It's all about the sum of the digits

Question

The sequence

Given an integer \$n>0\$, we define \$a(n)\$ as the lowest positive integer such that there exists exactly \$n\$ positive integers smaller than \$a(n)\$ whose sum of digits is equal to the sum of the digits of \$a(n)\$.

Edit: this sequence has since been published as A332046

First terms

  n | a(n) | sum of digits | matching integers
----+------+---------------+------------------------------------------------------
  1 |  10  |       1       | [1]
  2 |  20  |       2       | [2, 11]
  3 |  30  |       3       | [3, 12, 21]
  4 |  40  |       4       | [4, 13, 22, 31]
  5 |  50  |       5       | [5, 14, 23, 32, 41]
  6 |  60  |       6       | [6, 15, 24, 33, 42, 51]
  7 |  70  |       7       | [7, 16, 25, 34, 43, 52, 61]
  8 |  80  |       8       | [8, 17, 26, 35, 44, 53, 62, 71]
  9 |  90  |       9       | [9, 18, 27, 36, 45, 54, 63, 72, 81]
 10 | 108  |       9       | [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]
 .. |  ..  |      ..       | ..
 20 | 216  |       9       | [9, 18, 27, 36, 45, 54, 63, 72, 81, 90, ..., 207]
 .. |  ..  |      ..       | ..
 30 | 325  |      10       | [19, 28, 37, 46, 55, 64, 73, 82, 91, 109, ..., 316]
 .. |  ..  |      ..       | ..
 40 | 442  |      10       | [19, 28, 37, 46, 55, 64, 73, 82, 91, 109, ..., 433]
 .. |  ..  |      ..       | ..
 50 | 560  |      11       | [29, 38, 47, 56, 65, 74, 83, 92, 119, 128, ..., 551]

More examples

a(13) = 135
a(17) = 171
a(42) = 460
a(57) = 660
a(81) = 1093
a(82) = 1128
a(118) = 1507
a(669) = 9900
a(670) = 10089
a(1000) = 14552
a(5000) = 80292
a(10000) = 162085

As a side note, you may not assume that the sum of the digits may only increase. For instance, the sum of the digits for \$a(81)\$ is \$1+0+9+3=13\$, while the sum of the digits for \$a(82)\$ is \$1+1+2+8=12\$.

Rules

You may either:
- take a 1-indexed integer \$n\$ and return \$a(n)\$
- take a 0-indexed integer \$n\$ and return \$a(n+1)\$
- take a positive integer \$n\$ and return the \$n\$ first terms
- print the sequence indefinitely
This is code-golf!

A simple proof that \$a_n\$ is defined for any \$n\$: \$10^n\$ has digit sum of 1, and has exactly \$n\$ positive integers smaller than itself whose sum of digits is 1, namely \$10^0 \cdots 10^{n-1}\$. — Bubbler, Feb 6, 2020 at 1:51
Did you invent a(n)? I couldn't find it in oeis.org. That's an interesting function in order to troll "Continue the sequence" questions. What comes after 10, 20, 30, 40, 50, 60, 70, 80, 90? 108. — Eric Duminil, Feb 6, 2020 at 11:05
@EricDuminil Yes I invented it and I'm actually half-tempted to submit it on OEIS. — Arnauld, Feb 6, 2020 at 11:08

isaacg · Accepted Answer · 2020-02-05 16:37:03Z

10

Pyth, 12 bytes

f}hQ_XsjT;H1

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f: Count up from 1 to find the first number such that:
X ... H1: Increment or insert the value 1 into the dictionary H at index
sjT;: sum of base ten digits of current number
_: values of dictionary
}hQ: check whether input + 1 is contained within.

The first time this is true, exactly input numbers must have the same digit sum as the current number, for the first time.

answered Feb 5, 2020 at 16:37

isaacg

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

4

Gaia, 13 10 bytes

1⟨┅Σ¦ṅC=⟩#

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Straightforward implementation.

