Nobody knows GAU Numbers

Question

Let me introduce you to GAU numbers

GAU(1) = 1  
GAU(2) = 1122  
GAU(3) = 1122122333  
GAU(4) = 11221223331223334444  
GAU(6) = 11221223331223334444122333444455555122333444455555666666  
...  
GAU(10) = 11221223331223334444122333444455555122333444455555666666122333444455555666666777777712233344445555566666677777778888888812233344445555566666677777778888888899999999912233344445555566666677777778888888899999999910101010101010101010

This challenge is pretty simple!

Given an integer n>0, find the number of digits of GAU(n)

Example

Let's make GAU(4)
we take the following steps (until we get to 4) and concatenate them

[1][122][122333][1223334444]

you must write every number as many times as its value, but you have to count every time from 1

Let's try to make GAU(5)
we will have to count from 1 to 1

[1]

then from 1 to 2 (but repeating every number as many times as its value)

[122]

then from 1 to 3

[122333]

then from 1 to 4

[1223334444]

and finally from 1 to 5 (this is the last step because we want to find GAU(5))

[122333444455555]

Now we take all these steps and concatenate them
the result is GAU(5)

11221223331223334444122333444455555

We are interested in the number of digits of these GAU numbers.

Test cases

Input⟼Output

n   ⟼ Length(GAU(n))

1   ⟼ 1  
2   ⟼ 4  
3   ⟼ 10  
10  ⟼ 230   
50  ⟼ 42190  
100 ⟼ 339240  
150 ⟼ 1295790

This is a code-golf challenge.
Shortest code in bytes will win.

If you still have any questions please let me know.
I really want everyone here to understand this magic-hidden-complex pattern

Up until n=9, the lengths are tetrahedral numbers, but beyond that the multi-digit numbers get in the way of a simple closed form — Miff, Commented Sep 25, 2017 at 13:54
@numbermaniac thanks for spotting this. GUA numbers are totally different. They haven't been invented yet! — user72253, Commented Sep 30, 2017 at 19:00

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

14

SOGL V0.12, 11 10 8 7 5 bytes

∫∫l*+

Try it Here! - this expects to be called as a function with the input on the stack and the input box empty.
7 byte alternative taking the input from the input box:

0.∫∫l*+

Try it Here!

0      push 0
 .     push the input
  ∫    iterate over a range 1..POP (input) inclusive, pusing the current number
   ∫    iterate over 1..POP (above loops number) inclusive, pusing the current number
    l    push that numbers length without popping the number
     *   multiply the length by the number
      +  add to the zero, or whatever it is now

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2017 at 11:58

dzaima

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\$\begingroup\$ push that numbers length without popping the number nice \$\endgroup\$
– Erik the Outgolfer
Commented Sep 23, 2017 at 12:16

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jferard · Accepted Answer · 2017-09-23 16:26:18Z

8

Haskell, 45 bytes

f n=sum[j*length(show j)|i<-[1..n],j<-[1..i]]

Try it online!

answered Sep 23, 2017 at 16:26

jferard

1,7947 silver badges7 bronze badges

1

\$\begingroup\$ A bit more unreadable but one byte cheaper: TIO \$\endgroup\$
– ბიმო
Commented Sep 23, 2017 at 16:59

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Nitrodon · Accepted Answer · 2017-09-23 19:34:22Z

Brain-Flak, 166 bytes

<>(((()()())({}){})())<>{({}[()]<({}({}<<>({}[()])((){[()](<()>)}{}){{}((((({})({})){}{}){}))<>(({}<({}())>){()<({}[({})])>}{})(<>)}{}<>>({}({})())))>)}{}({}<{}{}{}>)

Try it online!

