9
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Starting at 1-TET, give equal temperaments that have better and better approximation of the perfect fifth(just ratio 3/2). (OEIS sequence A060528)

The formal description of the sequence, copied from the OEIS:

A list of equal temperaments (equal divisions of the octave) whose nearest scale steps are closer and closer approximations to the ratios of two tones of musical harmony: the perfect 4th, 4/3 and its complement the perfect 5th, 3/2.

Note that by symmetry, the perfect fourth doesn't matter.

Let's say we know that 3 is in the sequence. The frequencies in 3-TET are:

2^0, 2^⅓, 2^⅔

Where 2^⅔ is the closest logarithmic approximation of 3/2.

Is 4 in the sequence? The frequencies in 4-TET are:

2^0, 2^¼, 2^½, 2^¾

Where 2^½ is the closest approximation of 3/2. This is not better than 2^⅔, so 4 is not in the sequence.

By similar method, we confirm that 5 is in the sequence, and so on.

When given an integer n as input, the output must be the first N numbers of the sequence in order. For example, when n = 7, the output should be:

1 2 3 5 7 12 29

Sequence description by xnor

The irrational constant \$ \log_2(3) \approx 1.5849625007211563\dots\$ can be approximated by a sequence of rational fractions

$$ \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \frac{11}{7}, \frac{19}{12}, \frac{46}{29}, \dots$$

A fraction is included in the sequence if it's the new closest one by absolute distance \$ \left| \frac{p}{q} - \log_2(3)\ \right|\$, that is, closer than any other fraction with a smaller or equal denominator.

Your goal is to output the first \$n\$ denominators in order. These are sequence A060528 (table). The numerators (not required) are given by A254351 (table)

Rules:

  1. Do not import the sequence A060528 directly.
  2. The format doesn't matter as long as the numbers are distinguishable. In the example above, the output can also be:

    [1,2,3,5,7,12,29]

  3. As this is a code-golf, the shortest code in bytes wins.

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  • 5
    \$\begingroup\$ Hi and welcome to Code Golf SE! We require that all challenges be self-contained, so a description here of the sequence would be a great help. \$\endgroup\$ – AdmBorkBork Sep 24 at 21:54
  • 5
    \$\begingroup\$ I'm confused by the description from OEIS. It mentions perfect 4th as well (ratio 4/3), but the challenge is about perfect 5ths (ratio 3/2). I think it also needs explanation that the sequence values are the denominators of the rational approximations. \$\endgroup\$ – xnor Sep 24 at 21:58
  • 5
    \$\begingroup\$ I like the challenge, but I find the stuff added to the description still confusing, not knowing much about music. For instance, I don't know what 1-TET or 4-TET are, and nothing seems to show up on Google. I'll try writing an explanation of how I'd describe this sequence. \$\endgroup\$ – xnor Sep 24 at 23:06
  • 3
    \$\begingroup\$ @DannyuNDos Ah yes, the 12-tone equal temperament. That's my favourite instrument \$\endgroup\$ – Jo King Sep 25 at 1:53
  • 2
    \$\begingroup\$ @DannyuNDos Thanks. So the comparison is between 1/2 and log2(1.5), not between 2^(1/2) and 1.5. That should be made clearer in the text \$\endgroup\$ – Luis Mendo Sep 25 at 9:48
5
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05AB1E, 19 18 bytes

µ¯ßNLN/3.²<αßDˆ›D–

Try it online!

µ                      # repeat until counter == input
 ¯                     #  push the global array
  ß                    #  get the minimum (let's call it M)
   N                   #  1-based iteration count
    L                  #  range 1..N
     N/                #  divide each by N
       3.²             #  log2(3)
          <            #  -1
           α           #  absolute difference with each element of the range
            ß          #  get the minimum
             Dˆ        #  add a copy to the global array
               ›       #  is M strictly greater than this new minimum?
                D–     #  if true, print N
                       #  implicit: if true, add 1 to the counter
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  • 1
    \$\begingroup\$ Nice answer, but all I'm wondering right now is why the while-loop has 1-based indices.. :S \$\endgroup\$ – Kevin Cruijssen Sep 25 at 12:48
4
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Wolfram Language (Mathematica), 62 60 bytes

Denominator@NestList[Rationalize[r=Log2@3,Abs[#-r]]&,2,#-1]&

Try it online!

