Triple countdown sequence

Question

Let's start with the natural numbers

[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,100...

Now we will make a new list, replacing each natural number n with a countdown from n to 1.

[1,2,1,3,2,1,4,3,2,1,5,4,3,2,1,6,5,4,3,2,1,7,6,5,4,3,2,1,8,7,6,5,4,3,2,1,9,8,7,6,5,4,3,2,1,10,9,8,7,6,5,4,3,2,1,11,10,9,8,7,6,5,4,3,2,1,12,11,10,9,8,7,6,5,4,3,2,1,13,12,11,10,9,8,7,6,5,4,3,2,1,14,13,12,11,10,9,8,7,6..

Now we will repeat the process.

[1,2,1,1,3,2,1,2,1,1,4,3,2,1,3,2,1,2,1,1,5,4,3,2,1,4,3,2,1,3,2,1,2,1,1,6,5,4,3,2,1,5,4,3,2,1,4,3,2,1,3,2,1,2,1,1,7,6,5,4,3,2,1,6,5,4,3,2,1,5,4,3,2,1,4,3,2,1,3,2,1,2,1,1,8,7,6,5,4,3,2,1,7,6,5,4,3,2,1,6..

And now we will do it a third time:

[1,2,1,1,1,3,2,1,2,1,1,2,1,1,1,4,3,2,1,3,2,1,2,1,1,3,2,1,2,1,1,2,1,1,1,5,4,3,2,1,4,3,2,1,3,2,1,2,1,1,4,3,2,1,3,2,1,2,1,1,3,2,1,2,1,1,2,1,1,1,6,5,4,3,2,1,5,4,3,2,1,4,3,2,1,3,2,1,2,1,1,5,4,3,2,1,4,3,2,1...

This is the triple countdown sequence.

Your task is to implement the triple countdown sequence. You may either take input and give the value of the sequence at that index, or you may simply output the terms starting from the beginning without stop.¹

This is code-golf so answers will be scored in bytes with fewer bytes being the goal.

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^{1: Both zero and one indexing are permitted. Outputting the terms without halt includes lazy structures which can produce an infinite number of terms.}

Answer 1

APL (Dyalog Extended), 12 bytes

Returns the nth number in the series.

⊢⊃1(∊…¨⍨)⍣3⍳

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⊢ the argument

⊃ picks the element from

1(…)⍣3⍳ repeat thrice with 1 as constant left argument and indices 1 through n as initial right argument:

 ∊ enlist (flatten)

 … the sequence

  ¨ for each

   ⍨ with swapped arguments, i.e. counting down from that number to 1

Answer 2

05AB1E, 6 bytes

∞3FLí˜

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Infinitely outputs, thanks to Kevin Crujisen

05AB1E, 6 bytes

L3FLí˜

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I doubt it could be beaten

Answer 3

Haskell, 34 33 32 bytes

-1 byte thanks to Wheat Wizard!
-1 byte thanks to pxeger!

[1..]>>=f>>=f>>=f
f x=[x,x-1..1]

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Answer 4

jq, 50 bytes

def f:[range(.)+1]|reverse[];
1|while(1;.+1)|f|f|f

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funny how similar the second line is to the Zsh answer.

-1 from ovs.

Answer 5

Zsh, 37 bytes

c()xargs -i seq {} -1 1
seq inf|c|c|c

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Full program which outputs infinitely, one per line (ATO link includes a wrapper to halt after N terms).

"Implicit" "list" "flattening" FTW! (Zsh doesn't actually have any of these features - it's just text!)

c is a function that outputs the sequence from {} to 1 (with a step of -1) for each {} in the input (xargs -i). Then output "all" integers and pipe that to c three times.

Answer 6

JavaScript (V8), 60 bytes

for(a=1;;a++)for(i=a;--i;)for(j=i;--j;)for(k=j;--k;)print(k)

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Boring trivial solution.

Answer 7

Husk, 6 bytes

!4¡ṁṫN

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outputs an infinite list.

Answer 8

Wolfram Language (Mathematica), 36 bytes

Do[Print[0;],{,∞},#,#,#]&@{,,1,-1}

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Prints elements indefinitely.

