Kite's thoughts on pergens: Difference between revisions

Wikispaces>TallKite
**Imported revision 624772747 - Original comment: **
Wikispaces>TallKite
**Imported revision 624773549 - Original comment: **
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<h2>IMPORTED REVISION FROM WIKISPACES</h2>
<h2>IMPORTED REVISION FROM WIKISPACES</h2>
This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
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: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-12 05:03:54 UTC</tt>.<br>
: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-12 05:43:33 UTC</tt>.<br>
: The original revision id was <tt>624772747</tt>.<br>
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Heptatonic 5th-based notation is only possible if the 5th ranges from 600¢ to 720¢. In practice, the lower limit of this range is ~646¢, for 13b-edo. For every bare enharmonic, the following table shows in what parts of this range this interval should be upped or downed. The tipping point edo is simply the 3-exponent of the bare enharmonic.
Heptatonic 5th-based notation is only possible if the 5th ranges from 600¢ to 720¢. In practice, the lower limit of this range is ~646¢, for 13b-edo. For every bare enharmonic, the following table shows in what parts of this range this interval should be upped or downed. The tipping point edo is simply the 3-exponent of the bare enharmonic.
||||~ bare enharmonic interval ||~ 3-exponent ||~ tipping  
||||~ bare enharmonic interval ||~ 3-exponent ||~ tipping
point edo ||~ edo's 5th ||~ upping range ||~ downing range ||~ if the 5th is just ||
point edo ||~ edo's 5th ||~ upping range ||~ downing range ||~ if the 5th is just ||
||= M2 ||= C - D ||= 2 ||= 2-edo ||= 600¢ ||= none ||= all ||= downed ||
||= M2 ||= C - D ||= 2 ||= 2-edo ||= 600¢ ||= none ||= all ||= downed ||
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A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).
A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).


Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M3 = [1,2] = dd3. So E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: E' = \\A1 and 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: E + E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;\\d3 = 2·vv\m2, and E - E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;//ddd3 = 2·^\\d2. Thus vv\m2 and vv/d2 are also enharmonics, and v\4 and v/d4 are also generators. Here is the genchain:
Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. So E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2⋅G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2⋅m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: E' = \\A1 and 2⋅G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: E + E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;\\d3 = 2·vv\m2, and E - E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;``//``ddd3 = 2·vv/d2. Thus vv\m2 and vv/d2 are equivalent enharmonics, and v\4 and v/d4 are equivalent generators. Here is the genchain:
&lt;span style="display: block; text-align: center;"&gt;P1 -- ^M3=v\4 -- /m6=\M6 -- ^/8=vm9 -- P11&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E^=Fv\ -- Ab/=A\ -- C^/=Dbv -- F&lt;/span&gt;
&lt;span style="display: block; text-align: center;"&gt;P1 -- ^M3=v\4 -- /m6=\M6 -- ^/8=vm9 -- P11&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E^=Fv\ -- Ab/=A\ -- C^/=Dbv -- F&lt;/span&gt;
One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E' = v12m3. //Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E" = ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d2, and 4·G' = ^m2.//  
One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E = v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E' = /&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d2, and 4·G' = /m2.
 
The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - /m2 = \M3.
 
From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = \M3 + ^1 = ^\M3.
Equivalent enharmonics: E + E' = v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;/&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;dd4 = 3·v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;/m2, and E - 2·E' = v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;\&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;AA1 = 3·v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;\\A1.
