New Order #5: where Fibonacci and Beatty meet at Wythoff Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) The PPCG Site design is on its way - help us make it awesome! Sandbox for Proposed ChallengesNew order #4: WorldNew Order #2: Turn My WayNew Order #1: How does this feel?New Order #3: 5 8 6Print the intersection of sequencesFibonacci ExponentsFind the Fibonacci KernelSum my Fibonaccified divisors!Reverse FibonacciUpper or Lower Wythoff?New Order #1: How does this feel?New Order #2: Turn My WayNew Order #3: 5 8 6New order #4: World
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New Order #5: where Fibonacci and Beatty meet at Wythoff
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
The PPCG Site design is on its way - help us make it awesome!
Sandbox for Proposed ChallengesNew order #4: WorldNew Order #2: Turn My WayNew Order #1: How does this feel?New Order #3: 5 8 6Print the intersection of sequencesFibonacci ExponentsFind the Fibonacci KernelSum my Fibonaccified divisors!Reverse FibonacciUpper or Lower Wythoff?New Order #1: How does this feel?New Order #2: Turn My WayNew Order #3: 5 8 6New order #4: World
$begingroup$
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 fifth challenge in this series (links to the first, second, third and fourth challenge).
In this challenge, we will meet the Wythoff array, which is a intertwined avalanche of Fibonacci sequences and Beatty sequences!
The Fibonacci numbers are probably for most of you a well known sequence. Given two starting numbers $F_0$ and $F_1$, the following $F_n$ are given by: $F_n = F_(n-1) + F_(n-2)$ for $n>2$.
The Beatty sequence, given a parameter $r$ is: $B^r_n = lfloor rn rfloor$ for $n ge 1$. One of the properties of the Beatty sequence is that for every parameter $r$, there is exactly one parameter $s=r/(r-1)$, such that the Beatty sequences for those parameters are disjunct and joined together, they span all natural numbers excluding 0 (e.g.: $B^r cup B^r/(r-1) = BbbN setminus 0$).
Now here comes the mindblowing part: you can create an array, where each row is a Fibonacci sequence and each column is a Beatty sequence. This array is the Wythoff array. The best part is: every positive number appears exactly once in this array! The array looks like this:
1 2 3 5 8 13 21 34 55 89 144 ...
4 7 11 18 29 47 76 123 199 322 521 ...
6 10 16 26 42 68 110 178 288 466 754 ...
9 15 24 39 63 102 165 267 432 699 1131 ...
12 20 32 52 84 136 220 356 576 932 1508 ...
14 23 37 60 97 157 254 411 665 1076 1741 ...
17 28 45 73 118 191 309 500 809 1309 2118 ...
19 31 50 81 131 212 343 555 898 1453 2351 ...
22 36 58 94 152 246 398 644 1042 1686 2728 ...
25 41 66 107 173 280 453 733 1186 1919 3105 ...
27 44 71 115 186 301 487 788 1275 2063 3338 ...
...
An element at row $m$ and column $n$ is defined as:
$A_m,n = begincases
A_m,1 = leftlfloor lfloor mvarphi rfloor varphi rightrfloor\
A_m,2 = leftlfloor lfloor mvarphi rfloor varphi^2 rightrfloor\
A_m,n = A_m,n-2+A_m,n-1 text for n > 2
endcases$
where $varphi$ is the golden ratio: $varphi=frac1+sqrt52$.
If we follow the anti-diagonals of this array, we get A035513, which is the target sequence for this challenge (note that this sequence is added to the OEIS by Neil Sloane himself!). Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A035513.
There are different strategies you can follow to get to $a(n)$, which makes this challenge (in my opinion) really interesting.
Task
Given an integer input $n$, output $a(n)$ in integer format, where $a(n)$ is A035513.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 2$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 7
20 | 20
50 | 136
78 | 30
123 | 3194
1234 | 8212236486
3000 | 814
9999 | 740496902
29890 | 637
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767). Note that $a(n)$ goes up to 30 digit numbers in this range...
- 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 2 hours ago
agtoeveragtoever
1,354424
$endgroup$
|
show 3 more comments
$begingroup$
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 fifth challenge in this series (links to the first, second, third and fourth challenge).
In this challenge, we will meet the Wythoff array, which is a intertwined avalanche of Fibonacci sequences and Beatty sequences!
The Fibonacci numbers are probably for most of you a well known sequence. Given two starting numbers $F_0$ and $F_1$, the following $F_n$ are given by: $F_n = F_(n-1) + F_(n-2)$ for $n>2$.