Explanation:

1⟨	⟩#	| find the first 1 positive integers N where:
      C		| the count of
     ṅ		| the digital sum d(N)
  ┅Σ¦		| in the list [d(1)..d(N-1)]
       =	| is equal to the (implicit) input

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Feb 5, 2020 at 16:35

Giuseppe

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Nick Kennedy · Accepted Answer · 2020-02-05 20:58:54Z

3

Jelly, 10 bytes

D€§ċṪ$=ʋ1#

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A monadic link taking an integer \$n\$ and returning \$a(n)\$.

Explanation

       ʋ1# | Find the first integer x where the following is true;
D€         | - Digits of 1..x
  §        | - Sum each
   ċṪ$     | - Count the number equal to the tail (after popping tail)
      =    | - Equal to (implicit original argument to link)

answered Feb 5, 2020 at 20:58

Nick Kennedy

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Grimmy · Accepted Answer · 2020-02-05 21:39:29Z

3

05AB1E, 10 bytes

∞.Δ1Ÿ1öć¢Q

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answered Feb 5, 2020 at 21:39

Grimmy

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\$\begingroup\$ Ah, I keep forgetting about 1ö being the non-vectorizing digit-sum! Apart from that it's similar as what I had before I saw your answer. \$\endgroup\$
– Kevin Cruijssen
Feb 7, 2020 at 13:14

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Bubbler · Accepted Answer · 2020-02-06 04:35:49Z

APL (Dyalog Unicode), 27 bytes^SBCS

{⌈⌿+\∘.=⍨1⊥10⊥⍣¯1⍳20×⍵}⍳+∘1

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How it works

{⌈⌿+\∘.=⍨1⊥10⊥⍣¯1⍳20×⍵}⍳+∘1  ⍝ Right argument: n
{                     }⍳+∘1  ⍝ Find the first index of n+1 from...
 ⌈⌿+\∘.=⍨1⊥10⊥⍣¯1⍳20×⍵       ⍝ ... the list of cumulative count of own digit sum (sort of)
                  20×⍵         ⍝ Practical upper bound 20×n (could also use theoretical 10*⍵)
                 ⍳             ⍝ Range: 1..20×n
           10⊥⍣¯1  ⍝ Extract decimal digits from each
         1⊥        ⍝ Sum the digits of each number
     ∘.=⍨    ⍝ Self outer product by equality
   +\        ⍝ Cumulative sum in row direction; cumulative counts of own digit sum
 ⌈⌿          ⍝ Maximum in column direction; cumulative maximum of cumulative count

ovs · Accepted Answer · 2020-02-06 13:11:51Z

3

Python 2, 71 bytes

f=lambda n,x=[0]:x.count(x[-1])+~n and-~f(n,x+[sum(map(int,`len(x)`))])

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answered Feb 6, 2020 at 13:11

ovs

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Neil · Accepted Answer · 2020-02-06 00:46:44Z

2

Charcoal, 16 bytes

Ｗ⁻№υΣＬυθ⊞υΣＬυＩＬυ

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

Ｗ⁻№υΣＬυθ

Count the number of entries in the predefined list that are equal to the digital sum of the length of the list and repeat while that does not equal the input...

⊞υΣＬυ

... append the digital sum of the length of the list to the list.

ＩＬυ

Output the length of the list.

answered Feb 6, 2020 at 0:46

Neil

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G B · Accepted Answer · 2020-02-07 12:08:58Z

2

Ruby, 55 bytes

->n,r=0,*w{w<<k=(r+=1).digits.sum until w.count(k)>n;r}

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edited Feb 7, 2020 at 12:08

answered Feb 7, 2020 at 11:17

G B

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Shieru Asakoto · Accepted Answer · 2020-02-06 07:30:09Z