Equivalent generators: G = ^\M3 =
 
P1 -- \M3 -- \\A5=/m6-- P8
C -- E\ -- Ab/ -- C
P1 -- ^\M3 -- ^^\\A5=^^/m6=vv/M6=vv``//``d7 -- v/m9 -- F
C -- E^\ -- Ab^^/=Avv/ -- Dbv/ -- F//
 
P1 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3 -- v&lt;span style="vertical-align: super;"&gt;8&lt;/span&gt;A5=^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m6-- P8
C -- Ev&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Ab^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C
P1 -- v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/M3 -- v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``A5=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``m6=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\d7 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\m9 -- F
C -- Ev&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/ -- G#v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Ab^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Bbb^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\ -- Db^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\ -- F//


//Before we can divide by 4, we must quadruple all ups and downs: E = ^&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;d2 and 4·G' = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2. The bare alt-generator is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 - 4·(^1) = m2. Thus E' = \&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 and G' = ^/1.The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3.//
&lt;span style="display: block; text-align: center;"&gt;//P1 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3 -- v&lt;span style="vertical-align: super;"&gt;8&lt;/span&gt;A5=^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m6-- P8//
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;//C -- Ev&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Ab^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C//&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;//P1 -- v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/M3 -- v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``A5=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``m6=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\d7 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\m9 -- F//&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;//C -- Ev&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/ -- G#v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Ab^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Bbb^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\ -- Db^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\ -- F//&lt;/span&gt;


==Alternate enharmonics==  
==Alternate enharmonics==  
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[//Question: what if there are highs and lows?//]</pre></div>
[//Question: what if there are highs and lows?//]</pre></div>
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Untempered just intonation has a pergen of the octave, the fifth, and a list of commas, each containing only one higher prime: (P8, P5, 81/80, 64/63, ...). The higher prime's exponent in the monzo must be 1 or -1. The commas can be replaced with microtonal accidentals: (P8, P5, ^1, /1,...).&lt;br /&gt;
Untempered just intonation has a pergen of the octave, the fifth, and a list of commas, each containing only one higher prime: (P8, P5, 81/80, 64/63, ...). The higher prime's exponent in the monzo must be 1 or -1. The commas can be replaced with microtonal accidentals: (P8, P5, ^1, /1,...).&lt;br /&gt;
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For any comma containing primes 2 and 3, let m = the GCD of all the monzo's exponents other than the 2-exponent, and let n = the GCD of all its higher-prime exponents. The comma will split the octave into m parts, and if n &amp;gt; m, it will split some 3-limit interval into n parts.&lt;br /&gt;
For any comma containing primes 2 and 3, let m = the GCD of all the monzo's exponents other than the 2-exponent, and let n = the GCD of all its higher-prime exponents. The comma will split the octave into m parts, and if n &amp;gt; m, it will split some 3-limit interval into n parts.&lt;br /&gt;
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Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. Let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &amp;gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.&lt;br /&gt;
Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. Let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &amp;gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.&lt;br /&gt;
