The Squaring Sequence

Question

Each term in the squaring sequence, \$x_n\$, is created by taking \$x_{n-1}\$, squaring it, and removing all but the first four digits.

The sequence always begins with \$x_1 = 1111\$. Squaring this yields \$1234321\$, so \$x_2 = 1234\$

The first few terms are:

1111
1234
1522
2316
5363
...

---

The Challenge

Your task is to, given a non-negative integer \$n\$, calculate \$x_n\$. You may submit a full program which performs I/O, or a function which takes \$n\$ as a parameter.

Your solution can be zero or one indexed, as long as you specify which.

Because all the terms in this sequence are shorter than 5 digits, your code should be as short as possible too. Standard code-golf loopholes apply.

May the best golfer win!

---

Test Cases

Note: These are 1-indexed.

1   -> 1111
8   -> 6840
15  -> 7584
20  -> 1425
80  -> 4717

Answer 1

Japt -mh, 9 bytes

T=Â4î1wT²

Try it

T=Â4î1wT²     :Implicit map of range [0,input)
T=            :Assign to variable T (initially 0)
  Â           :Bitwise NOT of bitwise NOT of (i.e., cast to an integer)
   4î         :Mould to length 4 string
     1w       :  Maximum of 1 and
       T²     :  T squared
              :Implicit output of final result

No Flags, 12 10 bytes

1-indexed, returning a string

@4îX*X}g1ì

Try it

@4îX*X}g1ì     :Implicit input of input of integer U
@              :Function taking an integer X as argument
 4îX*X         :  Mould X*X to length 4
      }        :End function
       g       :Repeatedly take the last element of the following array, pass it through that function, push the result and return the Uth element
        1ì     :  Convert 1 to digit array

Answer 2

CJam, 15 bytes

'14*ri{i2#s4<}*

Uses 0-based indexing.

Try it online!

Answer 3

Perl 6, 36 bytes

{(1111,{+$_².substr(0,4)}...*)[$_]}

Explanation:

{                                 } # bare block lambda
  1111,                  ...        # sequence generator
                            *       # without a limit
       {                }           # lambda used to generate the next value
         $_²                        # start by squaring the previous value
            .substr(0,4)            # take only the first four digits
        +                           # make it numeric ( not necessary )
 (                           )[$_]  # return the requested value

Test:

say {(1111,{+$_².substr(0,4)}...*)[$_]}( 1,8,15,20,80 X- 1 ).perl
# (1111, 6840, 7584, 1425, 4717)

Answer 4

GNU Awk, 41 bytes

Pretty straightforward since you can easily mix math and string manipulations.

{for(x=1111;--$1;x=substr(x*x,1,4));}$1=x

Accepts zero or more positive numbers n (1-indexed), one number per line, on stdin. Outputs results x_n, one result per line, on stdout.

Sample input/output:

% awk -f sqrseq.awk
1
1111
8
6840
15
7584
20
1425
80
4717
%

Answer 5

Matlab, 79, 78 Bytes

function a=s(n)
if n<2;a=1111; else f=s(n-1);a=fix(f^2/10^(3+(f>1e7^.5)));end

Test cases:

s(79) = 2172
s(49) = 8059
s(6)  = 2876

Not an amazing solution. I'm sure there must be a better way to truncate to 4 digits but I don't know today.

Edit: Shaved a byte by setting 0.5 -> .5

Answer 6

Java 8, 77 93 74 71 69 78 bytes

int n=1111;int m=1;while(x>m++){n=Integer.parseInt((n*n+"").substring(0,4));}

x->{int n=1111;int m=1;while(x>m++){n=Integer.parseInt((n*n+"").substring(0,4))‌​;}return n;}

x->{int n=1111;for(;--x>0;){n=Integer.parseInt((n*n+"").substring(0,4));}}

x->{long n=1111;for(;--x>0;){n=Long.valueOf((n*n+"").substring(0,4));}}

x->{long n=1111;for(;--x>0;)n=Long.valueOf((n*n+"").substring(0,4));return n;}

Each repetition makes n the first 4 characters of n*n.