Explanation

<>(((()()())({}){})())<>           # Initialize second stack with 9 and 10
{({}[()]<                          # Do main loop n times:
  ({}
    ({}
      <
        <>({}[()])                 # Subtract 1 from counter to next power of 10
        ((){[()](<()>)}{}){        # If reached a power of 10 (say, 10^k):
          {}((((({})({})){}{}){})) # Multiply existing (10^k*0.9) by 10 and push twice
          <>                       # On first stack
          (
            ({}<({}())>)           # Increment length of numbers
            {()<({}[({})])>}{}     # Divide length of new set of numbers by this length
          )                        # Add together to get new set of numbers length
        (<>)}  
      {}<>>  
      ({}({})())                   # Add number length to number set length
    )                              # Add number set length to new segment length
  )                                # Add new segment length to total length
>)}                                # End main loop
{}({}<{}{}{}>)                     # Put result on stack by itself

H.PWiz · Accepted Answer · 2017-09-25 19:33:01Z

7

Husk, 5 bytes

ṁLΣQḣ

Try it online!

Explanation

ṁ  map over
 L  length
  Σ  sum list
   Q  all sublists
    ḣ  range 1 .. N

edited Sep 25, 2017 at 19:33

H.PWiz

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answered Sep 25, 2017 at 6:33

Icer Wild

1111 bronze badge

\$\begingroup\$ Nice algorithm! \$\endgroup\$
– H.PWiz
Commented Sep 25, 2017 at 19:33

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Leaky Nun · Accepted Answer · 2017-09-23 12:10:29Z

4

Jelly, 7 bytes

Rx`ƤDFL

Try it online!

answered Sep 23, 2017 at 12:10

Leaky Nun

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Icer Wild · Accepted Answer · 2017-09-25 04:58:59Z

4

05AB1E, 5 bytes

LŒJJg

Try it online!

Explanation

L      # push [1 .. a]
 Œ     # all substrings (a)
  J    # numbers to strings (inner)
   J   # numbers to strings (outer)
    g  # length

answered Sep 25, 2017 at 4:58

Icer Wild

1111 bronze badge

\$\begingroup\$ Welcome to the site :) \$\endgroup\$
– DJMcMayhem
Commented Sep 25, 2017 at 5:01

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Halvard Hummel · Accepted Answer · 2017-09-23 12:08:28Z

3

Python 2, 53 bytes

lambda x:sum(len(`i`)*i*(x-i+1)for i in range(1,x+1))

Try it online!

answered Sep 23, 2017 at 12:08

Halvard Hummel

3,2411 gold badge11 silver badges25 bronze badges

\$\begingroup\$ 51 bytes. \$\endgroup\$
– Jonathan Frech
Commented Sep 30, 2017 at 18:59

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ბიმო · Accepted Answer · 2017-09-23 12:29:57Z

3

Husk, 7 bytes

Σ∫mS*Lḣ

Try it online!

Ungolfed/Explanation

         -- implicit input N                        | 10
  m   ḣ  -- map the following function over [1..N]  | [1,2,3,4]
   S*L   --   multiply length of number by itself   | [1,2,3,4] (only important for numbers ≥ 10)
 ∫       -- prefix sums                             | [0,1,3,6,10]
Σ        -- sum                                     | 20

edited Sep 23, 2017 at 12:29

answered Sep 23, 2017 at 12:24

ბიმო

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

3

Husk, 7 bytes

ṁLṁṘNḣḣ

Try it online!

Explanation

          Implicit input, e.g 4
      ḣ   Range from 1 to n                               [1,2,3,4]
     ḣ    Prefixes                                        [[],[1],[1,2],[1,2,3],[1,2,3,4]]
  ṁ       Map and then concatenate
   ṘN     Repeat each number in each list by its index    [[],[1],[1,2,2],[1,2,2,3,3,3],[1,2,2,3,3,3,4,4,4,4]]
                                                          [1,1,2,2,1,2,2,3,3,3,1,2,2,3,3,3,4,4,4,4]
ṁ         Map and then sum
 L        Length (of number: 10 -> 2)                     26

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2017 at 12:14

H.PWiz

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\$\begingroup\$ Oh another Husk solution :) Didn't see your submission when posting mine, same bytecount but they're sufficiently different, so I'll leave mine here too. \$\endgroup\$
– ბიმო
Commented Sep 23, 2017 at 12:34