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One obvious application is to name regular temperaments in a logical, consistent manner, avoiding the need to memorize many arbitrary names. Many temperaments have pergen-like names: Hemififths is (P8, P5/2), semihemi is (P8/2, P4/2), triforce is (P8/3, P4/2), both tetracot and semihemififths are (P8, P5/4), fourfives is (P8/4, P5/5), pental is (P8/5, P5), and fifive is (P8/2, P5/5). Pergen names are an improvement over these because they specify more exactly what is split. Some temperament names are what might be called pseudo-pergens, because the multigen isn't 3-limit, and might not even actually be a generator. Meantone (mean = average, tone = major 2nd) implies (5/4)/2.&lt;br /&gt;
One obvious application is to name regular temperaments in a logical, consistent manner, avoiding the need to memorize many arbitrary names. Many temperaments have pergen-like names: Hemififths is (P8, P5/2), semihemi is (P8/2, P4/2), triforce is (P8/3, P4/2), both tetracot and semihemififths are (P8, P5/4), fourfives is (P8/4, P5/5), pental is (P8/5, P5), and fifive is (P8/2, P5/5). Pergen names are an improvement over these because they specify more exactly what is split. Some temperament names are what might be called pseudo-pergens, because the multigen isn't 3-limit, and might not even actually be a generator. Meantone (mean = average, tone = major 2nd) implies (5/4)/2.&lt;br /&gt;
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         &lt;th&gt;3-exponent&lt;br /&gt;
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point edo&lt;br /&gt;
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We can assign cents to each accidental symbol. First let c = the cents of the tuning's 5th from 700¢, the 12edo 5th. Thus P5 = 700¢ + c. From this we can calculate the cents of any 3-limit interval. Since the enharmonic = 0¢, we can derive the cents of the accidental. If the enharmonic is vvA1, then vvA1 = 0¢ and ^1 = (A1)/2. But A1 = 100¢ + 7c, so ^1 = 50¢ + 3.5c. If the 5th is 696¢, c = -4 and the up symbol equals 36¢. /1 can be similarly derived from its enharmonic. #1 always equals 100¢ + c.&lt;br /&gt;
We can assign cents to each accidental symbol. First let c = the cents of the tuning's 5th from 700¢, the 12edo 5th. Thus P5 = 700¢ + c. From this we can calculate the cents of any 3-limit interval. Since the enharmonic = 0¢, we can derive the cents of the accidental. If the enharmonic is vvA1, then vvA1 = 0¢ and ^1 = (A1)/2. But A1 = 100¢ + 7c, so ^1 = 50¢ + 3.5c. If the 5th is 696¢, c = -4 and the up symbol equals 36¢. /1 can be similarly derived from its enharmonic. #1 always equals 100¢ + c.&lt;br /&gt;
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So far, the largest multigen has been a 12th. As the multigen fractions get larger, the multigen gets quite wide. To avoid cumbersome degree names like 16th or 22nd, for degrees above 12, the widening is indicated with one &amp;quot;W&amp;quot; per octave. Thus 32/9 = Wm7, 9/2 = WWM2 or WM9, etc. For (P8, M/n), valid multigens are any voicing of the fifth that is less than n/2 octaves. For (P8, M/6), the multigen can be P4, P5, P11, P12, WWP4 or WWP5.&lt;br /&gt;
So far, the largest multigen has been a 12th. As the multigen fractions get larger, the multigen gets quite wide. To avoid cumbersome degree names like 16th or 22nd, for degrees above 12, the widening is indicated with one &amp;quot;W&amp;quot; per octave. Thus 32/9 = Wm7, 9/2 = WWM2 or WM9, etc. For (P8, M/n), valid multigens are any voicing of the fifth that is less than n/2 octaves. For (P8, M/6), the multigen can be P4, P5, P11, P12, WWP4 or WWP5.&lt;br /&gt;
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If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a &lt;strong&gt;single-split&lt;/strong&gt; pergen. If it has two fractions, it's a &lt;strong&gt;double-split&lt;/strong&gt; pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called &lt;strong&gt;single-pair&lt;/strong&gt; notation because it adds only a single pair of accidentals to conventional notation. &lt;strong&gt;Double-pair&lt;/strong&gt; notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.&lt;br /&gt;
If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a &lt;strong&gt;single-split&lt;/strong&gt; pergen. If it has two fractions, it's a &lt;strong&gt;double-split&lt;/strong&gt; pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called &lt;strong&gt;single-pair&lt;/strong&gt; notation because it adds only a single pair of accidentals to conventional notation. &lt;strong&gt;Double-pair&lt;/strong&gt; notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.&lt;br /&gt;
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A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.&lt;br /&gt;
A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.&lt;br /&gt;
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To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P and P8. If P is 6/5, the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P - P8 = (6/5)&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P = (2/1) · (7/6)&lt;span style="vertical-align: super;"&gt;-4&lt;/span&gt;. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - P4 = (10/9)^3 / (4/3) = 250/243.&lt;br /&gt;
