# The first n numbers without consecutive equal binary digits

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.

• Given your own MATL solution, is it acceptable to output the result in reverse order? – Shaggy Jan 22 '18 at 12:34
• Yes, as long as it's sorted. @Shaggy – Stewie Griffin Jan 22 '18 at 12:42
• Pushing my luck here, but would this output format be acceptable [85,[42,[21,[10,[5,[2,[1,0]]]]]]]? – Shaggy Jan 22 '18 at 18:05

# 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.

# 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


# 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.

• ḶḂḄƤ prefix quick was made for this – miles Jan 22 '18 at 13:11
• No need for the prefix quick even - Ḷ€ḂḄ would also work. – Jonathan Allan Jan 22 '18 at 14:50

# MATL, 5 bytes

:WI/k


### 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


Try it online!

# 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


# Python 2, 4537 36 bytes

-3 bytes thanks to user202729
-1 byte thanks to mathmandan

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


Try it online!

• Doubling s is the same as adding s to itself, so I believe you could do s+=s+~s%2 to save a byte. – mathmandan Jan 22 '18 at 21:38

# Python 3, 68615448 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.

Try It Online

• Thanks for that -1 byte. And I think I can't replace if else with and or? – Manish Kundu Jan 22 '18 at 10:12
• Okay done that already. – Manish Kundu Jan 22 '18 at 10:13
• Because x == 0 is equivalent to not x if x is an integer, swap the operands (i.e., x if c else y = y if not c else x) will save some more bytes. – user202729 Jan 22 '18 at 10:16
• You can also drop i%2 and use 1-r%2 instead – Rod Jan 22 '18 at 10:25
• Then you don't need to keep track of i. – user202729 Jan 22 '18 at 10:29

## Husk, 7 bytes

mḋḣ↑Θİ¬


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

### Explanation

     İ¬   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.


# APL (Dyalog Unicode), 11 bytesSBCS

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

# 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  • for the scoring i would say: 27 + 1 (-n) = 28 bytes, because to run a perl one-liner, one should use -e and to use 5.10 you just need to use -E, which is the same length – Nahuel Fouilleul Jan 22 '18 at 12:18 • can save 3 bytes using $|-- instead of ($.^=1) – Nahuel Fouilleul Jan 22 '18 at 12:22 # Haskell, 33 bytes l=1:0:l ($scanl((+).(2*))0l).take


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# APL (Dyalog), 7 bytes

⌊3÷⍨2*⍳


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# APL (Dyalog), 11 bytes

(2⊥2|⍳)¨1+⍳


Try it online!

Uses ⎕IO←0.

• Returns one-too-many terms. Run under ⎕IO←0 (but claim it is 1-indexed!) and change 0, to 1+: (2⊥2|⍳)¨1+⍳ – Adám Jan 22 '18 at 11:28

# 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));}


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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

• Welcome to PPCG! :) I'm not a C guy but you may be able to glean some hints from Steadybox's solution. – Shaggy Jan 22 '18 at 12:30
• Ok that makes more sense now, I've included the entire program when all I need is a function and the rest can be done in a footer. I guess this can be significantly improved then. – Minerscale Jan 22 '18 at 12:33
• Welcome to PPCG! You can save a byte by removing i++ and changing i&1 to i++&1. Also, although as global variables i and j are initialized to zero initially, they need to be initialized inside the function, because function submissions have to be reusable. – Steadybox Jan 22 '18 at 12:51
• Even better, it's possible to save 2 more bytes, eliminating the ternary entirely. – user202729 Jan 22 '18 at 13:39
• 50 bytes: i,j;f(c){for(i=j=0;i<c;)printf("%d ",j+=j+i++%2);} Try it online! – Steadybox Jan 22 '18 at 14:04

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

(takel)
l=0:[2*a+1-amod2|a<-l]


Try it online!