1

JavaScript (Node.js), 68 67 63 bytes

-1 -5 byte thanks Arnauld

n=>(F=z=>F[g=eval([...z+''].join`+`)]^n?F(-~z,F[g]=-~F[g]):z)``

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edited Feb 6, 2020 at 7:30

answered Feb 6, 2020 at 7:06

Shieru Asakoto

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mabel · Accepted Answer · 2020-02-06 08:50:32Z

1

Python 2, 92 93 bytes

f=lambda n,x=1:n==sum(sum(map(int,`i`))==sum(map(int,`x`))for i in range(x))and x or f(n,x+1)

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All I can get for now but I suspect someone much more competent than I am could do much better.

edited Feb 6, 2020 at 8:50

answered Feb 5, 2020 at 20:20

mabel

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2

\$\begingroup\$ Because you define the function recursively, you have to give it a name, so you have 94 bytes :/ at least for now :) \$\endgroup\$
– RGS
Feb 6, 2020 at 0:34
\$\begingroup\$ And then 93 bytes if you put the n== in the beginning, which allows for the next keyword to come closer to the ). \$\endgroup\$
– RGS
Feb 6, 2020 at 0:53
\$\begingroup\$ I see and thanks @RGS \$\endgroup\$
– mabel
Feb 6, 2020 at 8:50

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

Java 8, 138 bytes

n->{for(int a=9,c,i;;a++){for(c=i=0;++i<a;)c+=(i+"").chars().map(q->q-48).sum()==(a+"").chars().map(q->q-48).sum()?1:0;if(c==n)return a;}}

Try it online.

Explanation:

n->{               // Method with integer as both parameter and return-type
  for(int a=9,     //  Integer `a_i`, starting at 9 (since a(1)=10)
          c,       //  Count-integer
          i;       //  Loop-integer
      ;            //  Loop indefinitely:
       a++){       //    After every iteration: increase `a_i` by 1
    for(c=i=0;     //   Reset both the count and loop integers to 0
        ++i<a;)    //   Inner loop `i` in the range (0,a):
      c+=          //    Increase the count by:
         (i+"").chars().map(q->q-48).sum()
                   //     If the digit-sum of `i`
           ==(a+"").chars().map(q->q-48).sum()?
                   //     equals the digit-sum of `a_i`:
            1      //      Increase the count by 1
           :       //     Else:
            0;     //      Leave the count the same by increasing with 0
    if(c==n)       //   If the count equals the input integer `n` after the inner loop:
      return a;}}  //    Return `a_i` as our `a(n)` result

mazzy · Accepted Answer · 2020-02-07 17:02:49Z

1

PowerShell, 74 72 bytes

$a=@{}
1.."$args"|%{for(;$_-$a.(++$i|%{"$_"[0..9]-join'+'|iex})++){}}
$i

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2^31 = 2147483647, so the expression "$_"[0..9] works with any Powershell [int] > 0.

edited Feb 7, 2020 at 17:02

answered Feb 7, 2020 at 16:19

mazzy

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lesobrod · Accepted Answer · 2023-05-07 08:50:50Z

Wolfram Language (Mathematica), 92 or 113 bytes

Brute force short version:

(t=Total@IntegerDigits@#&;Catch@Do[If[Count[Range@(k-1),x_/;t@x==t@k]==#,Throw@k],{k,∞}])&

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Time- and memory-optimized version. Print table of values up to given number:

(n=0;a=<||>;r={};Do[While[FreeQ[a,i+1],a[k]=Lookup[a,k=Total@IntegerDigits@++n,0]+1];AppendTo[r,{i,n}],{i,#}];r)&

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Ungolfed self-explained:

next = 0; assoc = <||>;
max = 30; result = {};
Do[
 While[FreeQ[assoc, n + 1],
  next += 1;
  key = Total@IntegerDigits@next;
  assoc[key] = Lookup[assoc, key, 0] + 1
  ];
 AppendTo[result, {n, next}]
 , {n, max}]; result

138 Aspen · Accepted Answer · 2023-05-07 08:57:05Z

Scala, 115 bytes

Golfed version. Try it online!