To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P and P8. If P is 6/5, the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P - P8 = (6/5)&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P = (2/1) · (7/6)&lt;span style="vertical-align: super;"&gt;-4&lt;/span&gt;. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - P4 = (10/9)^3 / (4/3) = 250/243.&lt;br /&gt;
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There are also alternate enharmonics, see below.&lt;br /&gt;
There are also alternate enharmonics, see below.&lt;br /&gt;
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There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2 or a ddd2. There is an easy method for finding such a pergen, if one exists. First, some terminology and basic concepts:&lt;br /&gt;
There are multiple notations for a given pergen, depending on the enharmonic interval(s). Preferably, the enharmonic's degree will be a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting. If it's a unison, it will always be an A1. (P1 would be pointless, d1 would be inverted to A1, and AA1 would be split into two A1's.) If it's a 2nd, preferably it will be a m2 or a d2 or a dd2, and not a M2 or an A2 or a ddd2. There is an easy method for finding such a pergen, if one exists. First, some terminology and basic concepts:&lt;br /&gt;
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A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).&lt;br /&gt;
A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).&lt;br /&gt;
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Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M3 = [1,2] = dd3. So E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: E' = \\A1 and 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: E + E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;\\d3 = 2·vv\m2, and E - E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;&lt;em&gt;ddd3 = 2·^\\d2. Thus vv\m2 and vv/d2 are also enharmonics, and v\4 and v/d4 are also generators. Here is the genchain:&lt;br /&gt;
Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. So E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. But by using double-pair notation, we can avoid that. We find P11/2, which equals two generators: P11/2 = 2⋅G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2⋅m6 = [1,0] = A1. For this second enharmonic, we use the second pair of accidentals: E' = \\A1 and 2⋅G = /m6 or \M6. The sum or difference of two enharmonic intervals is also an enharmonic: E + E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;\\d3 = 2·vv\m2, and E - E' = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:025:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:025 --&gt;ddd3 = 2·vv/d2. Thus vv\m2 and vv/d2 are equivalent enharmonics, and v\4 and v/d4 are equivalent generators. Here is the genchain:&lt;br /&gt;
&lt;span style="display: block; text-align: center;"&gt;P1 -- ^M3=v\4 -- /m6=\M6 -- ^/8=vm9 -- P11&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E^=Fv\ -- Ab/=A\ -- C^/=Dbv -- F&lt;/span&gt;&lt;br /&gt;
&lt;span style="display: block; text-align: center;"&gt;P1 -- ^M3=v\4 -- /m6=\M6 -- ^/8=vm9 -- P11&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E^=Fv\ -- Ab/=A\ -- C^/=Dbv -- F&lt;/span&gt;&lt;br /&gt;
One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E' = v12m3. &lt;/em&gt;Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E&amp;quot; = ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d2, and 4·G' = ^m2.&lt;em&gt; &lt;br /&gt;
One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. We have [3,2]/12 = [0,0] = P1, and G' = ^1 and E = v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;m3. Next find 4·G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3·m2 = [0,-1] = descending d2. Thus E' = /&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d2, and 4·G' = /m2.&lt;br /&gt;
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The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - /m2 = \M3. &lt;br /&gt;
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From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = \M3 + ^1 = ^\M3.&lt;br /&gt;
Equivalent enharmonics: E + E' = v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;/&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;dd4 = 3·v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;/m2, and E - 2·E' = v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;\&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;AA1 = 3·v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;\\A1.&lt;br /&gt;
Equivalent generators: G = ^\M3 =&lt;br /&gt;
&lt;br /&gt;
P1 -- \M3 -- \\A5=/m6-- P8&lt;br /&gt;
C -- E\ -- Ab/ -- C&lt;br /&gt;
P1 -- ^\M3 -- ^^\\A5=^^/m6=vv/M6=vv&lt;!-- ws:start:WikiTextRawRule:026:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:026 --&gt;d7 -- v/m9 -- F&lt;br /&gt;