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  • \$\begingroup\$ How many precision? \$\endgroup\$ – Dannyu NDos Sep 25 at 1:38
  • \$\begingroup\$ @DannyuNDos This function uses exact values, so computations can be done to arbitrary precision. \$\endgroup\$ – attinat Sep 25 at 1:39
  • \$\begingroup\$ You win the challenge. \$\endgroup\$ – Dannyu NDos Sep 25 at 1:48
  • 5
    \$\begingroup\$ @DannyuNDos why accept an answer this quickly? It's also arguably better not to accept an answer at all.. \$\endgroup\$ – attinat Sep 25 at 2:03
  • \$\begingroup\$ Regarding the floating-point errors other languages are suffering, I'd like to present an alternative method of alloting score. So hold on. \$\endgroup\$ – Dannyu NDos Sep 25 at 2:10
4
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JavaScript (V8), 81 80 78 bytes

-2 bytes thanks Arnauld!

n=>{for(d=g=1;w=Math.log2(3),w+=~(w*g-.5)/g,n--;g++)w*w<d?d=print(g)||w*w:n++}

Try it online!

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2
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Python 2, 92 bytes

E=k=input()
n=0
while k:
 n+=1;e=abs((3.169925001442312*n-1)%2-1)/n
 if e<E:print n;E=e;k-=1

Try it online!

Uses the constant 3.169925001442312 for \$2 \log_2(3)\$. I wasn't sure how many digits of accuracy are required, since the inaccuracy will break the sequence eventually, so I used the full float precision of 2 * numpy.log2(3).

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  • 1
    \$\begingroup\$ This gives two extra terms after 665: ..., 665, (1995), (4655), 8286, ... Try it online! \$\endgroup\$ – Οurous Sep 25 at 1:31
  • \$\begingroup\$ @Οurous Yeah, that's pretty much inevitable sooner or later for any language without infinite precision, though I'm surprised it popped up so early with 32-bit floats as Python uses. I'll wait for the challenge writer to clarify on how far answers need to work. \$\endgroup\$ – xnor Sep 25 at 7:03
  • \$\begingroup\$ wouldn't it be fewer characters to use 2 * numpy.log2(3) rather than the fully spelled out number? (Or even better, numpy.log2(9)) \$\endgroup\$ – JDL Sep 25 at 13:28
  • \$\begingroup\$ @JDL that would require this code: from numpy import* and log2(9). \$\endgroup\$ – Jonathan Allan Sep 25 at 13:34
  • \$\begingroup\$ ah, that is what I get for assuming python works like R and you can write package::function without loading package first! \$\endgroup\$ – JDL Sep 25 at 13:35
2
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Clean, 128 111 108 bytes

import StdEnv
c=ln 3.0/ln 2.0
?d=abs(toReal(toInt(c*d))/d-c)
$i=take i(iterate(\d=until((>)(?d)o?)inc d)1.0)

Try it online!

Should work up to the limits of Real's 64-bit double precision type.

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2
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MATL, 27 25 bytes

1`@:@/Q3Zl-|X<hY<tdzG-}df

Try it online!

Explanation

1       % Push 1. This initiallizes the vector of distances
  `     % Do...while
  @:    %   Range [1, 2, ..., k], where k is the iteration index, staring at 1
  @/    %   Divide by k, element-wise. Gives [1/k, 2/k, ..., 1]
  Q     %   Add 1, element-wise. Gives [(k+1/k, (k+2)/k, ..., 2]
  3Zl   %   Push log2(3)
  -|    %   Absolute difference, element-wise
  X<    %   Minimum
  h     %   Concatenate with vector of previous distances
  Y<    %   Cumulative minimum
  t     %   Duplicate
  dz    %   Consecutive differences, number of nonzeros. This tells how many
        %   times the cumulative minimum has decreased
  G-    %   Subtract input n. This is the loop condition. 0 means we are done
}       % Finally (execute on loop exit)
  d     %   Consecutive differences (of the vector of cumulative differences)
  f     %   Indices of nonzeros. This is the final result
        % End. A new iteration is executed if the top of the stack is nonzero
        % Implicit display
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2
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Perl 5 (-MPOSIX=log2 -M5.01 -n), 73, 78, 71 bytes

Fixed following comment, may be improved...

-7 bytes thanks to Grimy

$o=abs$d-(0|.5+($d=log2 3)*++$;)/$;;$@=$o,$_-=say$;if!$@|$o<$@;$_&&redo

Try it online!

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