16
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Introduction

The sign of a number is either a +, or a - for every non-zero integer. Zero itself is signless (+0 is the same as -0). In the following sequence, we are going to alternate between the positive sign, the zero and the negative sign. The sequence starts with 1, so we write 1 with a positive sign, with zero (this one is weird, but we just multiply the number by 0) and the negative sign:

1, 0, -1

The next number is 2, and we do the same thing again:

2, 0, -2

The sequence eventually is:

1, 0, -1, 2, 0, -2, 3, 0, -3, 4, 0, -4, 5, 0, -5, 6, 0, -6, 7, 0, -7, ...

Or a more readable form:

a(0) = 1
a(1) = 0
a(2) = -1
a(3) = 2
a(4) = 0
a(5) = -2
a(6) = 3
a(7) = 0
a(8) = -3
a(9) = 4
...

The Task

Given a non-negative integer n, output the nth term of the above sequence. You can choose if you use the zero-indexed or one-indexed version.

Test cases:

Zero-indexed:

a(0) = 1
a(11) = -4
a(76) = 0
a(134) = -45
a(296) = -99

Or if you prefer one-indexed:

a(1) = 1
a(12) = -4
a(77) = 0
a(135) = -45
a(297) = -99

This is , so the submission with the smallest number of bytes wins!

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  • \$\begingroup\$ Is it Ok if you start with [0, 0, 0, -1, 0, 1... \$\endgroup\$ – Blue May 28 '16 at 16:54
  • \$\begingroup\$ @muddyfish no sorry, it has to start with 1. \$\endgroup\$ – Adnan May 28 '16 at 16:59

47 Answers 47

6
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Jelly, 7 bytes

+6d3’PN

Zero-indexed. Test cases here.

Explanation:

+6      Add 6:     x+6
d3      Divmod:    [(x+6)/3, (x+6)%3]
’       Decrement: [(x+6)/3-1, (x+6)%3-1]
P       Product    ((x+6)/3-1) * ((x+6)%3-1)
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6
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JavaScript ES6, 18 bytes

n=>-~(n/3)*(1-n%3)

Turned out very similar to @LeakyNun's answer but I didn't see his until after I posted mine.

Explanation and Ungolfed

-~ is shorthand for Math.ceil, or rounding up:

n =>               // input in var `n`
    Math.ceil(n/3) // Get every 3rd number 1,1,1,2,2,2, etc.
    *
    (1-n%3)        // 1, 0, -1, 1, 0, -1, ...

function f(n){n=i.value;o.value=-~(n/3)*(1-n%3);}
Input: <input id=i oninput="f()"/><br /><br />
Output: <input id=o readable/>

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  • 1
    \$\begingroup\$ (I hereby attest that he did not see my solution before he posted his solution) \$\endgroup\$ – Leaky Nun May 28 '16 at 16:53
  • \$\begingroup\$ Math.ceil and -~ are different; Math.ceil(1) == 1 whereas -~1 == 2 \$\endgroup\$ – Cyoce May 30 '16 at 23:26
  • 1
    \$\begingroup\$ 1 byte shorter: n=>~(n/3)*~-(n%3) \$\endgroup\$ – Cyoce May 30 '16 at 23:31
6
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MarioLANG, 93 81 bytes

one-indexed

Try It Online

;(-))+(-
"============<
>:(![<:![<:)![
 !=#="!#="!=#=
!  < !-< !- <
#==" #=" #=="

Explanation :

we begin by taking the imput

;

wich give us

          v
... 0 0 input 0 0 ...

we then decrement the left byte and increment the right byte with

;(-))+(
=======

we end up with

           v
... 0 -1 input +1 0 ...

we then set up the loop

;(-))+(-
"============<
>  ![< ![<  ![
   #=" #="  #=
!  < !-< !- <
#==" #=" #=="

the loop will go until the memory look like

         v 
... 0 -X 0 +X 0 ...

we then only need to output the result

;(-))+(-
"============<
>:(![<:![<:)![
 !=#="!#="!=#=
!  < !-< !- <
#==" #=" #=="
\$\endgroup\$
  • 2
    \$\begingroup\$ Nice! You seem to like MarioLang. \$\endgroup\$ – Rɪᴋᴇʀ May 29 '16 at 3:41
  • \$\begingroup\$ @EasterlyIrk The feeling doesn't seem mutual from MarioLang to EtherFrog, though: ;( and >:(. Although, two times [<: could be considered slightly happy. ;P \$\endgroup\$ – Kevin Cruijssen Aug 31 '16 at 14:34
4
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Python 2, 24 bytes

lambda n:(n/3+1)*(1-n%3)

