# The Squaring Sequence

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 loopholes apply.

May the best golfer win!

# Test Cases

Note: These are 1-indexed.

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

• Here's a related link :) Dec 2, 2016 at 18:56
• This sequence could find use as a pseudo-random number generator (although it's not very secure).
– user100690
Mar 11, 2021 at 8:51
• May we output as a string?
– user100690
Mar 17, 2021 at 16:18
• @ophact IO is flexible. Mar 18, 2021 at 13:14
• Any OEIS link for this? Mar 21, 2021 at 8:14

# JavaScript (ES7), 4443 36 bytes

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


This is a great example of abusing type coercion: ** converts both its arguments to numbers, and + converts both its arguments to strings unless they're both numbers. This means that f(n)**2+f first converts f(n) to a number and squares it, then concatenates the result with the string representation of f. We can then use .slice to retrieve the first 4 chars of the string.

Here are a few alternate approaches that don't use strings:

f=(n,x=1111)=>x<1e4?--n?f(n,x*x):x:f(n,x/10|0)
f=n=>--n?(x=f(n))*x/(x>3162?1e4:1e3)|0:1111


### Test snippet

let f=n=>--n?(Math.pow(f(n),2)+f).slice(0,4):1111
<input id=I type="number" step="1" min="1" value="1"><button onclick="console.log(f(I.value))">Run</button>

Note: this uses Math.pow because ** isn't supported in all browsers.

• If we can output as a string, then n=>n?(''+f(n-1)**2).slice(0,4):1111 should work for 35 bytes
– user100690
Mar 17, 2021 at 16:20

# 05AB1E, 8 7 bytes

Code:

$Fn4×4£  Explanation: $        # Push 1 and the input
F       # Input times do...
n      #   Square the number
4×    #   Repeat that string 4 times
4£  #   Take the first four characters
# Output the last computed number


Uses the CP-1252 encoding. Try it online!

• Beat me by 20 seconds :(. Dec 2, 2016 at 19:01
• ×4£ can be ∍ now for -2 bytes. Nov 23, 2022 at 8:27

# Python 2, 5146 44 Bytes

I'd like to get rid of the clunky if if possible, but I think an exec could be shorter.. Turns out for the moment that exec is shorter. Wrong again! The recursive function returns. This is one-indexed.

f=lambda n:1111*(n<2)or int(f(n-1)**2[:4])


An aleternative 46-byte solution with exec:

s=1111;exec's=int(s*s[:4]);'*input();print s


An alternative 49-byte recursive solution:

f=lambda n,s=1111:s*0**n or f(n-1,int(s*2[:4]))


Thanks to Flp.Tkc for saving a byte by reminding me that squaring doesn't need exponentiation :)

• Another 46 bytes solution: f=lambda n:1111if n<2else int(f(n-1)**2[:4]) Dec 2, 2016 at 19:36
• @daHugLenny that can actually be 45: repl.it/EejD Dec 2, 2016 at 19:47
• @Flp.Tkc And that can actually be 44 ;)
Dec 5, 2016 at 14:00

((iterate(read.take 4.show.(^2))1111)!!)


It's a 0-based sequence. Usage example: ((iterate(read.take 4.show.(^2))1111)!!) 79 -> 4717.

How it works:

iterate (   ) 1111               -- repeatedly apply a function starting
-- with 1111 and collect the results in a list
-- the function is
(^2)                  -- square
show                     -- turn into string
take 4                      -- take the first 4 chars
read                           -- turn back to number
!!          -- finally pick the nth element from the list


## Perl, 37 bytes

36 bytes of code + -p flag.

$\=1x4;$\=substr$\*$\,0,4while--$_}{  To run it: perl -pe '$\=1x4;$\=substr$\*$\,0,4while--$_}{' <<< 80


# V, 19 bytes

4é1Àñ|C="*"
5|D


Try it online!

This uses 0-based indexing.

Of course, since numbers aren't exactly V's forte, this isn't very golfy. However, it does show one nice advantage V has over vim. You can run a macro 0 times, which is not possible in vim since '0' is a command not a count.

