Pergen names: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

This is an imported revision from Wikispaces. The revision metadata is included below for reference:

: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-17 18:13:46 UTC</tt>.

: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-17 18:27:51 UTC</tt>.

: The original revision id was <tt>~~623960639~~</tt>.

: The original revision id was <tt>623960989</tt>.

: The revision comment was: <tt></tt>

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To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the alternate generator from the enharmonic.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^5M2, which happens to be P + G.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^5M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.

~~(P8~~/~~5, P4~~/2): P = ^~~4M2,~~ G = ~~^1, G'~~ = ~~^5M2, E~~ = ~~v10m2~~

P1 - ^4M2=v6m3 - vvP4 - ^^P5 - ^6M6=v4m7 - P8C - D^4=Ebv6 - Fvv - G^^ - A^6=Bbv4 - C

P1 - ^4M2~~=v6m3 - vvP4 - ^^P5 - ^6M6~~=v4m7 - P8C - D^4=~~Ebv6 - Fvv - G^^ - A^6=Bbv~~4 - C

P1 - ^5M2=v5m3 - P4C - D^5=Ebv5 - F

~~P1 - ^5<~~/~~span>M2~~=~~v5m3 -~~ P4~~<~~/~~span>C - D^5<~~/~~span>~~=~~Ebv5 - F~~

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For example, (P8/2, P4/2) has from P8/2 P = vA4 and E = ^^d2, and from P4/2 G = /M2 and E' = \\m2.

~~To find the~~ double-pair notation ~~for~~ a true double ~~pergen, find each~~ pair ~~from each half of~~ the ~~pergen~~.

A false double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).

==Alternate enharmonics==

For single-comma pergens, the enharmonic should equal the comma's mapping.

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Not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.

~~Each enharmonic interval implies a different notation. If every pergen could use every enharmonic, there would be an overwhelming choice of notations! Fortunately, not~~ all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.

For {P8/M, multi-gen/N}, an octave = M periods ± some number of enharmonics, and a multi-gen = N generators ± some number of enharmonics.

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For {P8/M, multi-gen/N}, there are two conditions on the enharmonic. If T is the 3-exponent of the multi-gen, the conditions are edo = Mx and edo = Ny ± T. For {P8/2, P4/2}, the two conditions are mutually exclusive: the edo must be both even and odd. Therefore there must be two accidental pairs, each with its own enharmonic interval. In the main table, this pergen is notated with both ups/downs and highs/lows. Since the 8ve and 4th are split, the 5th is too. Each interval has its own genchain. One of these is notated with ups/downs, another with highs/lows, and the third with both. The 3 possible ways of allocating the two accidental pairs are all listed. Furthermore, ups/downs can be exchanged for highs/lows.

For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^6d32, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d34. The octave genchain would be C - E#^3=Abbv3 - C. ~~Seeing the same pitch represented as both E and A is rather disconcerting. For this reason, enharmonics that are unisons or 2nds are preferred.~~

For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^6d32, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d34. The octave genchain would be C - E#^3=Abbv3 - C.

~~[//Question: if the edo is 14, is the enharmonic 2 A1's = AA1?//]~~

Even if a pergen with two fractions __can__ be notated with a single accidental pair, a second pair may be preferred. Again, analogous to 22-edo, a notation that causes familiar chords to be misspelled is not very welcoming.

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(to be continued)</pre></div>

<h4>Original HTML content:</h4>

<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>pergen names</title></head><body><a href="#Definition">Definition</a> | <a href="#Derivation">Derivation</a> | <a href="#Applications">Applications</a> | <a href="#Further Discussion">Further Discussion</a>

<div style="width:100%; max-height:400pt; overflow:auto; background-color:#f8f9fa; border: 1px solid #eaecf0; padding:0em"><pre style="margin:0px;border:none;background:none;word-wrap:break-word;width:200%;white-space: pre-wrap ! important" class="old-revision-html"><html><head><title>pergen names</title></head><body><a href="#Definition">Definition</a> | <a href="#Derivation">Derivation</a> | <a href="#Applications">Applications</a> | <a href="#Further Discussion">Further Discussion</a>

<h1 id="toc0"><a name="Definition"></a>Definition</h1>

<h1 id="toc0"><a name="Definition"></a>Definition</h1>

A pergen (pronounced &quot;peer-gen&quot;) is a way of identifying a regular temperament solely by its period and generator(s). For any temperament, there are many possible periods and generators. For the pergen, they are chosen to use the fewest, and smallest, prime factors possible. Fractions are allowed, e.g. half-octave, but avoided if possible.

