The first n numbers without consecutive equal binary digits

Question

The sequence contains the decimal representation of the binary numbers of the form: 10101..., where the n-th term has n bits.

The sequence is probably easiest to explain by just showing the relationships between the binary and decimal representations of the numbers:

0       ->  0
1       ->  1
10      ->  2
101     ->  5
1010    ->  10
10101   ->  21
101010  ->  42

Challenge:

Take an input integer n, and return the first n numbers in the sequence. You may choose to have the sequence 0-indexed or 1-indexed.

Test cases:

n = 1   <- 1-indexed
0

n = 18
0, 1, 2, 5, 10, 21, 42, 85, 170, 341, 682, 1365, 2730, 5461, 10922, 21845, 43690, 87381

Explanations are encouraged, as always.

This is OEIS A000975.

Answer 1

Python 2, 36 bytes

lambda n:[2**i*2/3for i in range(n)]

Try it online! Explanation: The binary representation of \$\frac{2}3\$ is 0.101010101... so it simply remains to multiply it by an appropriate power of 2 and take the integer portion.

Answer 2

05AB1E, 4 bytes

2 bytes saved using Neil's 2/3 trick

Lo3÷

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Explanation

L      # push range [1 ... input]
 o     # raise 2 to the power of each
  3÷   # integer division of each by 3

---

05AB1E, 6 bytes

TRI∍ηC

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Explanation

T        # push 10
 R       # reverse it
  I∍     # extend to the lenght of the input
    η    # compute prefixes
     C   # convert each from base-2 to base-10

Answer 3

Jelly, ... 4 bytes

Thanks miles for -1 byte!

ḶḂḄƤ

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Explanation:

Ḷ     Ḷowered range, or UnḶength. Get [0, 1, 2, 3, ..., n-1]
 Ḃ    Ḃit. Get the last bit of each number. [0, 1, 0, 1, ...]
   Ƥ  for each Ƥrefixes [0], [0, 1], [0, 1, 0], [0, 1, 0, 1], ...
  Ḅ   convert it from Ḅinary to integer.

---

Jelly, 4 bytes

Jonathan Allan's version.

Ḷ€ḂḄ

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Ḷ     Ḷowered range, or UnḶength.
 €    Apply for €each. Automatically convert the number n
      to the range [1,2,..,n]. Get [[0],[0,1],[0,1,2],..].
  Ḃ   Ḃit. Get the last bit from each number.
      Current value: [[0],[0,1],[0,1,0],..]
   Ḅ  Convert each list from Ḅinary to integer.

---

A version based on Neil's 2/3 trick gives 5 bytes, see revision history.

Answer 4

MATL, 5 bytes

:WI/k

Based on Neil's answer.

Explanation

:       % Implicit input, n. Push range [1 2 ... n]
W       % 2 raised to that, element-wise. Gives [2 4 ...2^n] 
I       % Push 3
/       % Divide, element-wise
k       % Round down, element-wise. Implicit display

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---

MATL, 9 bytes

:q"@:oXBs

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Explanation

:       % Implicit input n. Range [1 2 ... n]
q       % Subtract 1, element-wise: gives [0 1 ... n-1]
"       % For each k in [0 1 ... n-1]
  @     %   Push k
  :     %   Range [1 2 ... k]
  o     %   Modulo 2, element-wise: gives [1 0 1 ...]
  XB    %   Convert from binary to decimal
  s     %   Sum. This is needed for k=0, to transform the empty array into 0
        % Implicit end. Implicit display

Answer 5

Python 3, 68 61 54 48 43 bytes

c=lambda x,r=0:x and[r]+c(x-1,2*r+~r%2)or[]  

Thanks to user202729 for helping save 19 bytes and ovs for helping save 6 bytes.

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Answer 6

Python 2, 45 37 36 bytes

^{-3 bytes thanks to user202729
-1 byte thanks to mathmandan}

s=0
exec"print s;s+=s+~s%2;"*input()

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Answer 7

Husk, 7 bytes

mḋḣ↑Θݬ

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1-based, so input n gives the first n results.

