New order #4: World
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Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
$endgroup$
add a comment |
$begingroup$
Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
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I would answer this using TI-BASIC, but the input would be limited to $0<N<1000$ since lists are limited to 999 elements. Great challenge nonetheless!
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– Tau
yesterday
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@Tau : although out-of-spec (and this non-competing), I'd be interested in your solution. Do your have one you can post?
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– agtoever
yesterday
1
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I deleted the program, but I should be able to recreate it. I'll post it as non-competing once I have redone it.
$endgroup$
– Tau
yesterday
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@agtoever, "non-competing" doesn't cover invalid solutions; it was for solutions using languages or language features that were created after a challenge was posted.
$endgroup$
– Shaggy
yesterday
$begingroup$
PP&CG meta is indeed very clear on this. I wasn't award of such a strict interpretation of "non-competing"... @Tau : it seems you can't post your TI-BASIC solution under these rules. Sorry.
$endgroup$
– agtoever
19 hours ago
add a comment |
$begingroup$
Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
$endgroup$
Introduction (may be ignored)
Putting all positive numbers in its regular order (1, 2, 3, ...) is a bit boring, isn't it? So here is a series of challenges around permutations (reshuffelings) of all positive numbers. This is the fourth challenge in this series (links to the first, second and third challenge).
In this challenge, we will explore not one permutation of the natural numbers, but an entire world of permutations!
In 2000, Clark Kimberling posed a problem in the 26th issue of Crux Mathematicorum, a scientific journal of mathematics published by the Canadian Mathematical Society. The problem was:
$text{Sequence }a = begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{2} rfloortext{ if }lfloor frac{a_{n-1}}{2} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = 3 a_{n-1}text{ otherwise}
end{cases}$
Does every positive integer occur exactly once in this sequence?
In 2004, Mateusz Kwasnicki provided positive proof in the same journal and in 2008, he published a more formal and (compared to the original question) a more general proof. He formulated the sequence with parameters $p$ and $q$:
$begin{cases}
a_1 = 1\
a_n = lfloor frac{a_{n-1}}{q} rfloortext{ if }lfloor frac{a_{n-1}}{q} rfloor notin {0, a_1, ... , a_{n-1}}\
a_n = p a_{n-1}text{ otherwise}
end{cases}$
He proved that for any $p, q>1$ such that $log_p(q)$ is irrational, the sequence is a permutation of the natural numbers. Since there are an infinite number of $p$ and $q$ values for which this is true, this is truly an entire world of permutations of the natural numbers. We will stick with the original $(p, q)=(3, 2)$, and for these paramters, the sequence can be found as A050000 in the OEIS. Its first 20 elements are:
1, 3, 9, 4, 2, 6, 18, 54, 27, 13, 39, 19, 57, 28, 14, 7, 21, 10, 5, 15
Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A050000.
Task
Given an integer input $n$, output $a(n)$ in integer format, where:
$begin{cases}
a(1) = 1\
a(n) = lfloor frac{a(n-1)}{2} rfloortext{ if }lfloor frac{a(n-1)}{2} rfloor notin {0, a_1, ... , a(n-1)}\
a(n) = 3 a(n-1)text{ otherwise}
end{cases}$
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 3$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 2
20 | 15
50 | 165
78 | 207
123 | 94
1234 | 3537
3000 | 2245
9999 | 4065
29890 | 149853
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767)
- Invalid input (0, floats, strings, negative values, etc.) may lead to unpredicted output, errors or (un)defined behaviour.
- Default I/O rules apply.
Default loopholes are forbidden.- This is code-golf, so the shortest answers in bytes wins
code-golf sequence
code-golf sequence
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
edited Apr 7 at 22:00
agtoever
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
asked Apr 7 at 21:27
agtoeveragtoever
1,327424
1,327424
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I would answer this using TI-BASIC, but the input would be limited to $0<N<1000$ since lists are limited to 999 elements. Great challenge nonetheless!
$endgroup$
– Tau
yesterday
$begingroup$
@Tau : although out-of-spec (and this non-competing), I'd be interested in your solution. Do your have one you can post?
$endgroup$
– agtoever
yesterday
1
$begingroup$
I deleted the program, but I should be able to recreate it. I'll post it as non-competing once I have redone it.
