Up go the bits!

Question

Given an integer N perform the following steps: (using 9 as an example).

1. Receive input N. (9)
2. Convert N from base10 to base2. (1001)
3. Increase every bit by 1. (2112)
4. Treat the result as base3 and convert it back to base10. (68)
5. Return/Output the result.

---

Input

May be received in any reasonable number format.
You only need to handle cases where N > 0.

---

Output

Either return as a number or string, or print to stdout.

---

Rules

• This is code-golf, the shortest code in bytes wins.
• Default loopholes are forbidden.

---

Test Cases

1 -> 2
2 -> 7
5 -> 23
9 -> 68
10 -> 70
20 -> 211
1235 -> 150623
93825 -> 114252161

Answer 1

Python 2, 31 bytes

f=lambda n:n and 3*f(n/2)+n%2+1

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Answer 2

JavaScript (Node.js), 23 bytes

f=x=>x&&x%2+1+3*f(x>>1)

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Answer 3

Java (JDK 10), 44 bytes

long n(long x){return x<1?0:x%2+1+3*n(x/2);}

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Answer 4

Jelly, 4 bytes

B‘ḅ3

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Answer 5

Java 10, 81 52 bytes (Base conversion)

n->n.toString(n,2).chars().reduce(0,(r,c)->r*3+c-47)

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-29 bytes thanks to @Holger.

Explanation:

n->{                         // Method with Long as both parameter and return-type
  n.toString(n,2)            //  Convert the input to a Base-2 String
  .chars().reduce(0,(r,c)->  //  Loop over its digits as bytes
    r*3+c-47)                //  Multiply the current result by 3, and add the digit + 1
                             //  (which is equal to increasing each digit by 1,
                             //  and then converting from Base-3 to Base-10)

---

Java 10, 171 167 151 150 149 bytes (Sequence)

n->{int t=31-n.numberOfLeadingZeros(n);return a(t+1)+b(n-(1<<t));};int a(int n){return--n<1?n+2:3*a(n)+1;}int b(int n){return n<1?0:n+3*b(n/=2)+n*2;}

-16 bytes thanks to @musicman523, changing (int)Math.pow(2,t) to (1<<t).
-1 byte thanks to @Holger, changing (int)(Math.log(n)/Math.log(2)) to 31-n.numberOfLeadingZeros(n).

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Explanation:

n->{                         // Method with Integer as both parameter and return-type
  int t=31-n.numberOfLeadingZeros(n);
                             //  2_log(n)
  return a(t+1)              //  Return A060816(2_log(n)+1)
         +b(n-(1<<t));}      //   + A005836(n-2^2_log(n))

// A060816: a(n) = 3*a(n-1) + 1; a(0)=1, a(1)=2
int a(int n){return--n<1?n+2:3*a(n)+1;}

// A005836: a(n+1) = 3*a(floor(n/2)) + n - 2*floor(n/2).
int b(int n){return n<1?0:n+3*b(n/=2)+n*2;}

When we look at the sequence:

2,  7,8,  22,23,25,26,  67,68,70,71,76,77,79,80,  202,203,205,206,211,212,214,215,229,230,232,233,238,239,241,242, ...

We can see multiple subsequences:

A053645(n):
0,  0,1,  0,1,2,3,  0,1,2,3,4,5,6,7,  0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,  ...

A060816(A053645(n)):
2,  7,7,  22,22,22,22,  67,67,67,67,67,67,67,67,  202,202,202,202,202,202,202,202,202,202,202,202,202,202,202,  ...

A005836(A053645(n)+1)
0,  0,1,  0,1,3,4,  0,1,3,4,9,10,12,13,  0,1,3,4,9,10,12,13,27,28,30,31,36,37,39,40,  ...

So the sequence being asked is:

A060816(A053645(n)) + A005836(A053645(n)+1)

---

I suck at finding patterns, so I'm proud of what I found above.. Having said that, @user202729 found a better and shorter approach in Java within a few minutes.. :'(

Answer 6

05AB1E, 5 bytes

b€>3β

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b       binary
 €>     increment each
   3β   base 3

---

05AB1E, 5 bytes

2в>3β

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Answer 7

J, 7 bytes

3#.1+#:

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Thanks Galen Ivanov for -4 bytes! I really need to improve my J golfing skill...

