# Half, Half Half, and, Half

Consider the following number sequence:

$0, \frac{1}{2}, \frac{1}{4}, \frac{3}{4}, \frac{1}{8}, \frac{3}{8}, \frac{5}{8}, \frac{7}{8}, \frac{1}{16}, \frac{3}{16}, \frac{5}{16}, \frac{7}{16}, \frac{9}{16}, \frac{11}{16}, \frac{13}{16}, \frac{15}{16}, \frac{1}{32}, \frac{3}{32}, \frac{5}{32}, \dots$

It enumerates all binary fractions in the unit interval $[0, 1)$.

(To make this challenge easier, the first element is optional: You may skip it and consider the sequence starts with 1/2.)

Write a program (complete program or a function) which...

Choose one of these behaviors:

• Input n, output nth element of the sequence (0-indexed or 1-indexed);
• Input n, output first n elements of the sequence;
• Input nothing, output the infinite number sequence which you can take from one by one;

### Rule

• Your program should at least support first 1000 items;
• You may choose to output decimals, or fractions (built-in, integer pair, strings) as you like;
• Input / Output as binary digits is not allowed in this question;
• This is , shortest codes win;
• Standard loopholes disallowed.

### Testcases

input output
1     1/2     0.5
2     1/4     0.25
3     3/4     0.75
4     1/8     0.125
10    5/16    0.3125
100   73/128  0.5703125
511   511/512 0.998046875
512   1/1024  0.0009765625


These examples are based on 0-indexed sequence with the leading 0 included. You would need to adjust the input for fitting your solution.

• OEIS A006257
• Josephus problem: $a_{2n} = 2a_n-1, a_{2n+1} = 2a_n+1$. (Formerly M2216)
• 0, 1, 1, 3, 1, 3, 5, 7, 1, 3, 5, 7, 9, 11, 13, 15, 1, 3, 5, ...
• OEIS A062383
• $a_0 = 1$: for $n>0$, $a_n = 2^{\lfloor log_2n+1 \rfloor}$ or $a_n = 2a_{\lfloor \frac{n}{2} \rfloor}$.
• 1, 2, 4, 4, 8, 8, 8, 8, 16, 16, 16, 16, 16, 16, 16, 16, 32, 32, 32, ...
• A006257(n)/A062383(n) = (0, 0.1, 0.01, 0.11, 0.001, ...) enumerates all binary fractions in the unit interval [0, 1). - Fredrik Johansson, Aug 14 2006

• "Input nothing, output the infinite number sequence one by one" Does it have to be one-by-one, or are we also allowed to output an infinite list (possible in Haskell, Elixir, 05AB1E, etc.)? Sep 14, 2018 at 15:14
• Can I output a list of strings? e.g. "1/2" "1/4" "1/8"... Sep 14, 2018 at 15:21
• @KevinCruijssen Infinite list is fine as long as you can take n elements from it later.
– tsh
Sep 15, 2018 at 5:45
• @Barranka I think it is acceptable. That is nothing different to print fractions to stdout.
– tsh
Sep 15, 2018 at 5:46
• When you say Input / Output as binary numbers is not allowed, you mean we can't write a function that returns a pair if ints, or a double in a language / implementation where double uses IEEE binary64 format? I hope you don't mean was have to parse an ASCII string if we want to take an integer input? Normal integer types are binary in languages like C. Or do you mean the input/output can't be an array or string of integer or ASCII zeros/ones? Sep 17, 2018 at 5:41

pred.until(<2)(/2).(+0.5)


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Outputs decimals, one-indexed without the initial zero term.

Adds 0.5 to the input, then halves until the results is below 2, then subtracts 1. Using a pointfree expression saves 1 bytes over

f n=until(<2)(/2)(n+0.5)-1


# Java 10, 68 64 bytes

First try at code golf!

Option 1: find the n-th element (1-indexed)

-4 bytes thanks to @Kevin Cruijssen

n->{int x=0;for(;n>>++x!=1;);return((~(1<<x)&n)*2.+1)/(1<<x+1);}


This is an anonymous method that finds the n-th term by removing the most significant bit from n, doubling it and adding one, then dividing by the next highest power of 2.