Full program:

a=lambda n:(n/3+1)*(1-n%3)

print(a(0))   #   1
print(a(11))  #  -4
print(a(76))  #   0
print(a(134)) # -45
print(a(296)) # -99
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4
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MATL, 15 12 bytes

3/XkG3X\2-*_

This uses one based indexing.

Try it online! or verify test cases

Explanation:

    G          #Input
     3X\       #Modulus, except multiples of 3 give 3 instead of 0
        2-     #Subtract 2, giving -1, 0 or 1
3/Xk           #Ceiling of input divided by 3.
          *    #Multiply 
           _   #Negate
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  • \$\begingroup\$ To take care of most of the issues something like Q3/Xk-1:1G_)* may work better. It can probably be modified ever further for 1-based indexing instead. \$\endgroup\$ – Suever May 28 '16 at 17:55
4
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Haskell, 27 bytes

f x=div(x+3)3*(1-mod(x+3)3)

Slightly more interesting 28 byte solution:

(((\i->[i,0,-i])=<<[1..])!!)

(Both are 0-indexed)

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3
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MATL, 8 bytes

:t~y_vG)

The result is 1-based.

Try it online!

Explanation

This builds the 2D array

 1  2  3  4  5 ...
 0  0  0  0  0 ...
-1 -2 -3 -4 -5 ...

and then uses linear indexing to extract the desired term. Linear indexing means index down, then across (so in the above array the first entries in linear order are 1, 0, -1, 2, 0, ...)

:     % Vector [1 2 ... N], where N is implicit input
t~    % Duplicate and logical negate: vector of zeros
y_    % Duplicate array below the top and negate: vector [-1 -2 ... -N]
v     % Concatenate all stack contents vertically
G)    % Index with input. Implicit display
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3
\$\begingroup\$

Perl 5, 22 bytes

21 plus one for -p:

$_=(-$_,$_+2)[$_%3]/3

Uses 1-based indexing.

Explanation:

-p sets the variable $_ equal to the input. The code then sets it equal to the $_%3th element, divided by 3, of the 0-based list (-$_,$_+2) (where % is modulo). Note that if $_%3 is two, then there is no such element, and the subsequent division by 3 numifies the undefined to 0. -p then prints $_.

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3
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Bash, 28 25 Bytes

echo $[(1+$1/3)*(1-$1%3)]
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  • \$\begingroup\$ @DigitalTrauma, tkx, didn't know this... \$\endgroup\$ – rexkogitans May 30 '16 at 18:38
2
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Perl 6,  26  23 bytes

{({|(++$,0,--$)}...*)[$_]}
{($_ div 3+1)*(1-$_%3)}

( The shorter one was translated from other answers )

Explanation (of the first one):

{ # bare block with implicit parameter 「$_」
  (

    # start of sequence generator

    { # bare block
      |(  # slip ( so that it flattens into the outer sequence )
        ++$, # incrementing anon state var =>  1, 2, 3, 4, 5, 6
        0,   # 0                           =>  0, 0, 0, 0, 0, 0
        --$  # decrementing anon state var => -1,-2,-3,-4,-5,-6
      )
    }
    ...  # repeat
    *    # indefinitely

    # end of sequence generator

  )[ $_ ] # get the nth one (zero based)
}

Test:

#! /usr/bin/env perl6
use v6.c;
use Test;