Answer 8

R, 55 43 bytes

function(n)(n%/%2^(x=0:log2(n))%%2+1)%*%3^x

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Uses the standard base conversion trick in R, increments, and then uses a dot product with powers of 3 to convert back to an integer.

Thanks to @user2390246 for dropping 12 bytes!

Answer 9

APL (Dyalog), 10 bytes

3⊥1+2⊥⍣¯1⊢

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    2⊥⍣¯1  binary
  1+       go guess
3⊥         base 3

Answer 10

Brachylog, 7 bytes

ḃ+₁ᵐ~ḃ₃

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Explanation

Not that you really need one, but…

ḃ            To binary
 +₁ᵐ         Map increment
    ~ḃ₃      From ternary

Answer 11

Ruby, 27 bytes

f=->x{x>0?x%2+1+3*f[x/2]:0}

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Answer 12

Python 2, 56 55 bytes

lambda n:int(''.join('12'[c>'0']for c in bin(n)[2:]),3)

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Answer 13

Attache, 19 bytes

FromBase&3@1&`+@Bin

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This is a composition of three functions:

• FromBase&3
• 1&`+
• Bin

This first converts to binary (Bin), increments it (1&`+), then converts to ternary (FromBase&3).

Alternatives

Non-pointfree, 21 bytes: {FromBase[Bin!_+1,3]}

Without builtins, 57 bytes: Sum@{_*3^(#_-Iota!_-1)}@{If[_>0,$[_/2|Floor]'(1+_%2),[]]}

Answer 14

Retina 0.8.2, 36 bytes

.+
$*
+`^(1+)\1
$1;1
^
1
+`1;
;111
1

Try it online! Explanation:

.+
$*

Convert from decimal to unary.

+`^(1+)\1
$1;1

Repeatedly divmod by 2, and add 1 to the result of the modulo.

^
1

Add 1 to the first digit too.

+`1;
;111

Convert from unary-encoded base 3 to unary.

1

Convert to decimal.

Answer 15

Japt, 6 bytes

¤cÄ n3
¤      // Convert the input to a base-2 string,
 c     // then map over it as charcodes.
  Ä    // For each item, add one to its charcode
       // and when that's done,
    n3 // parse the string as a base 3 number.

Takes input as a number, outputs a number.

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Answer 16

MATL, 12 7 6 bytes

BQ3_ZA

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Saved 5 bytes thanks to Giuseppe and another one thanks to Luis Mendo.

Old 7 byte answer:

YBQc3ZA

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Explanation:

YB        % Convert to binary string
  Q       % Increment each element
   c      % Convert ASCII values to characters
    3     % Push 3
     ZA   % Convert from base 3 to decimal.

---

Old one for 12 bytes:

BQtz:q3w^!Y*

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Oh my, that was messy... So is this: `BQ3GBn:q^!Y*.

Explanation:

               % Implicit input
B              % Convert to binary vector
 Q             % Increment all numbers
  t            % Duplicate
   z           % Number of element in vector
    :          % Range from 1 to that number
     q         % Decrement to get the range from 0 instead of 1
      3        % Push 3
       w       % Swap order of stack
        ^      % Raise 3 to the power of 0, 1, ...
         !     % Transpose
          Y*   % Matrix multiplication
               % Implicit output

Answer 17

C# (Visual C# Compiler), 128 bytes

using System;using System.Linq;i=>{int z=0;return Convert.ToString(i,2).Reverse().Select(a=>(a-47)*(int)Math.Pow(3,z++)).Sum();}

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I am counting System because i use Convert and Math.

Answer 18

Python 2, 56 54 bytes

lambda i:int(''.join(`int(x)+1`for x in bin(i)[2:]),3)

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Answer 19

PHP, 84 64 Bytes

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ORIGINAL Code

function f($n){$b=decbin($n);echo base_convert($b+str_repeat('1',strlen($b)),3,10);}

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Thanks to Cristoph, less bytes if ran with php -R

function f($n){echo base_convert(strtr(decbin($n),10,21),3,10);}

Explanation

function f($n){
$b=decbin($n);                    #Convert the iteger to base 2
echo base_convert(                  #base conversion PHP function
    $b+str_repeat('1',strlen($b)),  #It adds to our base 2 number
    3,                              #a number of the same digits length
    10);                            #with purely '1's
}

Answer 20

C, 32 27 bytes

n(x){x=x?x%2+1+3*n(x/2):0;}

Based on user202729's Java answer. Try it online here. Thanks to Kevin Cruijssen for golfing 5 bytes.