• You can replace otherwise with e.g. 1>0 (otherwise == True) – flawr Jan 22 '18 at 9:57
• To golf it even more, you can use the guard for assigning something, e.g. like this: Try it online! – flawr Jan 22 '18 at 9:59
• PS: Also check out tips for golfing in haskell as well as our haskell-chatroom of monads and men. – flawr Jan 22 '18 at 10:01
• You need to make a function that returns the first n elements of the list where n is the argument. – totallyhuman Jan 22 '18 at 10:59
• Yes, exactly. I can recommend to have a look at our Guide to Golfing Rules in Haskell, which tries to capture the current consensus on what is allowed and what isn't. – Laikoni Jan 22 '18 at 16:58

# Ruby, 26 bytes

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


Try it online!

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.

# 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!

## Retina, 28 bytes

)K0
"$+"+¶<.+$.(*__2*$-1*  Try it online! 0-based, so input n gives the first n+1 results. ### Explanation Uses the recursion from OEIS: a(n) = a(n-1) + 2*a(n-2) + 1  Let's go through the program: )K0  This is a constant stage: it discards the input and sets the working string to 0, the initial value of the sequence. The ) wraps this stage in a group. That group itself does nothing, but almost every stage (including group stages) records its result in a log, and we'll need two copies of the 0 on that log for the program to work. "$+"+¶<.+
$.(*__2*$-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. # Brain-Flak, 36 bytes {([()]{}<((({}<>)<>){}([{}]()))>)}<>  Try it online! ## 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  # Ruby42 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}}  Try it online! • Good work! I like your formula ^^ – Unihedron Jan 22 '18 at 11:16 • save 3 bytes: ->x{a=0;x.times{a-=~a+p(a)%2}} – G B Jan 24 '18 at 9:52 # ><>, 22 + 3 (-v flag) bytes 0:nao::1+2%++$1-:?!;$!  Try it online! ### 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.  # Java 8, 1158180 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  # 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 }  # 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)  Try it online! Returns in reverse order. # 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


# Japt, 1097 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


(takemap a[0..])
a 0=0
a 1=1
a n=a(n-1)+2*a(n-2)+1


Try it online!

# MATL, 7 bytes

:&+oRXB


Try it online!

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.

• Good idea, &+ – Luis Mendo Jan 22 '18 at 11:13

# 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|$.^=1}  Try it online! ## Explanation # while gets(); -- assumed by -n v=0 # First element of the sequence ?1.upto($_){   # 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  • Interesting. It's possible in 27 bytes, though. – Eric Duminil Jan 22 '18 at 15:38 • Nice! Seems that we all got outgolfed by 26b though. – Unihedron Jan 23 '18 at 0:44 # AWKa=0, 31 bytes {for(;$1--;a=a*2+1-a%2)print a}


Try it online!

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.

• I like your formula ^^ – Asone Tuhid Jan 22 '18 at 16:42

# C, 52 bytes

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


1-indexed

Try it online!

# 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

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


Try it online!

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.

• When I run it with  (0d018) as par the test case, your code prints  *UªUªUªUªUªUª (0x01 02 05 0a 15 2a 55 aa 55 aa 55 aa 55 aa 55 aa 55 aa; 0d001 002 005 010 021 042 085 170 085 170 085 170 085 170 085 170 085 170) :( tio.run/##SypKzMxLK03O/… – Unihedron Jan 22 '18 at 11:51
• Ok, seems it is a cell size problem. I think either your code should adapt to big integers or you need to specify the implementation that would run your code properly, but the default of 8-bit cells isn't enough – Unihedron Jan 22 '18 at 11:53
• Forgot about that, thanks @Unihedron! I'll have a think about an 8-bit version, probably outputting in unary. – Jo King Jan 22 '18 at 12:38
• Using an interpreter with 32-bit cells, it works. Though I think I might have a try at a bitinteger (8bit) version myself if you haven't by the weekend :D – Unihedron Jan 22 '18 at 13:25