$endgroup$
– Tau
yesterday
$begingroup$
@agtoever, "non-competing" doesn't cover invalid solutions; it was for solutions using languages or language features that were created after a challenge was posted.
$endgroup$
– Shaggy
yesterday
$begingroup$
PP&CG meta is indeed very clear on this. I wasn't award of such a strict interpretation of "non-competing"... @Tau : it seems you can't post your TI-BASIC solution under these rules. Sorry.
$endgroup$
– agtoever
19 hours ago
add a comment |
$begingroup$
I would answer this using TI-BASIC, but the input would be limited to $0<N<1000$ since lists are limited to 999 elements. Great challenge nonetheless!
$endgroup$
– Tau
yesterday
$begingroup$
@Tau : although out-of-spec (and this non-competing), I'd be interested in your solution. Do your have one you can post?
$endgroup$
– agtoever
yesterday
1
$begingroup$
I deleted the program, but I should be able to recreate it. I'll post it as non-competing once I have redone it.
$endgroup$
– Tau
yesterday
$begingroup$
@agtoever, "non-competing" doesn't cover invalid solutions; it was for solutions using languages or language features that were created after a challenge was posted.
$endgroup$
– Shaggy
yesterday
$begingroup$
PP&CG meta is indeed very clear on this. I wasn't award of such a strict interpretation of "non-competing"... @Tau : it seems you can't post your TI-BASIC solution under these rules. Sorry.
$endgroup$
– agtoever
19 hours ago
$begingroup$
I would answer this using TI-BASIC, but the input would be limited to $0<N<1000$ since lists are limited to 999 elements. Great challenge nonetheless!
$endgroup$
– Tau
yesterday
$begingroup$
I would answer this using TI-BASIC, but the input would be limited to $0<N<1000$ since lists are limited to 999 elements. Great challenge nonetheless!
$endgroup$
– Tau
yesterday
$begingroup$
@Tau : although out-of-spec (and this non-competing), I'd be interested in your solution. Do your have one you can post?
$endgroup$
– agtoever
yesterday
$begingroup$
@Tau : although out-of-spec (and this non-competing), I'd be interested in your solution. Do your have one you can post?
$endgroup$
– agtoever
yesterday
1
1
$begingroup$
I deleted the program, but I should be able to recreate it. I'll post it as non-competing once I have redone it.
$endgroup$
– Tau
yesterday
$begingroup$
I deleted the program, but I should be able to recreate it. I'll post it as non-competing once I have redone it.
$endgroup$
– Tau
yesterday
$begingroup$
@agtoever, "non-competing" doesn't cover invalid solutions; it was for solutions using languages or language features that were created after a challenge was posted.
$endgroup$
– Shaggy
yesterday
$begingroup$
@agtoever, "non-competing" doesn't cover invalid solutions; it was for solutions using languages or language features that were created after a challenge was posted.
$endgroup$
– Shaggy
yesterday
$begingroup$
PP&CG meta is indeed very clear on this. I wasn't award of such a strict interpretation of "non-competing"... @Tau : it seems you can't post your TI-BASIC solution under these rules. Sorry.
$endgroup$
– agtoever
19 hours ago
$begingroup$
PP&CG meta is indeed very clear on this. I wasn't award of such a strict interpretation of "non-competing"... @Tau : it seems you can't post your TI-BASIC solution under these rules. Sorry.
$endgroup$
– agtoever
19 hours ago
add a comment |
19 Answers
19
active
oldest
votes
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JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
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55 bytes
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– Embodiment of Ignorance
Apr 7 at 22:09
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@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
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– Arnauld
Apr 7 at 22:16
2
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But this is code-golf, we don't care about speed, as long as it gets the job done
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– Embodiment of Ignorance
Apr 7 at 22:22
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n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
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– tsh
2 days ago
add a comment |
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Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
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add a comment |
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05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 2 days ago
EmignaEmigna
47.9k434145
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15 bytes
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– Kevin Cruijssen
2 days ago
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@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
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– Emigna
2 days ago
add a comment |
$begingroup$
Perl 6, 49 bytes
-2 bytes thanks to nwellnof
{(1,3,{(3*@_[*-1]Xdiv 6,1).max(*∉@_)}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
( ).max( ) # The first of
@_[*-1]X # The previous element
3* div 6 # Halved and floored
3* div ,1 # Or tripled
*∉@_ # That hasn't appeared in the sequence yet
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
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add a comment |
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J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered Apr 7 at 22:17
JonahJonah
2,6711017
$endgroup$
add a comment |
$begingroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
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add a comment |
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Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 2 days ago
ShaggyShaggy
18.9k21768
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add a comment |
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Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 2 days ago
user1472751user1472751
1,26126
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add a comment |
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Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
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add a comment |
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Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 2 days ago
Kirill L.Kirill L.