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

3

JavaScript (ES6), 57 55 bytes

n=>[...Array(n)].reduce(x=>x+(++y+"").length*y*n--,y=0)

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2017 at 15:06

Shaggy

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totallyhuman · Accepted Answer · 2017-09-23 18:27:49Z

3

Python 2, 59 58 bytes

Another one bytes the dust thanks to Jonathan Frech.

f=lambda n:n and sum(i*len(`i`)for i in range(n+1))+f(n-1)

Try it online!

Not short but eh... what the heck.

edited Sep 23, 2017 at 18:27

answered Sep 23, 2017 at 14:32

totallyhuman

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\$\begingroup\$ len(`i`)*i for -> i*len(`i`)for. \$\endgroup\$
– Jonathan Frech
Commented Sep 23, 2017 at 17:54
\$\begingroup\$ 53 bytes, non-recursive. \$\endgroup\$
– Jonathan Frech
Commented Sep 23, 2017 at 22:46

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Iaka Noe · Accepted Answer · 2017-09-24 05:30:01Z

CJam, 20 bytes

q~),(\{),{_s,*+}*+}%

Try it online!

The number is passed in the "input" field.

Ungolfed explanation: (example input = 2)

q~),(\{),{_s,*+}*+}%                                             | Stack:
q                     read input as string                       | "2"
 ~                    eval input (add it to stack as integer)    | 2
  )                   add 1                                      | 3
   ,                  range (convert to array with values 0...N) | [0, 1, 2]
    (                 pop first item of array                    | [1, 2] 0
     \                swap top two values of stack               | 0 [1, 2]
      {           }   for each item in array...                  | 0 1
       )              add 1                                      | 0 2
        ,             range (convert to array with values 0...N) | 0 [0, 1]
         {     }      for every element in the array...          | 0 0
          _           duplicate                                  | 0 0 0
           s          convert to string                          | 0 0 "0"
            ,         get length of string                       | 0 0 1
             *        multiply                                   | 0 0
              +       add                                        | 0 1
                *     fold                                       | 0 1
                 +    add                                        | 1
                   %  repeat                                     | 4

It seems hard when explained lol.

Arnauld · Accepted Answer · 2017-09-23 12:46:51Z

2

JavaScript (ES6), 50 42 bytes

Updated: now basically a port of what other answers are doing.

f=(n,i=1)=>n&&`${n}`.length*n*i+f(n-1,i+1)

Test cases

f=(n,i=1)=>n&&`${n}`.length*n*i+f(n-1,i+1)

console.log(f(1))   // 1
console.log(f(2))   // 4
console.log(f(3))   // 10
console.log(f(10))  // 230
console.log(f(50))  // 42190
console.log(f(100)) // 339240
console.log(f(150)) // 1295790

edited Sep 23, 2017 at 12:46

answered Sep 23, 2017 at 12:18

Arnauld

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

2

J, 24 bytes

[:+/[:+/\[:(*#@":"0)1+i.

Similar high-level approach to dzaima's APL answer, translated into J, except we calculate the number's length by turning it into a string first instead of taking logs, and we get to use J's hook to multiply that length by the number itself: (*#@":"0). After that it's just the sum of the scan sum.

Try it online!

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2017 at 22:03

Jonah

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1

\$\begingroup\$ 1(#.]*#\*#\.)1#@":@+i. also works for 22 bytes \$\endgroup\$
– miles
Commented Sep 26, 2017 at 2:31
\$\begingroup\$ @miles That's clever -- it took me a bit to figure it out. How long have been programming in J for? \$\endgroup\$
– Jonah
Commented Sep 26, 2017 at 5:11
\$\begingroup\$ A bit after I joined code-golf. I don't actually use it to write any real programs as no I know would be able to read it, but I do use it as an advanced desktop calculator now, and usually always have a window open to calculate something. \$\endgroup\$
– miles
Commented Sep 26, 2017 at 5:47

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Giuseppe · Accepted Answer · 2017-09-25 13:27:40Z

2

R, 39 bytes

function(n)sum(nchar(rep(1:n,n:1*1:n)))

Verify all test cases!