This contains many unprintable characters, so here is a hexdump:

0000000: 34e9 31c0 f17c 4312 3d12 222a 1222 0a1b  4.1..|C.=."*."..
0000010: 357c 44                                  5|D


And here is a readable version:

4é1Àñ|C<C-r>=<C-r>"*<C-r>"
<esc>5|D


Explanation:

4                           " 4 times:
é1                         " Insert a '1'
Àñ                       " Arg1 times:
|                      "   Move to the first character on this line
C                     "   Delete this whole line and enter insert mode
<C-r>=               "   Insert the following evaluated as vimscript:
<C-r>"         "     Insert what we just deleted
*        "     Times
<C-r>"  "     What we just deleted
<esc>                       "   Escape to normal mode
5|                     "   Move to the fifth column on this line
D                    "   And delete until the end of this line
" The second 'ñ' is added implicitly


# Mathematica, 48 bytes

Nest[⌊10^(3-⌊t=2Log[10,#]⌋+t)⌋&,1111,#]&


Unnamed function taking an integer argument; 0-indexed. Uses four three-byte characters ⌊⌊⌋⌋: Mathematica uses either Floor[x] or ⌊x⌋ to round a real number down to an integer, and the latter is generally one fewer byte. The command names in Mathematica for converting integers to strings are too long, so instead we do a mathematical calculation to find the first four digits of x^2: we take the base-10 logarithm of x^2, subtract its integer part, raise 10 back to that power, and multiply by 1000 and round down.

tl;dr: logarithms ftw

# Jelly, 12 9 bytes

-3 bytes thanks to Dennis using 1-based indexing and the ṁ mold/reshape atom. Golfing suggestions welcome! Try it online!

²Dṁ4Ḍ
1Ç¡


Ungolfing

Helper link
²       Square.
D      Integer to decimal (a list of digits).
ṁ4    Mold/reshape list_of_digits to be 4 digits long.
Ḍ   Decimal to integer.

Main link: implicit left argument n
Ç¡   Call the helper link n times.

• This saves 3 bytes with 1-based indexing. Dec 2, 2016 at 19:31
• Uh, I don't think you can use 1 instead of ⁽¡n. Jan 28, 2018 at 10:35
• @EriktheOutgolfer How come? Jan 28, 2018 at 16:09
• @Sherlock9 Oh, you seem to 1-index this sequence? Hm, looks like the code is a bit tricky to understand... Jan 28, 2018 at 16:18

# APL (Dyalog Extended), 13 bytes

-2 (changing the input method) thanks to @Razetime

{⍎4⍴⍕⍵*2}⍣⎕⊢1


Explanation:

{⍎4⍴⍕⍵*2}⍣⎕⊢1     ⍝ apply the dfn input times
1     ⍝ .. with initial argument 1
⍵*2          ⍝ current value squared
⍎4⍴⍕             ⍝ take first 4 digits


Try it online!

• -2 Dec 22, 2020 at 9:36

## Powershell, 73 55 bytes

Huge thanks to TimmyD for shaving off 18 bytes!

Code:

for($A=1111;$args[0]---1;$A=-join"$(+$A*$A)"[0..3]){}$A $A=1111;1..($n=2)|%{[string]$B=[math]::pow($A,2);$A=$B.substring(0,4)};$A

$n is n in xn-1 Explanation and exploded code: $A=1111                            #starting number
$n=4 #n in formula for($i=0; $i -lt$n;$i++) #loop n times { [string]$B=[math]::pow($A,2) #create a new string$B and set it to $A raised to the power of 2$A=$B.substring(0,4) #set$A to the first 4 characters of $B }$A                             #print $A  Some notes: • Powershell lets you assign variables in the same statements where you reference them. For example, 1..($n=4)|% will set $n to 4 and then start a loop that runs$n times. 1 can be changed to any integer and it will loop $n-[your integer]+1 times. • The default data type when using [math]:: in Powershell is a double. In the code above, we have to explicitly cast $B to a string so that we can call .substring() on it because there is no .substring() function for doubles in Powershell.

# Python 2,  44  41 bytes

-3 bytes thanks to xnor (use an integer division to avoid and)

f=lambda n:int(1/n*1111orf(n-1)**2[:4])


repl.it

1-based recursive function.

When n>1 the integer division, 1/n, results in 0, then 0*1111=0 which is falsey, so the right of the or is evaluated, which takes the first four characters of the representation of the square of the n-1th result; this is then cast to an int.

When n=1 the integer division, 1/n, results in 1, then 1*1111=1111, which is truthy, and the int 1111 cast to an int is 1111.