6,1081527
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add a comment |
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C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered 2 days ago
Neil A.Neil A.
1,358120
$endgroup$
$begingroup$
@ceilingcat Unfortunately that affects operator precedence and changes the output of the function.
$endgroup$
– Neil A.
yesterday
add a comment |
$begingroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 2 days ago
ArBoArBo
38115
$endgroup$
add a comment |
$begingroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 2 days ago
recursiverecursive
5,7491322
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
$endgroup$
add a comment |
$begingroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 2 days ago
Noodle9Noodle9
30137
$endgroup$
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
2 days ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
2 days ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
2 days ago
add a comment |
$begingroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
$endgroup$
add a comment |
$begingroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 2 days ago
xnorxnor
94k18192451
$endgroup$
add a comment |
$begingroup$
Lua, 78 bytes
x,y=1,3 u={}for _=2,...do
u[x]=0
x,y=y,y//2
if u[y]then y=3*x end
end
print(x)
Try it online!
answered yesterday
wastlwastl
2,289526
$endgroup$
$begingroup$
68 bytes through removing some whitespace, removing thezvariable and changing the if statement to ternary
$endgroup$
– Jo King
11 hours ago
add a comment |
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19 Answers
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19 Answers
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$begingroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
$endgroup$
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:09
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
Apr 7 at 22:16
2
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:22
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
2 days ago
add a comment |
$begingroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
$endgroup$
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:09
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
Apr 7 at 22:16
2
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:22
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
2 days ago
add a comment |
$begingroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
$endgroup$
JavaScript (ES6), 55 51 50 bytes
Saved 1 byte thanks to @EmbodimentofIgnorance
Saved 1 byte thanks to @tsh
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
Try it online!
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
edited 2 days ago
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
answered Apr 7 at 22:01
ArnauldArnauld
80.7k797334
80.7k797334
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:09
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
Apr 7 at 22:16
2
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:22
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
2 days ago
add a comment |
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:09
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
Apr 7 at 22:16
2
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:22
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")
$endgroup$
– tsh
2 days ago
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:09
$begingroup$
55 bytes
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:09
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
Apr 7 at 22:16
$begingroup$
@EmbodimentofIgnorance I usually avoid that trick, as the eval'ed code is much slower. But the difference is barely noticeable for that one, so I guess that's fine.
$endgroup$
– Arnauld
Apr 7 at 22:16
2
2
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:22
$begingroup$
But this is code-golf, we don't care about speed, as long as it gets the job done
$endgroup$
– Embodiment of Ignorance
Apr 7 at 22:22
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")$endgroup$
– tsh
2 days ago
$begingroup$
n=>eval("for(o=[p=2];n--;)o[p=o[q=p>>1]?3*p:q]=p")$endgroup$
– tsh
2 days ago
add a comment |
$begingroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
$endgroup$
add a comment |
$begingroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
$endgroup$
Jelly, 15 bytes
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ
A full program accepting the integer, n (1-based), from STDIN which prints the result.
Try it online!
How?