# store it lexically
my &alt-seq-sign = {({|(++$,0,--$)}...*)[$_]}
my &short-one = {($_ div 3+1)*(1-$_%3)}

my @tests = (
    0 =>   1,
   11 =>  -4,
   76 =>   0,
  134 => -45,
  296 => -99,
  15..^30  => (6,0,-6,7,0,-7,8,0,-8,9,0,-9,10,0,-10)
);

plan @tests * 2 - 1;

for @tests {
  is alt-seq-sign( .key ), .value, 'alt-seq-sign  ' ~ .gist;

  next if .key ~~ Range; # doesn't support Range as an input
  is short-one(    .key ), .value, 'short-one     ' ~ .gist;
}
1..11
ok 1 - alt-seq-sign  0 => 1
ok 2 - short-one     0 => 1
ok 3 - alt-seq-sign  11 => -4
ok 4 - short-one     11 => -4
ok 5 - alt-seq-sign  76 => 0
ok 6 - short-one     76 => 0
ok 7 - alt-seq-sign  134 => -45
ok 8 - short-one     134 => -45
ok 9 - alt-seq-sign  296 => -99
ok 10 - short-one     296 => -99
ok 11 - alt-seq-sign  15..^30 => (6 0 -6 7 0 -7 8 0 -8 9 0 -9 10 0 -10)
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2
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J, 19 15 bytes

>.@(%&3)*1-3|<:

Probably need to golf this further...

1-indexed.

Ungolfed:

>> choose_sign      =: 1-3|<:      NB. 1-((n-1)%3)
>> choose_magnitude =: >.@(%&3)    NB. ceil(n/3)
>> f                =: choose_sign * choose_magnitude
>> f 1 12 77
<< 1 _4 0

Where >> means input (STDIN) and << means output (STDOUT).

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2
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Pyke, 8 7 bytes (old version)

3.DeRt*

Try it here! - Note that link probably won't last for long

3.D      - a,b = divmod(input, 3)
   e     - a = ~a -(a+1)
     t   - b -= 1
      *  - a = a*b
         - implicit output a

Newest version

3.DhRt*_

Try it here!

3.D      - a,b = divmod(input, 3)
   h     - a+=1
     t   - b-=1
      *  - a = a*b
       _ - a = -a
         - implicit output a
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  • \$\begingroup\$ Can you provide a link to the (old version) \$\endgroup\$ – Downgoat May 28 '16 at 17:09
  • \$\begingroup\$ Latest commit where the old code works here (this is earlier today) \$\endgroup\$ – Blue May 28 '16 at 17:44
2
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J, 27 bytes

Whilst not the golfiest, I like it better, as it uses an agenda.

>.@(>:%3:)*1:`0:`_1:@.(3|])

Here is the tree decomposition of it:

         ┌─ >.      
  ┌─ @ ──┤    ┌─ >: 
  │      └────┼─ %  
  │           └─ 3: 
  ├─ *              
──┤           ┌─ 1: 
  │      ┌────┼─ 0: 
  │      │    └─ _1:
  └─ @. ─┤          
         │    ┌─ 3  
         └────┼─ |  
              └─ ]  

This is very similar to Kenny's J answer, in that it chooses the magnitude and sign, but it's different in that I use an agenda to choose the sign.

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2
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MATL, 8 bytes

_3&\wq*_

This solution uses 1-based indexing into the sequence.

Try it Online

Modified version showing all test cases

Explanation

        % Implicitly grab the input
_       % Negate the input
3&\     % Compute the modulus with 3. The second output is floor(N/3). Because we negated
        % the input, this is the equivalent of ceil(input/3)
w       % Flip the order of the outputs
q       % Subtract 1 from the result of mod to turn [0 1 2] into [-1 0 1]
*       % Take the product with ceil(input/3)
_       % Negate the result so that the sequence goes [N 0 -N] instead of [-N 0 N]
        % Implicitly display the result
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2
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Pyth, 10 bytes

*h/Q3-1%Q3

Try it online!

Explanation:

*     : Multiply following two arguments
h/Q3  : 1 + Input/3
-1%Q3 : 1 - Input%3

Note: I've assumed the zero-indexed sequence.

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2
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Actually, 10 bytes

3@│\u)%1-*

Try it online!

Explanation:

3@│\u)%1-*
3@│         push 3, swap, duplicate entire stack ([n 3 n 3])
   \u)      floor division, increment, move to bottom ([n 3 n//3+1])
      %1-   mod, subtract from 1 ([1-n%3 n//3+1])
         *  multiply ([(1-n%3)*(n//3+1)])
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2
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05AB1E, 7 bytes

Code:

(3‰`<*(

Explained:

(           # negate input: 12 -> -12
 3‰         # divmod by 3: [-4, 0]
   `        # flatten array: 0, -4
    <       # decrease the mod-result by 1: -1, -4
     *      # multiply: 4
      (     # negate -4
\$\endgroup\$
2
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GeoGebra, 44 bytes

Element[Flatten[Sequence[{t,0,-t},t,1,n]],n]

where n is one-indexed.

