Let's converge to 9!

Question

Given an integer n > 2, print or return the smallest non-negative integer k such that a(n, k) = 9, where a(n, k) is defined by:

• a(n, 0) = n
• a(n, k+1) =
  • a(n, k) / 2 + 1 if a(n, k) is even
  • the sum of the digits of a(n, k)² (in base 10) if a(n, k) is odd

Examples

For n = 5, the expected output is k = 4:

a(5, 0) = 5
a(5, 1) = 7  (5² = 25 and 2 + 5 = 7)
a(5, 2) = 13 (7² = 49 and 4 + 9 = 13)
a(5, 3) = 16 (13² = 169 and 1 + 6 + 9 = 16)
a(5, 4) = 9  (16 / 2 + 1)

For n = 40, the expected output is k = 2:

a(40, 0) = 40
a(40, 1) = 21 (40 / 2 + 1)
a(40, 2) = 9  (21² = 441 and 4 + 4 + 1 = 9)

Clarifications and rules

• The input is guaranteed to be greater than 2.
• Your program should theoretically work for any value of n. (In practice, it may be limited by the maximum integer size supported by your language.)
• k may be either 0-indexed or 1-indexed. Please state it in your answer.
• This is code-golf, so the shortest answer in bytes wins!

First values

Below are the first values from n = 3 to n = 422, with k 0-indexed. (For 1-indexing, just add 1 to these values.)

 1  2  4  3  3  5  0  4  3  4  2  6  1  1  6  5  5  4  1  5  2  3  3  7  6  2  3  2  2  7
 6  6  5  6  6  5  1  2  2  6  6  3  1  4  3  4  4  8  1  7  6  3  5  4  6  3  2  3  3  8
 7  7  3  7  4  6  6  7  5  7  6  6  6  2  4  3  3  3  6  7  3  7  2  4  7  2  6  5  6  4
 7  5  2  5  6  9  6  2  3  8  2  7  1  4  6  6  6  5  1  7  4  4  3  3  7  4  3  4  2  9
 6  8  6  8  6  4  6  8  2  5  3  7  6  7  3  8  2  6  7  8  6  7  5  7  6  7  4  3  3  5
 6  4  3  4  4  4  6  7  6  8  3  4  6  8  7  3  6  5  6  8  3  3  2  7  6  6  5  7  6  5
 7  8  2  6  3  3  6  6  6  7  4 10  6  7  3  3  6  4  1  9  2  3  3  8  7  2  6  5  2  7
 7  7  6  7  3  6  7  2  4  8  3  5  6  5  6  4  2  4  6  8  3  5  6  4  7  5  2  3  6 10
 7  7  3  9  2  7  1  9  5  7  6  5  6  7  4  9  6  3  6  6  3  4  2  8  7  7  6  8  6  4
 7  9  4  3  3  7  7  8  3  9  4  7  6  8  3  6  6  8  7  7  7  8  6  5  7  4  6  4  2  6
 7  7  6  5  3  4  7  5  4  5  3  5  7  7  6  8  2  7  1  9  6  4  6  5  7  7  2  9  6  8
 7  4  3  7  4  6  6  7  6  9  3  4  6  4  2  3  3  8  1  7  6  7  2  6  7  8  3  7  5  6
 7  8  2  9  3  3  6  7  6  4  4  4  6  7  6  7  6  7  6  8  7  5  6 11  7  7  3  8  4  4
 7  4  6  7  3  5  6  2  2 10  6  3  6  4  3  4  4  9  7  8  3  3  6  7  7  6  4  3  6  8

Answer 1

Perl 6, 41 bytes (40 chars)

{+($_,{$_%2??[+] $_².comb!!$_/2+1}...9)}

Try it online!

This uses 1-indexing of k's, so it gives 1 higher answers than the examples in OP. If this is not what the 1-indexing means, I'll have to add 1 byte more.

Explanation: It's an anonymous function. We just use the Perl 6's facility for generating lists using recursion :—). It looks like this: (first element),(block that takes the previous element and gives the next)...(end condition). In this case, the first element is $_ (argument of the main function) and the end condition is 9 (fulfilled when we generate a 9). In the middle block, we use $_ to refer to its argument ( = the previous element of the sequence). The ?? !! is the old ternary operator (better known as ? :). Finally, we take the length of this list by forcing numerical context by +(...).