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A somewhat more interesting solution which generalizes to \$n\$-countdown sequences:

Wolfram Language (Mathematica), 48 bytes

Nest[gArray[g,#,{#,1}]&,Print,3]@i~Do~{i,∞}

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Answer 9

PowerShell Core, 33 bytes

Outputs terms without stopping!

filter s{$_..1}
for(){++$x|s|s|s}

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Answer 10

Jelly, 8 bytes

r1)⁺⁺Fị@

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Because it is extremely cumbersome to produce infinite output in Jelly, it instead applies countdown 3 times to the input's range and outputs the input-th item. The part without indexing (6 bytes) ties with 05AB1E's equivalent.

r1)⁺⁺Fị@    Monadic main link. Input = n
r1)         [x,x-1..1] for each number x (autorange)
   ⁺⁺       Do the above; do the above (no autorange)
     Fị@    Flatten and get the nth item

Answer 11

Scala, 60 bytes

Stream.from(1)flatMap>flatMap>flatMap>
def> =(_:Int)to(1,-1)

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A less trivial solution should be coming soon.

> is a function that takes an Int and returns a range going from that number to 1. We then take an infinite Stream starting at 1, and apply > to each of those numbers thrice, flattening each time.

Answer 12

Raku, 33 bytes

(1..*,{.flatmap:{$_...1}}...*)[3]

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This is a lazy list of countdown sequences, where the first element is the list of natural numbers, the second is the countdown sequence, the third is the double countdown sequence, and the fourth is the triple countdown sequence, et cetera. The triple countdown sequence is the one at index 3, which is returned.

Answer 13

Wolfram Language (Mathematica), 45 bytes

Flatten[r@r@r@Range@#][[#]]&
r=Range[#,1,-1]&

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-23 bytes from @att

Answer 14

R, 59 51 bytes

`[`=`for`;while(F<-F+1)i[F:1,j[i:1,k[j:1,show(k)]]]

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Prints the resulting sequence infinitely.

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Less boring:

R, 68 bytes

`!`=unlist;s=sapply;while(F<-F+1)cat(!s(!s(F:1,seq,,-1),seq,,-1),"")

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Answer 15

C++ (gcc), 101 bytes

#include<cstdio>
int a,x,y,z;main(){for(;;)for(x=a++;x--;)for(y=x;y--;)for(z=y;z;)printf("%d,",z--);}

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Answer 16

Factor + combinators.extras lists lists.lazy, 58 bytes

[ 1 lfrom [ [ 1 [a,b] >list ] lmap-lazy lconcat ] thrice ]

Running in the listener, as lmap-lazy postdates build 1525, the one TIO uses. Note that output has been limited to 30 so I could actually take the screenshot. This limitation is not present in the code itself.

Explanation

It's a quotation (anonymous function) that returns an infinite lazy list of the triple countdown sequence.

• 1 lfrom An infinite lazy list of the natural numbers.
• [ ... ] thrice Apply a quotation (to the natural numbers) three times.
• [ ... ] lmap-lazy Use a quotation to map over a lazy list.
• 1 [a,b] >list Create a list from an input number to one. e.g. 5 -> L{ 5 4 3 2 1 }
• lconcat Concatenate a list of lists into a single list.

Answer 17

C (gcc), 69 bytes

Nested Ternary operator solution.

a, b and c are the triple countdown state from fastest to slowest, and d is the increasing count. a is printed for every step in the countdown. (make that a+1 thanks to AZTECCO

Thanks to AZTECCO for -10 bytes!! And -2 more!!! It took me an embarrassingly long time to figure out what's going on with the last suggestion, that's some serious golfing.

a,b,c,d;main(){for(;;a=!a?b=!b?c-=c?1:--d:b-1:a-1)printf("%d,",a+1);}

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Answer 18

Swift, 103 bytes

let a:([Int])->[Int]={$0.flatMap{(1...$0).reversed()}};(1...).forEach{a(a(a([$0]))).forEach{print($0)}}

Functional programming in Swift, excepting the print call, which obviously is non-pure.

a is a closure that converts an array of integers to an array of their countdown, which is called three times for every number.