C -- E^\ -- Ab^^/=Avv/ -- Dbv/ -- F&lt;em&gt;&lt;br /&gt;
&lt;br /&gt;
P1 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3 -- v&lt;span style="vertical-align: super;"&gt;8&lt;/span&gt;A5=^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m6-- P8&lt;br /&gt;
C -- Ev&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Ab^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;br /&gt;
P1 -- v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/M3 -- v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:027:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:027 --&gt;A5=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:028:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:028 --&gt;m6=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\d7 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\m9 -- F&lt;br /&gt;
C -- Ev&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/ -- G#v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:029:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:029 --&gt;=Ab^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:030:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:030 --&gt;=Bbb^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\ -- Db^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\ -- F&lt;/em&gt;&lt;br /&gt;
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&lt;/em&gt;Before we can divide by 4, we must quadruple all ups and downs: E = ^&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;d2 and 4·G' = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2. The bare alt-generator is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 - 4·(^1) = m2. Thus E' = \&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 and G' = ^/1.The period can be deduced from 4·G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4·G' = P4 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3.&lt;em&gt;&lt;br /&gt;
&lt;span style="display: block; text-align: center;"&gt;&lt;/em&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3 -- v&lt;span style="vertical-align: super;"&gt;8&lt;/span&gt;A5=^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m6-- P8&lt;em&gt;&lt;br /&gt;
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;&lt;/em&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Ab^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;em&gt;&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;&lt;/em&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/M3 -- v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:025:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:025 --&gt;A5=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:026:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:026 --&gt;m6=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\d7 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\m9 -- F&lt;em&gt;&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;&lt;/em&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/ -- G#v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:027:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:027 --&gt;=Ab^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:028:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:028 --&gt;=Bbb^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\ -- Db^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\ -- F//&lt;/span&gt;&lt;br /&gt;
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&lt;!-- ws:start:WikiTextHeadingRule:51:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc10"&gt;&lt;a name="Further Discussion-Alternate enharmonics"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:51 --&gt;Alternate enharmonics&lt;/h2&gt;
&lt;!-- ws:start:WikiTextHeadingRule:53:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc10"&gt;&lt;a name="Further Discussion-Alternate enharmonics"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:53 --&gt;Alternate enharmonics&lt;/h2&gt;
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Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;A2, which is an improvement but still awkward. The period is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 and the generator is v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M2.&lt;br /&gt;
Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;A2, which is an improvement but still awkward. The period is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 and the generator is v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M2.&lt;br /&gt;
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&lt;span style="display: block; text-align: center;"&gt;P1 -- vM3 -- ^m6 -- P8&lt;br /&gt;
&lt;span style="display: block; text-align: center;"&gt;P1 -- vM3 -- ^m6 -- P8&lt;br /&gt;
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev -- Ab^ -- C&lt;br /&gt;
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev -- Ab^ -- C&lt;br /&gt;