µ×3żHḞḢḟȯ1Ṫ;µ¡Ḣ - Main Link: no arguments (implicit left argument = 0)
µ µ¡ - repeat this monadic chain STDIN times (starting with x=0)
- e.g. x = ... 0 [1,0] [9,3,1,0]
×3 - multiply by 3 0 [3,0] [27,9,3,0]
H - halve 0 [1.5,0] [4.5,1.5,0.5,0]
ż - zip together [0,0] [[3,1.5],[0,0]] [[27,4.5],[9,1.5],[3,0.5],[0,0]]
Ḟ - floor [0,0] [[3,1],[0,0]] [[27,4],[9,1],[3,0],[0,0]]
Ḣ - head 0 [3,1] [27,4]
ḟ - filter discard if in x [3] [27,4]
ȯ1 - logical OR with 1 1 [3] [27,4]
Ṫ - tail 1 3 4
; - concatenate with x [1,0] [3,1,0] [4,9,3,1,0]
Ḣ - head 1 3 4
- implicit print
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
answered Apr 7 at 22:46
Jonathan AllanJonathan Allan
54.1k537174
54.1k537174
add a comment |
add a comment |
$begingroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 2 days ago
EmignaEmigna
47.9k434145
$endgroup$
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
2 days ago
add a comment |
$begingroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 2 days ago
EmignaEmigna
47.9k434145
$endgroup$
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
2 days ago
add a comment |
$begingroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 2 days ago
EmignaEmigna
47.9k434145
$endgroup$
05AB1E, 16 15 bytes
Saved 1 byte thanks to Kevin Cruijssen.
0-indexed.
¾ˆ$FDˆx3*‚;ï¯Kн
Try it online!
Explanation
Using n=1 as example
¾ˆ # initialize global array as [0]
$ # initialize stack with 1, input
F # input times do:
Dˆ # duplicate current item (initially 1) and add one copy to global array
# STACK: 1, GLOBAL_ARRAY: [0, 1]
x # push Top_of_stack*2
# STACK: 1, 2, GLOBAL_ARRAY: [0, 1]
3* # multiply by 3
# STACK: 1, 6, GLOBAL_ARRAY: [0, 1]
‚;ï # pair and integer divide both by 2
# STACK: [0, 3], GLOBAL_ARRAY: [0, 1]
¯K # remove any numbers already in the global array
# STACK: [3], GLOBAL_ARRAY: [0, 1]
н # and take the head
# STACK: 3
answered 2 days ago
EmignaEmigna
47.9k434145
edited 2 days ago
answered 2 days ago
EmignaEmigna
47.9k434145
answered 2 days ago
EmignaEmigna
47.9k434145
answered 2 days ago
EmignaEmigna
47.9k434145
47.9k434145
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
2 days ago
add a comment |
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
2 days ago
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
15 bytes
$endgroup$
– Kevin Cruijssen
2 days ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
2 days ago
$begingroup$
@KevinCruijssen: Thanks! I thought of using the global array, but assumed it would be the same length as a list on the stack and never tried it :/
$endgroup$
– Emigna
2 days ago
add a comment |
$begingroup$
Perl 6, 49 bytes
-2 bytes thanks to nwellnof
{(1,3,{(3*@_[*-1]Xdiv 6,1).max(*∉@_)}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
( ).max( ) # The first of
@_[*-1]X # The previous element
3* div 6 # Halved and floored
3* div ,1 # Or tripled
*∉@_ # That hasn't appeared in the sequence yet
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
$endgroup$
add a comment |
$begingroup$
Perl 6, 49 bytes
-2 bytes thanks to nwellnof
{(1,3,{(3*@_[*-1]Xdiv 6,1).max(*∉@_)}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
( ).max( ) # The first of
@_[*-1]X # The previous element
3* div 6 # Halved and floored
3* div ,1 # Or tripled
*∉@_ # That hasn't appeared in the sequence yet
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
$endgroup$
add a comment |
$begingroup$
Perl 6, 49 bytes
-2 bytes thanks to nwellnof
{(1,3,{(3*@_[*-1]Xdiv 6,1).max(*∉@_)}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
( ).max( ) # The first of
@_[*-1]X # The previous element
3* div 6 # Halved and floored
3* div ,1 # Or tripled
*∉@_ # That hasn't appeared in the sequence yet
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
$endgroup$
Perl 6, 49 bytes
-2 bytes thanks to nwellnof
{(1,3,{(3*@_[*-1]Xdiv 6,1).max(*∉@_)}...*)[$_]}
Try it online!