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To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the alternate generator from the enharmonic.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^5M2, which happens to be P + G.~~ ~~

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^5M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.

~~(P8/5, P4/2): P = ^4M2, G = ^1, G' = ^5M2, E = v10m2 ~~

P1 - ^4M2=~~v6m3~~ - vvP4 - ^^P5 - ^~~6M6~~=v4m7 - P8C - D^4=~~Ebv6~~ - Fvv - G^^ - A^6=Bbv4 - C

P1 - ^4M2=v6m3 - vvP4 - ^^P5 - ^6M6=v4m7 - P8C - D^4=Ebv6 - Fvv - G^^ - A^6=Bbv4 - C

P1 - ^5M2=v5m3 - P4C - D^5=Ebv5 - F

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen.

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For example, (P8/2, P4/2) has from P8/2 P = vA4 and E = ^^d2, and from P4/2 G = /M2 and E' = \\m2.

A false double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).

<h2 id="toc8"><a name="Further Discussion-Alternate enharmonics"></a>Alternate enharmonics</h2>

For single-comma pergens, the enharmonic should equal the comma's mapping.

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~~ ~~

Not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.

~~Each enharmonic interval implies a different notation. If every pergen could use every enharmonic, there would be an overwhelming choice of notations! Fortunately, not~~ all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.

For {P8/M, multi-gen/N}, an octave = M periods ± some number of enharmonics, and a multi-gen = N generators ± some number of enharmonics.

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For {P8/M, multi-gen/N}, there are two conditions on the enharmonic. If T is the 3-exponent of the multi-gen, the conditions are edo = Mx and edo = Ny ± T. For {P8/2, P4/2}, the two conditions are mutually exclusive: the edo must be both even and odd. Therefore there must be two accidental pairs, each with its own enharmonic interval. In the main table, this pergen is notated with both ups/downs and highs/lows. Since the 8ve and 4th are split, the 5th is too. Each interval has its own genchain. One of these is notated with ups/downs, another with highs/lows, and the third with both. The 3 possible ways of allocating the two accidental pairs are all listed. Furthermore, ups/downs can be exchanged for highs/lows.

For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^6d32, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d34. The octave genchain would be C - E#^3=Abbv3 - C. ~~Seeing the same pitch represented as both E and A is rather disconcerting. For this reason, enharmonics that are unisons or 2nds are preferred. ~~

For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^6d32, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d34. The octave genchain would be C - E#^3=Abbv3 - C.

~~ ~~

~~[Question: if the edo is 14, is the enharmonic 2 A1's = AA1?]~~

Even if a pergen with two fractions can be notated with a single accidental pair, a second pair may be preferred. Again, analogous to 22-edo, a notation that causes familiar chords to be misspelled is not very welcoming.