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- /m2 -- &lt;!-- ws:start:WikiTextRawRule:029:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:029 --&gt;d3=\\A2 -- \M3 -- P4&lt;br /&gt;
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- /m2 -- &lt;!-- ws:start:WikiTextRawRule:031:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:031 --&gt;d3=\\A2 -- \M3 -- P4&lt;br /&gt;
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Db/ -- Ebb&lt;!-- ws:start:WikiTextRawRule:030:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:030 --&gt;=D#\\ -- E\ -- F&lt;/span&gt;&lt;br /&gt;
&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Db/ -- Ebb&lt;!-- ws:start:WikiTextRawRule:032:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:032 --&gt;=D#\\ -- E\ -- F&lt;/span&gt;&lt;br /&gt;
Sometimes the temperament being notated implies a certain enharmonic. Specifically, the comma tempered out should map to the enharmonic, or some multiple of it.&lt;br /&gt;
Sometimes the temperament being notated implies a certain enharmonic. Specifically, the comma tempered out should map to the enharmonic, or some multiple of it.&lt;br /&gt;
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&lt;em&gt;(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4·G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.&lt;/em&gt;&lt;br /&gt;
&lt;em&gt;(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4·G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.&lt;/em&gt;&lt;br /&gt;
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&lt;!-- ws:start:WikiTextHeadingRule:53:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc11"&gt;&lt;a name="Further Discussion-Alternate keyspans and stepspans"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:53 --&gt;Alternate keyspans and stepspans&lt;/h2&gt;
&lt;!-- ws:start:WikiTextHeadingRule:55:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc11"&gt;&lt;a name="Further Discussion-Alternate keyspans and stepspans"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:55 --&gt;Alternate keyspans and stepspans&lt;/h2&gt;
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One might wonder, when using gedras, why 12-edo keyspans? Why heptatonic stepspans? Heptatonic is best because conventional notation is heptatonic, and we want to minimize the heptatonic stepspan of the enharmonic. In order for the matrix to be invertible, the edo must be connected to the 4\7 kite on the scale tree. The only choices are 5-edo, 12-edo, 19-edo, 26-edo, 33-edo, 40-edo and 47-edo, and on the other side, 2-edo, 9-edo, 16-edo and 23-edo. Any of these edos would also work. 12-edo is merely the most convenient choice, because of its familiarity. Dividing the gedra directly only gives you an estimate of the best period or generator. As noted in the previous section, to improve the enharmonic, this initial estimate must often be revised. So the choice of estimating edo isn't very important.&lt;br /&gt;
One might wonder, when using gedras, why 12-edo keyspans? Why heptatonic stepspans? Heptatonic is best because conventional notation is heptatonic, and we want to minimize the heptatonic stepspan of the enharmonic. In order for the matrix to be invertible, the edo must be connected to the 4\7 kite on the scale tree. The only choices are 5-edo, 12-edo, 19-edo, 26-edo, 33-edo, 40-edo and 47-edo, and on the other side, 2-edo, 9-edo, 16-edo and 23-edo. Any of these edos would also work. 12-edo is merely the most convenient choice, because of its familiarity. Dividing the gedra directly only gives you an estimate of the best period or generator. As noted in the previous section, to improve the enharmonic, this initial estimate must often be revised. So the choice of estimating edo isn't very important.&lt;br /&gt;
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&lt;!-- ws:start:WikiTextHeadingRule:55:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc12"&gt;&lt;a name="Further Discussion-Chord names"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:55 --&gt;Chord names&lt;/h2&gt;
&lt;!-- ws:start:WikiTextHeadingRule:57:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc12"&gt;&lt;a name="Further Discussion-Chord names"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:57 --&gt;Chord names&lt;/h2&gt;
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Using pergens, all rank-2 chords can be named using ups and downs, and if needed highs and lows as well. See the &lt;a class="wiki_link" href="/Ups%20and%20Downs%20Notation"&gt;ups and downs&lt;/a&gt; page for chord naming conventions. The genchain and/or the perchain creates a lattice in which each note and each interval has its own name. The many enharmonic equivalents allow proper chord spelling.&lt;br /&gt;