Ungolfed version:

n(x) { // recursive function; both argument and return type are implicitly int
    x = // implicit return
    x ? x % 2 + 1 + 3*n(x/2) // if x != 0 return x % 2 + 1 + 3*n(x/2) (recursive call)
    : 0; // else return 0
}

Answer 21

Octave with the communication toolbox, 33 32 bytes

@(x)(de2bi(x)+1)*3.^(0:log2(x))'

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Converts the input to a binary vector using de2bi, and incrementing all numbers. Does matrix multiplication with a vertical vector of 3 raised to the appropriate powers: 1, 3, 9, ..., thus getting the sum without an explicit call to sum.

Answer 22

Husk, 5 bytes

B3m→ḋ

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Explanation

B3m→ḋ
    ḋ  Convert to base 2
  m→   Map increment
B3     Convert from base 3

Answer 23

CJam, 8 bytes

ri2b:)3b

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Explanation

ri   e# Read input as an integer
2b   e# Convert to base 2. Gives a list containing 0 and 1
:)   e# Add 1 to each number in that list
3b   e# Convert list from base 3 to decimal. Implicitly display

Answer 24

Brain-Flak, 74 bytes

({<>(())<>({<({}[()])><>([{}]())<>})}(<>)){{}((({})()){}{}[{}])([][()])}{}

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"Readable" version

({<>(())<>
  ({
    <({}[()])>
    <>
    ([{}]())
    <>
  })
}
# At this point we have a inverted binary string on the stack
(<>)
)
{
  {}
  (
    (({})()){}{}[{}]
  )
  ([][()])
}{}

Answer 25

Whitespace, 117 bytes

[S S S N
_Push_0][S N
S _Duplicate_0][S N
S _Duplicate_0][T   N
T   T   _Read_STDIN_as_number][T    T   T   _Retrieve][N
S S S N
_Create_Label_OUTER_LOOP][S N
S _Duplicate][S S S T   S N
_Push_2][T  S T T   _Modulo][S S S T    N
_Push_1][T  S S S _Add][S N
T   _Swap][S S S T  S N
_Push_2][T  S T S _Integer_division][S N
S _Duplicate][N
T   S N
_If_0_jump_to_Label_INNER_LOOP][N
S N
S N
_Jump_to_Label_OUTER_LOOP][N
S S N
_Create_Label_INNER_LOOP][S S S T   T   N
_Push_3][T  S S N
_Multiply][T    S S S _Add][S N
T   _Swap][S N
S _Duplicate][N
T   S T N
_If_0_jump_to_Label_PRINT_AND_EXIT][S N
T   _Swap][N
S N
N
_Jump_to_Label_INNER_LOOP][N
S S T   N
_Create_Label_PRINT_AND_EXIT][S N
T   _Swap][T    N
S T _Output_integer_to_STDOUT]

Letters S (space), T (tab), and N (new-line) added as highlighting only.
[..._some_action] added as explanation only.

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Explanation in pseudo-code:

I first converted the recursive function int f(int n){return n<1?0:n%2+1+3*f(n/2);} to its iterative form (in pseudo-code):

Integer n = STDIN as integer
Add starting_value 0 to the stack
function OUTER_LOOP:
  while(true){
    Add n%2+1 to the stack
    n = n/2
    if(n == 0):
      Jump to INNER_LOOP
    Else:
      Jump to next iteration OUTER_LOOP

function INNER_LOOP:
  while(true){
    n = 3*n
    n = n + Value at the top of the stack (the ones we calculated with n%2+1)
    Swap top two items
    Check if the top is now 0 (starting value):
      Jump to PRINT_AND_EXIT
    Else:
      Swap top two items back
      Jump to next iteration INNER_LOOP

function PRINT_AND_EXIT:
  Swap top two items back
  Print top to STDOUT as integer
  Exit program with error: Exit not defined

And I then implemented this iterative approach in the stack-based language Whitespace, using it's default stack.