Try it online!

Code walkthrough:

n->{                      // builds anonymous function with input n
int x=0;                  // stores floor of log(n) (base 2) for most significant digit
for(;n>>++x!=1;);         // calculates floor of log(n) by counting right shifts until 1
return((~(1<<x)&n)        // removes most significant digit of n
*2.+1)                     // multiplies 2 and adds 1 to get the odd numerator
/(1<<x+1);}               // divides by the next highest power of 2 and returns


Will edit if it's necessary to print the final value instead of returning it.

• Welcome to PPCG, good to have you with us :) Sep 14, 2018 at 19:39
• Hi, welcome to PPCG! Great first answer, +1 from me. Currently it's the same byte-count as my Java answer, but you can still golf some parts of your answer to make it shorter than mine: The {} after the loop can be a ; instead; you can remove the space after the return; 2.0 can be 2.; And changing the n>>x!=1;x++, 1<<x, and 1<<x+1 to n>>x++!=1;, 1<<x-1, 1<<x respectively also saves a byte. Try it online: 64 bytes. Enjoy your stay! Sep 14, 2018 at 20:43
• Oh, and if you haven't seen it yet: Tips for golfing in Java and Tips for golfing in <all languages> are both pretty interesting to read through. :) Sep 14, 2018 at 20:44
• See my 30 bytes answer, originally based on yours but golfed and golfed and golfed. Sep 17, 2018 at 9:36

# MathGolf, 5 4 bytes

╫\╨]


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## How it would look like with the operator working correctly

╫\)╨]   (")" adds 1 to TOS, making rounding behave as expected)


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## Explanation

╫     Left-rotate all bits in input
\    Swap top two elements on stack, pushing the input to the top
╨   Round up to nearest power of 2
]  Wrap in array (just for pretty printing)


I took my inspiration from this question to solve the problem, my "own" solution was around 10-12 bytes I think.

I had intended for the round up to closest power of 2 to return the number itself if it was a number of two, but due to a mistake it rounds to the next power of two (e.g. 4 -> 8 instead of 4 -> 4). This will have to be fixed later, but now it saves me one byte.

• I don't know MathGolf but if the ] serves no other purpose than formatting the output, I'd say you don't need to include it in your byte count. Sep 14, 2018 at 17:17
• I was unsure about it. Since the stack is printed on output as a joined string, it outputs a stack with the numbers 1 and 2 as 12. If that still counts I'll remove a byte
– maxb
Sep 14, 2018 at 17:21
• I think you should leave it in. It sometimes saves a byte to output the stack as one string, sometimes it will cost you a byte. Sep 14, 2018 at 17:40
• @H.PWiz that was my original thinking, since it seems fair to use the strengths of your language. Some languages only print the top of the stack when finished, some print it as a list. Usually it's a 1 byte difference, but it's part of the challenge.
– maxb
Sep 14, 2018 at 18:52

# Java 10, 89857069 68 bytes

v->{for(float j,t=2;;t*=2)for(j=1;j<t;j+=2)System.out.println(j/t);}


Port of @Emigma's 05AB1E answer, so outputs decimals indefinitely as well.
-15 bytes thanks to @Arnauld.

Try it online.

Explanation:

v->{                      // Method with empty unused parameter and no return-type
for(float j,t=2;;       //  Loop t from 2 upwards indefinitely,
t*=2)  //  doubling t after every iteration
for(j=1;j<t;          //   Inner loop j in the range [1, t),
j+=2)     //   in steps of 2 (so only the odd numbers)
System.out.println( //    Print with trailing new-line:
j/t);}            //     j divided by t

• How often can I say that I half your byte count? Well, I think this is the first time ;-) Sep 17, 2018 at 9:41
• @OlivierGrégoire Dang, now that is an impressive Java answer. :) I saw your 37-byte version as comment on TCFP's answer, but you even stripped off more bytes. It looks so extremely simple now in your 30-byte version, but it's still ingenious how you've golfed it from the initial version. Well done! Sep 17, 2018 at 10:03