Simple algorithm; I observed, as most did, that for i in 1:n, i is repeated i*(n-i+1) times. So I create that vector, count the number of characters in each, and sum them.

edited Sep 25, 2017 at 13:27

answered Sep 23, 2017 at 12:18

Giuseppe

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Brad Gilbert b2gills · Accepted Answer · 2017-09-26 13:53:53Z

2

Perl 6, 36 bytes

{[+] 1..*Z*($_...1).map:{.chars*$_}}

Test it

Expanded:

{  # bare block lambda with implicit parameter ｢$_｣

  [+]               # reduce the following using &infix:«+»

    1 .. *          # Range from 1 to infinity

    Z*              # zip using &infix:«*»

    ( $_ ... 1 )    # sequence from the input down to 1
    .map:           # for each one
    { .chars * $_ } # multiply the number of digits with itself
}

answered Sep 26, 2017 at 13:53

Brad Gilbert b2gills

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fergusq · Accepted Answer · 2017-10-08 12:11:19Z

2

Röda, 31 bytes

{f={seq 1,_}f|f|[#`$_`*_1]|sum}

Try it online!

answered Oct 8, 2017 at 12:11

fergusq

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orlp · Accepted Answer · 2017-09-23 12:15:43Z

1

Python 2, 51 50 bytes

lambda n:sum(~k*(k-n)*len(`k+1`)for k in range(n))

edited Sep 23, 2017 at 12:15

answered Sep 23, 2017 at 12:09

orlp

39.2k7 gold badges81 silver badges195 bronze badges

\$\begingroup\$ @LeakyNun Why? I developed this answer myself. I didn't even check the other answers. \$\endgroup\$
– orlp
Commented Sep 23, 2017 at 12:12
1

\$\begingroup\$ This doesn't even output the right answer, gives 0 for n=1, 3 for n=2 and 14 for n=3 \$\endgroup\$
– Halvard Hummel
Commented Sep 23, 2017 at 12:12
\$\begingroup\$ @HalvardHummel Oops, messed up a sign and forgot a +1. Fixed now. \$\endgroup\$
– orlp
Commented Sep 23, 2017 at 12:16
\$\begingroup\$ I see that you finally understood the patern! Is there a way to test your code online or the other Python 2 answer covers this, too? \$\endgroup\$
– user72253
Commented Sep 23, 2017 at 13:28

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

1

Mathematica, 66 bytes

Tr[1^(f=Flatten)[IntegerDigits/@f@(a=Array)[a[#~Table~#&,#]&,#]]]&

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2017 at 12:59

ZaMoC

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steenbergh · Accepted Answer · 2017-09-23 18:19:48Z

1

QBIC, 21 bytes

[:|[a|[c|A=A+!c$}?_lA

answered Sep 23, 2017 at 18:19

steenbergh

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user45941user45941 · Accepted Answer · 2017-09-23 20:10:19Z

1

Actually, 13 bytes

R♂R♂i⌠;n⌡Mεjl

Try it online!

Explanation:

R♂R♂i⌠;n⌡Mεjl
R              range(1, n+1)
 ♂R            range(1, x+1) for x in previous
   ♂i          flatten into 1D list
     ⌠;n⌡M     for x in list:
      ;n       repeat x x times
          εj   concatenate to string
            l  length

answered Sep 23, 2017 at 20:10

user45941

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

1

Japt, 12 11 10 9 bytes

õõÈ*sÊÃxx

Try it or test all numbers from 1 to 150.

Explanation

Implicit input of integer U.

õõ

Generate an array of integers from 1 to U and then generate sub-arrays from 1 to each integer.

È   Ã

Pass the elements of each sub-array through a function.

