaScript, 35 bytes

J, 24 bytes

Python 2, 42 bytes

Ruby, 39 bytes

Charcoal, 12 bytes

Burlesque, 19 bytes

Pari/GP, 30 bytes

C (gcc), 62 50 49 bytes

Desmos, 49 bytes

48 bytes, port of @tsh's answer

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