{($_+.5)/2**.msb-1}  Try it online! # Java (JDK 10), 30 bytes n->(n+.5)/n.highestOneBit(n)-1  Try it online! Returns the nth item in the sequence. This answer is originally a succession of golfs of TCFP's Java answer. In the end, the golfs didn't look like the original answer anymore (though the math used is the same) so I decided to post the golfs as a separate answer instead of simply commenting on the TCFP's answer. So if you like this answer, go upvote TCFP's answer as well! ;-) Intermediate golfs were: n->{int x=0;for(;n>>++x!=1;);return((~(1<<x)&n)*2.+1)/(1<<x+1);} // 64 bytes (TCFP's answer when I started golfing) n->{int x=0;for(;n>>++x!=1;);x=1<<x;return((~x&n)*2.+1)/x/2;} // 61 bytes n->{int x=n.highestOneBit(n);return((~x&n)*2.+1)/x/2;} // 54 bytes n->{int x=n.highestOneBit(n);return((~x&n)+.5)/x;} // 50 bytes n->((n&~(n=n.highestOneBit(n)))+.5)/n // 37 bytes n->(n-(n=n.highestOneBit(n))+.5)/n // 34 bytes n->(n+.5)/n.highestOneBit(n)-1 // 30 bytes, current score  • And I was sitting here thinking my answer was as short as I could make it, you come along and cut it by more than half! Amazing stuff, definitely deserves a +1 from me. – TCFP Sep 17, 2018 at 13:25 • @TCFP It was an iterative process over the span of several hours. I actually posted each intermediate golfing as a comment to your answer, but deleted them as I found better golfs. Thanks for the praise ;-) Sep 17, 2018 at 13:32 # Python 3, 33 bytes lambda n:(8*n+4)/2**len(bin(n))-1  Try it online! Outputs decimals, one-indexed without the initial zero term. # 05AB1E, 11 8 bytes Saved 3 bytes thanks to Kevin Cruijssen. ∞oDÅÉs/˜  Try it online! Explanation ∞ # start an infinite list [1... o # calculate 2**N D # duplicate ÅÉ # get a list of odd numbers up to 2**N s/ # divide each by 2**N ˜ # flatten  • -1 byte by using ∞ (infinite list starting at 1): ∞oεDÅÉs/}˜ Sep 14, 2018 at 15:12 • @KevinCruijssen: Cool! That's a command I hadn't seen before. Thanks :) Sep 15, 2018 at 8:55 • Ah, and nice way of saving two more bytes due to the implicit mapping.. Hadn't even thought about that, lol.. Sep 15, 2018 at 9:41 • :O how is this possible. You condensed the ~2 page question into 8 bytes. Sep 17, 2018 at 2:21 • I was thinking using the prime numbers and a list of [1,2,4,4,8,8,8,8,16,16,...,2**n] and prepending the correct indexed prime followed by a /... But that wasn't working so well. Well, but not 8-bytes well. Something like 9LoDÅP)ζ. Oct 26, 2018 at 16:02 # Jelly, 9 bytes Bṙ1Ḅ,æċ2$


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# PowerShell, 40 bytes

for($i=2;;$i*=2){1..$i|?{$_%2}|%{$_/$i}}


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Outputs the infinite sequence as decimal values. Given language limitations, will eventually run into precision problems, but easily handles the first 1000 entries.

Starts by setting $i=2, then enters a for loop. Each iteration, we construct a range from 1..$i and pull out the odd values with |?{$_%2}. Those are fed into their own inner loop, where we divide each to get the decimal |%{$_/$i}. Those are left on the pipeline and output when the pipeline is flushed after every for iteration. Each iteration we're simply incrementing $i by \$i*=2 to get the next go-round.

Edit: -3 bytes thanks to @Delfad0r.