Explanation

     ݬ   The infinite list [1, 0, 1, 0, 1, ...]
    Θ     Prepend a zero.
   ↑      Take the first n elements.
  ḣ       Get the prefixes of that list.
mḋ        Interpret each prefix as base 2.

Answer 8

Haskell, 47 40 53 49 44 40 34 bytes

-4 bytes thanks to user202729
-6 bytes thanks to Laikoni

(`take`l)
l=0:[2*a+1-a`mod`2|a<-l]

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Answer 9

APL (Dyalog), 7 bytes

⌊3÷⍨2*⍳

Try it online!

---

APL (Dyalog), 11 bytes

(2⊥2|⍳)¨1+⍳

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Uses ⎕IO←0.

Answer 10

Perl v5.10 -n, 24+1 bytes

-3 bytes thanks to Nahuel Fouilleul!

say$v=$v*2|$|--while$_--

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Same logic as my Ruby version, but shorter because perl is more concise. For some odd reason, print wouldn't do a seperator (dammit!), so I had to use say from v5.10; in order for this to run, I'm not sure how to score this, so I'm leaving it out for now?...

Explanation

say    # Like shouting, but milder.
  $v = $v*2 | $|-- # Next element is this element times 2 bitwise-OR
                   # with alternating 0 1 0 1..., so 0b0, 0b1, 0b10, 0b101...
                   # $. is OUTPUT_AUTOFLUSH, which is initially 0 and
                   #   setting all non-zero values seem to be treated as 1
  while $_-- # do for [input] times

Answer 11

Haskell, 33 bytes

l=1:0:l
($scanl((+).(2*))0l).take

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Answer 12

APL (Dyalog Unicode), 11 bytes^SBCS

Assumes ⎕IO (Index Origin) to be 0, which is default on many systems. Anonymous tacit prefix function. 1-indexed.

(2⊥⍴∘1 0)¨⍳

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⍳ ɩndices 0…n−1

(…)¨ apply the following tacit function to each

 ⍴∘1 0 cyclically reshape the list [1,0] to that length

 2⊥ convert from base-2 (binary) to normal number

Answer 13

C, 81 55 59 bytes

1 indexed.

i,j;f(c){for(i=j=0;i<c;)printf("%d ",i++&1?j+=j+1:(j+=j));}

Full program, less golfed:

i;j;main(c,v)char**v;{c=atoi(*++v);for(;i<c;i++)printf("%d ",i&1?j+=j+1:(j+=j));}

Try it online!

EDIT 2: I was under the assumption that functions didn't need to be reusable now that I think of it, it makes perfect sense that they would have to be reusable :P

EDIT: I was under the misconception that I had to include the entire program in the answer, turns out I only needed the function that does it. That's nice.

I'm decently sure I can shave off a few bytes here and there. I've already employed a few tricks. A large chunk of the program is dedicated to getting the argument and turning it into an int. This is my first code golf. If I'm doing anything wrong tell me :P

Answer 14

J, 9 bytes

[:#.\2|i.

How it works?

i. - list 0..n-1

2| - the list items mod 2

\ - all prefixes

#. - to decimal

[: - caps the fork (as I have even number (4) of verbs)

Try it online!

Answer 15

Ruby, 26 bytes

->n{(1..n).map{|i|2**i/3}}

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Beats all the older ruby answers.

Explanation

1/3 in binary looks like 0.01010101..., so If you multiply it by powers of two, you get:

n| 2^n/3
-+---------
1|0.1010101...
2|01.010101...
3|010.10101...
4|0101.0101...
5|01010.101...
6|010101.01...

But Ruby floors the numbers on int division, giving me the sequence I need.

Answer 16

Retina, 28 bytes

)K`0
"$+"+¶<`.+
$.(*__2*$-1*

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0-based, so input n gives the first n+1 results.