The Beatty sequence, given a parameter $r$ is: $B^r_n = lfloor rn rfloor$ for $n ge 1$. One of the properties of the Beatty sequence is that for every parameter $r$, there is exactly one parameter $s=r/(r-1)$, such that the Beatty sequences for those parameters are disjunct and joined together, they span all natural numbers excluding 0 (e.g.: $B^r cup B^r/(r-1) = BbbN setminus 0$).
Now here comes the mindblowing part: you can create an array, where each row is a Fibonacci sequence and each column is a Beatty sequence. This array is the Wythoff array. The best part is: every positive number appears exactly once in this array! The array looks like this:
1 2 3 5 8 13 21 34 55 89 144 ...
4 7 11 18 29 47 76 123 199 322 521 ...
6 10 16 26 42 68 110 178 288 466 754 ...
9 15 24 39 63 102 165 267 432 699 1131 ...
12 20 32 52 84 136 220 356 576 932 1508 ...
14 23 37 60 97 157 254 411 665 1076 1741 ...
17 28 45 73 118 191 309 500 809 1309 2118 ...
19 31 50 81 131 212 343 555 898 1453 2351 ...
22 36 58 94 152 246 398 644 1042 1686 2728 ...
25 41 66 107 173 280 453 733 1186 1919 3105 ...
27 44 71 115 186 301 487 788 1275 2063 3338 ...
...
An element at row $m$ and column $n$ is defined as:
$A_m,n = begincases
A_m,1 = leftlfloor lfloor mvarphi rfloor varphi rightrfloor\
A_m,2 = leftlfloor lfloor mvarphi rfloor varphi^2 rightrfloor\
A_m,n = A_m,n-2+A_m,n-1 text for n > 2
endcases$
where $varphi$ is the golden ratio: $varphi=frac1+sqrt52$.
If we follow the anti-diagonals of this array, we get A035513, which is the target sequence for this challenge (note that this sequence is added to the OEIS by Neil Sloane himself!). Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A035513.
There are different strategies you can follow to get to $a(n)$, which makes this challenge (in my opinion) really interesting.
Task
Given an integer input $n$, output $a(n)$ in integer format, where $a(n)$ is A035513.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 2$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 7
20 | 20
50 | 136
78 | 30
123 | 3194
1234 | 8212236486
3000 | 814
9999 | 740496902
29890 | 637
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767). Note that $a(n)$ goes up to 30 digit numbers in this range...
- 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 2 hours ago
agtoeveragtoever
1,354424
$endgroup$
$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
1 hour ago
$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
Is a floating point representation of phi (or rt(5)) and application of the recurrence going to satisfy the range requirement?
$endgroup$
– Jonathan Allan
1 hour ago
$begingroup$
@JonathanAllan : good point... I'll look into that later. For now: let's pose that if some code passes the test cases, then it works sufficiently.
$endgroup$
– agtoever
1 hour ago
|
show 3 more comments
$begingroup$
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 fifth challenge in this series (links to the first, second, third and fourth challenge).
In this challenge, we will meet the Wythoff array, which is a intertwined avalanche of Fibonacci sequences and Beatty sequences!
The Fibonacci numbers are probably for most of you a well known sequence. Given two starting numbers $F_0$ and $F_1$, the following $F_n$ are given by: $F_n = F_(n-1) + F_(n-2)$ for $n>2$.
The Beatty sequence, given a parameter $r$ is: $B^r_n = lfloor rn rfloor$ for $n ge 1$. One of the properties of the Beatty sequence is that for every parameter $r$, there is exactly one parameter $s=r/(r-1)$, such that the Beatty sequences for those parameters are disjunct and joined together, they span all natural numbers excluding 0 (e.g.: $B^r cup B^r/(r-1) = BbbN setminus 0$).
Now here comes the mindblowing part: you can create an array, where each row is a Fibonacci sequence and each column is a Beatty sequence. This array is the Wythoff array. The best part is: every positive number appears exactly once in this array! The array looks like this:
1 2 3 5 8 13 21 34 55 89 144 ...
4 7 11 18 29 47 76 123 199 322 521 ...
6 10 16 26 42 68 110 178 288 466 754 ...
9 15 24 39 63 102 165 267 432 699 1131 ...
12 20 32 52 84 136 220 356 576 932 1508 ...
14 23 37 60 97 157 254 411 665 1076 1741 ...
17 28 45 73 118 191 309 500 809 1309 2118 ...
19 31 50 81 131 212 343 555 898 1453 2351 ...
22 36 58 94 152 246 398 644 1042 1686 2728 ...
25 41 66 107 173 280 453 733 1186 1919 3105 ...
27 44 71 115 186 301 487 788 1275 2063 3338 ...
...