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Answer 19

Nibbles, 6 bytes

<$/, 3,~.@\, 

explanation

<$      # first int input elements
  /     # fold
   ,3   # [1,2,3] (any list of length 3 would work)
   ,~   # [1,2,...inf] (initial value to fold)
   .@   # map on accumulator
     \, # reverse of list from 1 to
        # implicit $ (the element of list to map)

Answer 20

Python 3, 72 bytes

def c(x,i):i>0<x!=(print(x),c(x-1,i),c(x-1,i-1))
i=1
while 1:c(i,2);i+=1

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-19 bytes thanks to Wheat Wizard by using even more recursion
-1 byte thanks to ovs

Answer 21

JavaScript (V8), 51 bytes

e=f=>i=>{for(;--i;)f(i)};e(_=>e(e(e(print)))(-_))``

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Answer 22

MathGolf, 9 bytes

)╒3æ╒mx─§

Outputs the 0-based value.

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Explanation:

)          # Increase the (implicit) input-integer by 1 (work-around for input 0)
 ╒         # Pop and push a list in the range [1,input+1]
  3æ       # Loop 3 times, using four characters as inner code-block:
    ╒      #  Convert each integer in the list to a [1,n] ranged list
     m     #  Map over each list:
      x    #   Reverse the list
       ─   #  Flatten the list of lists
        §  # After the loop, get the value at the (implicit) input as index
           # (after which the entire stack is output implicitly as result)

Answer 23

C++ (gcc), 132 123 114 113 112 bytes

#include<argp.h>
main(int a,int b){if(a<2)for(;;)main(2,a++);while(--b)if(a<4)main(a+1,b);else printf("%d,",b);}

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Thanks to @wheat-wizard for tip on removing all whitespace

Answer 24

Julia 1.0, 56 bytes

~i=i:-1:1
!i=[.~i...;]
for i=1:-1%UInt println.(!!!i)end

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Answer 25

Japt, 13 bytes

There's gotta be a better way than this.

Outputs the nth term, 0-indexed.

gÈc!õ1}g3NËôÄ

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Answer 26

Perl 5, 48 bytes

sub d{sub c{map{reverse 1..$_}@_}sub{c c c++$i}}

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Returns an anonymous sub that can be called repeatedly to obtain the next list of items in the sequence. Ungolfed and commented:

sub d {
    sub c {
        # This is where the magic happens. c(1) returns (1), c(2) returns (2, 1).
        # c(1, 2, 3) returns (1, 2, 1, 3, 2, 1)
        return map {reverse 1 .. $_} @_;
    }
    # 3 nested calls to c() to get the triple countdown
    return sub{c(c(c(++$i)))};
}

Answer 27

Ruby, 48 bytes

x=0
loop{x-=1;eval"(x...0).map{|x|"*3+"p -x}}}"}

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• outputs infinitely

We eval "(x...0).map{|x|"*3 which becomes a depth 3 loop

x is decremented thus negative so that we can use range x..0

Then we put -x

Answer 28

Vyxal, 9 bytes

ɾṘ3(vɾf)Ṙ

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Answer 29

Charcoal, 36 bytes

Ｎθ≔⁰ηＷ›θＬυ«≦⊖ηＦ…η⁰Ｆ…κ⁰Ｆ…λ⁰⊞υμ»Ｉ±§υ⊖θ

Try it online! Link is to verbose version of code. Outputs the 1-indexed nth value. Explanation: Based on @AZTECCO's answer, generates the triple countup sequence for the negated natural numbers, and then negates the final value to produce the desired result.

Ｎθ

Input n.

≔⁰η

Start with no terms.

Ｗ›θＬυ«

Repeat until we have at least n terms.

≦⊖η

Get the next negated natural number.

Ｆ…η⁰Ｆ…κ⁰Ｆ…λ⁰⊞υμ

Generate the triple countup for that number.

»Ｉ±§υ⊖θ

Output the nth term, negating to give the desired result.

Answer 30

Burlesque, 14 bytes

r1{m{ro^p}}3E!

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r1    # Range 1..inf
{
 m{   # Apply to each
  ro  # Range from 1..N
  ^p  # Push in reverse
 }
}
3E!   # Run 3 times