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










5












$begingroup$


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









share|improve this question









$endgroup$











  • $begingroup$
    So what's the New Order reference here?
    $endgroup$
    – Luis Mendo
    2 hours ago










  • $begingroup$
    @LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
    $endgroup$
    – agtoever
    2 hours ago










  • $begingroup$
    Ah, I completely missed that! Now I feel regret...
    $endgroup$
    – Luis Mendo
    2 hours 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
    2 hours 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
    2 hours ago
















5












$begingroup$


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









share|improve this question









$endgroup$











  • $begingroup$
    So what's the New Order reference here?
    $endgroup$
    – Luis Mendo
    2 hours ago










  • $begingroup$
    @LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
    $endgroup$
    – agtoever
    2 hours ago










  • $begingroup$
    Ah, I completely missed that! Now I feel regret...
    $endgroup$
    – Luis Mendo
    2 hours 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
    2 hours 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
    2 hours ago














5












5








5





$begingroup$


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









share|improve this question









$endgroup$




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






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 2 hours ago









agtoeveragtoever

1,354424




1,354424











  • $begingroup$
    So what's the New Order reference here?
    $endgroup$
    – Luis Mendo
    2 hours ago










  • $begingroup$
    @LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
    $endgroup$
    – agtoever
    2 hours ago










  • $begingroup$
    Ah, I completely missed that! Now I feel regret...
    $endgroup$
    – Luis Mendo
    2 hours 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
    2 hours 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
    2 hours ago

















  • $begingroup$
    So what's the New Order reference here?
    $endgroup$
    – Luis Mendo
    2 hours ago










  • $begingroup$
    @LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
    $endgroup$
    – agtoever
    2 hours ago










  • $begingroup$
    Ah, I completely missed that! Now I feel regret...
    $endgroup$
    – Luis Mendo
    2 hours 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
    2 hours 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
    2 hours ago
















$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
2 hours ago




$begingroup$
So what's the New Order reference here?
$endgroup$
– Luis Mendo
2 hours ago












$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
2 hours ago




$begingroup$
@LuisMendo: the avalanche of Fibonacci and Beatty sequences, which form the Wythoff array...
$endgroup$
– agtoever
2 hours ago












$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
2 hours ago




$begingroup$
Ah, I completely missed that! Now I feel regret...
$endgroup$
– Luis Mendo
2 hours 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
2 hours 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
2 hours 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
2 hours 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
2 hours ago











3 Answers
3






active

oldest

votes


















1












$begingroup$


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.






share|improve this answer









$endgroup$












  • $begingroup$
    ...×⁹r‘ÆḞ¤Sð/ saves one.
    $endgroup$
    – Jonathan Allan
    2 mins ago


















1












$begingroup$


Jelly, 30 bytes



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.






share|improve this answer











$endgroup$








  • 1




    $begingroup$
    you are right! 740496902 is the result for 999
    $endgroup$
    – J42161217
    39 mins ago










  • $begingroup$
    Combining the first part of yours and second part of mine gives 25 bytes. Not sure which of us should have the combined version!
    $endgroup$
    – Nick Kennedy
    29 mins ago










  • $begingroup$
    @NickKennedy - nice, go for it!
    $endgroup$
    – Jonathan Allan
    27 mins ago


















0












$begingroup$


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!






share|improve this answer









$endgroup$













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$


    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.






    share|improve this answer









    $endgroup$












    • $begingroup$
      ...×⁹r‘ÆḞ¤Sð/ saves one.
      $endgroup$
      – Jonathan Allan
      2 mins ago















    1












    $begingroup$


    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.






    share|improve this answer









    $endgroup$












    • $begingroup$
      ...×⁹r‘ÆḞ¤Sð/ saves one.
      $endgroup$
      – Jonathan Allan
      2 mins ago













    1












    1








    1





    $begingroup$


    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.






    share|improve this answer









    $endgroup$




    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.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 40 mins ago









    Nick KennedyNick Kennedy

    1,56649




    1,56649











    • $begingroup$
      ...×⁹r‘ÆḞ¤Sð/ saves one.
      $endgroup$
      – Jonathan Allan
      2 mins ago
















    • $begingroup$
      ...×⁹r‘ÆḞ¤Sð/ saves one.
      $endgroup$
      – Jonathan Allan
      2 mins ago