Explanation

Uses the recursion from OEIS:

a(n) = a(n-1) + 2*a(n-2) + 1

Let's go through the program:

)K`0

This is a constant stage: it discards the input and sets the working string to 0, the initial value of the sequence. The ) wraps this stage in a group. That group itself does nothing, but almost every stage (including group stages) records its result in a log, and we'll need two copies of the 0 on that log for the program to work.

"$+"+¶<`.+
$.(*__2*$-1*

There's a bunch of configuration here: "$+"+ wraps the stage in a loop. The "$+" is treated as a substitution, and $+ refers to the program's input, i.e. n. This means that the loop is run n times.

Then ¶< wraps each iteration in an output stage, which prints the stage's input with a trailing linefeed (so the first iteration prints the zero, the second iteration prints the first iteration's result and so on).

The stage itself replaces the entire working string with the substitution on the last line. That one makes use of an implicit closing parenthesis and implicit arguments for the repetition operator *, so it's actually short for:

$.($&*__2*$-1*_)

The stuff inside the parentheses can be broken up into three parts:

• $&*_: gives a string of a(n-1) _s.
• _: gives a single _.
• 2*$-1*_: gives a string of 2*a(n-1) _. The $-1 refers to the penultimate result in the result log, i.e. the loop iteration before the last. That's why we needed to copies of the zero on the log to begin with, otherwise this would refer to the program's input on the first iteration.

Then $.(…) measures the length of the resulting string. In other words, we've computed a(n) = a(n-1) + 1 + 2*a(n-2) by going through unary (not really though: $.(…) is lazy and doesn't actually evaluate its content if it can determine the resulting length directly through arithmetic, so this is even quite efficient).

The result of the final loop iteration (the n+1th element of the sequence) is printed due to Retina's implicit output at the end of the program.

Answer 17

Ruby 42 41 43 41 37 35 31 33 30 bytes

-2 bytes thanks to Unihedron

-3 bytes thanks to G B

->x{a=0;x.times{a-=~a+p(a)%2}}

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Answer 18

Brain-Flak, 36 bytes

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

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Explanation:

The next number in the sequence is obtained by n*2+1 or n*2+0.

{([()]{}< Loop input times
  (
   (({}<>)<>){} Copy n to other stack; n*2
   ([{}]())  i = 1-i
  ) push n*2 + i
>)} End loop
<> Output other stack

Answer 19

Japt, 10 9 7 6 bytes

All derived independently from other solutions.

1-indexed.

õ!²mz3

Try it

---

Explanation

õ        :[1,input]
 !²      :Raise 2 to the power of each
   m     :Map
    z3   :Floor divide by 3

Try it

---

7 byte version

õ_ou ì2

Try it

õ            :[1,input]
 _           :Pass each through a function
   o         :[0,current element)
    u        :Modulo 2 on above
      ì2     :Convert above from base-2 array to base-10

---

9 byte version

õ_îA¤w)n2

Try it

õ            :[1,input]
 _           :Pass each through a function
   A         :10
    ¤        :Convert to binary
     w       :Reverse
  î          :Repeat the above until it's length equals the current element
      )      :Close nested methods
       n2    :Convert from binary to base-10

Answer 20

Java 8, 115 81 80 52 bytes

n->{for(int i=2;n-->0;i*=2)System.out.println(i/3);}

Port of @Neil's Python 2 answer.
1-indexed and outputted directly, each value on a separated line.

Explanation:

Try it online.

n->{                           // Method with integer parameter and no return-type
  for(int i=2;                 //  Start integer `i` at 2
      n-->0;                   //  Loop `n` times:
      i*=2)                    //    Multiply `i` by 2 after every iteration
    System.out.println(i/3);}  //   Print `i` integer-divided by 3 and a new-line

---

Old 80 bytes answer:

n->{String t="",r=t;for(Long i=0L;i<n;)r+=i.parseLong(t+=i++%2,2)+" ";return r;}

1-indexed input and space-delimited String output

Explanation:

Try it online.

n->{                             // Method with integer parameter and String return-type
  String t="",r=t;               //  Temp and result-Strings, both starting empty
  for(Long i=0L;i<n;)            //  Loop from 0 to `n` (exclusive)
    r+=                          //   Append the result-String with:
       i.parseLong(        ,2);  //    Binary to integer conversion
                   t+=           //     append the temp-String with:
                      i  %2      //      current index `i` modulo-2
                       ++        //      and increase `i` by one afterwards
       +" ";                     //    + a space
  return r;}                     //  Return the result-String

Answer 21

><>, 22 + 3 (-v flag) bytes

0:nao::1+2%++$1-:?!;$!