def f(n: Int)={var r,k=0;var w=List[Int]();do{r+=1;k=r.toString.map(_.asDigit).sum;w:+=k}while(w.count(_==k)<=n);r}

Ungolfed version. Try it online!

object RubyToScala {
  def main(args: Array[String]): Unit = {
    println(f(1))
    println(f(2))
    println(f(3))
    println(f(30))
    println(f(50))
    println(f(1000))
  }

  def f(n: Int): Int = {
    var r = 0
    var w = List[Int]()
    var k = 0
    do {
      r += 1
      k = r.toString.map(_.asDigit).sum
      w :+= k
    } while (w.count(_ == k) <= n)
    r
  }
}

Shaggy · Accepted Answer · 2023-05-07 19:05:52Z

0

Japt, 14 bytes

@¶XÇìxÃè¶Xìx}a

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answered May 7 at 19:05

Shaggy

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bluswimmer · Accepted Answer · 2023-05-09 00:34:03Z

0

Lua, 93 bytes

t={}n=0repeat m=0n=n+1(n..''):gsub('.',load'm=m+...')t[m]=(t[m]or 0)+1until...+0<t[m]print(n)

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Ungolfed version:

t={}n=0
repeat 
    m=0 n=n+1                        --Increment n, set m to 0
    (n..''):gsub('.',load'm=m+...')  --Sum digits of n, store in m
    t[m]=(t[m]or 0)+1                --Increment t[m], or set to 1 if null
until...+0<t[m]                      --If input < t[m], stop loop
print(n)

answered May 9 at 0:34

bluswimmer

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The Thonnu · Accepted Answer · 2023-08-25 15:39:19Z

0

Thunno 2, 6 bytes

ƘRʂṪc=

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Explanation

ƘRʂṪc=  # Implicit input
Ƙ       # Find the first positive integer, n, where:
 R      #  Push [1..n]
  ʂ     #  Get the digital sum of each number
   Ṫ    #  Remove the last item and push it separately
    c   #  Count the occurrences of this in the list
     =  #  Does this equal the input?

edited Aug 25 at 15:39

answered May 7 at 19:50

The Thonnu

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Stack Exchange Network

It's all about the sum of the digits

The sequence

First terms

More examples

Rules

17 Answers 17

Pyth, 12 bytes

Gaia, 13 10 bytes

Explanation:

Jelly, 10 bytes

Explanation

05AB1E, 10 bytes

APL (Dyalog Unicode), 27 bytes^SBCS

How it works

Python 2, 71 bytes

Charcoal, 16 bytes

Ruby, 55 bytes

JavaScript (Node.js), 68 67 63 bytes

Python 2, 92 93 bytes

Java 8, 138 bytes

PowerShell, 74 72 bytes

Wolfram Language (Mathematica), 92 or 113 bytes

Scala, 115 bytes

Japt, 14 bytes

Lua, 93 bytes

Thunno 2, 6 bytes

Explanation

Your Answer

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code-golf
sequence
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Hot Network Questions

It's all about the sum of the digits

The sequence

First terms

More examples

Rules

17 Answers 17

Pyth, 12 bytes

Gaia, 13 10 bytes

Explanation:

Jelly, 10 bytes

Explanation

05AB1E, 10 bytes

APL (Dyalog Unicode), 27 bytesSBCS

How it works

Python 2, 71 bytes

Charcoal, 16 bytes

Ruby, 55 bytes

JavaScript (Node.js), 68 67 63 bytes

Python 2, 92 93 bytes

Java 8, 138 bytes

PowerShell, 74 72 bytes

Wolfram Language (Mathematica), 92 or 113 bytes

Scala, 115 bytes

Japt, 14 bytes

Lua, 93 bytes

Thunno 2, 6 bytes

Explanation

Your Answer

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Hot Network Questions

APL (Dyalog Unicode), 27 bytes^SBCS

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