The last weird thing here is the sum of digits. Numbers are Cool (behave both like strings and numbers), so we use a string method .comb on $_² (give list of characters = digits), then adding the characters up (that converts them back to numbers).

Answer 2

Jelly, 17 bytes

²DSµH‘$Ḃ?ßµ-n9$?‘

Try it online!

Straight-forward approach. Uses 0-based indexing.

Explanation

²DSµH‘$Ḃ?ßµ-n9$?‘  Input: n
               ?   If
            n9$      n != 9
          µ        Then
        ?            If
       Ḃ               n % 2 == 1
   µ                 Then
²                      Square
 D                     Decimal digits
  S                    Sum
      $              Else
    H                  Halve
     ‘                 Increment
         ß           Call recursively
                   Else
           -         The constant -1
                ‘  Increment

Answer 3

Python 2, 129 126 76 68 67 64 54 53 bytes

-3 bytes thanks to Jonathan Frech. -8 bytes thanks to Maltysen. -7 bytes thanks to Jonathan Allan. -1 byte thanks to Mr. Xcoder.

f=lambda n:n-9and-~f(n%2*sum(map(int,`n*n`))or 1+n/2)

Try it online!

From somebody who probably doesn't know enough math, this seems completely arbitrary. :P

Answer 4

Husk, 13 bytes

€9¡?o→½ȯΣd□¦2

This is 1-indexed. Try it online!

Explanation

Nothing too fancy here.

€9¡?o→½ȯΣd□¦2  Implicit input, say n = 5
  ¡            Iterate the following function:
   ?       ¦2   If divisible by 2,
    o→½         then halve and increment,
       ȯΣd□     else square, take digits and get their sum.
               This gives an infinite sequence: [5,7,13,16,9,9,9,9,9..
€9             1-based index of 9; print implicitly.

Answer 5

Mathematica, 58 bytes

1-indexed

If[#!=9,#0@If[OddQ@#,Total@IntegerDigits[#^2],#/2+1]+1,0]&

Try it online! (in order to work on Mathics, Tr is replaced with Total)

here is -1 byte version by @JungHwanMin (but it doesn't work on mathics so I kept both)

Mathematica, 57 bytes

If[#!=9,#0@If[2∣#,#/2+1,Total@IntegerDigits[#^2]]+1,0]&

Answer 6

JavaScript (ES6), 59 50 bytes

0-indexed.

f=n=>n-9&&f(n%2?eval([...""+n*n].join`+`):n/2+1)+1

---

Try it

o.innerText=(
f=n=>n-9&&f(n%2?eval([...""+n*n].join`+`):n/2+1)+1
)(i.value=5);oninput=_=>o.innerText=f(+i.value)

<input id=i min=3 type=number><pre id=o>

---

Explanation

The first thing we do is calculate n-9. If n==9 then that, obviously, gives 0 and things stop there. If n!=9 then n-9 will give a non-zero value which, being truthy, means we can continue on through the logical AND. We call the function again, passing a new n to it, calculated as follows:

n%2?

If n modulo 2 is truthy - i.e., n is odd.

[...""+n*n]

Multiply n by itself, convert it to a string and destructure that string to an array of individual characters (digits).

 .join`+`

Rejoin the characters to a string using +, giving us a mathematical expression.

eval(                   )

Evaluate that expression, giving us the sum of the digits of n*n.

:n/2+1

If n%2 is falsey (i.e., n is even) then we simply divide n by 2 and add 1.

To the result of calling the function again, we then add 1. So, using an initial input of 5, the process goes as follows:

f(5)
= -4&&f(7)+1
= -2&&(f(13)+1)+1
=  4&&((f(16)+1)+1)+1
=  7&&(((f(9)+1)+1)+1)+1
=     (((0+1)+1)+1)+1
= 4

Answer 7

Jelly,  16  15 bytes

-1 byte thanks to miles (use of ternary if)

²DSµH‘µḂ?_9$пL

A monadic link taking and returning numbers.
1-indexed

Try it online! or see a test-suite (coerces results to be 0-indexed and formats like OP code-block)

How?