Example runs:

Input: 1

Command    Explanation                   Stack           Heap    STDIN   STDOUT   STDERR

SSSN       Push 0                        [0]
SNS        Duplicate top (0)             [0,0]
SNS        Duplicate top (0)             [0,0,0]
TNTT       Read STDIN as integer         [0,0]           {0:1}   1
TTT        Retrieve                      [0,1]           {0:1}
NSSSN      Create Label OUTER_LOOP       [0,1]           {0:1}
 SNS       Duplicate top (1)             [0,1,1]         {0:1}
 SSSTSN    Push 2                        [0,1,1,2]       {0:1}
 TSTT      Modulo top two (1%2)          [0,1,1]         {0:1}
 SSSTN     Push 1                        [0,1,1,1]       {0:1}
 TSSS      Add top two (1+1)             [0,1,2]         {0:1}
 SNT       Swap top two                  [0,2,1]         {0:1}
 SSSTSN    Push 2                        [0,2,1,2]       {0:1}
 TSTS      Int-divide top two (1/2)      [0,2,0]         {0:1}
 SNS       Duplicate top (0)             [0,2,0,0]       {0:1}
 NTSN      If 0: Go to Label INNER_LOOP  [0,2,0]         {0:1}
 NSSN      Create Label INNER_LOOP       [0,2,0]         {0:1}
  SSSTTN   Push 3                        [0,2,0,3]       {0:1}
  TSSN     Multiply top two (0*3)        [0,2,0]         {0:1}
  TSSS     Add top two (2+0)             [0,2]           {0:1}
  SNT      Swap top two                  [2,0]           {0:1}
  SNS      Duplicate top (0)             [2,0,0]         {0:1}
  NTSTN    If 0: Jump to Label PRINT     [2,0]           {0:1}
  NSSTN    Create Label PRINT            [2,0]           {0:1}
   SNT     Swap top two                  [0,2]           {0:1}
   TNST    Print top to STDOUT           [0]             {0:1}           2
                                                                                  error

Try it online (with raw spaces, tabs and new-lines only).
Stops with error: Exit not defined.

Input: 4

Command    Explanation                   Stack           Heap    STDIN   STDOUT   STDERR

SSSN       Push 0                        [0]
SNS        Duplicate top (0)             [0,0]
SNS        Duplicate top (0)             [0,0,0]
TNTT       Read STDIN as integer         [0,0]           {0:4}   4
TTT        Retrieve                      [0,4]           {0:4}
NSSSN      Create Label OUTER_LOOP       [0,4]           {0:4}
 SNS       Duplicate top (4)             [0,4,4]         {0:4}
 SSSTSN    Push 2                        [0,4,4,2]       {0:4}
 TSTT      Modulo top two (4%2)          [0,4,0]         {0:4}
 SSSTN     Push 1                        [0,4,0,1]       {0:4}
 TSSS      Add top two (0+1)             [0,4,1]         {0:4}
 SNT       Swap top two                  [0,1,4]         {0:4}
 SSSTSN    Push 2                        [0,1,4,2]       {0:4}
 TSTS      Int-divide top two (4/2)      [0,1,2]         {0:4}
 SNS       Duplicate top (2)             [0,1,2,2]       {0:4}
 NTSN      If 0: Go to Label INNER_LOOP  [0,1,2]         {0:4}
 NSNSN     Jump to Label OUTER_LOOP      [0,1,2]         {0:4}
 SNS       Duplicate top (2)             [0,1,2,2]       {0:4}
 SSSTSN    Push 2                        [0,1,2,2,2]     {0:4}
 TSTT      Modulo top two (2%2)          [0,1,2,0]       {0:4}
 SSSTN     Push 1                        [0,1,2,0,1]     {0:4}
 TSSS      Add top two (0+1)             [0,1,2,1]       {0:4}
 SNT       Swap top two                  [0,1,1,2]       {0:4}
 SSSTSN    Push 2                        [0,1,1,2,2]     {0:4}
 TSTS      Int-divide top two (2/2)      [0,1,1,1]       {0:4}
 SNS       Duplicate top (1)             [0,1,1,1,1]     {0:4}
 NTSN      If 0: Go to Label INNER_LOOP  [0,1,1,1]       {0:4}
 NSNSN     Jump to Label OUTER_LOOP      [0,1,1,1]       {0:4}
 SNS       Duplicate top (1)             [0,1,1,1,1]     {0:4}
 SSSTSN    Push 2                        [0,1,1,1,1,2]   {0:4}
 TSTT      Modulo top two (1%2)          [0,1,1,1,1]     {0:4}
 SSSTN     Push 1                        [0,1,1,1,1,1]   {0:4}
 TSSS      Add top two (1+1)             [0,1,1,1,2]     {0:4}
 SNT       Swap top two                  [0,1,1,2,1]     {0:4}
 SSSTSN    Push 2                        [0,1,1,2,1,2]   {0:4}
 TSTS      Int-divide top two (1/2)      [0,1,1,2,0]     {0:4}
 SNS       Duplicate top (0)             [0,1,1,2,0,0]   {0:4}
 NTSN      If 0: Go to Label INNER_LOOP  [0,1,1,2,0]     {0:4}
 NSSN      Create Label INNER_LOOP       [0,1,1,2,0]     {0:4}
  SSSTTN   Push 3                        [0,1,1,2,0,3]   {0:4}
  TSSN     Multiply top two (0*3)        [0,1,1,2,0]     {0:4}
  TSSS     Add top two (2+0)             [0,1,1,2]       {0:4}
  SNT      Swap top two                  [0,1,2,1]       {0:4}
  SNS      Duplicate top (1)             [0,1,2,1,1]     {0:4}
  NTSTN    If 0: Jump to Label PRINT     [0,1,2,1]       {0:4}
  SNT      Swap top two                  [0,1,1,2]       {0:4}
  NSNN     Jump to Label INNER_LOOP      [0,1,1,2]       {0:4}
  SSSTTN   Push 3                        [0,1,1,2,3]     {0:4}
  TSSN     Multiply top two (2*3)        [0,1,1,6]       {0:4}
  TSSS     Add top two (1+6)             [0,1,7]         {0:4}
  SNT      Swap top two                  [0,7,1]         {0:4}
  SNS      Duplicate top (1)             [0,7,1,1]       {0:4}
  NTSTN    If 0: Jump to Label PRINT     [0,7,1]         {0:4}
  SNT      Swap top two                  [0,1,7]         {0:4}
  NSNN     Jump to Label INNER_LOOP      [0,1,7]         {0:4}
  SSSTTN   Push 3                        [0,1,7,3]       {0:4}
  TSSN     Multiply top two (7*3)        [0,1,21]        {0:4}
  TSSS     Add top two (1+21)            [0,22]          {0:4}
  SNT      Swap top two                  [22,0]          {0:4}
  SNS      Duplicate top (0)             [22,0,0]        {0:4}
  NTSTN    If 0: Jump to Label PRINT     [22,0]          {0:4}
  NSSTN    Create Label PRINT            [22,0]          {0:4}
   SNT     Swap top two                  [0,22]          {0:4}
   TNST    Print top to STDOUT           [0]             {0:4}           22
                                                                                  error