Try Java online!

Post history:

• 77 bytes: initial code (incomplete)

• +16 bytes, by Olivier Grégoire: completed code by making it a Lambda function.

• -19 bytes: replace while with for cycle.

• -4 bytes: used longs instead of ints

• -2 bytes, by Roman Gräf: removed unnecessary brackets

• +9 bytes, missing return statement

Thanks to @OlivierGrégoire and @RomanGräf for pointing out some issues!

Wait, Java beats... (drumroll) Clojure and Matlab here! A big applause to Java please!

Answer 7

Perl, 36 bytes

A different approach from the other Perl solution, leading to slightly shorter code. No command-line arguments are needed (other than the usual version selection argument, -M5.010, which doesn't count against the byte count), meaning that this is the same amount of code but with fewer penalties, giving a better overall score.

say+eval'($&*$&||1x4)=~/(....)/;'x<>

We create a loop Underload-style via repeating and eval-ing a string; I experimented with starting the string in the middle, but starting it at the start turns out to be shortest. We multiply $& (the result of the last regex match) by itself to square it; if the result's zero, we use 1x4 (i.e. 1111; Perl has an operator for repeating things, including digits of a number) instead of the result. Then we regex the first four characters. The whole thing runs in list context due to being inside say, thus the final result of the eval will be the contents of the parentheses of the final match.

Answer 8

Java, 79 67 66 64 bytes

• Version 2.2/64 bytes:

Thanks to @Oliver Grégoire.

int a(int i){i=i<2?1111:a(--i);for(i*=i;i>1e4;)i/=10;return i;}

• Version 2.1/66 bytes:

Thanks to @ETHProduction.

long a(long i){i=i<2?1111:a(--i);for(i*=i;i>1e4;)i/=10;return i;}

• Version 2.0/67 bytes:

Replaced substring and stuff with the idea from @Xanderhall

long a(long i){i=i<2?1111:a(--i);i*=i;for(;i>1e4;)i/=10;return i;}

• Version 1.0/79 bytes:

Although there are shorter solutions I wanted to post one recursive:). And I am the shortest "real" function:). Edit: Seems like I am the shortest now:)))

long a(long i){i=i<2?1111:a(--i);return Long.valueOf((i*i+"").substring(0,4));}

Answer 9

Java 8, 82 59 bytes

Saved 23 bytes thanks to help from commenters @adrianmp and @OlivierGrégoire

n->{int x=1111;while(--n>0)for(x*=x;x>1e4;x/=10);return x;}

Each iteration, we square the number, then loop and shrink it by factors of 10 until it's a 4 digit number. Repeat this process n times and return.

Answer 10

Husk, 11 bytes

¡öd↑4d□1111

Try it online!

Answer 11

Stax, 8 bytes

å\S∩♪öH╠

Run and debug it

Answer 12

JavaScript, 35 bytes

n=>n?(''+f(n-1)**2).slice(0,4):1111

Zero-indexed. Outputs a string if n >= 0 and the number 1111 if n == 0.

Answer 13

><> (Fish), 59 bytes

1e4f*+f*+$v /a,:1%-11!.|
:*:$v!?:-1<$^?)+*f+*f+*f2e6a
 ;n$<

Try it

Explanation

Red loop repeatedly squares. Blue subloop divides by 10 while the value is greater than or equal to 10000.

1e4 is 1111 encoded as base 15. Then repeated f*+ sums it.

a6e2 is 10000 encoded as base 15.

I'm not sure if my base 15 encoding is optimal. Maybe some string encoding with base 255 would be better? But then how to properly add the elements? Hmm this could be a metagolf challenge.

Answer 14

MathGolf, 8 bytes

1kƲ░4<i

Try it online.