²DSµH‘µḂ?_9$пL - Link: number, n
            п  - collect results in a list while:
           $    -   last two links as a monad:
         _9     -     subtract nine
        ?       -   if:
       Ḃ        -     bit - current loop input modulo by 2 (1 if odd, 0 if even)
   µ            -   ...then:
²               -     square the current loop input
 D              -     cast to a list of its decimal digits
  S             -     sum
      µ         -   ...else:
    H           -     halve current loop input
     ‘          -     increment
              L - length (get the number of results collected
                -         - this includes the 9, so is 1-indexed w.r.t. k)

Answer 8

Haskell, 62 59 bytes

f 9=0
f a=1+f(cycle[div a 2+1,sum[read[d]|d<-show$a^2]]!!a)

Try it online!

Edit: -3 bytes thanks to @Ørjan Johansen.

Answer 9

Perl 5, 56 + 1 (-n) = 57 bytes

$|++,$_=$_%2?eval$_**2=~s/./+$&/gr:1+$_/2while$_-9;say$|

Try it online!

Answer 10

05AB1E, 16 bytes

[Ð9Q#Èi2÷>ënSO]N

Try it online!

Explanation

[                  # start a loop
 Ð                 # triplicate current number
  9Q#              # if it equals 9, break
     Èi            # if even
       2÷>         # divide by 2 and increment
          ë        # else
           n       # square
            SO     # sum digits
              ]    # end loop
               N   # push the iteration counter N

Answer 11

VB.NET (.NET 4.5.2), 107 + 20 (imports) = 117 bytes

Requires Imports System.Linq

Function A(n)
While n<>9
n=If(n Mod 2=0,n/2+1,CStr(n^2).Sum(Function(c)Val(c)))
A+=1
End While
End Function

Function which takes n as an integer input and returns a 0-based k.

Ungolfed:

Function A(n) ' input/output types are Object, but we will be casting to integer
    'A = 0 ' VB will create an implicit variable with the same name as the function

    ' loop until a(n, k) = 9
    ' using n as the variable to store a(n, k)
    While n <> 9

        n = If(n Mod 2 = 0, ' equivalent to c# ternary ?: operator

            n / 2 + 1, ' even case

            CStr(n ^ 2).Sum(Function(c) Val(c)))
            ' odd case
            ' cast number to string
            ' then convert each char to the number it represents
            ' and do a linq sum

        A += 1 ' Object + Integer will coerce to an integer
    End While

    ' Where's the return?
    ' That implicit variable with the matching name will get returned if there's no explicit return
End Function

Answer 12

Golfscript, 34 bytes

Try it online!

I really need a better way than I have to add up the digits of a number.

~{9-}{.2%{.*`{+48-}*48-}{2/)}if}/,

Answer 13

Pyth,  23  22 bytes

For now, this is a recursive function but I'll try to switch to .W (functional while) to save bytes instead.

L&-b9hy|*%b2sj^b2Th/b2

Try it here! (with additional code to call the function - use y<your_number> - without spaces)

Answer 14

Java 8, 110 98 bytes

n->{int k=0,s;for(;n!=9;k++){s=0;for(int c:(n*n+"").getBytes())s+=c-48;n=n%2<1?n/2+1:s;}return k;}

0-indexed

Explanation:

Try it here.

 n->             // Method with integer as both input and return-type
   int k=0,      //  Result-integer `k` starting at 0
       s;        //  Sum-integer
   for(;n!=9;    //  Loop (1) as long as `n` is not 9
        k++){    //    And increase `k` by 1 after every iteration
     s=0;        //   Reset sum `s` to 0
     for(int c:(n*n+"").getBytes())
                 //   Do `n*n` and inner loop (2) over the digits as characters
       s+=c-48;  //    And increase the sum `s` with these digits
                 //   End of inner loop (2) (implicit / single-line body)
     n=n%2<1?    //   If `n` is even:
        n/2+1    //    Change `n` to `n/2+1`
       :         //   Else:
        s;       //    Change `n` to sum `s`
  }              //  End of loop (1)
  return k;      //  Return the result `k`
}                // End of separated method (2)

Answer 15

Clojure v1.8, 124 113 112 bytes

0-indexed

(fn[n](loop[a n k 0](if(= a 9)k(recur(if(even? a)(+(/ a 2)1)(apply +(map #(-(int %)48)(str(* a a)))))(inc k)))))

Try it online!