• Good one, ninja'd me by one byte! Dec 2, 2016 at 20:51
• I just looked for your answer and then realised you wrote the challenge! Nice job. Dec 2, 2016 at 20:56
• Nice idea with taking the int outside. If you 1-index, you can do the base case shorter with g=lambda n:int(1/n*1111org(n-1)**2[:4]).
– xnor
Dec 2, 2016 at 21:51
• "Crossed out 44 still looks like 44 :(" Dec 2, 2016 at 22:23
• @Flp.Tkc not as much as it does without the &nbsp;s! Dec 2, 2016 at 22:25

# Groovy, 49 bytes

{x=1111;(it-1).times{x="${x**2}"[0..3] as int};x}  ## Pyke, 10 bytes 1RVX4*4<b  Try it here! # MATL, 14, 13 bytes 1111G:"UV4:)U  Try it online! Explanation: 1111 % Push 1111 G % Push input :" % Input times: U % Square the top of the stack V % Convert it to a string 4:) % Take the first four digits U % Convert it back to a number % Implictly display  • You can use U (square, for numeric input) insted of t* Dec 2, 2016 at 20:12 • @LuisMendo Thanks for reminding of the function I recommended! :P Dec 2, 2016 at 21:21 # R, 585655 53 bytes x=3334;for(e in N<-scan():1)x=x^2%/%10^(3+(x>3162));x  Takes N from stdin. 3334 is practically X_0, which is needed because the for-loop needs to be executed at least once (it would be longer to skip). R really is a terrible language for taking the first four digits of a number, but since the number of cases are limited, we only have to worry about the squares of x<3163 and x>3162, the former yield a 6 digit number, the latter a 7 digit number. The rest is pretty straightforward, %/% divides and ignores the remainder. x is printed to stdout. Saved 2 bytes thanks to @ETHproductions • This is so notrivial. Brilliant! Dec 3, 2016 at 2:02 • Nice one! What would happen if you started with 3334 (or perhaps 3333)? Dec 3, 2016 at 2:16 • @ETHproductions 3333^2 = 11108889 so would yield 1110, and .... as im checking this I see 3334 would work :| . Not sure why I didn't check that anymore. – JAD Dec 3, 2016 at 9:44 # Javagony - 153 bytes Javagony is a restricted version of Java, that doesn't allow any control flow except recursion and try-catch, no for loops, while loops, or if's. Coding in it is a pretty fun exercise, but frustrating. Not that regular Java isn't nearly as frustrating by itself. int a(int i){return a(i-1,1111);}int a(int i,int n){try{int x=1/i;return a(i-1,Integer.parseInt((n*n+"").substring(0,4)));}catch(Exception e){return n;}}  # PHP, 55 52 bytes Saved 3 bytes thanks to @user59178 for($i=1111;$argv[1]--;)$i=substr($i**2,0,4);echo$i;


Run from command line, zero-indexed.

Thanks for not caring about what type my variables are, PHP! Here we simply square the number and trim off everything past the first 4 digits, casually alternating between number and string without a care in the world.