Returns the 0-indexed element in the sequence. You can change this to 1-indexed by changing the starting elements to 0,1 instead of 1,3
Explanation:
{ } # Anonymous code block
( ...*)[$_] # Index into the infinite sequence
1,3 # That starts with 1,3
,{ } # And each element is
( ).max( ) # The first of
@_[*-1]X # The previous element
3* div 6 # Halved and floored
3* div ,1 # Or tripled
*∉@_ # That hasn't appeared in the sequence yet
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
edited 12 hours ago
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
answered Apr 7 at 23:11
Jo KingJo King
26.7k365132
26.7k365132
add a comment |
add a comment |
$begingroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered Apr 7 at 22:17
JonahJonah
2,6711017
$endgroup$
add a comment |
$begingroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered Apr 7 at 22:17
JonahJonah
2,6711017
$endgroup$
add a comment |
$begingroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered Apr 7 at 22:17
JonahJonah
2,6711017
$endgroup$
J, 47 40 bytes
[:{:0 1(],<.@-:@{:@](e.{[,3*{:@])])^:[~]
Try it online!
ungolfed
[: {: 0 1 (] , <.@-:@{:@] (e. { [ , 3 * {:@]) ])^:[~ ]
Direct translation of the definition into J. It builds bottom up by using ^: to iterate from the starting value the required number of times.
answered Apr 7 at 22:17
JonahJonah
2,6711017
edited 2 days ago
answered Apr 7 at 22:17
JonahJonah
2,6711017
answered Apr 7 at 22:17
JonahJonah
2,6711017
answered Apr 7 at 22:17
JonahJonah
2,6711017
2,6711017
add a comment |
add a comment |
$begingroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
$endgroup$
add a comment |
$begingroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
$endgroup$
add a comment |
$begingroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
$endgroup$
Java 10, 120 99 bytes
n->{var L=" 1 0 ";int r=1,t;for(;n-->0;L+=r+" ")if(L.contains(" "+(r=(t=r)/2)+" "))r=t*3;return r;}
Try it online.
Explanation:
n->{ // Method with integer as both parameter and return-type
var L=" 1 0 "; // Create a String that acts as 'List', starting at [1,0]
int r=1, // Result-integer, starting at 1
t; // Temp-integer, uninitialized
for(;n-->0; // Loop the input amount of times:
L+=r+" ")) // After every iteration: add the result to the 'List'
t=r // Create a copy of the result in `t`
r=(...)/2 // Then integer-divide the result by 2
if(L.contains(" "+(...)+" ")) // If the 'List' contains this result//2:
r=t*3; // Set the result to `t` multiplied by 3 instead
return r;} // Return the result
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
edited 2 days ago
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
answered 2 days ago
Kevin CruijssenKevin Cruijssen
42.6k571217
42.6k571217
add a comment |
add a comment |
$begingroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 2 days ago
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 2 days ago
ShaggyShaggy
18.9k21768
$endgroup$
add a comment |
$begingroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 2 days ago
ShaggyShaggy
18.9k21768
$endgroup$
Japt, 15 14 bytes
1-indexed.
@[X*3Xz]kZ Ì}g
Try it
@[X*3Xz]kZ Ì}g :Implicit input of integer U
g :Starting with the array [0,1] do the following U times, pushing the result to the array each time
@ : Pass the last element X in the array Z through the following function
[ : Build an array containing
X*3 : X multiplied by 3
Xz : X floor divided by 2
] : Close array
kZ : Remove all elements contained in Z
Ì : Get the last element
} : End function
:Implicit output of the last element in the array
answered 2 days ago
ShaggyShaggy
18.9k21768
edited 2 days ago
answered 2 days ago
ShaggyShaggy
18.9k21768
answered 2 days ago
ShaggyShaggy
18.9k21768
answered 2 days ago
ShaggyShaggy
18.9k21768
18.9k21768
add a comment |
add a comment |
$begingroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 2 days ago
user1472751user1472751
1,26126
$endgroup$
add a comment |
$begingroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 2 days ago
user1472751user1472751
1,26126
$endgroup$
add a comment |
$begingroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 2 days ago
user1472751user1472751
1,26126
$endgroup$
Haskell, 67 65 bytes
(h[1,0]!!)
h l@(a:o)|elem(div a 2)o=a:h(3*a:l)|1>0=a:h(div a 2:l)
Try it online!
Uses 0-based indexing.