*sÊ

Convert the current element to a string (s), get it's length (Ê) and multiply it by the element.

xx

Reduce the main array by addition after first doing the same to each sub-array.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 23, 2017 at 13:04

Shaggy

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jq170727 · Accepted Answer · 2017-09-24 00:24:10Z

1

Jq 1.5, 82 49 43 bytes

[range(.)+1|range(.)+1|"\(.)"*.|length]|add

Expanded

[   range(.)+1        # for i=1 to N
  | range(.)+1        # for j=1 to i
  | "\(.)"*.          # "j" copied j times
  | length            # convert to length
] | add               # add lengths

Sample Run

$ jq -Mr '[range(.)+1|range(.)+1|"\(.)"*.|length]|add' <<< "150"
1295790

Try it online! also jqplay.org

edited Sep 24, 2017 at 0:24

answered Sep 23, 2017 at 23:30

jq170727

5312 silver badges5 bronze badges

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Conor O'Brien · Accepted Answer · 2017-09-24 21:11:13Z

Stacked, 28 bytes

[~>[~>[:rep]"!]"!flat''#`#']

Try it online!

Some might ask, "At which point are aliases unreadable?" If this isn't close, you have a very liberal definition of "readability".

Explanation

[~>[~>[:rep]"!]"!flat''#`#']    input: N
 ~>[          ]"!               for each number K from 1 to N
    ~>[    ]"!                  for each number J from 1 to K
       :rep                     repeat J J times
                 flat           flatten the resultant array
                     ''#`       join by the empty string
                         #'     get the length of said string

G B · Accepted Answer · 2017-09-25 07:15:25Z

1

Ruby, 41 40 bytes

->n{(1..n).map{|x|[x]*x*=-x-~n}*''=~/$/}

Try it online!

edited Sep 25, 2017 at 7:15

answered Sep 25, 2017 at 6:54

G B

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

1

C# (.NET Core), 94 80 74 bytes

n=>{int b=0,a=0,i;while(a++<n)for(i=0;i++<a;)b+=(i+"").Length*i;return b;}

Try it online!

I was hoping to find a direct solution like what @kamoroso94's answer started, but gave up as I was spending too much time on it. There probably is a way of doing it, but the formula needs to adjust for every magnitude step.

Acknowledgements

14 bytes saved thanks to @someone

6 bytes saved thanks to @Kevin Cruijssen

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Sep 24, 2017 at 1:54

Ayb4btu

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1

\$\begingroup\$ n=>{int b=0,a=0,i;for(;a++<n;)for(i=0;i++<a;)b+=i.ToString().Length*i;return b;} Try it online! for 80 bytes and performance. \$\endgroup\$
– the default.
Commented Sep 24, 2017 at 10:54
1

\$\begingroup\$ i.ToString() can be (i+"") to save some more bytes. \$\endgroup\$
– Kevin Cruijssen
Commented Sep 25, 2017 at 7:42

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Giuseppe · Accepted Answer · 2017-09-25 13:46:04Z

1

MATL, 15 bytes

:ttP*Y"10&YlQks

Try it online!

Explanation:

:                range 1:input (implicit input)
 tt              duplicate twice
   P             reverse
    *            multiply elementwise
     Y"          runlength decoding
       10&Yl     log10
            Qk   increment and floor
              s  sum (implicit output)

answered Sep 25, 2017 at 13:46

Giuseppe

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\$\begingroup\$ That logarithm is expensive :-) You can replace it by converting to string, removing spaces, length: :ttP*Y"VXzn \$\endgroup\$
– Luis Mendo
Commented Oct 6, 2017 at 23:33

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Neil · Accepted Answer · 2017-10-04 19:27:41Z

1

Charcoal, 18 14 bytes

ＩΣＥ⊕ＮΣＥ⊕ι×λＬＩλ

Try it online! Link is to verbose version of code. Edit: Using Sum saved me 4 bytes. Explanation:

