The Nine Pattern

Question

Introduction

I stumbled across this (useless) pattern the other day while I was watching TV. I named it "the 9 pattern" because the first number to use it was 9. The gist of it is, you enter a number (let's say x), and then you get back:

• x
• x + (x / 3) [let's call this y]
• two-thirds of y [let's call this z]
• z + 1

So, if I put inside this pattern the number 9 as x, this is what would come out:

• 9 (9)
• 12 (9 + 9 / 3) [9 over 3 is 3, and 9 + 3 is 12]
• 8 (12 times two-thirds) [a third of 12 is 4, and 4 * 2 is 8]
• 9 (8 + 1 is 9)

Challenge

Write me a function (in any programming language) that takes in a number, and outputs an integer array using the pattern.
Somewhat like this psuedo-code:

function ninePattern(int myInt) returns IntegerArray {  
    int iterationA = myInt + (myInt / 3);  
    int iterationB = iterationA * (2 / 3); 
    int iterationC = iterationB + 1;  
    IntegerArray x = [myInt, iterationA, iterationB, iterationC];  
    return x;  
}

Clarifications

Discussions have been arousing in comments regarding the specifications of the question. This section is meant to clarify some of those.

"better to count in bytes than characters"

I picked characters because (for me, at least) it would be easier to judge. Of course, I can't change that now. (lots of answers are already posted)

"rounding"

Rounding follows this rhyme:

If it's 5 or more, raise the score
If it's 4 or less, let it rest

Simply, put, if it is something like 4.7 or 3.85, round them to 5 and 4 respectively.

Examples

Input => Result
9 => [9, 12, 8, 9]
8 => [8, 11, 7, 8]
6 => [6, 8, 5, 6]
23 => [23, 31, 21, 22]
159 => [159, 212, 141, 142]

If, however, the numbers are something like 2.3 or 10.435446, round them to 2 and 10 respectively.

"language support"

You are free to not use functions and/or arrays IF AND ONLY IF the language of your choice does not support them. If it does (even if it will increase your characters count), you must use them.

Answer 1

CJam, 21 bytes

qi__3d/+mo_2d3/*i_)]`

Feedback is welcome

Explanation

• qi__ - Read the input as an integer and duplicate it twice
• 3D/+mo - Divide one instance of the input by 3, then add it to the second instance to make y
• _2d3/*i - Duplicate y, then multiply it by .6
• _)]` - Dupe, increment, wrap in array, print as array (not in code because of the ` operator :( )

Edit: Forgot to make the first three a double, so the program was broken. Fixed.

Answer 2

Axiom, 59 bytes

g(x)==round(x)::INT;f(x)==[x,a:=g(x+x/3.),b:=g(a*2./3),b+1]

test

(3) -> [[i,f(i)] for i in [9,8,6,23,159]]
   (3)
   [[9,[9,12,8,9]], [8,[8,11,7,8]], [6,[6,8,5,6]], [23,[23,31,21,22]],
    [159,[159,212,141,142]]]
                                                      Type: List List Any

Answer 3

PHP, 60 Bytes

print_r([$a=$argn,$b=round($a+$a/3),$c=round($b*2/3),$c+1]);

Try it online!

Answer 4

PHP, 67 bytes

function f($x){return [$x,$y=round($x*4/3),$z=round($y*2/3),$z+1];}

Answer 5

Zsh*, 58 bytes

r()<<<${$(($1+.5))%.*}
r $1;r $[_+_/3];r $[2*_/3];r $[_+1]

Try it online!

(*) Using force_float and extendedglob options.