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Explanation

The stack gets initialized with the loop counter.

0:nao                  : Push 0 to the stack, duplicate and print with a new line.
                         [7] -> [7, 0]
     ::1+              : Duplicate the stack top twice more then add 1 to it.
                         [7, 0] -> [7, 0, 0, 1]
         2%++          : Mod the stack top by 2 then add all values on the stack bar the loop counter.
                         [7, 0, 0, 1] -> [7, 1]
             $1-:?!;$! : Swap the loop counter to the top, minus 1 from it and check if zero, if zero stop the program else continue.

Answer 22

Perl 6,  35 30 27 25  20 bytes

{[\~](0,+!*...*)[^$_]».&{:2(~$_)}}

Try it (35)

{(0,{$_*2+|($+^=1)}…*)[^$_]}

Try it (30)

{(⅓X*(2,4,8…2**$_))».Int}

Try it (30)

{(⅔,* *2…*)[^$_]».Int}

Try it (27)

{((2 X**1..$_)X/3)».Int}

Try it (25)

{(2 X**1..$_)Xdiv 3}

Try it (20)

Expanded:

{
 (
  2                  # 2
    X**              # cross to the power of
       1..$_         # Range from 1 to the input (inclusive)
            )

             Xdiv    # cross using integer divide
                  3  # by 3
}

Answer 23

C, 47 46 bytes

a;f(n){for(a=0;n--;a+=a-~a%2)printf("%d ",a);}

The accumulator a begins with zero. At each step, we double it (a+=a) and add one if the previous least-significant bit was zero (!(a%2), or equivalently, -(~a)%2).

Test program

#include <stdlib.h>
#include <stdio.h>

int main(int argc, char **argv)
{
    while (*++argv) {
        f(atoi(*argv));
        puts("");
    }
}

Results

$ ./153783 1 2 3 4 5 6
0 
0 1 
0 1 2 
0 1 2 5 
0 1 2 5 10 
0 1 2 5 10 21 

Answer 24

Python 2, 33 bytes

s=2
exec"print s/3;s*=2;"*input()

Try it online!

---

Python 2, 34 bytes

f=lambda n:n*[f]and[2**n/3]+f(n-1)

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Returns in reverse order.

Answer 25

Ruby -n, 32 30+1 bytes

Since we have exactly 1 line of input, $. is godly convenient!

EDIT: I'm amazed that I managed to outgolf myself, but it seems using -n which counts as 1 (by rule 2 in default special conditions, since Ruby can be run with ruby -e 'full program' (thus -n is 1) all instances of gets which is only used once can be golfed down 1 char this way; I believe this is a milestone for ruby, please speak up if you disagree with this train of thought before I repeatedly reuse it in the future)

v=0
?1.upto($_){p v=v*2|$.^=1}

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Explanation

# while gets(); -- assumed by -n
v=0            # First element of the sequence
?1.upto($_){   # Do from "1" to "$LAST_READ_LINE" aka: Repeat [input] times
  p            # print expression
  v=v*2|$.^=1  # Next element is current element times two
               # bitwise-or 0 or 1 alternating
               # $. = lines of input read so far = 1 (initially)
}
# end           -- assumed by -n

Answer 26

JavaScript (ES7), 39 35 31 30 bytes

1-indexed with output in reverse order.

f=n=>n?[2**n/3|0,...f(--n)]:[]

---

Try it

o.innerText=(
f=n=>n?[2**n/3|0,...f(--n)]:[]
)(i.value=8);oninput=_=>o.innerText=f(+i.value)

