Digitangular numbers

Question

A triangular number is a number that can be expressed as the sum of consecutive positive integers, starting at 1. They can also be expressed with the formula n(n + 1) / 2, where n is some positive integer.

A number's digitangular counterpart is calculated in the following way:

1. Split a number into an array of its digits e.g. 613 => [6 1 3]
2. For each number in the array, calculate the nth triangular number; [6 1 3] => [21 1 6]
3. Sum the resultant array; [21 1 6] => 28

Your task is, given an integer n, repeatedly calculate n's digitangular counterpart, until the result is equal to 1, then output all values that were calculated. You may output the values in any order, and with an optional inclusion of the original number at the start of the array. This is a code-golf so the shortest code wins.

Test cases

23 => 9 45 25 18 37 34 16 22 6 21 4 10 1
72 => 31 7 28 39 51 16 22 6 21 4 10 1
55 => 30 6 21 4 10 1
78 => 64 31 7 28 39 51 16 22 6 21 4 10 1
613 => 28 39 51 16 22 6 21 4 10 1
8392 => 90 45 25 18 37 34 16 22 6 21 4 10 1
11111 => 5 15 16 22 6 21 4 10 1
8592025 => 117 30 6 21 4 10 1
999999999 => 405 25 18 37 34 16 22 6 21 4 10 1

Answer 1

Husk, 6 bytes

U¡(ṁΣd

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Explanation

U¡(ṁΣd
 ¡(       Iterate the following function on the input:
     d       Split the number into digits
   ṁΣ        Map each digit to its triangular number, then sum the results
U         Take the results of iteration until before the first repeated one

Answer 2

05AB1E, 6 5 bytes

Δ=SLO

Try it online! Edit: Saved 1 byte thanks to @Emigna. Explanation:

Δ       Repeat until value doesn't change
 =      Print current value
  S     Split into characters
   L    Turn each character into range from 1 to N
    O   Sum

Answer 3

APL (Dyalog), 23 20 17 bytes

3 bytes saved thanks to @ngn

{⍵∪⍨+/∊⍳¨⍎¨⍕⊃⍵}⍣≡

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How?

⍵∪⍨ - prepend current array to the

+/ - sum of

∊ - flattened

⍳¨ - ranges of each

⍎¨⍕ - digit of the

⊃⍵ - previous value

⍣≡ until convergence. The usage of ∪ (union) makes sure after the first 1 is joined, the next will be excluded due to set uniqueness, and the array will converge.

Answer 4

J, ₂₀ 19 bytes

(1#.2!1+,.&.":)^:a:

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Outputs the original number, too.

Explanation

(1#.2!1+,.&.":)^:a:
               ^:a:  Apply until input converges, storing all results in an array
(1#.2!1+,.&.":)      Digitangular sum
        ,.&.":         Split to digits
          &.":           Convert to string, apply left function, then undo conversion
        ,.               Ravel items (make array of singleton arrays of items)
                         This ensures that when cast back to integers, the digits are split.
      1+               Add 1 to each
    2!                 Compute (n choose 2) on each (nth triangular number)
 1#.                   Sum (debase 1)

Answer 5

Haskell, 51 47 46 bytes

f 1=[1]
f x=x:f(sum$do d<-show x;[1..read[d]])

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f 1=[1]                     -- if x == 1, return the singleton list [1]
f x=                        -- else
         do d<-show x       --  for each char 'd' in the string representation of x
                            --    (using the list monad)
           [1..read[d]]     --  make a list of [1..d]
                            --    (the list monad concatenates all those lists into a single list)
        sum                 --  sum those numbers
      f                     --  call f on it
    x:                      --  and put x in front of it 

Edit: @H.PWiz saved a byte. Thanks!

Answer 6

Python 2, 62 bytes

f=lambda x:x<2and[1]or[x]+f(sum(-~int(i)*int(i)/2for i in`x`))

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Answer 7

Wolfram Language (Mathematica), 43 41 bytes

Echo@#>1&&#0[#.(#+1)/2&@IntegerDigits@#]&

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How it works

The expression #.(#+1)/2&@IntegerDigits@# gives the digitangular counterpart of #. We Echo the input, use short-circuit evaluation with && to stop if we've reached 1, and otherwise recurse on the digitangular counterpart.

---

-2 bytes thanks to Martin Ender for the . trick: we don't have to use Tr to sum the digits if we replace the multiplication #(#+1)/2 by the dot product #.(#+1)/2.

Answer 8

Wolfram Language (Mathematica), 49 42 39 bytes

Thanks to Misha Lavrov for saving 3 bytes.

#//.x_:>(y=IntegerDigits@Echo@x).++y/2&

Try it online! (TIO needs parentheses around the ++y for some reason. In my local Mathematica installation it works without them, as it should.)

Prints each value on its own line, preceded by >>, and includes the starting number.

Answer 9

Ruby, 60 47 42 bytes

-13 bytes by @JustinMariner

-5 bytes by @GB

->x{p x=x.digits.sum{|k|k*-~k/2}while x>1}

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Answer 10

Japt, 19 17 bytes

Takes input as a single element array.

_Ì¥1}a@pUÌì mò xx

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Answer 11

Retina, 21 bytes

;{:G`
.
$*1¶
1
$%`1
1

Try it online! (The outputs of the individual cases aren't well separated, but each output ends with a 1.)

Prints each number on its own line, in order, including the starting number.

Explanation

;{:G`

This is just some configuration of the program. { makes the program loop until it fails to change the result (which happens once we get to 1), : prints number before each iteration, and ; prevents the final result from being printed twice at the end of the program. The G is just my usual way of creating a no-op stage.

.
$*1¶

Convert each digit to unary and put it on its own line.

1
$%`1

Compute the triangular number on each line, by replacing each 1 with its prefix. We could also use M!&`1+ here, which gives us all suffixes of each line.

1

Count all 1s, which sums up all the triangular numbers and converts the result back to decimal.

Answer 12

Ohm v2,  9  7 bytes

·Ω}#ΣΣu

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Explanation

          Implicit input as a string
·Ω        Evaluate until the result has already been seen, pushing intermediate results
  }       Split digits
   #      Range from 0 to N
    ΣΣ    Sum
      u   Convert to string

Answer 13

Befunge-93, 51 bytes

p&>>:25*%:1+*2/v
  |:/*52p00+g00<
00<vp000_@#-1.::g

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James Holderness cleverly reshaped my progarm into a 51-byte form. Thanks!

Answer 14

Pushy, 24 22 21 17 bytes

[sL:R{;Svc^#&1=?i

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Explanation

[sL:R{;Svc^#&1=?i

[           &1=?i   \ Loop until result == 1:
 s                  \   Split last result into digits
  L:  ;             \   For each digit n:
    R{              \       Push the range (1, n) inclusive
       S            \   Sum the ranges
        vc^         \   Delete all stack items, except the sum
           #        \   Print result

Answer 15

R, 70 bytes

f=function(n)"if"(n-1,c(n,f((d=n%/%10^(nchar(n):0)%%10)%*%(d+1)/2)),n)

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Returns the original value as well.

R, 80 bytes

function(n){o={}
while(n-1){n=(d=n%/%10^(nchar(n):0)%%10)%*%(d+1)/2
o=c(o,n)}
o}

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Does not return the original value.

Answer 16

Lua, 91 bytes

function f(n)x=0 for d in((0|n)..""):gmatch"."do x=x+d*(d+1)/2 end print(x)c=x==1or f(x)end

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Answer 17

05AB1E, 20 12 bytes

Saved 2 bytes thanks to caird coinheringaahing

ΔD,þ€iLO}O}}

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Explanation

(old version)

Δþ€iD>*;}OD1›iD,}}1,  Main program
Δ                }    Repeat until there is no changes
 þ                    Push digits of the input number
  €i    }             Apply for each digit
    D>*;              Calculate the triangular number for given digit
         O            Sum all numbers
          D1›iD,}     Print if greater than 1
                  1,  Print 1 at the end

Answer 18

JavaScript, 61 57 bytes

f=a=>a-1?([...a+[]].map(b=>a+=b++*b/2,a=0),a)+' '+f(a):''

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Answer 19

Python 2, 69 bytes

def f(x):
 while x>1:
  x=sum(-~int(z)*int(z)/2for z in`x`)
  print x

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Answer 20

k, 19 bytes

{+/(+/1+!"I"$)'$x}\

Unsurprisingly works similarly to the already posted APL and J solutions

               $x    cast x (implicit var of lambda) to string
   (         )'      apply composition (function train) to each character of string
    +/1+!"I"$        cast character to number, create list from 1 to n, sum it
 +/                  sum triangular numbers
{                }\  iterate lambda until result converges, appending each result

Answer 21

Charcoal, 18 bytes

Ｗ⊖Ｉθ«≔ＩΣ⭆θΣ…·0κθθ⸿

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

   θ                Input
  Ｉ                 Cast to integer
 ⊖                  Decrement
Ｗ   «               Loop while not zero
         θ          Current value
        ⭆           Map over characters and concatenate
             0      Literal character 0
              κ     Current character
           …·       Inclusive range
          Σ         Concatenate
       Σ            Sum of digits
      Ｉ             Cast to string
     ≔         θ    Assign back to value
                θ   Output latest value
                 ⸿  Move to start of next line

Answer 22

dc, 62 bytes

?sj[0soljZ1-[dljr10r^/10%d1+*2/lo+sod1-r0<i]dsixlopdsj1<b]dsbx

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Answer 23

Jelly, 7 bytes

DRFSµÐĿ

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• DRFSµÐĿ: Full program / monadic link.

• ÐĿ: Loop until results are no longer unique (if something other than 1 would occur twice, then the given input does not have a defined result, since it would never reach 1).

• D: Convert from integer to decimal.

• R: Range (1-indexed). Vectorizes.