[(y,2^x)|x<-[1..],y<-[1,3..2^x]]


This is an infinite list of integer pairs.

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s=(1,2):[(i*2+u,j*2)|(i,j)<-s,u<-[-1,1]]


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Infinite sequence as pairs of integers (starting from (1,2)).

Quite a bit longer than @nimi's answer, but the approach is completely different, so I decided to post it anyway.

This solution is based on the following observation.

Consider the infinite sequence $$\left\{\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{1}{16},\frac{3}{16},\ldots\right\}$$ and apply the following steps.

• Replace every number $$\\frac{i}{j}\$$ in the sequence with the list $$\\left\{\frac{2i-1}{2j},\frac{2i+1}{2j}\right\}\$$: $$\left\{\left\{\frac{1}{4},\frac{3}{4}\right\},\left\{\frac{1}{8},\frac{3}{8}\right\},\left\{\frac{5}{8},\frac{7}{8}\right\},\left\{\frac{1}{16},\frac{3}{16}\right\},\ldots\right\}$$
• Join all the lists into a single sequence: $$\left\{\frac{1}{4},\frac{3}{4},\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{1}{16},\frac{3}{16},\ldots\right\}$$
• Add $$\\frac{1}{2}\$$ at the beginning of the sequence: $$\left\{\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{1}{8},\frac{3}{8},\frac{5}{8},\frac{7}{8},\frac{1}{16},\frac{3}{16},\ldots\right\}$$

Notice how you get back to the sequence you started with!

The solution exploits this fact (together with Haskell's laziness) to compute the sequence s.

# R, 42 41 bytes

function(n)2*c(y<-2^(log2(n)%/%1),n-y+.5)


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-1 byte thanks to Robin Ryder

Returns a pair Denominator,Numerator. Uses the formula $$\$$N = 2\times\left(n-2^{\lfloor \log_2(n)\rfloor}\right)+1$$\$$ from the Josephus sequence and $$\$$D = 2^{\lfloor \log_2(n)\rfloor+1}$$\$$ from the other sequence. Happily we are able to re-use the denominator as the two formulas have quite a lot in common!

• 41 bytes by distributing the multiplication. Mar 29, 2021 at 20:33
• @RobinRyder thanks! Always nice to see an improvement to an old golf :-) Mar 30, 2021 at 14:17

# Python 2 - 68 66 bytes

-2 bytes thanks to Kevin

from math import*
def g(n):a=2**floor(log(n,2));print(n-a)*2+1,2*a


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• You can golf 1 byte by changing return 2*(n-a) to return(n-a)*2. And you could save an additional byte by using Python 2 instead of 3, so return can be print (with parenthesis). Sep 14, 2018 at 15:27
• @KevinCruijssen For someone who doesn't use Python, you certainly are a better Python programmer than me. Sep 14, 2018 at 15:27
• Hehe. :D Simple things like this comes with experience I guess. I'm on this site for about two years now I think (EDIT: Since April 2016). I sometimes even suggests things to golf for answers that are written in languages I had never seen before.. Some basic things work in most languages. For example, last week I suggested a golf for a T-SQL answer, and I once suggested a golf in a Red answer. xD Sep 14, 2018 at 15:33
• 44 bytes using len and bin instead of log.
– ovs
Sep 15, 2018 at 6:16
• @ovs 42 bytes. Sep 15, 2018 at 10:54

# Python 3, 53 51 bytes

• Saved two bytes thanks to mypetlion; reusing default parameters to reset n.
def f(m=2,n=1):n<m and print(n/m)&f(m,2+n)or f(m+m)


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• Swap the parameters to save 2 bytes: def f(m=2,n=1):n<m and print(n/m)&f(m,n+2)or f(m+m) Sep 14, 2018 at 16:14
• @mypetlion Cool, thank you! Sep 14, 2018 at 16:17

# Racket, 92 91 bytes

(define(f k(m 2)(n 1))(if(> k 0)(if(=(+ n 1)m)(f(- k 1)(+ m m))(f(- k 1)m(+ n 2)))(/ n m)))


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• Saved a byte thanks to Giuseppe -- removing superfluous whitespace.