Explanation

(loop[a n k 0](if(= a 9)...))  Loop until a=9
(if(even? a)(+(/ a 2)1)...)    If even, a(n, k) / 2 + 1 if a(n, k)
(if(even? a)...(apply +(map #(-(int %)48)(str(* a a)))))  If odd, calculate the sum of digits of a(n, k)²
#(-(int %)48)                  Convert character to number

Answer 16

Pyth, 18 bytes

tl.u?%N2sj*NNTh/N2

Try it online: Demonstration

Explanation:

tl.u?%N2sj*NNTh/N2
  .u                 apply the following function to the input, 
                     until it runs into a fixed point
    ?%N2                if value % 2 == 1:
          *NN               value * value
         j   T              convert to digits
        s                   sum
                        else:
               /N2          value / 2
              h              + 1
 l                   get the length of all visited values
t                     - 1

Answer 17

Japt, 22 21 17 bytes

-9©ÒßUg[U/2ÄU²ìx]

Try it

-9©ÒßUg[U/2ÄU²ìx]     :Implicit input of integer U
-9                    :Subtract 9
  ©                   :Logical AND with
   Ò                  :  Negate the bitwise NOT of
    ß                 :    Recursive call with argument
     Ug               :      Index U into (0-based & modular)
       [              :        Array containing
        U/2Ä          :         1. U/2+1
            U²        :         2. U squared
              ì       :            To digit array
               x      :            Reduce by addition
                ]     :        End array

Answer 18

AWK, 73 bytes

{for(j=$1;j!=9;++k){l=split(j%2?j^2:j/2+1,g,X);for(j=0;l;)j+=g[l--]}}$0=k

Zero index as it saves a few bytes.

{for(j=$1;j!=9;++k){     # main loop until 9
l=split(                 # number into digits
j%2?j^2:j/2+1,g,X);      # maths, split into array g
for(j=0;l;)j+=g[l--]     # add digits together
}}$0=k                   # set output to iterations

Answer 19

iogii 18 bytes

i;2%$*\_$2/)if?wz{

Try it! (0 indexed)

If you want to see all test cases we must change \ to \, otherwise it would mean transpose instead of digits: All test cases

I don't really know how to explain it beyond saying what each character does and how the parse would look in something like Haskell:

i iterate
; mdup (put orig value after putting function applied to value)
2
%
$ arg
*
\ digits
_ sum
$ arg
2
/
) inc
if
? isDebut?
w takeWhile
z last
{ whole numbers

last (takeWhile wholeNumbers (isDebut (iterate implicitInput (\i ->
    if (i%2) then (i/2+1) else sum (digits (i*i)) end
))))

The mdup is used instead of arg again so that the end block of iterate can be implicit (first time stack size goes back to 1 is after the if).

Using is debut instead of =9 saves a byte

Answer 20

Uiua, 29 bytes

⧻[⍢(⨬(+1÷2|/+⍜°⋕≡¤ⁿ2)⊸◿2.)≠₉]

Try it!

1-indexed (if 2-indexing was allowed, we could replace ⍢ with ⍥ and ≠₉ with ∞ for -1 byte)

⧻[⍢(⨬(+1÷2|/+⍜°⋕≡¤ⁿ2)⊸◿2.)≠₉]
  ⍢(                       )≠₉   # while n≠9: 
 [                        .   ]  #   push n into an array
     ⨬(    |         )⊸◿2       #   switch on n%2:
      (    |                     #     0: 
       +1÷2                      #       n ← n/2+1
           |          )          #     1:
                    ⁿ2           #       square
              ⍜°⋕≡¤             #       get digits
            /+                   #       and sum
⧻                                # get the length of the array

Answer 21

Vyxal, 10 bytes

½›n²∑"i)İL

Try it Online!

        cL # Count the unique values from...
-------)   # Applying the following function
      i    # Index, modularly, into
     "     # The pair of
½›         # n/2+1 if n is even
  n²∑      # And the sum of the digits of n^2 if n is odd

Since İ excludes the initial value, all we have to do is collect values until we reach a fixed point, with no explicit check for 9.