  Ｅ⊕Ｎ           Map from 0 to the input (loop variable i)
      Ｅ⊕ι       Map from 0 to i (loop variable l)
            Ｉλ  Cast l to string
           Ｌ    Take the length
         ×λ     Multiply by l
     Σ          Sum the results
 Σ              Sum the results
Ｉ               Cast to string
                Implicitly print

edited Oct 4, 2017 at 19:27

answered Sep 23, 2017 at 19:33

Neil

177k12 gold badges74 silver badges281 bronze badges

\$\begingroup\$ :| Sum sums numbers in the strings when given string arguments \$\endgroup\$
– ASCII-only
Commented Oct 4, 2017 at 11:23
\$\begingroup\$ @ASCII-only It wasn't that, it was just printing a Σ instead... \$\endgroup\$
– Neil
Commented Oct 4, 2017 at 12:36
\$\begingroup\$ @ASCII-only Also, best I can do with Sum is still 18 bytes: Print(Cast(Sum(Map(InclusiveRange(1, InputNumber()), Sum(Map(InclusiveRange(1, i), Times(l, Length(Cast(l))))))))); \$\endgroup\$
– Neil
Commented Oct 4, 2017 at 12:46
\$\begingroup\$ wait you keep forgetting Incremented :P \$\endgroup\$
– ASCII-only
Commented Oct 4, 2017 at 12:50
\$\begingroup\$ @ASCII-only I tried sum of product but that was 17 bytes: ≔⊕ＮθＩΣＥθ×⁻θι×ιＬＩι. However, using Incremented instead of InclusiveRange shaves 4 bytes off my previous comment! \$\endgroup\$
– Neil
Commented Oct 4, 2017 at 14:08

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Cinaski · Accepted Answer · 2017-10-05 07:32:14Z

1

Ohm v2, 7 bytes

@@D×JJl

Try it online!

answered Oct 5, 2017 at 7:32

Cinaski

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dzaima · Accepted Answer · 2017-10-08 08:58:16Z

[Dyalog APL], 22 20 bytes

{+/≢¨⍕¨↑,/(/⍨¨⍳¨⍳⍵)}

Try it online!

Explanation:

{+/≢¨⍕¨↑,/(/⍨¨⍳¨⍳⍵)}
{                  } anonymous function with right argument named ⍵
                ⍳⍵   range 1 to right arg
              ⍳¨     for each, range 1 to it
             ¨       for each
           /⍨          for each item, repeat right arg left arg times
          (       )  take that and
        ,/           join the sub-arrays together
       ↑             convert from a nested array to a simple array (or something like that, I don't quite understand it :p)
     ⍕¨              convert each number to a char-array (aka string version)
   ≢¨                get length of each
 +/                  sum that together

Nobody knows GAU Numbers

Let me introduce you to GAU numbers

This challenge is pretty simple!

Example

Test cases

47 Answers 47

SOGL V0.12, 11 10 8 7 5 bytes

Haskell, 45 bytes

Brain-Flak, 166 bytes

Explanation

Husk, 5 bytes

Jelly, 7 bytes

05AB1E, 5 bytes

Python 2, 53 bytes

Husk, 7 bytes

Ungolfed/Explanation

Husk, 7 bytes

Explanation

JavaScript (ES6), 57 55 bytes

Python 2, 59 58 bytes

CJam, 20 bytes

JavaScript (ES6), 50 42 bytes

Test cases

J, 24 bytes

R, 39 bytes

Perl 6, 36 bytes

Expanded:

Röda, 31 bytes

Python 2, 51 50 bytes

Mathematica, 66 bytes

QBIC, 21 bytes

Actually, 13 bytes

Japt, 12 11 10 9 bytes

Explanation

Jq 1.5, 82 49 43 bytes

Stacked, 28 bytes

Explanation

Ruby, 41 40 bytes

C# (.NET Core), 94 80 74 bytes

Acknowledgements

MATL, 15 bytes

Charcoal, 18 14 bytes

Ohm v2, 7 bytes

[Dyalog APL], 22 20 bytes

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