Try it online (with raw spaces, tabs and new-lines only).
Stops with error: Exit not defined.

Answer 26

Add++, 14 bytes

L,BBu1€+B]3$Bb

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Answer 27

Japt, 7 bytes

¤£°XÃn3

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Answer 28

Stax, 7 bytes

É¥ê4¼⌐○

Run and debug it

This one is pretty straightforward. After unpacking, and commenting, the program looks like this.

:B  Convert to bits
{^m Increment each
3|b Base-3 convert

Answer 29

Haskell, 32 bytes

f 0=0
f a=1+mod a 2+3*f(div a 2)

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Answer 30

Pyth, 8

ihMjQ2 3

How to eliminate the space and make the Q implicit?

Pyth online.

Answer 31

Perl 5, 36 bytes

$\+=(1+$_%2)*3**$b++,$_>>=1while$_}{

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Answer 32

dc, 66 bytes

[2~rd0<B]dsBxz1-si[1+z:tz0<T]dsTx0dsj[ljd1+dsj;tr3r^*+ljli>M]dsMxp

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[2~rd0<B]dsBx is a macro that uses dc's quotient/remainder integer division, ~ to continuously break divide by two, breaking n down to individual binary bits on the stack. It will always leave one extra zero, so we subtract one from the stack depth and store this total length in i with z1-si.

[1+z:tz0<T]dsTx is a macro that adds one to whatever is on the stack, and then pops that value, storing it at index(stack depth) in array t. This basically means that if we started with the binary 1101, t now holds 2, 1, 2, 2, assuming 1-indexing. 0dsj puts a zero on the stack so we don't get a stack empty error when we do our first addition, and it stores a zero in register j as well.