# MATL, 8 bytes

BnWGEy-Q


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Returns Numerator, then Denominator. Uses the same method as my R answer, although it's a bit more efficient.

Explanation, with input 5:

           # implicit input 5
B          # convert to array of bits
# STACK: [[1 0 1]]
n          # length (place of Most Significant Bit)
# STACK: [3]
W          # elementwise raise 2^x
# STACK: [8]
G          # paste input
# STACK: [8, 5]
E          # double
# STACK: [8, 10]
y          # copy from below
# STACK: [8, 10, 8]
-          # subtract
# STACK: [8, 2]
Q          # increment
# STACK: [8, 3]
# implicit end of program, display stack contents


# Factor, 52 bytes

{ 1/2 } [ dup stack. dup 1 v+n append 2 v/n t ] loop


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Prints each fraction on its own line forever. Keeps track of the array of all terms with equal denominator.

{ 1/2 }         ! Push the initial array
[ ... t ] loop  ! Loop forever:
dup stack.    !   Print all fractions separated by a newline
dup 1 v+n     !   Copy and add 1 to all fractions
append 2 v/n  !   Join and halve all fractions


# Shakespeare Programming Language, 426 bytes

,.Ajax,.Ford,.Act I:.Scene I:.[Exeunt][Enter Ajax and Ford]Ajax:You be the sum ofyou a cat.Ford:You cat.Scene V:.Ford:Is twice you nicer I?If solet usScene X.You be twice you.Let usScene V.Scene X:.Ford:Remember twice you.You be the sum oftwice the remainder of the quotient betweenI you a cat.Open heart.You big big big big big cat.Speak thy.Recall.Open heart.You be twice the sum ofa cat a big big cat.Speak thy.Let usAct I.


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Outputs the sequence infinitely as both numbers separated by a space, with each item being separated by a newline.

• Love it. Lol You be twice the sum of a cat Sep 17, 2018 at 2:24
• Actually, it's "twice the sum ofa cat a big big cat" (i.e. 10 for some reason). Sep 17, 2018 at 2:39

# Python 2, 44 bytes

def f(n):m=2**len(bin(n))/4;return 2*n-m+1,m


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Function returns a tuple of (numerator, denominator). An input of 0 is not handled (it was optional).

• return 2*n-m+1,m can be print-~n+n-m,m to save 2 bytes. Sep 14, 2018 at 20:56

# Excel 48 28 Bytes

Saved 20 bytes (!) thanks to tsh

=(A1+0.5)/2^INT(LOG(A1,2))-1


=MOD(A1+0.5,2^(INT(LOG(A1,2))))/2^INT(LOG(A1,2))

Assumes value in A1, output is in decimal. If you want the output to be in fraction, you can create a custom format for the output cell as "0/###0" and it will show it as fraction.

Explanation: Difficult to explain, since there is a shortcut taken to get to this formula. Basically the numerator is a bit shift left of the input, and the denominator is the next power of 2 higher than the number input.

I originally started with Excel built in functions for BITLSHIFT and BITRSHIFT, but they will shift the entire 48 bits which is not what you want. The functions DEC2BIN (and BIN2DEC) have a limit of -512 to 511 (10 bits) so this wouldn't work. Instead I had to rebuild the number with a modulus of the original number, then times two, then add 1 (since the left digit would always be 1 before a shift).

=MOD(A1                        Use MOD for finding what the right digits are
+0.5                    this will later add the left "1" to the right digits
,2^INT(LOG(A1,2)))) Take Log base 2 number of digits on the right
this creates the numerator divided by 2 (explained later)
/ 2^INT(LOG(A1,2))             The denominator should be 2^ (Log2 + 1) but instead of
adding a 1 here, we cause the numerator to be divided by 2 instead
This gives us a fraction.  In the numerator, we also added .5
instead of 1 so that we wouldn't need to divide it in both the
numerator and denominator
Then tsh showed how I could take the int/log out of the mod and remove it from numerator/denominator.