• F: Flatten and S: Sum (µ just creates a new monadic chain)

Answer 24

dc, 31 bytes

[[I~d1+*2/rd0<m+]dsmxpd1<f]dsfx

The function m computes the digitangular function of its input; f repeats this until the result reaches 1.

Note that we use the input radix to extract digits - this means it will work in any base system, not only decimal.

Demo

for i in 23 72 55 78 613 8392 11111 8592025 999999999
    do echo $i '=>' $(dc -e $i'[[I~d1+*2/rd0<m+]dsmxpd1<f]dsfx')
done

23 => 9 45 25 18 37 34 16 22 6 21 4 10 1
72 => 31 7 28 39 51 16 22 6 21 4 10 1
55 => 30 6 21 4 10 1
78 => 64 31 7 28 39 51 16 22 6 21 4 10 1
613 => 28 39 51 16 22 6 21 4 10 1
8392 => 90 45 25 18 37 34 16 22 6 21 4 10 1
11111 => 5 15 16 22 6 21 4 10 1
8592025 => 117 30 6 21 4 10 1
999999999 => 405 25 18 37 34 16 22 6 21 4 10 1

Answer 25

Python 2, 76 bytes

f=lambda k,s=0:k+s<2and[1]or[k]*(s<1)+f(*[k/10,s,s+k%10*(k%10+1)/2][k<1::2])

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Answer 26

Neim, 8 bytes

ͻtD÷D

Explanation:

ͻ             Start infinite loop
             Split top of stack into each of its characters
  t           Push infinite list of triangular numbers
             For each of the characters, get the nth element in the above list.
              Sum.
     D         Duplicate.
      ÷        If top of stack is equal to 1, break.
       D       Duplicate.
               Implicitly print all elements in the stack, concatenated.

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Formatted output

Answer 27

C, 72 bytes

s;F(n){for(s=0;n;n/=10)for(;n%10;n--)s+=n%10;printf("%d ",s);s<2?:F(s);}

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Answer 28

D, 140 bytes

import std.algorithm,std.range,std.conv,std.stdio;void f(T)(T n){n=n.text.map!(u=>u.to!T-48+(u.to!T-48).iota.sum).sum;n.writeln;if(n-1)n.f;}

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Answer 29

PHP, 71+1 bytes

for(;1<$n="$s"?:$argn;print$s._)for($i=$s=0;$p--||~$p=$n[$i++];)$s+=$p;

Run as pipe with -nR or try it online. (requires PHP 5.3 or later for the Elvis operator)

Answer 30

Add++, 32 bytes

D,f,@,EDBFEREss
+?
y:1
W!,$f>x,O

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Doesn't output the first value

How it works

D,f,@,   - Create a monadic function, f.
         - Example argument: [613]
      ED - Digits;   STACK = [[6 1 3]]
      BF - Flatten;  STACK = [6 1 3]
      ER - Range;    STACK = [[1 2 3 4 5 6] [1] [1 2 3]]
      Es - Sum each; STACK = [21 1 6]
      s  - Sum;      STACK = [28]

           f calculates n's digitangular counterpart

+?       - Take input;     x = 613; y = 0
y:1      - Set y to 1;     x = 613; y = 1
W!,      - While x != y...
   $f>x, -   Call f;       x =  28; y = 1
   O     -   Print x;

Answer 31

Perl 6, 36 bytes

{$_,{sum map {sum 1..$_},.comb}...1}

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Includes the input number in the output list.

Answer 32

K (oK) / K4, 17 bytes

Solution:

{+/,/!:'1+.:'$x}\

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Examples:

{+/,/!:'1+.:'$x}\23
23 9 45 25 18 37 34 16 22 6 21 4 10 1
{+/,/!:'1+.:'$x}\72
72 31 7 28 39 51 16 22 6 21 4 10 1
{+/,/!:'1+.:'$x}\999999999
999999999 405 25 18 37 34 16 22 6 21 4 10 1

Explanation:

Converge over lambda returning all results along the way:

{+/,/!:'1+.:'$x}\ / the solution
{              }\ / lambda {}, converge \
             $x   / string input, 613 => "613"
          .:'     / value (.:) each (') "613" => 6 1 3
        1+        / add 1, 6 1 3 => 7 1 4
     !:'          / til (!:) each (') to generate ranges
   ,/             / flatten list
 +/               / sum list

Answer 33

Pyt, 17 bytes

`Đą△ƩĐ1≠?ŕł:;ŕ↔ŕ↔

Explanation:

                        Implicit input
`         ł             Loop ... while top of stack is not zero
 Đ                      Duplicate the top of the stack
  ą△Ʃ                   Get the digitangular sum of the top of the stack
     Đ1≠                Is it not equal to 1?
        ?  :;           If top of stack is 1, Then continue; Else jump to ':'
         ŕ              Remove top of stack
          ł             Loop back to ` if top of stack is 1 
           :;           If top of stack isn't 1, then continue from ':'
             ŕ↔ŕ↔       Remove top and bottom of stack
                        Implicit print

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