EDIT: saved 2 bytes by using elem instead of notElem and switching conditions
answered 2 days ago
user1472751user1472751
1,26126
edited 2 days ago
answered 2 days ago
user1472751user1472751
1,26126
answered 2 days ago
user1472751user1472751
1,26126
answered 2 days ago
user1472751user1472751
1,26126
1,26126
add a comment |
add a comment |
$begingroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
$endgroup$
add a comment |
$begingroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
$endgroup$
add a comment |
$begingroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
$endgroup$
Jelly, 21 bytes
Ø.;0ị×3$:2$:2eɗ?Ɗ$⁸¡Ṫ
Try it online!
A monadic link that takes zero-indexed $n$ as the argument and returns $a(n)$.
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
answered Apr 7 at 22:08
Nick KennedyNick Kennedy
1,50649
1,50649
add a comment |
add a comment |
$begingroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 2 days ago
Kirill L.Kirill L.
6,1081527
$endgroup$
add a comment |
$begingroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 2 days ago
Kirill L.Kirill L.
6,1081527
$endgroup$
add a comment |
$begingroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 2 days ago
Kirill L.Kirill L.
6,1081527
$endgroup$
Ruby, 54 52 48 bytes
->n{*s=0;j=2;n.times{s<<j=s==s-[j/2]?j/2:j*3};j}
Try it online!
answered 2 days ago
Kirill L.Kirill L.
6,1081527
edited 2 days ago
answered 2 days ago
Kirill L.Kirill L.
6,1081527
answered 2 days ago
Kirill L.Kirill L.
6,1081527
answered 2 days ago
Kirill L.Kirill L.
6,1081527
6,1081527
add a comment |
add a comment |
$begingroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered 2 days ago
Neil A.Neil A.
1,358120
$endgroup$
$begingroup$
@ceilingcat Unfortunately that affects operator precedence and changes the output of the function.
$endgroup$
– Neil A.
yesterday
add a comment |
$begingroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered 2 days ago
Neil A.Neil A.
1,358120
$endgroup$
$begingroup$
@ceilingcat Unfortunately that affects operator precedence and changes the output of the function.
$endgroup$
– Neil A.
yesterday
add a comment |
$begingroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered 2 days ago
Neil A.Neil A.
1,358120
$endgroup$
C++ (gcc), 189 180 bytes
-9 bytes to small golfing
#import<vector>
#import<algorithm>
int a(int n){std::vector<int>s={1};for(int i=0;i<n;++i)s.push_back(i&&std::find(s.begin(),s.end(),s[i]/2)==s.end()?s[i]/2:3*s[i]);return s[n-1];}
Try it online!
Computes the sequence up to n, then returns the desired element. Slow for larger indices.
answered 2 days ago
Neil A.Neil A.
1,358120
edited 2 days ago
answered 2 days ago
Neil A.Neil A.
1,358120
answered 2 days ago
Neil A.Neil A.
1,358120
answered 2 days ago
Neil A.Neil A.
1,358120
1,358120
$begingroup$
@ceilingcat Unfortunately that affects operator precedence and changes the output of the function.
$endgroup$
– Neil A.
yesterday
add a comment |
$begingroup$
@ceilingcat Unfortunately that affects operator precedence and changes the output of the function.
$endgroup$
– Neil A.
yesterday
$begingroup$
@ceilingcat Unfortunately that affects operator precedence and changes the output of the function.
$endgroup$
– Neil A.
yesterday
$begingroup$
@ceilingcat Unfortunately that affects operator precedence and changes the output of the function.
$endgroup$
– Neil A.
yesterday
add a comment |
$begingroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 2 days ago
ArBoArBo
38115
$endgroup$
add a comment |
$begingroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 2 days ago
ArBoArBo
38115
$endgroup$
add a comment |
$begingroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 2 days ago
ArBoArBo
38115
$endgroup$
Python 2, 66 bytes
l=lambda n,p=1,s=[0]:p*(n<len(s))or l(n,3*p*(p/2in s)or p/2,[p]+s)
Try it online!