Explanation:

1         # Push 1
 k        # Push the input-integer
  Æ       # Pop and loop that many times, using 5 characters as inner code-block:
   ²      #  Square the current integer
    ░     #  Convert it to a string
     4<   #  Pop and leave the first 4 characters/digits
       i  #  Convert it back to an integer
          # (after the loop, the entire stack is joined together and output implicitly)

Answer 15

Go, 79 bytes

func f(n int)int{i,a:=0,1111
for;i<n;i++{for s:=a*a;s>999;s/=10{a=s}}
return a}

Attempt This Online!

Answer 16

Thunno 2, 8 bytes

«¥[s{²4ɱ

Attempt This Online!

0-indexed.

Explanation

«¥[s{²4ɱ  # Implicit input
«¥[       # Push compressed integer 1111
   s{     # Repeat input number of times:
     ²    #  Square top of stack
      4ɱ  #  Take the first four characters
          # Implicit output

Answer 17

Julia 1.0, 42 bytes

!x=parse(Int,("$(1(x<2||!~-x)^2)"^4)[1:4])

Try it online!

Answer 18

QBIC, 43 41 bytes

#1111|:[a-1|f=!A! A=$LEFT$|(!f*f/z$,5)]?A

Way too much going on in here... Time to implement Substring in QBIC.

Explanation:

#1111|                     Define string constant A$ as '1111'
:                          Get CMD line param for N, called 'a'
[a-1|                      FOR 'b' = 1 to 'a'-1
  f=!A!                    Make 'f' be A$ cast to num

             f*f/z         Square 'f', and since we only want the first 4 digits,
                           divide it by 10 (z=10). This prevents that QBASIC 
                           switches to E^ notation because f^2 is too large
    $LEFT$|(!       $|,5)  Cast it to string, take the first 5 positions (QBASIC 
                           adds a leading space on cast-to-num...)
  A=                       And assign to A$
]                          END FOR
?A                         Print the latest value of A$, end

Verified for all test cases.

Answer 19

Wonder, 28 bytes

@:^#0(genc@><""tk4^#0 2)1111

Usage:

(@:^#0(genc@><""tk4^#0 2)1111)79

Simply gets the n th item from an infinite list of squaring sequence numbers. Zero-indexed.

More readable:

@
  iget #0
    (
      genc@
        join "" tk 4 ^ #0 2
    ) 1111

Answer 20

Pip, 13 bytes

Lao:(o*o)@<4o

1-indexed. Takes input as a command-line argument. Try it online!

Explanation

Same algorithm as most of the 1-indexed golflang answers.

               Implicit: a is 1st cmdline arg; o is 1
La             Loop a times:
  o:             Assign this expression to o:
         @<4     Leftmost four characters of
    (o*o)        o squared (parens necessary for precedence)
            o  Print final value of o

The first step, from 1 to 1111, works without any explicit string repetition: string slicing in Pip repeats the string as necessary until the slice index is valid. So "leftmost four characters of 1" gives 1111.

Answer 21

Python 3, 67 Bytes

Indexed at 1

a=1111
for _ in range(int(input())-1):a=str(int(a)**2)[:4]
print(a)

I was originally going to do this as a list comprehension, but I realised that I couldn't receive the last item in the list in order to square it.

Answer 22

Minkolang v0.15, 43 40 33 bytes

'1111'nd?C[2;6ZrI4-[x]r6$Z]N.CxN.

0-indexed

Try it online! (for n=2)

Explanation

'1111'                            push this number
      n                           push input as number
       d                          duplicate
        ?                         check if it is 0:

If input is 0: (the reason 0 has to be handled specially is that otherwise Minkolang gets confused with the for-loops and 0iterations that you get an error)

         C                        start comment
           ... C                  end comment
                x                 delete value at top of stack (ie 0)
                 N.               output as number and end program

If input is > 0:

         C                        gets jumped over (since input is true)
          [               ]       start for-loop with the input as the number of iterations
           2;                      pushes 2, and raises the number by 2 (ie squares it)
             6Z                    converts number to string
               r                   reverses it
                I4-                pushes the length of the stack (ie the string value of the number) and subtracts it by 4 (so we now how many of the string to discard of)
                   [x]             discard the top of the stack by that many times
                      r            reverses stack
                       6$Z         converts it to a number
                                   for-loop ends
                           N.     output stack as number and end program

Answer 23

C, 53 52 bytes

a;s(n){return n?(a=s(n-1))*a/(a>3162?1e4:1e3):1111;}

Zero based. First timer - This is a recursive variation on the earlier solution (expressed in the same way), as requested.