<input id=i type=number><pre id=o>

---

35 byte version, without recursion

n=>[...Array(n)].map(_=>2**n--/3|0)

o.innerText=(f=
n=>[...Array(n)].map(_=>2**n--/3|0)
)(i.value=8);oninput=_=>o.innerText=f(+i.value)

<input id=i type=number><pre id=o>

Answer 27

Haskell, 52 bytes

(`take`map a[0..])
a 0=0
a 1=1
a n=a(n-1)+2*a(n-2)+1

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Answer 28

MATL, 7 bytes

:&+oRXB

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Explanation:

         % Implicitly grab input, n
:        % Range: 1 2 ... n

 &+      % Add the range to itself, transposed
         % 2 3 4 5 ...
         % 3 4 5 6 ...
         % 4 5 6 7 ...
         % 5 6 7 8 ...

   o     % Parity (or modulus 2)
         % 0 1 0 1 ...
         % 1 0 1 0 ...
         % 0 1 0 1 ...
         % 1 0 1 0 ...

    R    % Upper triangular matrix:
         % 0 1 0 1
         % 0 0 1 0
         % 0 0 0 1
         % 0 0 0 0

    XB   % Convert rows to decimal:
         % [5, 2, 1, 0]
         % Implicitly output

The output would be 0, 1, 2, 5 ... if P was added to the end (flip), making it 8 bytes.

Answer 29

brainfuck, 40 bytes

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

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0-indexed. Input as char code, output as unary with null bytes separating series of char code 1s. Assumes 8-bit cells unless you want to input over 255. Assumes negative cells, though this could be fixed at the expense of several bytes.

Previously, 50 bytes

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

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Inputs as char code, outputs as char code. 1-indexed. Probably could be golfed a little.

@Unihedron points out I forgot to specify that this needs infinite sized cells, otherwise it tops out at the 8th number.

Answer 30

AWK a=0, 31 bytes

{for(;$1--;a=a*2+1-a%2)print a}

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Uses the formula shamelessly stolen from this other Ruby answer.

While not having a=0 would work (awk treats "empty" as 0), the first element of 0 won't get printed and instead be an empty line, which while I would argue is a valid output probably won't pass, so there's a=0 which can be inserted as command line argument.

Answer 31

C, 52 bytes

i,k;f(n){for(i=k=0;i<n;k=++i%2+2*k)printf("%d ",k);}

1-indexed

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Answer 32

R, 21 bytes

cat(2^(1:scan())%/%3)

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Based on the same algorithm as many here. 1-indexed.

R, 37 bytes

for(i in 0:scan())cat(F<-2*F+i%%2,"")

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0-indexed. Doubling and adding n mod 2 at each iteration yields the correct result. F is initialized to zero.

Answer 33

Julia 0.6, 15 14 bytes

!n=2.^(1:n)÷3

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Using the 2/3 method. ÷ does integer division in Julia and . is element-wise function application.

-1 Byte thanks to Dennis.

Answer 34

Ruby, 27 bytes

->n{n.times{|i|p 2**i*2/3}}

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It's just a Ruby port of this awesome Python answer.

Answer 35

Actually, 14 bytes

r⌠;"10"*H2@¿⌡M

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Explanation:

r⌠;"10"*H2@¿⌡M
r               range(0, input)
 ⌠;"10"*H2@¿⌡M  map (for n in range):
   "10"*          repeat "10" n times
  ;     H         first n characters
         2@¿      interpret as binary integer

Answer 36

Pyt, 5 bytes

1←ř«₃

Explanation:

1        Pushes 1
 ←       Gets input
  ř      Pushes [1,2,...,input]
   «     Bit-shift 1 to the left by each element in the array
    ₃    Python 2-style division by 3 (2^k/3)

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Answer 37

Octave, 20 bytes

@(x)fix(2.^(1:x)./3)

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Using @Neils Python method (+1 to him) saves a heck of a lot of bytes.

