Equality in the sum of digits

Question

Introduction

Let's take the number 180. This is an interesting number because the sum of digits of this number is equal to:

1 + 8 + 0 = 9

And the squared version of this number, or:

180² = 32400 > 3 + 2 + 4 + 0 + 0 = 9

These are both 9. The sum of digits of the original number and the squared number are the same. Of course, this is also found at OEIS: A058369.

Task

Given a non-negative integer n, output the nth positive number with this condition.

Test cases (zero-indexed)

Input > Output

0 > 1
1 > 9
2 > 10
3 > 18
4 > 19
5 > 45
6 > 46
7 > 55
8 > 90
9 > 99
10 > 100
11 > 145
12 > 180
13 > 189
14 > 190
15 > 198
16 > 199
17 > 289
18 > 351
19 > 361

The input can also be 1-indexed if that fits you better.

This is code-golf, so the submission with the least amount of bytes wins!

Answer 1

Jelly, 13 bytes

,²DS€=/
1dz#Ṫ

Input is 1-indexed. Try it online!

How it works

1dz#Ṫ    Main link. Argument: n (index)

1        Set the return value to 1.
   #     Execute ... until ... matches have been found.
 Ç         the helper link
  ³        n
    Ṫ    Extract the last match.


,²DS€=/  Helper link. Argument: k (integer)

,²       Pair k with k².
  D      Convert each to decimal.
   S€    Compute the sum of each list of base 10 digits.
     =/  Reduce by equality.

Answer 2

05AB1E, 10 9 8 bytes

µNÐn‚1öË

1-indexed.

-1 byte thanks to @Emigna by removing the implicit ½ (increase counter_variable after every iteration) at the end
-1 byte thanks to @Grimy removing the duplicated SO by using ‚1ö

Try it online.

Explanation:

µ         # Loop while the counter_variable is not equal to the (implicit) input yet:
 NÐ       #  Push the 0-based loop index three times
   n      #  Take the square of this index
          #   i.e. 180 → 32400
    ‚     #  Pair it with the index
          #   i.e. 180 and 32400 → [180,32400]
     1ö   #  Convert both numbers from base-1 to base-10, which basically sums the digits
          #   i.e. [180,32400] → [9,9]
       Ë  #  Check if both sums are equal
          #   i.e. [9,9] → 1 (truthy)
          #  (if they are: implicitly increase the counter_variable by 1)
          # (after the loop: implicitly print the top of the stack, which is the remaining
          #  copy of the index from the triplicate we've used)

Answer 3

Haskell, 54 bytes

s=sum.map(read.pure).show
([x|x<-[1..],s x==s(x^2)]!!)

Usage example: ([x|x<-[1..],s x==s(x^2)]!!) 17 -> 289.

s calculates the digit sum:

                    show     -- turn number into a string
     map(read.pure)          -- turn every character (the digits) in to a
                             -- one element string and convert back to integer
sum                          -- sum those integers

main function:

[x|x<-[1..]            ]     -- make a list of all x starting from 1
           ,s x==s(x^2)      -- where s x == s (x^2)
                        !!   -- pick nth element from that list

Answer 4

JavaScript (ES6), 76 73 72 bytes

n=>eval("for(q=s=>eval([...s+''].join`+`),i=1;q(i)!=q(i*i)||n--;i++);i")

I spent 30 minutes trying to get this to work until I realized I was outputting the wrong variable :|

This is zero-indexed.

Answer 5

Perl 6, 47 46 bytes

{(grep {$_.comb.sum==$_².comb.sum},1..*)[$_]}

Answer 6

Mathematica, 64 bytes

a=Tr@*IntegerDigits;Nest[NestWhile[#+1&,#+1,a@#!=a[#^2]&]&,1,#]&

Simple anonymous function. Zero-indexed.

Answer 7

Pyth, 15

e.fqsjZTsj^Z2TQ

1 byte thanks to DenkerAffe!

Try it here or run a Test Suite.

Uses the 1-indexed option.