An element at row $m$ and column $n$ is defined as:
$A_m,n = begincases
A_m,1 = leftlfloor lfloor mvarphi rfloor varphi rightrfloor\
A_m,2 = leftlfloor lfloor mvarphi rfloor varphi^2 rightrfloor\
A_m,n = A_m,n-2+A_m,n-1 text for n > 2
endcases$
where $varphi$ is the golden ratio: $varphi=frac1+sqrt52$.
If we follow the anti-diagonals of this array, we get A035513, which is the target sequence for this challenge (note that this sequence is added to the OEIS by Neil Sloane himself!). Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A035513.
There are different strategies you can follow to get to $a(n)$, which makes this challenge (in my opinion) really interesting.
Task
Given an integer input $n$, output $a(n)$ in integer format, where $a(n)$ is A035513.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 2$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 7
20 | 20
50 | 136
78 | 30
123 | 3194
1234 | 8212236486
3000 | 814
9999 | 740496902
29890 | 637
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767). Note that $a(n)$ goes up to 30 digit numbers in this range...
- 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 2 hours ago
agtoeveragtoever
1,354424
$endgroup$
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 fifth challenge in this series (links to the first, second, third and fourth challenge).
In this challenge, we will meet the Wythoff array, which is a intertwined avalanche of Fibonacci sequences and Beatty sequences!
The Fibonacci numbers are probably for most of you a well known sequence. Given two starting numbers $F_0$ and $F_1$, the following $F_n$ are given by: $F_n = F_(n-1) + F_(n-2)$ for $n>2$.
The Beatty sequence, given a parameter $r$ is: $B^r_n = lfloor rn rfloor$ for $n ge 1$. One of the properties of the Beatty sequence is that for every parameter $r$, there is exactly one parameter $s=r/(r-1)$, such that the Beatty sequences for those parameters are disjunct and joined together, they span all natural numbers excluding 0 (e.g.: $B^r cup B^r/(r-1) = BbbN setminus 0$).
Now here comes the mindblowing part: you can create an array, where each row is a Fibonacci sequence and each column is a Beatty sequence. This array is the Wythoff array. The best part is: every positive number appears exactly once in this array! The array looks like this:
1 2 3 5 8 13 21 34 55 89 144 ...
4 7 11 18 29 47 76 123 199 322 521 ...
6 10 16 26 42 68 110 178 288 466 754 ...
9 15 24 39 63 102 165 267 432 699 1131 ...
12 20 32 52 84 136 220 356 576 932 1508 ...
14 23 37 60 97 157 254 411 665 1076 1741 ...
17 28 45 73 118 191 309 500 809 1309 2118 ...
19 31 50 81 131 212 343 555 898 1453 2351 ...
22 36 58 94 152 246 398 644 1042 1686 2728 ...
25 41 66 107 173 280 453 733 1186 1919 3105 ...
27 44 71 115 186 301 487 788 1275 2063 3338 ...
...
An element at row $m$ and column $n$ is defined as:
$A_m,n = begincases
A_m,1 = leftlfloor lfloor mvarphi rfloor varphi rightrfloor\
A_m,2 = leftlfloor lfloor mvarphi rfloor varphi^2 rightrfloor\
A_m,n = A_m,n-2+A_m,n-1 text for n > 2
endcases$
where $varphi$ is the golden ratio: $varphi=frac1+sqrt52$.
If we follow the anti-diagonals of this array, we get A035513, which is the target sequence for this challenge (note that this sequence is added to the OEIS by Neil Sloane himself!). Since this is a "pure sequence" challenge, the task is to output $a(n)$ for a given $n$ as input, where $a(n)$ is A035513.
There are different strategies you can follow to get to $a(n)$, which makes this challenge (in my opinion) really interesting.
Task
Given an integer input $n$, output $a(n)$ in integer format, where $a(n)$ is A035513.
Note: 1-based indexing is assumed here; you may use 0-based indexing, so $a(0) = 1; a(1) = 2$, etc. Please mention this in your answer if you choose to use this.
Test cases
Input | Output
---------------
1 | 1
5 | 7
20 | 20
50 | 136
78 | 30
123 | 3194
1234 | 8212236486
3000 | 814
9999 | 740496902
29890 | 637
Rules
- Input and output are integers (your program should at least support input and output in the range of 1 up to 32767). Note that $a(n)$ goes up to 30 digit numbers in this range...
- 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 2 hours ago
agtoeveragtoever
1,354424
asked 2 hours ago
agtoeveragtoever
1,354424
asked 2 hours ago
agtoeveragtoever
1,354424
asked 2 hours ago
agtoeveragtoever
1,354424
asked 2 hours ago
agtoeveragtoever
1,354424
1,354424
$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
1 hour ago
$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
Is a floating point representation of phi (or rt(5)) and application of the recurrence going to satisfy the range requirement?