Explanation:

Element[                      , n] # Return the nth element of the list                  .
 Flatten[                    ]     # Strip all the unnecessary braces from the list     /|\
  Sequence[{t,0,-t}, t, 1, n]      # Generate a list of lists of the form {t, 0, -t}     |
                             # This list will start with {1,0,-1} and end with {n,0,-n}  |

It is not necessary to generate all triplets through {n, 0, -n}, but it's shorter than writing ceil(n/3) or something to that effect. Note that n must be defined to create this object (if it isn't defined at the time this is run, GeoGebra will prompt you to create a slider for n).

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  • \$\begingroup\$ Hi and welcome to PPCG! Do you have a link I can test this (preferably online)? \$\endgroup\$ – Rɪᴋᴇʀ Jun 2 '16 at 2:54
  • \$\begingroup\$ @EᴀsᴛᴇʀʟʏIʀᴋ, thanks! Here's a link to an online applet thingamabob. The page looked blank for a little while, but then it showed up. \$\endgroup\$ – Joe Jun 2 '16 at 21:30
  • \$\begingroup\$ Oh cool. But how where do I put in the formula? >_> I tried pasting it into the blank, and it prompted to create a slider, but nothing else happened. \$\endgroup\$ – Rɪᴋᴇʀ Jun 2 '16 at 22:37
  • \$\begingroup\$ @EᴀsᴛᴇʀʟʏIʀᴋ: Over on the left-hand side, where it says "Input..." First, to initialise n, enter something like n=297 (this will give you a slider that's configured nicely). Then paste the formula into the Input box, which should now be below the n. (Make sure to hit return ;) The formula should evaluate to the nth term of the sequence, and it should change when you move the slider. \$\endgroup\$ – Joe Jun 2 '16 at 22:50
2
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Labyrinth, 17 15 14 bytes

Saved 3 bytes using Sok's idea of using 1-(n%3) instead of ~(n%3-2).

1?:#/)}_3%-{*!

The program terminates with an error (division by zero), but the error message goes to STDERR.

Try it online!

Explanation

The program is completely linear, although some code is executed in reverse at the end.

1     Turn top of stack into 1.
?:    Read input as integer and duplicate.
#     Push stack depth (3).
/)    Divide and increment.
}     Move over to auxiliary stack.
_3%   Take other copy modulo 3.
-     Subtract from 1. This turns 0, 1, 2 into 1, 0, -1, respectively.
{*    Move other value back onto main stack and multiply.
!     Output as integer.

The instruction pointer now hits a dead end and turns around, so it starts to execute the code from the end:

*     Multiply two (implicit) zeros.
{     Pull an (implicit) zero from the auxiliary to the main stack.
-     Subtract two (implicit) zeros from one another.
      Note that these were all effectively no-ops due to the stacks which are
      implicitly filled with zeros.
%     Attempt modulo, which terminates the program due to a division-by-zero error.
\$\endgroup\$
2
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Erlang, 40 bytes

F=fun(N)->trunc((N/3+1)*(1-N rem 3))end.

Sadly Erlang has no '%' modulo operator and 'rem' requires the spaces, even before the 3.

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2
\$\begingroup\$

Hexagony, 25 bytes

?'+}@/)${':/3$~{3'.%(/'*!

Or, in non-minified format:

    ? ' + }
   @ / ) $ {
  ' : / 3 $ ~
 { 3 ' . % ( /
  ' * ! . . .
   . . . . .
    . . . .

Try it online!

My first foray into Hexagony, so I'm certain I've not done this anywhere near as efficiently as it could be done...

Calculates -(n%3 - 1) on one memory edge, n/3 + 1 on an adjacent one, then multiplies them together.

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  • \$\begingroup\$ Wow, very interesting to see this! :) \$\endgroup\$ – Adnan Jun 2 '16 at 20:17
2
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R, 28 bytes

-((n=scan())%%3-1)*(n%/%3+1)

Looks like this is a variation of most of the answers here. Zero based.

   n=scan()                  # get input from STDIN
  (        )%%3-1            # mod by 3 and shift down (0,1,2) -> (-1,0,1)
-(               )           # negate result (1,0,-1), this handles the alternating signs
                  *(n%/%3+1) # integer division of n by 3, add 1, multiply by previous

The nice thing about it is that it handles multiple inputs

> -((n=scan())%%3-1)*(n%/%3+1)
1: 0 3 6 9 1 4 7 10 2 5 8 11
13: 
Read 12 items
 [1]  1  2  3  4  0  0  0  0 -1 -2 -3 -4
> 

Originally I wanted to do the following, but couldn't trim off the extra bytes.

rbind(I<-1:(n=scan()),0,-I)[n]

Uses rbind to add 0's and negatives to a range of 1 to n then return the n'th term (one based).

# for n = 5
rbind(                    )    # bind rows 
            n=scan()           # get input from STDIN and assign to n
      I<-1:(        )          # build range 1 to n and assign to I
                     ,0        # add a row of zeros (expanded automatically)
                       ,-I     # add a row of negatives
                           [n] # return the n'th term
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2
\$\begingroup\$

Batch (Windows), 86 bytes

Alternate.bat

SET /A r=%1%%3
SET /A d=(%1-r)/3+1
IF %r%==0 ECHO %d%
IF %r%==1 ECHO 0
IF %r%==2 ECHO -%d%

This program is run as Alternate.bat n where n is the number you wish to call the function on.

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2
\$\begingroup\$

APL, 12 chars

-×/1-0 3⊤6+⎕

0 3⊤ is APL's divmod 3.

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2
\$\begingroup\$

Java 7, 38 37 36 bytes

My first golf, be gentle

int a(int i){return(1+i/3)*(1-i%3);}

Try it here! (test cases included)

Edit: I miscounted, and also golfed off one more character by replacing (-i%3+1) with (1-i%3).

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  • 1
    \$\begingroup\$ Hello, and welcome to PPCG! You can remove the space after return, and use a Java 8 lambda. \$\endgroup\$ – NoOneIsHere Jun 23 '16 at 17:43
  • \$\begingroup\$ I should specify that this was Java 7. I'll remove that space, though. Thanks! \$\endgroup\$ – Steven H. Jun 23 '16 at 17:47
1
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Retina, 45 bytes

.+
11$&$*
(111)+(1)*
$#2$#1
T`d`+0-`^.
^0.+
0

Try it online!

Test suite.

Takes input/output in base-ten. 1-indexed.

Unary input, base-ten output, 1-indexed: 40 bytes

$
11
(111)+(1)*
$#2$#1
T`d`+0-`^.
^0.+
0

Try it online!

Test suite.

\$\endgroup\$
1
\$\begingroup\$

MATLAB / Octave, 27 bytes

@(n)ceil(n/3)*(mod(-n,3)-1)

This creates an anonymous function that can be called using ans(n). This solution uses 1-based indexing.

All test cases

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1
\$\begingroup\$

Mathematica 26 bytes

With 4 bytes saved thanks to Martin Ender.

⎡#/3⎤(-#~Mod~3-1)&

Uses the same approach as Suever.

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1
\$\begingroup\$

Octave, 23 bytes

With no mod cons...

@(n)(-[-1:1]'*[1:n])(n)

Uses 1-based indexing magic.


Explanation

Creates an anonymous function that will:

(-[-1:1]'*[1:n])(n)
  [-1:1]              % make a row vector [-1 0 1]
 -      '             % negate and take its transpose making a column vector
          [1:n]       % make a row vector [1..n], where n is the input
         *            % multiply with singleton expansion
               (n)    % use linear indexing to get the nth value

After the multiplication step we'll have a 3xn matrix like so (for n=12):

 1    2    3    4    5    6    7    8    9   10   11   12
 0    0    0    0    0    0    0    0    0    0    0    0
-1   -2   -3   -4   -5   -6   -7   -8   -9  -10  -11  -12

Making n columns is overkill, but it's a convenient number that is guaranteed to be large enough. Linear indexing counts down each column from left to right, so the element at linear index 4 would be 2.

All test cases on ideone.

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1
\$\begingroup\$

dc, 10

?2+3~1r-*p

Uses 1-based indexing.

?              # Push input to stack
 2+            # Add 2
   3~          # divmod by 3
     1r-       # subtract remainder from 1
        *      # multiply by quotient
         p     # print
\$\endgroup\$

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