• You could save 3 bytes by using $argv[1]-- as the loop counter. Dec 5, 2016 at 9:36 # cQuents, 14 bytes =1111:(ZZ)[:4]  Try it online! 1-indexed. ## Explanation =1111 first term is 1111 : given n, output nth term, otherwise output full sequence each term equals (ZZ) previous term * previous term [:4] [:4] (python slicing)  • Really nice showcase of this language's abilities! Aug 25, 2022 at 18:18 # Pushy, 20 bytes 1111@:2esL4-:.;Kjk;#  Try it online! Note this is 1-indexed.  % Implicit: N is on stack 1111@ % Push 1111, and then reverse stack to get [1111, n] : % N times do: (this consumes N) 2e % Square last term s % Split into individual digits L4-:.; % Get stack length -4, pop that many times Kj % Join remaining digits (Uses flag "K" for whole stack) k % Set "K" flag to false, so operations only affect last item ; % End loop. # % Output final calculated term  # Brachylog, 18 bytes ,1111:?:{^@[.l4,}i  Try it online! This answer is 0-indexed. ### Explanation ,1111:?:{ }i Iteratively call Input times the predicate in brackets starting with input 1111: ^ Square @[. Output is a prefix of the square .l4, Its length is 4  ## Batch, 82 bytes @set n=1111 @for /l %%i in (1,1,%1)do @set/an*=n&call set n=%%n:~0,4%% @echo %n%  Like Perl, integers are strings, but unlike Perl I can only take the substring of a variable, and taking substrings inside a loop is somewhat awkward. • I think you can leave out the space after @for. Dec 5, 2016 at 13:20 • @YourDeathIsComing 'for' is not recognised as an internal or external command, operable program or batch file. – Neil Dec 5, 2016 at 13:34 # C, 56 bytes a;s(n){for(a=1111;--n;)a=a*a/(a>3162?1e4:1e3);return a;}  One-indexed. • I have a feeling that you can go for recursion... Dec 3, 2016 at 9:05 # Clojure, 76 bytes (defn s[n](if(= n 1)1111(read-string(subs(str(*(s(dec n))(s(dec n))))0 4))))  First Clojure golf (seems like a nice language). This is 1-indexed. Will explain the code later. # C#, 64 60 bytes Saved 4 bytes by following Olivier Grégoire's comment on a Java answer! n=>{int x=1111;for(;n-->1;)for(x*=x;x>1e4;x/=10);return x;};  Previous version (64 bytes): n=>{int x=1111;while(n-->1){x*=x;while(x>9999)x/=10;}return x;};  Full program with ungolfed method and test cases: using System; namespace SquaringSequence { class Program { static void Main(string[] args) { Func<int, int> f = n => { int x = 1111; while (n-- > 1) { x *= x; while (x > 9999) x /= 10; } return x; }; // test cases: Console.WriteLine(f(1)); // 1111 Console.WriteLine(f(8)); // 6840 Console.WriteLine(f(15)); // 7584 Console.WriteLine(f(20)); // 1425 Console.WriteLine(f(80)); // 4717 } } }  • +1 for the dark goes-to-operator n-->1 Dec 3, 2016 at 0:43 # Ruby, 47 bytes First golf! Saves bytes with -n option (but still count as 1! :)). a=1111;$_.to_i.times{a="#{a*a}"[0,4].to_i};p a


0-indexed. To run it:

ruby -ne 'a=1111;\$_.to_i.times{a="#{a*a}"[0,4].to_i};p a' <<< 80

• Welcome to the site, and nice first answer! One nitpick though, technically this is 47 bytes because of our policy of counting command line flags towards the byte count. Other than that, it looks good to me! Dec 3, 2016 at 18:26
• Thanks! Didn't know the rules, answer changed! Dec 3, 2016 at 18:29

# Pyth, 13 12 bytes

Thanks to @Jakube for -1 byte

us<*4*GG4Q1


A program that takes input of a 1-indexed integer and prints the result.

Test suite

How it works

us<*4*GG4Q1  Program. Input: Q
u         Q1  Execute the following Q times, starting at 1, with variable G:
*GG      Yield G*G
Convert to string
*4           Repeat 4 times
<      4      Yield first 4 characters
s              Convert to integer
Implicitly print

• *GG instead of ^G2<space> Dec 4, 2016 at 20:30

# GolfScript, 15 bytes

~1111\{.*4<~}*


Try it online!

0-indexed

~                 # Parse n to a number
1111             # Push 1111
\{      }*   # Execute this block n times
.*         # Square the number
# Parse it to a string
4<      # Get the first 4 digits
~     # Parse it back to a number

• Nice answer. Is it not possible to take integer input directly? Oct 11, 2020 at 4:44
• @Razetime No, in the input and output section of golfscript.com/golfscript/tutorial.html , it says: "all input from stdin is read first and placed as a string onto the stack". But this site also doesn't mention that the numbers can be complex, so maybe there is a way. If there is, it cannot be shorter than ~. Oct 11, 2020 at 14:51

# Vyxal, 6 5 bytes

Ṫ(²4Ẏ


Explanation:

Ṫ      # Pop, push 1 then the input
(     # [Input] times:
²    #   Square
4Ẏ  #   Get the first 4 digits


# TI-Basic, 31 bytes

Input N
1111
For(I,2,N
int(Ans²/₁₀^(-3+int(log(Ans²
End
Ans


1-indexed input. - represents the negative symbol.

# Excel VBA, 42 bytes

v=1111:For i=2To[A1]:v=Left(v^2,4):Next:?v


Input is in cell A1 of the active sheet. Code is run in the immediate window. Output is to the immediate window. Here's the result for 80 as an input: