Half, Half Half, and, Half

Clash Royale CLAN TAG#URR8PPP

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8
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Consider the following number sequence:

$ 0, frac12, frac14, frac34, frac18, frac38, frac58, frac78, frac116, frac316, frac516, frac716, frac916, frac1116, frac1316, frac1516, frac132, frac332, frac532, 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.)

Task

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 one by one;

Rule

• Your program should at least support first 1000 items;


• You may choose to output decimal, built-in fraction format, or, a pair of decimal integers as you like;
  
  
  • Input / Output as binary numbers is not allowed in this question;
  
  


• This is code-golf, 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.

Read More

• 
  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 fracn2 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

  
  

code-golf number sequence integer rational-numbers

share|improve this question

asked 48 mins ago

tsh

7,1101941

add a comment | 

up vote
8
down vote

favorite

Consider the following number sequence:

$ 0, frac12, frac14, frac34, frac18, frac38, frac58, frac78, frac116, frac316, frac516, frac716, frac916, frac1116, frac1316, frac1516, frac132, frac332, frac532, 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.)

Task

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 one by one;

Rule

• Your program should at least support first 1000 items;


• You may choose to output decimal, built-in fraction format, or, a pair of decimal integers as you like;
  
  
  • Input / Output as binary numbers is not allowed in this question;
  
  


• This is code-golf, 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.

Read More

• 
  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 fracn2 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

  
  

code-golf number sequence integer rational-numbers

share|improve this question

asked 48 mins ago

tsh

7,1101941

add a comment | 

up vote
8
down vote

favorite

up vote
8
down vote

favorite

Consider the following number sequence:

$ 0, frac12, frac14, frac34, frac18, frac38, frac58, frac78, frac116, frac316, frac516, frac716, frac916, frac1116, frac1316, frac1516, frac132, frac332, frac532, 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.)

Task

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 one by one;

Rule

• Your program should at least support first 1000 items;


• You may choose to output decimal, built-in fraction format, or, a pair of decimal integers as you like;
  
  
  • Input / Output as binary numbers is not allowed in this question;
  
  


• This is code-golf, 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.

Read More

• 
  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 fracn2 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

  
  

code-golf number sequence integer rational-numbers

share|improve this question

asked 48 mins ago

tsh

7,1101941

Consider the following number sequence:

$ 0, frac12, frac14, frac34, frac18, frac38, frac58, frac78, frac116, frac316, frac516, frac716, frac916, frac1116, frac1316, frac1516, frac132, frac332, frac532, 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.)

Task

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 one by one;

Rule

• Your program should at least support first 1000 items;


• You may choose to output decimal, built-in fraction format, or, a pair of decimal integers as you like;
  
  
  • Input / Output as binary numbers is not allowed in this question;
  
  


• This is code-golf, 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.

Read More

• 
  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 fracn2 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

  
  

code-golf number sequence integer rational-numbers

code-golf number sequence integer rational-numbers

share|improve this question

asked 48 mins ago

tsh

7,1101941

share|improve this question

asked 48 mins ago

tsh

7,1101941

share|improve this question

share|improve this question

asked 48 mins ago

tsh

7,1101941

asked 48 mins ago

tsh

7,1101941

asked 48 mins ago

tsh

7,1101941

7,1101941

add a comment | 

add a comment | 

6 Answers
6

active

oldest

votes

up vote
4
down vote

MathGolf, 5 bytes

k╫k╨]

Assuming implicit input (does not work on TIO currently)

_╫╨] ("_" duplicates TOS, "" swaps the top two elements)

Try it online!

Explanation

k Read input as int
 ╫ Left-rotate all bits in input
 k Read input as int
 ╨ 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.

share|improve this answer

edited 1 min ago

answered 36 mins ago

maxb

1,209519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user202729 Yes, thanks for noticing. I updated the explanation.
  – maxb
  8 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

05AB1E, 11 bytes

[N>oDÅÉs/»,

Try it online!

share|improve this answer

answered 43 mins ago

Emigna

43k431131

add a comment | 

up vote
2
down vote

Jelly, 9 bytes

Bṙ1Ḅ,æċ2$

Try it online!

share|improve this answer

answered 42 mins ago

Dennis♦

182k32291722

add a comment | 

up vote
2
down vote

PowerShell, 40 bytes

for($i=2;;$i*=2)?$_%2

Try it online!

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.

share|improve this answer

answered 22 mins ago

AdmBorkBork

24.4k359214

add a comment | 

up vote
2
down vote

Haskell, 35 bytes

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

This is an infinite list of integer pairs.

Try it online!

share|improve this answer

answered 20 mins ago

nimi

29.9k31881

add a comment | 

up vote
0
down vote

Java 10, 89 85 bytes

v->for(double i=1,j,t;;)for(t=Math.pow(2,i++),j=1;j<t;j+=2)System.out.println(j/t);

Port of @Emigma's 05AB1E answer, so outputs decimals indefinitely as well.

Try it online.

Explanation:

v-> // Method with empty unused parameter and no return-type
 for(double i=1,j,t;;) // Loop `i` from 1 upwards indefinitely
 for(t=Math.pow(2,i++), // Set `t` to 2 to the power `i`
 j=1;j<t;j+=2) // Inner loop `j` in the range [1, `t`) in steps of 2
 System.out.println( // Print with trailing new-line:
 j/t); // `j` divided by `t`

share

answered 5 mins ago

Kevin Cruijssen

29.9k552165

add a comment | 

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

MathGolf, 5 bytes

k╫k╨]

Assuming implicit input (does not work on TIO currently)

_╫╨] ("_" duplicates TOS, "" swaps the top two elements)

Try it online!

Explanation

k Read input as int
 ╫ Left-rotate all bits in input
 k Read input as int
 ╨ 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.

share|improve this answer

edited 1 min ago

answered 36 mins ago

maxb

1,209519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user202729 Yes, thanks for noticing. I updated the explanation.
  – maxb
  8 mins ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

MathGolf, 5 bytes

k╫k╨]

Assuming implicit input (does not work on TIO currently)

_╫╨] ("_" duplicates TOS, "" swaps the top two elements)

Try it online!

Explanation

k Read input as int
 ╫ Left-rotate all bits in input
 k Read input as int
 ╨ 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.

share|improve this answer

edited 1 min ago

answered 36 mins ago

maxb

1,209519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user202729 Yes, thanks for noticing. I updated the explanation.
  – maxb
  8 mins ago
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

MathGolf, 5 bytes

k╫k╨]

Assuming implicit input (does not work on TIO currently)

_╫╨] ("_" duplicates TOS, "" swaps the top two elements)

Try it online!

Explanation

k Read input as int
 ╫ Left-rotate all bits in input
 k Read input as int
 ╨ 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.

share|improve this answer

edited 1 min ago

answered 36 mins ago

maxb

1,209519

MathGolf, 5 bytes

k╫k╨]

Assuming implicit input (does not work on TIO currently)

_╫╨] ("_" duplicates TOS, "" swaps the top two elements)

Try it online!

Explanation

k Read input as int
 ╫ Left-rotate all bits in input
 k Read input as int
 ╨ 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.

share|improve this answer

edited 1 min ago

answered 36 mins ago

maxb

1,209519

share|improve this answer

share|improve this answer

edited 1 min ago

edited 1 min ago

edited 1 min ago

answered 36 mins ago

maxb

1,209519

answered 36 mins ago

maxb

1,209519

answered 36 mins ago

maxb

1,209519

1,209519

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user202729 Yes, thanks for noticing. I updated the explanation.
  – maxb
  8 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user202729 Yes, thanks for noticing. I updated the explanation.
  – maxb
  8 mins ago
  

  
  
  
  

@user202729 Yes, thanks for noticing. I updated the explanation.
– maxb
8 mins ago

@user202729 Yes, thanks for noticing. I updated the explanation.
– maxb
8 mins ago

add a comment | 

up vote
2
down vote

05AB1E, 11 bytes

[N>oDÅÉs/»,

Try it online!

share|improve this answer

answered 43 mins ago

Emigna

43k431131

add a comment | 

up vote
2
down vote

05AB1E, 11 bytes

[N>oDÅÉs/»,

Try it online!

share|improve this answer

answered 43 mins ago

Emigna

43k431131

add a comment | 

up vote
2
down vote

up vote
2
down vote

05AB1E, 11 bytes

[N>oDÅÉs/»,

Try it online!

share|improve this answer

answered 43 mins ago

Emigna

43k431131

05AB1E, 11 bytes

[N>oDÅÉs/»,

Try it online!

share|improve this answer

answered 43 mins ago

Emigna

43k431131

share|improve this answer

share|improve this answer

answered 43 mins ago

Emigna

43k431131

answered 43 mins ago

Emigna

43k431131

answered 43 mins ago

Emigna

43k431131

43k431131

add a comment | 

add a comment | 

up vote
2
down vote

Jelly, 9 bytes

Bṙ1Ḅ,æċ2$

Try it online!

share|improve this answer

answered 42 mins ago

Dennis♦

182k32291722

add a comment | 

up vote
2
down vote

Jelly, 9 bytes

Bṙ1Ḅ,æċ2$

Try it online!

share|improve this answer

answered 42 mins ago

Dennis♦

182k32291722

add a comment | 

up vote
2
down vote

up vote
2
down vote

Jelly, 9 bytes

Bṙ1Ḅ,æċ2$

Try it online!

share|improve this answer

answered 42 mins ago

Dennis♦

182k32291722

Jelly, 9 bytes

Bṙ1Ḅ,æċ2$

Try it online!

share|improve this answer

answered 42 mins ago

Dennis♦

182k32291722

share|improve this answer

share|improve this answer

answered 42 mins ago

Dennis♦

182k32291722

answered 42 mins ago

Dennis♦

182k32291722

answered 42 mins ago

Dennis♦

182k32291722

182k32291722

add a comment | 

add a comment | 

up vote
2
down vote

PowerShell, 40 bytes

for($i=2;;$i*=2)?$_%2

Try it online!

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.

share|improve this answer

answered 22 mins ago

AdmBorkBork

24.4k359214

add a comment | 

up vote
2
down vote

PowerShell, 40 bytes

for($i=2;;$i*=2)?$_%2

Try it online!

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.

share|improve this answer

answered 22 mins ago

AdmBorkBork

24.4k359214

add a comment | 

up vote
2
down vote

up vote
2
down vote

PowerShell, 40 bytes

for($i=2;;$i*=2)?$_%2

Try it online!

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.

share|improve this answer

answered 22 mins ago

AdmBorkBork

24.4k359214

PowerShell, 40 bytes

for($i=2;;$i*=2)?$_%2

Try it online!

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.

share|improve this answer

answered 22 mins ago

AdmBorkBork

24.4k359214

share|improve this answer

share|improve this answer

answered 22 mins ago

AdmBorkBork

24.4k359214

answered 22 mins ago

AdmBorkBork

24.4k359214

answered 22 mins ago

AdmBorkBork

24.4k359214

24.4k359214

add a comment | 

add a comment | 

up vote
2
down vote

Haskell, 35 bytes

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

This is an infinite list of integer pairs.

Try it online!

share|improve this answer

answered 20 mins ago

nimi

29.9k31881

add a comment | 

up vote
2
down vote

Haskell, 35 bytes

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

This is an infinite list of integer pairs.

Try it online!

share|improve this answer

answered 20 mins ago

nimi

29.9k31881

add a comment | 

up vote
2
down vote

up vote
2
down vote

Haskell, 35 bytes

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

This is an infinite list of integer pairs.

Try it online!

share|improve this answer

answered 20 mins ago

nimi

29.9k31881

Haskell, 35 bytes

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

This is an infinite list of integer pairs.

Try it online!

share|improve this answer

answered 20 mins ago

nimi

29.9k31881

share|improve this answer

share|improve this answer

answered 20 mins ago

nimi

29.9k31881

answered 20 mins ago

nimi

29.9k31881

answered 20 mins ago

nimi

29.9k31881

29.9k31881

add a comment | 

add a comment | 

up vote
0
down vote

Java 10, 89 85 bytes

v->for(double i=1,j,t;;)for(t=Math.pow(2,i++),j=1;j<t;j+=2)System.out.println(j/t);

Port of @Emigma's 05AB1E answer, so outputs decimals indefinitely as well.

Try it online.

Explanation:

v-> // Method with empty unused parameter and no return-type
 for(double i=1,j,t;;) // Loop `i` from 1 upwards indefinitely
 for(t=Math.pow(2,i++), // Set `t` to 2 to the power `i`
 j=1;j<t;j+=2) // Inner loop `j` in the range [1, `t`) in steps of 2
 System.out.println( // Print with trailing new-line:
 j/t); // `j` divided by `t`

share

answered 5 mins ago

Kevin Cruijssen

29.9k552165

add a comment | 

up vote
0
down vote

Java 10, 89 85 bytes

v->for(double i=1,j,t;;)for(t=Math.pow(2,i++),j=1;j<t;j+=2)System.out.println(j/t);

Port of @Emigma's 05AB1E answer, so outputs decimals indefinitely as well.

Try it online.

Explanation:

v-> // Method with empty unused parameter and no return-type
 for(double i=1,j,t;;) // Loop `i` from 1 upwards indefinitely
 for(t=Math.pow(2,i++), // Set `t` to 2 to the power `i`
 j=1;j<t;j+=2) // Inner loop `j` in the range [1, `t`) in steps of 2
 System.out.println( // Print with trailing new-line:
 j/t); // `j` divided by `t`

share

answered 5 mins ago

Kevin Cruijssen

29.9k552165

add a comment | 

up vote
0
down vote

up vote
0
down vote

Java 10, 89 85 bytes

v->for(double i=1,j,t;;)for(t=Math.pow(2,i++),j=1;j<t;j+=2)System.out.println(j/t);

Port of @Emigma's 05AB1E answer, so outputs decimals indefinitely as well.

Try it online.

Explanation:

v-> // Method with empty unused parameter and no return-type
 for(double i=1,j,t;;) // Loop `i` from 1 upwards indefinitely
 for(t=Math.pow(2,i++), // Set `t` to 2 to the power `i`
 j=1;j<t;j+=2) // Inner loop `j` in the range [1, `t`) in steps of 2
 System.out.println( // Print with trailing new-line:
 j/t); // `j` divided by `t`

share

answered 5 mins ago

Kevin Cruijssen

29.9k552165

Java 10, 89 85 bytes

v->for(double i=1,j,t;;)for(t=Math.pow(2,i++),j=1;j<t;j+=2)System.out.println(j/t);

Port of @Emigma's 05AB1E answer, so outputs decimals indefinitely as well.

Try it online.

Explanation:

v-> // Method with empty unused parameter and no return-type
 for(double i=1,j,t;;) // Loop `i` from 1 upwards indefinitely
 for(t=Math.pow(2,i++), // Set `t` to 2 to the power `i`
 j=1;j<t;j+=2) // Inner loop `j` in the range [1, `t`) in steps of 2
 System.out.println( // Print with trailing new-line:
 j/t); // `j` divided by `t`

share

answered 5 mins ago

Kevin Cruijssen

29.9k552165

share

share

answered 5 mins ago

Kevin Cruijssen

29.9k552165

answered 5 mins ago

Kevin Cruijssen

29.9k552165

answered 5 mins ago

Kevin Cruijssen

29.9k552165

29.9k552165

add a comment | 

add a comment | 

 

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