Naive implementation using .f which gets the first n numbers that match the given condition.

Answer 8

MATL, 24 23 bytes

x`@2:^"@V!Us]=?@]NG<]1$

Uses 1-based input.

Try it online!

x        % take inpout and delete it (gets copied into clipboard G)
`        %   do...while
  @      %   push loop iteration index: candidate number, n
  2:^    %   array [n n^2]
  "      %   for each element of that array 
    @    %     push that element 
    V!U  %     get its digits (to string, transpose, to number)
    Xs   %     compute their sum
  ]      %   end for each
  =      %   are the two sums equal?
  ?      %   if so
    @    %     the candidate number is valid: push it
  ]      %   end if
  NG<    %   is number of elements in stack less than input?
]        % if so, proceed with next iteration. End do...while. 
1$       % specify 1 input for implicit display: only top of stack

Answer 9

Mathematica, 63 60 61 59 bytes

Select[Range[9^#],Equal@@Tr/@IntegerDigits/@{#,#^2}&][[#]]&

While making this the other answer popped up but I'm beating them by a single byte and I'm posting this before that one gets golfed. One indexed.

Answer 10

Convex 0.2, 36 35 bytes

Convex is a new language that I am developing that is heavily based on CJam and Golfscript. The interpreter and IDE can be found here. Input is an integer into the command line arguments. Indexes are one-based. Uses the CP-1252 encoding.

1\{\__2#¶{s:~:+}%:={\(\)\}{)\}?}h;(

Answer 11

Julia, 79 66 bytes

f(n,x=0,i=1,s=c->sum(digits(c)))=x<n?f(n,x+(s(i)==s(i^2)),i+1):i-1

This is a recursive function that accepts an integer and returns an integer. It uses 1-based indexing.

We store a few things as function arguments:

• n : The input
• x : A counter for how many numbers with this condition we've found
• i : A number to check for the condition
• s : A function to compute the sum of the digits of its input

While x is less than the input, we recurse, incrementing x if i meets the condition and incrementing i. Once x == n, we return i, but we have to subtract 1 because it will have been incremented one too many times.

Answer 12

Retina, 103 bytes

\d+
$*1 x
{`x+
$.0$*x¶$.0$*a¶$.0$*b
%`b
$_
a+|b+
$.0
\d
$*
+`1¶1
¶
1(.*)¶¶$|¶[^d]+
$1x
}`^ ?x

x

Definitely golfable.

Uses the new Retina feature % for squaring (hence not working with the online version yet).

Answer 13

Mathcad, 70 50 bytes

Mathcad has no built in functions to convert a number to its digit string, so the user function d(a) does this job. A program then iterates through the positive integers, testing for equality of sums, until it has accumulated n numbers in the vector v. The program is evaluated using the = operator, which displays the result vector. (Note that the whole program appears exactly as displayed below on the Mathcad worksheet)

Updated program: Assumes default initialization of a to zero and makes use of fact that Mathcad returns the value of the last evaluated statement in a program.
Makes use of evaluation order of expressions to increment variable a in the first summation (and which is then available for use in the sum of square)

Original program: Returns a vector of all numbers up to n.

Answer 14

Japt, 15 bytes

1-indexed

_ìx ¶Z²ìx «U´}a

Try it

Answer 15

Java 8, 113 bytes

n->{int r=0;for(;n>=0;)if((++r+"").chars().map(c->c-48).sum()==(r*r+"").chars().map(c->c-48).sum())n--;return r;}

0-indexed

Explanation:

Try it online.

n->{           // Method with integer as both parameter and return-type
  int r=0;     //  Result-integer, starting at 0
  for(;n>=0;)  //  Loop as long as `n` is zero or positive
    if((++r    //   Increase `r` by 1 first
       +"").chars().map(c->c-48).sum()
               //   And if the sum of its digits
       ==(r*r+"").chars().map(c->c-48).sum())
               //   equals the sum of the digit of its square
      n--;     //    Decrease `n` by 1
  return r;}   //  Return the result

Answer 16

Perl 5 -p, 53 bytes

Includes +1 for -p

1 based

#!/usr/bin/perl -p
$.++while$_-=!eval+("-($.) ".$.**2)=~s/\B/+/gr;$_=$.