$endgroup$
– Jonathan Allan
1 hour ago
$begingroup$
@JonathanAllan : good point... I'll look into that later. For now: let's pose that if some code passes the test cases, then it works sufficiently.
$endgroup$
– agtoever
1 hour ago
|
show 3 more comments
$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
1 hour ago
$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
Is a floating point representation of phi (or rt(5)) and application of the recurrence going to satisfy the range requirement?
$endgroup$
– Jonathan Allan
1 hour ago
$begingroup$
@JonathanAllan : good point... I'll look into that later. For now: let's pose that if some code passes the test cases, then it works sufficiently.
$endgroup$
– agtoever
1 hour ago
$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
1 hour ago
$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
1 hour ago
$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
Is a floating point representation of phi (or rt(5)) and application of the recurrence going to satisfy the range requirement?
$endgroup$
– Jonathan Allan
1 hour ago
$begingroup$
Is a floating point representation of phi (or rt(5)) and application of the recurrence going to satisfy the range requirement?
$endgroup$
– Jonathan Allan
1 hour ago
$begingroup$
@JonathanAllan : good point... I'll look into that later. For now: let's pose that if some code passes the test cases, then it works sufficiently.
$endgroup$
– agtoever
1 hour ago
$begingroup$
@JonathanAllan : good point... I'll look into that later. For now: let's pose that if some code passes the test cases, then it works sufficiently.
$endgroup$
– agtoever
1 hour ago
|
show 3 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Wolfram Language (Mathematica), 90 bytes
Flatten[Table[(F=Fibonacci)[a+1]⌊(b-a+1)GoldenRatio⌋+(b-a)F@a,b,#,a,b,1,-1]][[#]]&
Try it online!
answered 1 hour ago
J42161217J42161217
14k21353
$endgroup$
add a comment |
$begingroup$
Jelly, 30 bytes
If 9999 is really meant to yield 740496902 and not 108240 then I got something wrong
p`SÞ⁸ịð;Øp,²;¤×Ḟ¥/;+ƝQƊ⁹¡ị@ð/
Try it online!
This is a little slow, but a huge improvement is made with a prefix of Ḥ½Ċ (double, square-root, ceiling) like in this test-suite.
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
$endgroup$
1
$begingroup$
you are right!740496902is the result for999
$endgroup$
– J42161217
7 mins ago
add a comment |
$begingroup$
Jelly, 27 bytes
RṁṬ€œið’;×ØpḞ¥×ạ‘+Ø.ÆḞʋSð@/
Try it online!
Monadic link using 1-based indexing. I’m sure there’s a better way of generating the row/column indices from n, but this works ok. In its shortest form it’s too slow for larger n on TIO, so the following Try it online! reduces the size initial triangular list at the cost of three bytes.
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Wolfram Language (Mathematica), 90 bytes
Flatten[Table[(F=Fibonacci)[a+1]⌊(b-a+1)GoldenRatio⌋+(b-a)F@a,b,#,a,b,1,-1]][[#]]&
Try it online!
answered 1 hour ago
J42161217J42161217
14k21353
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 90 bytes
Flatten[Table[(F=Fibonacci)[a+1]⌊(b-a+1)GoldenRatio⌋+(b-a)F@a,b,#,a,b,1,-1]][[#]]&
Try it online!
answered 1 hour ago
J42161217J42161217
14k21353
$endgroup$
add a comment |
$begingroup$
Wolfram Language (Mathematica), 90 bytes
Flatten[Table[(F=Fibonacci)[a+1]⌊(b-a+1)GoldenRatio⌋+(b-a)F@a,b,#,a,b,1,-1]][[#]]&
Try it online!
answered 1 hour ago
J42161217J42161217
14k21353
$endgroup$
Wolfram Language (Mathematica), 90 bytes
Flatten[Table[(F=Fibonacci)[a+1]⌊(b-a+1)GoldenRatio⌋+(b-a)F@a,b,#,a,b,1,-1]][[#]]&
Try it online!
answered 1 hour ago
J42161217J42161217
14k21353
answered 1 hour ago
J42161217J42161217
14k21353
answered 1 hour ago
J42161217J42161217
14k21353
answered 1 hour ago
J42161217J42161217
14k21353
14k21353
add a comment |
add a comment |
$begingroup$
Jelly, 30 bytes
If 9999 is really meant to yield 740496902 and not 108240 then I got something wrong
p`SÞ⁸ịð;Øp,²;¤×Ḟ¥/;+ƝQƊ⁹¡ị@ð/
Try it online!