Using pergens, all rank-2 chords can be named using ups and downs, and if needed highs and lows as well. See the &lt;a class="wiki_link" href="/Ups%20and%20Downs%20Notation"&gt;ups and downs&lt;/a&gt; page for chord naming conventions. The genchain and/or the perchain creates a lattice in which each note and each interval has its own name. The many enharmonic equivalents allow proper chord spelling.&lt;br /&gt;
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A different temperament may result in the same pergen with the same enharmonic, but may still produce a different name for the same chord. For example, injera (2.3.5.7 with 81/80 and 50/49, or rryy&amp;amp;gT) is also half-octave. However, the tipping point for the d2 enharmonic is at 700¢, and while pajara favors a fifth wider than that, injera favors a fifth narrower than that. Hence ups and downs are exchanged, and E = vvd2, and P = ^A4 = vd5. The mapping is [(2 2 0 1) (0 1 4 4)] = [(2 0) (2 1) (0 4) (1 4)]. The square mapping (the first two columns) are the same, hence the pergen is the same, but the other columns are different, hence the higher primes are mapped differently. 5/4 = M3 and 7/4 = M3 + vd5 = vm7, and 4:5:6:7 = C E G Bbv = C,v7.&lt;br /&gt;
A different temperament may result in the same pergen with the same enharmonic, but may still produce a different name for the same chord. For example, injera (2.3.5.7 with 81/80 and 50/49, or rryy&amp;amp;gT) is also half-octave. However, the tipping point for the d2 enharmonic is at 700¢, and while pajara favors a fifth wider than that, injera favors a fifth narrower than that. Hence ups and downs are exchanged, and E = vvd2, and P = ^A4 = vd5. The mapping is [(2 2 0 1) (0 1 4 4)] = [(2 0) (2 1) (0 4) (1 4)]. The square mapping (the first two columns) are the same, hence the pergen is the same, but the other columns are different, hence the higher primes are mapped differently. 5/4 = M3 and 7/4 = M3 + vd5 = vm7, and 4:5:6:7 = C E G Bbv = C,v7.&lt;br /&gt;
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  &lt;!-- ws:start:WikiTextHeadingRule:59:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc14"&gt;&lt;a name="Further Discussion-Combining pergens"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:59 --&gt;Combining pergens&lt;/h2&gt;
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Tempering out 250/243 creates third-fourth, and 49/48 creates half-fourth, and tempering out both commas creates sixth-fourth. Therefore (P8, P4/3) + (P8, P4/2) = (P8, P4/6).&lt;br /&gt;
Tempering out 250/243 creates third-fourth, and 49/48 creates half-fourth, and tempering out both commas creates sixth-fourth. Therefore (P8, P4/3) + (P8, P4/2) = (P8, P4/6).&lt;br /&gt;
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However, (P8/2, M2/4) + (P8, P4/2) = (P8/4, P4/2), so the sum isn't always obvious.&lt;br /&gt;
However, (P8/2, M2/4) + (P8, P4/2) = (P8/4, P4/2), so the sum isn't always obvious.&lt;br /&gt;
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&lt;!-- ws:start:WikiTextHeadingRule:61:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc15"&gt;&lt;a name="Further Discussion-Pergens and EDOs"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:61 --&gt;Pergens and EDOs&lt;/h2&gt;
&lt;!-- ws:start:WikiTextHeadingRule:63:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc15"&gt;&lt;a name="Further Discussion-Pergens and EDOs"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:63 --&gt;Pergens and EDOs&lt;/h2&gt;
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Pergens have much in common with edos. Pergens of rank-2 assume only primes 2 and 3, edos assume only prime 2. There are an infinite number of edos, but fewer than a hundred have been explored. There are an infinite number of pergens, but fewer than a hundred will suffice most composers.&lt;br /&gt;
Pergens have much in common with edos. Pergens of rank-2 assume only primes 2 and 3, edos assume only prime 2. There are an infinite number of edos, but fewer than a hundred have been explored. There are an infinite number of pergens, but fewer than a hundred will suffice most composers.&lt;br /&gt;
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Pergens were discovered by Kite Giedraitis in 2017, and developed with the help of Praveen Venkataramana. Earlier drafts of this article can be found at &lt;!-- ws:start:WikiTextUrlRule:3138:http://xenharmonic.wikispaces.com/pergen+names --&gt;&lt;a href="http://xenharmonic.wikispaces.com/pergen+names"&gt;http://xenharmonic.wikispaces.com/pergen+names&lt;/a&gt;&lt;!-- ws:end:WikiTextUrlRule:3138 --&gt;&lt;br /&gt;
Pergens were discovered by Kite Giedraitis in 2017, and developed with the help of Praveen Venkataramana. Earlier drafts of this article can be found at &lt;!-- ws:start:WikiTextUrlRule:3137:http://xenharmonic.wikispaces.com/pergen+names --&gt;&lt;a href="http://xenharmonic.wikispaces.com/pergen+names"&gt;http://xenharmonic.wikispaces.com/pergen+names&lt;/a&gt;&lt;!-- ws:end:WikiTextUrlRule:3137 --&gt;&lt;br /&gt;
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