Answer 24

Scala, 44 bytes

Stream.iterate(1111)(n=>n*n+"" take 4 toInt)

This is an expression which evaluates to a Stream[Int], which is an infinte sequence of integers. In scala, Stream[A] extends (Int => A), which means that a Stream is a function from an index to the element at that index.

Explantion:

Stream.iterate( //create an infinite Sequence,
    1111        //starting with 1111,
  )(            //by calculating the next element with the following function:
    n=>
      n*n       //square the previous element
      +""       //convert it to a string
      take 4    //take the first 4 characters and drop the rest
      toInt     //convert the string back to an int
  )

Answer 25

Swift, 80 Bytes

func s(i:Int)->Int{var c=1111;for _ in 0..<i{c*=c;while c>9999{c/=10}};return c}

---

Ungolfed

func squaringSequence(index: Int) -> Int {
    var currentNum = 1111

    for _ in 0 ..< index {
        currentNum *= currentNum

        while currentNum > 9999 {
            currentNum /= 10
        }
    }

    return currentNum
}

Due to the pretty-verbose nature of string manipulation in Swift, in which converting an Int to a String, getting the first four characters, and converting back might look something like this:

let str = String(num);
num = Int(str[str.startIndex ..< str.index(str.startIndex, offsetBy: 4)])!

I'm opting to get the first four digits by dividing by 10 as needed.

Answer 26

Retina, 77 bytes

Sequence is zero-indexed. Doesn't work for n > 1 in TIO because of memory constraints. Byte count assumes ISO 8859-1 encoding.

$
¶1111
{`(?!0¶).+
$*
(?<=1¶(1*))1(?=(1*))
1$1$2
^1

M%`1
}`(\d{4}).*$
$1
0¶

Try it online

Explanation:

$                           # Append a newline and the starting number
¶1111
{`(?!0¶).+                  # Begin loop. If counter is not zero, convert numbers to unary
$*
(?<=1¶(1*))1(?=(1*))        # If the counter >= 1, replace each digit with entire number
1$1$2                       #   (This squares the number)
^1                          # Counter -= 1

M%`1                        # Convert each line's numbers back to decimal
}`(\d{4}).*$                # Keep only first four digits. End loop
$1
0¶                          # Remove the counter

Answer 27

Clojure, 70 bytes

#(first(drop %(iterate(fn[i](read-string(subs(str(* i i))0 4)))1111)))

Zero-based indexing. Iterates a function which returns first 4 characters of the string representation of the squared integer value.

Answer 28

C++ (clang), 52 bytes

void s(int&n){n=--n?s(n),n*n/1e3/(n<3163?:10):1111;}

Port of Lynn's C answer using recursion, is 1 indexed and returns via out parameter.

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Answer 29

brev, 66 bytes

(fn(eif x(strse*((fn(* x x))(f(- x 1)))"^...."(m 0)0)1111))

There's no square unless I import the math lib so (fn(* x x)) had to do. Using a regex to yank the first four digits. This is zero-indexed.

Answer 30

RProgN, 19 Bytes

~{2^''.4m14¢}\*1\C

RProgN continues to be amazing average! Although it easily passes the score of verbose languages, it still chokes in comparison to other golfier languages.

Explained

~{2^''.4m14¢}\*1\C  # Main link
~                    # Zero Space Segment
 {          }       # Define anonymous function
  2^                # Square the top of the stack
    ''.             # Convert to a string.
       4m           # Repeated 4 times.
         14¢        # Push back only the first 4 letters.
             \*     # Multiplied by the implicit input, Creating another anonymous function that calls this one n times.
               1\C  # Push 1 to the stack, and call the chained function.

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