This is a little slow, but a huge improvement is made with a prefix of Ḥ½Ċ (double, square-root, ceiling) like in this test-suite.
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
$endgroup$
1
$begingroup$
you are right!740496902is the result for999
$endgroup$
– J42161217
7 mins ago
add a comment |
$begingroup$
Jelly, 30 bytes
If 9999 is really meant to yield 740496902 and not 108240 then I got something wrong
p`SÞ⁸ịð;Øp,²;¤×Ḟ¥/;+ƝQƊ⁹¡ị@ð/
Try it online!
This is a little slow, but a huge improvement is made with a prefix of Ḥ½Ċ (double, square-root, ceiling) like in this test-suite.
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
$endgroup$
1
$begingroup$
you are right!740496902is the result for999
$endgroup$
– J42161217
7 mins ago
add a comment |
$begingroup$
Jelly, 30 bytes
If 9999 is really meant to yield 740496902 and not 108240 then I got something wrong
p`SÞ⁸ịð;Øp,²;¤×Ḟ¥/;+ƝQƊ⁹¡ị@ð/
Try it online!
This is a little slow, but a huge improvement is made with a prefix of Ḥ½Ċ (double, square-root, ceiling) like in this test-suite.
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
$endgroup$
Jelly, 30 bytes
If 9999 is really meant to yield 740496902 and not 108240 then I got something wrong
p`SÞ⁸ịð;Øp,²;¤×Ḟ¥/;+ƝQƊ⁹¡ị@ð/
Try it online!
This is a little slow, but a huge improvement is made with a prefix of Ḥ½Ċ (double, square-root, ceiling) like in this test-suite.
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
answered 11 mins ago
Jonathan AllanJonathan Allan
54.3k537174
54.3k537174
1
$begingroup$
you are right!740496902is the result for999
$endgroup$
– J42161217
7 mins ago
add a comment |
1
$begingroup$
you are right!740496902is the result for999
$endgroup$
– J42161217
7 mins ago
1
1
$begingroup$
you are right!
740496902 is the result for 999$endgroup$
– J42161217
7 mins ago
$begingroup$
you are right!
740496902 is the result for 999$endgroup$
– J42161217
7 mins ago
add a comment |
$begingroup$
Jelly, 27 bytes
RṁṬ€œið’;×ØpḞ¥×ạ‘+Ø.ÆḞʋSð@/
Try it online!
Monadic link using 1-based indexing. I’m sure there’s a better way of generating the row/column indices from n, but this works ok. In its shortest form it’s too slow for larger n on TIO, so the following Try it online! reduces the size initial triangular list at the cost of three bytes.
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
$endgroup$
add a comment |
$begingroup$
Jelly, 27 bytes
RṁṬ€œið’;×ØpḞ¥×ạ‘+Ø.ÆḞʋSð@/
Try it online!
Monadic link using 1-based indexing. I’m sure there’s a better way of generating the row/column indices from n, but this works ok. In its shortest form it’s too slow for larger n on TIO, so the following Try it online! reduces the size initial triangular list at the cost of three bytes.
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
$endgroup$
add a comment |
$begingroup$
Jelly, 27 bytes
RṁṬ€œið’;×ØpḞ¥×ạ‘+Ø.ÆḞʋSð@/
Try it online!
Monadic link using 1-based indexing. I’m sure there’s a better way of generating the row/column indices from n, but this works ok. In its shortest form it’s too slow for larger n on TIO, so the following Try it online! reduces the size initial triangular list at the cost of three bytes.
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
$endgroup$
Jelly, 27 bytes
RṁṬ€œið’;×ØpḞ¥×ạ‘+Ø.ÆḞʋSð@/
Try it online!
Monadic link using 1-based indexing. I’m sure there’s a better way of generating the row/column indices from n, but this works ok. In its shortest form it’s too slow for larger n on TIO, so the following Try it online! reduces the size initial triangular list at the cost of three bytes.
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
answered 7 mins ago
Nick KennedyNick Kennedy
1,55649
1,55649
add a comment |
add a comment |
If this is an answer to a challenge…
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$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
1 hour ago
$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
1 hour ago
$begingroup$
Is a floating point representation of phi (or rt(5)) and application of the recurrence going to satisfy the range requirement?
$endgroup$
– Jonathan Allan
1 hour ago
$begingroup$
@JonathanAllan : good point... I'll look into that later. For now: let's pose that if some code passes the test cases, then it works sufficiently.
$endgroup$
– agtoever
1 hour ago