Uses zero-based indexing. The lambda does little more than recursively building up the sequence and returning as soon as the required index is reached.
answered 2 days ago
ArBoArBo
38115
answered 2 days ago
ArBoArBo
38115
answered 2 days ago
ArBoArBo
38115
answered 2 days ago
ArBoArBo
38115
38115
add a comment |
add a comment |
$begingroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 2 days ago
recursiverecursive
5,7491322
$endgroup$
add a comment |
$begingroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 2 days ago
recursiverecursive
5,7491322
$endgroup$
add a comment |
$begingroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 2 days ago
recursiverecursive
5,7491322
$endgroup$
Stax, 14 bytes
üÑα↕○Ü1∟¡f↑ô┬♥
Run and debug it
Zero-indexed.
answered 2 days ago
recursiverecursive
5,7491322
answered 2 days ago
recursiverecursive
5,7491322
answered 2 days ago
recursiverecursive
5,7491322
answered 2 days ago
recursiverecursive
5,7491322
5,7491322
add a comment |
add a comment |
$begingroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
$endgroup$
Wolfram Language (Mathematica), 63 bytes
(L=Last)@Nest[{##,If[FreeQ[#,x=⌊L@#/2⌋],x,3L@#]}&,{0,1},#]&
Try it online!
This is 0-indexed
(In TIO I added -1 in every test case)
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
edited Apr 8 at 0:17
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
answered Apr 7 at 22:05
J42161217J42161217
13.9k21353
13.9k21353
add a comment |
add a comment |
$begingroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
add a comment |
$begingroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
$endgroup$
Python 2, 62 bytes
a=lambda n:n<1or a(n-1)*6**(a(n-1)//2in[0]+map(a,range(n)))//2
Try it online!
Returns True for a(0). 0-indexed.
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
answered 2 days ago
Erik the OutgolferErik the Outgolfer
33k429106
33k429106
add a comment |
add a comment |
$begingroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 2 days ago
Noodle9Noodle9
30137
$endgroup$
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
2 days ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
2 days ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
2 days ago
add a comment |
$begingroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 2 days ago
Noodle9Noodle9
30137
$endgroup$
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
2 days ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
2 days ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
2 days ago
add a comment |
$begingroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 2 days ago
Noodle9Noodle9
30137
$endgroup$
Python 3, 105 103 100 95 83 bytes
-2 bytes thanks to agtoever
-12 bytes thanks to ArBo
def f(n):
s=0,1
while len(s)<=n:t=s[-1]//2;s+=(t in s)*3*s[-1]or t,
return s[-1]
Try it online!
answered 2 days ago
Noodle9Noodle9
30137
edited 2 days ago
answered 2 days ago
Noodle9Noodle9
30137
answered 2 days ago
Noodle9Noodle9
30137
answered 2 days ago
Noodle9Noodle9
30137
30137
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
2 days ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
2 days ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
2 days ago
add a comment |
$begingroup$
You can replace the for loop withwhile len(s)<=nand replace the i's with-1. This should shave off one of two characters.
$endgroup$
– agtoever
2 days ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing theiffrom thewhileloop to allow one-lining that loop
$endgroup$
– ArBo
2 days ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
You can replace the for loop with
while len(s)<=n and replace the i's with -1. This should shave off one of two characters.$endgroup$
– agtoever
2 days ago
$begingroup$
You can replace the for loop with
while len(s)<=n and replace the i's with -1. This should shave off one of two characters.$endgroup$
– agtoever
2 days ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
@agtoever that's so clever - thanks! :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing the
if from the while loop to allow one-lining that loop$endgroup$
– ArBo
2 days ago
$begingroup$
83 bytes by working with a tuple instead of a list, and removing the
if from the while loop to allow one-lining that loop$endgroup$
– ArBo
2 days ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
2 days ago
$begingroup$
@ArBo wow! absolutely brilliant - thanks :)
$endgroup$
– Noodle9
2 days ago
add a comment |
$begingroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
$endgroup$
add a comment |
$begingroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
$endgroup$
add a comment |
$begingroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
$endgroup$
Gaia, 22 20 bytes
2…@⟨:):3פḥ⌋,;D)+⟩ₓ)
Try it online!
0-based index.