Examples:

• What about =(A1+0.5)/2^INT(LOG(A1,2))-1?
– tsh
Sep 18, 2018 at 9:50

# Wolfram Language (Mathematica), 22 bytes

2^Mod[Log2[2#+1],1]-1&


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## C++, 9775 71 bytes

-26 bytes thanks to tsh, ceilingcat, Zacharý

float f(int i){float d=2,n=1;while(--i)n=d-n==1?d*=2,1:n+2;return n/d;}


Testing code :

std::cout << "1\t:\t" << f(1) << '\n';
std::cout << "2\t:\t" << f(2) << '\n';
std::cout << "3\t:\t" << f(3) << '\n';
std::cout << "4\t:\t" << f(4) << '\n';
std::cout << "10\t:\t" << f(10) << '\n';
std::cout << "100\t:\t" << f(100) << '\n';
std::cout << "511\t:\t" << f(511) << '\n';
std::cout << "512\t:\t" << f(512) << '\n';

• You may just omit if(!i)return 0; since 0 is not required in the challenge.
– tsh
Sep 17, 2018 at 7:58
• When trying to golf C-like language. You should avoid using while but try for. for(;exp;) is as same as while(exp) but you can write two more other statement into it. Prefer ?: instead of if else, which would be shorter in most cases.
– tsh
Sep 17, 2018 at 8:02
• I don't think you need the (...) around d-n-1. Sep 23, 2018 at 20:38

# Husk, 13 bytes

ṁz*İ1zR:1İ2İ½


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Abuses the built in sequences to the fullest.

Outputs an infinite list.

## Explanation

ṁz*İ1zR:1İ2İ½
İ½ powers of ½
:1İ2   powers of 2, with 1 prepended
zR       zip replicate the second by the first
z*           then zip multiply that with:
İ1         odd numbers
ṁ             map to each of the power arrays, and concatenate into single list


# Factor, 68 66 bytes

Saved 2 bytes thanks to @chunes

[ 1 2 [ 2dup / . [ 2 + ] dip 2dup > [ 2 * 1 swap ] when t ] loop ]


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Outputs the sequence infinitely.

[
1 2                    ! First numerator and denominator
[                      ! Loop body
2dup /               ! Divide to get next element of sequence
.                    ! Print the element
[ 2 + ] dip          ! Add 2 to the numerator
2dup >               ! If numerator > denominator
[
2 *                ! Multiply denominator by 2
1 swap             ! Set numerator to 1
]
when                 ! Functions like an if statement
t                    ! Push true on the stack so the loop doesn't stop
]
loop                   ! Run while top of stack is true (which is forever)
]

• when is shorter than [ ] if. Mar 29, 2021 at 22:42
• @chunes Sweet, thanks!
– user
Mar 29, 2021 at 22:44

# C (gcc), 63 bytes

No input, prints infinite sequence:

f(i,j){for(i=1,j=2;;i+=2,i>j&&(j*=2,i=1))printf("%d/%d ",i,j);}


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• Sep 14, 2018 at 15:38
• Building on @JonathanFrech 59 bytes Oct 27, 2019 at 4:10

# JavaScript (ES6), 44 bytes

Returns the $$\n\$$-th term, 1-indexed.

f=(n,p=q=1)=>n?f(n-1,p<q-2?p+2:!!(q*=2)):p/q


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# Ruby, 42 bytes

1.step{|i|(1..x=2**i).step(2){|j|p [j,x]}}


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Prints integer pairs infinitely, starting from 1/2.

# JavaScript (Node.js), 30 bytes

f=(n,d=.5)=>d>n?n/d:f(n-d,d*2)


Try it online! 0-indexed. Started out as a port of my Batch answer but I was able to calculate in multiples of $\frac{1}{2}$ which saved several bytes.

# Ruby, 31 bytes

->x{(2r*x+1)/2**x.bit_length-1}
`

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