---

Previous answer (independent creation):

Octave, 49 40 bytes

@(n)arrayfun(@(x)sum(2.^(x-1:-2:0)),0:n)

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Basically for each value x in 0:n where n is the input (0-indexed), we take a range of x-1:-2:0, and raise 2 to the power of each element in the range. The range results in alternating powers of 2, starting with an empty array [] for 0, then [],[1] for 0:1, then [],[1],[1 4] for 0:2, and so on.

If we then sum each of the produced alternating powers of two, we end up with the required sequence. This only works because in Octave the sum of an empty array is 0, so we can produce the first number 0 by producing no powers of two.

The resulting array, which contains all numbers in the pattern up to and including n is then returned.

Answer 38

JavaScript (Node.js), 44 bytes

In ascending order. Simple recursion. 1-indexed.

f=(n,i=0,a=[])=>n?f(n-1,~i&1+i*2,[...a,i]):a

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JavaScript (Node.js), 43 41 38 35 bytes

... or return as string. Still in ascending order. 0-indexed.

f=(n,i=0)=>n?i+[,f(n-1,~i&1+i*2)]:i

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JavaScript (Node.js), 40 bytes

In ascending order. 2**n/3 trick. 1-indexed.

n=>Array(n).fill(i=0).map(_=>2**++i/3|0)

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Answer 39

Ruby, 72 68 61 bytes

->n{a=[0]*n;n.times{|i|i.times{|j|a[i]|=1<<j if i%2!=j%2}};a}

Explained:

def f(n)
  a = [0] * n
  n.times do |i|
    i.times do |j|
      if i.even? != j.even?
        a[i] |= (1 << j)
      end
    end
  end
  a
end

This approach uses n'th bit installation using x | (1 << n). We start from the last bit and proceeding to the first, setting each 2'nd, alternating ones and zeros 'even?' check tells where to start.

Try Now!

I am new in both code golf and Ruby, so any comments will be appreciated!

Answer 40

Python 3 53 bytes

lambda n:[int(('0'+'10'*i)[:i+1],2)for i in range(n)]

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Answer 41

Red, 71 67 bytes

f: func[n][d: 0 loop n - 1[print d d: d * 2 + either odd? d[0][1]]]

1-indexed

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And here's the Red impementation of Neil's 2/3 trick:

Red, 51 bytes

f: func[n][repeat i n[print to-integer 2 ** i / 3]]

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Answer 42

SNOBOL4 (CSNOBOL4), 64 bytes

	N =INPUT
I	X =LT(X,N) X + 1	:F(END)
	OUTPUT =2 ^ X / 3	:(I)
END

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1-indexed. Uses the 2^i/3 method.

Answer 43

C 52 bytes

i,a;f(n){for(;i++<n;){printf("%d ",a);a=a*2+1-a%2;}}

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Without error messages (61 bytes):

int i,a;void f(n){for(;i++<n;){printf("%d ",a);a=a*2+1-a%2;}}

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Answer 44

Clean, 61 bytes

import StdEnv
$i=[sum[2^(n-p)\\p<-[1..n]|isOdd p]\\n<-[0..i]]

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Answer 45

Perl 5, 44 + 2 (-pa) = 46 bytes

$\+=(length sprintf'%b',$_)/$F[0]while$_--}{

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Answer 46

clojure, 61 bytes

(fn f[n r](if(> n 0)(cons r(f(- n 1)(+ r r 1(-(mod r 2)))))))

Usage:

user> (f 10 0)
(0 1 2 5 10 21 42 85 170 341)

Answer 47

Dart, 49 bytes

f(n,{a:0})=>new List.generate(n,(x)=>a=a<<1|x&1);

Use as

main() {
 print(f(31));
}

See DartPad

Answer 48

Jelly, 5 bytes

R2*:3

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Took Emigna's strategy and ported it to Jelly.

Answer 49

Scala, 64 bytes

val f=(n:Int)=>Stream from 1 map(i=>(1<<i)/3)take n mkString " "

1-indexed. A call to f(7) for example would return 0 1 2 5 10 21 42.

Answer 50

Jelly, 13 bytes

Rṁ@⁾10VDḄ
ḶÇ€

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