[ljd1+dsj;tr3r^*+ljli>M]dsMx is, unfortunately, a lot of counter nonsense. We need register j for two things - pull element (j+1) from array t, and multiply it by 3^j. We start macro M by putting j on the stack and duplicating it. We increment the new copy, duplicate it, and store it back into j. With the copy of (j+1) we left behind, we pull from array t. Swap so that our original j is at the top of the stack, do the necessary base-three math with 3r^*, add this 'bit' to our total, and then compare j with i to see if we still have more 'bits' to do. At the very end, we print.

Answer 33

Excel, 54 bytes

=DECIMAL(SUBSTITUTE(SUBSTITUTE(BASE(A1,2),1,2),0,1),3)

A fairly straightforward translation of the problem. Takes input from A1. I tried some cuter things with bit math like

=SUM(IF(ROW(1:31)<LOG(A1,2)+1,POWER(3,ROW(1:31)-1)*(1+MOD(BITRSHIFT(A1, ROW(1:31)-1),2))))

because nesting SUBSTITUTE always feels wasteful, but couldn't get nearly as short.

Answer 34

Ruby, 28 bytes

f=->x{x>0?x%2+3*f[x>>1]+1:0}

black magic, no idea how it works

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Answer 35

Jstx, 7 bytes

£?☺å♥£F

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Explanation

£? # Push the first stack value in binary.
☺  # Push literal 1
å  # Push the second stack value with all characters arithmetically shifted by the first stack value.
♥  # Push literal 3
£F # Push the second stack value in base first stack value converted to decimal.

Answer 36

x86, 26 bytes

Rewrite of the recursive formula as a tail-recursive stack function, since function calls are expensive. Uses __fastcall convention (input in ecx, output in eax).

The cmp %esp,%ebp/je sum1 branch is done to prevent an extra multiplication of 3 from occurring. It might save bytes to avoid this branch. Reordering the multiplication makes this unnecessary and saves 4 bytes.

.section .text
.globl main
main:
        mov     $10, %ecx

start:
        mov     %esp, %ebp          # Save sp 
build:
        mov     %ecx, %eax
        and     $1, %eax
        inc     %eax                # n%2 + 1
        push    %eax                # push n%2 + 1
        sar     %ecx                # n >>= 1
        jnz     build               # do while (n)

        xor     %eax, %eax          # sum = 0
sum0:
        lea     (%eax,%eax,2),%eax  # sum *= 3
        pop     %ebx
        add     %ebx, %eax          # pop stack, add to sum
        cmp     %esp, %ebp
        jnz     sum0                # do while stack non-empty


        ret

Hexdump:

00000039  89 e5 89 c8 83 e0 01 40  50 d1 f9 75 f5 31 c0 8d  |.......@P..u.1..|
00000049  04 40 5b 01 d8 39 e5 75  f6 c3                    |.@[..9.u..|

---

Assembly-friendly python:

stack = []
while n:
    stack.append(n%2 + 1)
    n //= 2

while stack:
    s *= 3
    s += stack.pop()

Answer 37

AWK, 43 bytes

func n(x){return x<1?0:x%2+1+3*n(int(x/2))}

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I discovered that AWK has rshift and lshift functions. They use more bytes than just doing a divide and cast, but might be useful sometime. :)

Answer 38

Bash + GNU utilities, 28

dc -e?2op|tr 01 12|dc -e3i?p

Input read from STDIN.

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Answer 39

Pari/GP, 25 bytes

f(n)=if(n,n%2+1+3*f(n\2))

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Answer 40

brainfuck, 50 bytes

+<,[[>-]++>[<]<[>+[<]<]>>-]>[>]<-<<[>[-<+++>]<<]>.

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Input and output as code points, and assumes arbitrary size cells otherwise it can't get past n=31.

Answer 41

Julia 0.6, 34 bytes

x->parse(Int,map(x->x+1,bin(x)),3)

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Tho it seems like this is the cool way

Julia 0.6, 25 bytes

f(n)=n<1?0:n%2+1+3f(n÷2)

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Answer 42

Zsh, 37 bytes

a=3#$[[##2]$1];echo $[a+${a//[01]/1}]

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• $[...] is another (deprecated?) form of $(( ))
• [#n] in arithmetic expansion sets the output base but includes the base in output (so you get 2#1001), [##n] omits the base in output: 1001.
• n# in arithmetic expansion sets the input base
• ${var//pat/rep} replaces all matches of pat with rep.

Answer 43

Go, 54 bytes

func f(n int)(o int){if n>0{o=n%2+1+3*f(n>>1)};return}

Base solution is the same as the one that everyone else uses.

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