Try it online!

Answer 17

TI-BASIC 66 62 bytes

Ans→N:While X<N:IS>(A,A::A:prgmA:Ans→B:A²:prgmA:If B=Ans:IS>(X,N:End:A
sum(int(10fPart(Ans₁₀^(seq(⁻X-1,X,0,log(Ans

Input is \$n\$ in Ans.
Output is the 1-indexed \$n\$th term in the sequence.

Helper function generates the sum of the digits of the value in Ans.

Examples:

3:prgmCDGF1E
             10
5:prgmCDGF1E
             19
8:prgmCDGF1E
             55
10:prgmCDGF1E
             99

Explanation:

Ans→N:While X<N:IS>(A,A::A:prgmA:Ans→B:A²:prgmA:If B=Ans:IS>(X,N:End:A ;prgmCDGF1E

Ans→N            ;store the input in N
While X<N        ;loop until the Nth term has been reached
IS>(A,A:         ;add 1 to A
                 ; (Increment A and skip the next statement if A>A)
A                ;leave A in Ans
prgmA            ;call the helper program below
Ans→B            ;store the result of the helper program in B
A²               ;square A and leave the result in Ans
prgmA            ;call the helper program below
                 ; (result is in Ans)
If B=Ans         ;if the two results are equal
IS>(X,N          ;add 1 to X
                 ; (Increment X and skip the next statement if X>N)
End
A                ;leave A in Ans
                 ;implicit print of Ans

sum(int(10fPart(Ans₁₀^(seq(⁻X-1,X,0,log(Ans   ;prgmA

                      seq(⁻X-1,X,0,log(Ans    ;generate a list...
                                              ; using X as the variable,
                                              ; starting at 0,
                                              ; ending at the log of Ans,
                                              ; and evaluating "⁻X-1" for each element
                                              ; (implicit increment of 1)
                   ₁₀^(                       ;raise 10 to the power of each element
                Ans                           ;multiply each element by the input
          fPart(                              ;remove the integer part from each element
        10                                    ;multiply each element by 10
    int(                                      ;round each element to the nearest integer
sum(                                          ;then sum the resulting list

---

Note: TI-BASIC is a tokenized language. Character count does not equal byte count.

Answer 18

J, 62 bytes

[:{:({.@](>:@[,],[#~(=&(1#."."0@":)*:)@[)}.@])^:(#@]<1+[)^:_&1

Try it online!

1 indexed. J once again not performing well on these "nth of" tasks, because of the excessive bookeeping mechanics.

Answer 19

APL(NARS), 49 chars, 98 bytes

r←h w;c
c←r←0
→2×⍳∼=/+/¨(⍎¨⍕)¨r,r×r+←1⋄→2×⍳w>c+←1

1-indexed, test:

  h¨⍳20
1 9 10 18 19 45 46 55 90 99 100 145 180 189 190 198 199 289 351 361 

Answer 20

MathGolf, 10 bytes

♪╒gÆ‼Σ²Σ=§

Try it online!

Explanation

To make this usable, I chose to present a version which calculates every number with this property below 1000, and fetches the correct item from that list. To have a solution which would work for any input size, the first byte could be replaced by ú (push 10**TOS). Since the n:th term in the sequence is always less than \$10^n\$, the script would always succeed. However, this puts a practical limit on calculation that's very low.

♪            push 1000
 ╒           range(1,n+1)
  gÆ         filter list using the next 5 operators
    ‼        apply next two commands to TOS
     Σ       sum(list), digit sum(int)
      ²      pop a : push(a*a) (square)
       Σ     sum(list), digit sum(int) (pushes the digit sum of the square)
        =    pop(a, b), push(a==b) (compares the two)
         §   get from array (returns the <input>th item from the filtered list