    $begingroup$
    ...×⁹r‘ÆḞ¤Sð/ saves one.
    $endgroup$
    – Jonathan Allan
    2 mins ago




    $begingroup$
    ...×⁹r‘ÆḞ¤Sð/ saves one.
    $endgroup$
    – Jonathan Allan
    2 mins ago











    1












    $begingroup$


    Jelly, 30 bytes



    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.






    share|improve this answer











    $endgroup$








    • 1




      $begingroup$
      you are right! 740496902 is the result for 999
      $endgroup$
      – J42161217
      39 mins ago










    • $begingroup$
      Combining the first part of yours and second part of mine gives 25 bytes. Not sure which of us should have the combined version!
      $endgroup$
      – Nick Kennedy
      29 mins ago










    • $begingroup$
      @NickKennedy - nice, go for it!
      $endgroup$
      – Jonathan Allan
      27 mins ago















    1












    $begingroup$


    Jelly, 30 bytes



    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.






    share|improve this answer











    $endgroup$








    • 1




      $begingroup$
      you are right! 740496902 is the result for 999
      $endgroup$
      – J42161217
      39 mins ago










    • $begingroup$
      Combining the first part of yours and second part of mine gives 25 bytes. Not sure which of us should have the combined version!
      $endgroup$
      – Nick Kennedy
      29 mins ago










    • $begingroup$
      @NickKennedy - nice, go for it!
      $endgroup$
      – Jonathan Allan
      27 mins ago













    1












    1








    1





    $begingroup$


    Jelly, 30 bytes



    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.






    share|improve this answer











    $endgroup$




    Jelly, 30 bytes



    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.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 17 mins ago

























    answered 44 mins ago









    Jonathan AllanJonathan Allan

    54.3k537174




    54.3k537174







    • 1




      $begingroup$
      you are right! 740496902 is the result for 999
      $endgroup$
      – J42161217
      39 mins ago










    • $begingroup$
      Combining the first part of yours and second part of mine gives 25 bytes. Not sure which of us should have the combined version!
      $endgroup$
      – Nick Kennedy
      29 mins ago










    • $begingroup$
      @NickKennedy - nice, go for it!
      $endgroup$
      – Jonathan Allan
      27 mins ago












    • 1




      $begingroup$
      you are right! 740496902 is the result for 999
      $endgroup$
      – J42161217
      39 mins ago










    • $begingroup$
      Combining the first part of yours and second part of mine gives 25 bytes. Not sure which of us should have the combined version!
      $endgroup$
      – Nick Kennedy
      29 mins ago










    • $begingroup$
      @NickKennedy - nice, go for it!
      $endgroup$
      – Jonathan Allan
      27 mins ago







    1




    1




    $begingroup$
    you are right! 740496902 is the result for 999
    $endgroup$
    – J42161217
    39 mins ago




    $begingroup$
    you are right! 740496902 is the result for 999
    $endgroup$
    – J42161217
    39 mins ago












    $begingroup$
    Combining the first part of yours and second part of mine gives 25 bytes. Not sure which of us should have the combined version!
    $endgroup$
    – Nick Kennedy
    29 mins ago




    $begingroup$
    Combining the first part of yours and second part of mine gives 25 bytes. Not sure which of us should have the combined version!
    $endgroup$
    – Nick Kennedy
    29 mins ago












    $begingroup$
    @NickKennedy - nice, go for it!
    $endgroup$
    – Jonathan Allan
    27 mins ago




    $begingroup$
    @NickKennedy - nice, go for it!
    $endgroup$
    – Jonathan Allan
    27 mins ago











    0












    $begingroup$


    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!






    share|improve this answer









    $endgroup$

















      0












      $begingroup$


      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!






      share|improve this answer









      $endgroup$















        0












        0








        0





        $begingroup$


        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!






        share|improve this answer









        $endgroup$




        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!







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 1 hour ago









        J42161217J42161217

        14k21353




        14k21353



























            draft saved

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            Кастелфранко ди Сопра Становништво Референце Спољашње везе Мени за навигацију43°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.5588543°37′18″ СГШ; 11°33′32″ ИГД / 43.62156° СГШ; 11.55885° ИГД / 43.62156; 11.558853179688„The GeoNames geographical database”„Istituto Nazionale di Statistica”проширитиууWorldCat156923403n850174324558639-1cb14643287r(подаци)