Credit to Shaggy for the approach
2… | push [0 1]
@⟨ ⟩ₓ | do the following n times:
:): | dup the list L, take the last element e, and dup that
3פḥ⌋, | push [3*e floor(e/2)]
;D | take the asymmetric set difference [3*e floor(e/2)] - L
)+ | take the last element of the difference and add it to the end of L (end of loop)
) | finally, take the last element and output it
;D
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
edited 2 days ago
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
answered 2 days ago
GiuseppeGiuseppe
17.7k31153
17.7k31153
add a comment |
add a comment |
$begingroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 2 days ago
xnorxnor
94k18192451
$endgroup$
add a comment |
$begingroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 2 days ago
xnorxnor
94k18192451
$endgroup$
add a comment |
$begingroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 2 days ago
xnorxnor
94k18192451
$endgroup$
Haskell, 55 bytes
(1%[0]!!)
a%o|b<-div a 2=a:last(b:[3*a|elem b o])%(a:o)
Try it online!
Golfing user1472751's slick list-generation method.
Same length:
(1%[0]!!)
a%o=a:[x|x<-[div a 2,a*3],all(/=x)o]!!0%(a:o)
Try it online!
answered 2 days ago
xnorxnor
94k18192451
edited 2 days ago
answered 2 days ago
xnorxnor
94k18192451
answered 2 days ago
xnorxnor
94k18192451
answered 2 days ago
xnorxnor
94k18192451
94k18192451
add a comment |
add a comment |
$begingroup$
Lua, 78 bytes
x,y=1,3 u={}for _=2,...do
u[x]=0
x,y=y,y//2
if u[y]then y=3*x end
end
print(x)
Try it online!
answered yesterday
wastlwastl
2,289526
$endgroup$
$begingroup$
68 bytes through removing some whitespace, removing thezvariable and changing the if statement to ternary
$endgroup$
– Jo King
11 hours ago
add a comment |
$begingroup$
Lua, 78 bytes
x,y=1,3 u={}for _=2,...do
u[x]=0
x,y=y,y//2
if u[y]then y=3*x end
end
print(x)
Try it online!
answered yesterday
wastlwastl
2,289526
$endgroup$
$begingroup$
68 bytes through removing some whitespace, removing thezvariable and changing the if statement to ternary
$endgroup$
– Jo King
11 hours ago
add a comment |
$begingroup$
Lua, 78 bytes
x,y=1,3 u={}for _=2,...do
u[x]=0
x,y=y,y//2
if u[y]then y=3*x end
end
print(x)
Try it online!
answered yesterday
wastlwastl
2,289526
$endgroup$
Lua, 78 bytes
x,y=1,3 u={}for _=2,...do
u[x]=0
x,y=y,y//2
if u[y]then y=3*x end
end
print(x)
Try it online!
answered yesterday
wastlwastl
2,289526
answered yesterday
wastlwastl
2,289526
answered yesterday
wastlwastl
2,289526
answered yesterday
wastlwastl
2,289526
2,289526
$begingroup$
68 bytes through removing some whitespace, removing thezvariable and changing the if statement to ternary
$endgroup$
– Jo King
11 hours ago
add a comment |
$begingroup$
68 bytes through removing some whitespace, removing thezvariable and changing the if statement to ternary
$endgroup$
– Jo King
11 hours ago
$begingroup$
68 bytes through removing some whitespace, removing the
z variable and changing the if statement to ternary$endgroup$
– Jo King
11 hours ago
$begingroup$
68 bytes through removing some whitespace, removing the
z variable and changing the if statement to ternary$endgroup$
– Jo King
11 hours ago
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
I would answer this using TI-BASIC, but the input would be limited to $0<N<1000$ since lists are limited to 999 elements. Great challenge nonetheless!
$endgroup$
– Tau
yesterday
$begingroup$
@Tau : although out-of-spec (and this non-competing), I'd be interested in your solution. Do your have one you can post?
$endgroup$
– agtoever
yesterday
1
$begingroup$
I deleted the program, but I should be able to recreate it. I'll post it as non-competing once I have redone it.
$endgroup$
– Tau
yesterday
$begingroup$
@agtoever, "non-competing" doesn't cover invalid solutions; it was for solutions using languages or language features that were created after a challenge was posted.
$endgroup$
– Shaggy
yesterday
$begingroup$
PP&CG meta is indeed very clear on this. I wasn't award of such a strict interpretation of "non-competing"... @Tau : it seems you can't post your TI-BASIC solution under these rules. Sorry.
$endgroup$
– agtoever
19 hours ago