@@ Line 1: / Line 1: @@
 <h2>IMPORTED REVISION FROM WIKISPACES</h2>
 This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
-: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-17 18:13:46 UTC</tt>.<br>
+: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2017-12-17 18:27:51 UTC</tt>.<br>
-: The original revision id was <tt>623960639</tt>.<br>
+: The original revision id was <tt>623960989</tt>.<br>
 : The revision comment was: <tt></tt><br>
 The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br>
@@ Line 320: / Line 320: @@
 To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the alternate generator from the enharmonic.
-For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2, which happens to be P + G.
+For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.
-&lt;span style="display: block; text-align: center;"&gt;(P8/5, P4/2): P = ^4M2, G = ^1, G' = ^5M2, E = v10m2&lt;/span&gt;
+&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 - vvP4 - ^^P5 - ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 - P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; - Fvv - G^^ - A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&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 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v6m3 - vvP4 - ^^P5 - ^6M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 - P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv6 - Fvv - G^^ - A^6=Bbv&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 - ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 - P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; - F&lt;/span&gt;
-&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 - P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; - F&lt;/span&gt;
+To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For example, (P8/2, P4/2) has from P8/2 P = vA4 and E = ^^d2, and from P4/2 G = /M2 and E' = \\m2.
-To find the double-pair notation for a true double pergen, find each pair from each half of the pergen.
+A false double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).
+==Alternate enharmonics==
+For single-comma pergens, the enharmonic should equal the comma's mapping.
@@ Line 340: / Line 342: @@
+Not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.
-Each enharmonic interval implies a different notation. If every pergen could use every enharmonic, there would be an overwhelming choice of notations! Fortunately, not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.
 For {P8/M, multi-gen/N}, an octave = M periods ± some number of enharmonics, and a multi-gen = N generators ± some number of enharmonics.
@@ Line 367: / Line 368: @@
 For {P8/M, multi-gen/N}, there are two conditions on the enharmonic. If T is the 3-exponent of the multi-gen, the conditions are edo = Mx and edo = Ny ± T. For {P8/2, P4/2}, the two conditions are mutually exclusive: the edo must be both even and odd. Therefore there must be two accidental pairs, each with its own enharmonic interval. In the main table, this pergen is notated with both ups/downs and highs/lows. Since the 8ve and 4th are split, the 5th is too. Each interval has its own genchain. One of these is notated with ups/downs, another with highs/lows, and the third with both. The 3 possible ways of allocating the two accidental pairs are all listed. Furthermore, ups/downs can be exchanged for highs/lows.
-For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;2, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;4. The octave genchain would be C - E#^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;=Abbv&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt; - C. Seeing the same pitch represented as both E and A is rather disconcerting. For this reason, enharmonics that are unisons or 2nds are preferred.
+For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;2, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;4. The octave genchain would be C - E#^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;=Abbv&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt; - C.
-[//Question: if the edo is 14, is the enharmonic 2 A1's = AA1?//]
 Even if a pergen with two fractions __can__ be notated with a single accidental pair, a second pair may be preferred. Again, analogous to 22-edo, a notation that causes familiar chords to be misspelled is not very welcoming.
@@ Line 393: / Line 392: @@
 (to be continued)</pre></div>
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 A &lt;strong&gt;pergen&lt;/strong&gt; (pronounced &amp;quot;peer-gen&amp;quot;) is a way of identifying a regular temperament solely by its period and generator(s). For any temperament, there are many possible periods and generators. For the pergen, they are chosen to use the fewest, and smallest, prime factors possible. Fractions are allowed, e.g. half-octave, but avoided if possible.&lt;br /&gt;
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 To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multi-gen as before. Then deduce the period from the enharmonic. If the multi-gen was changed by unreducing, deduce the alternate generator from the enharmonic.&lt;br /&gt;
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-For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2, which happens to be P + G.&lt;br /&gt;
+For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The generator is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G + E, G is ^1. This is much too small to be half a 4th, so we'll find an alternate generator later. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Deduce the alternate generator G' as we did for the period: P4 plus or minus some number of enharmonics must equal 2*G', and ([5,3] + y[1,1]) mod 2 must be 0. The smallest y is either 1 or -1. Choose -1, to get a smaller G', since the other one is equivalent. 2*G' = P4 - E, and G' = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2, which happens to be P + G. Thus P = ^4M2, G = ^1, G' = ^5M2, and E = v10m2.&lt;br /&gt;
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-&lt;span style="display: block; text-align: center;"&gt;(P8/5, P4/2): P = ^4M2, G = ^1, G' = ^5M2, E = v10m2&lt;/span&gt;&lt;br /&gt;
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-&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v6m3 - vvP4 - ^^P5 - ^6M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 - P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv6 - Fvv - G^^ - A^6=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; - C&lt;/span&gt;&lt;br /&gt;
+&lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 - vvP4 - ^^P5 - ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 - P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; - Fvv - G^^ - A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; - C&lt;/span&gt;&lt;br /&gt;
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 &lt;span style="display: block; text-align: center;"&gt;P1 - ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 - P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C - D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; - F&lt;/span&gt;&lt;br /&gt;
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-To find the double-pair notation for a true double pergen, find each pair from each half of the pergen.&lt;br /&gt;
+To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. For example, (P8/2, P4/2) has from P8/2 P = vA4 and E = ^^d2, and from P4/2 G = /M2 and E' = \\m2.&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 preferable when the single-pair enharmonic is a 3rd or a 4th, as with (P8/2, P4/3).&lt;br /&gt;
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+For single-comma pergens, the enharmonic should equal the comma's mapping.&lt;br /&gt;
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+Not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.&lt;br /&gt;
-Each enharmonic interval implies a different notation. If every pergen could use every enharmonic, there would be an overwhelming choice of notations! Fortunately, not all enharmonics work with all pergens. There are two logical restrictions on the enharmonic, one based on its degree and the other on its 3-exponent (implied edo). In addition, smaller degrees are always preferred. It should be either a unison or a 2nd, because equating two notes a 3rd or 4th apart is very disconcerting.&lt;br /&gt;
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 For {P8/M, multi-gen/N}, an octave = M periods ± some number of enharmonics, and a multi-gen = N generators ± some number of enharmonics.&lt;br /&gt;
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 For {P8/M, multi-gen/N}, there are two conditions on the enharmonic. If T is the 3-exponent of the multi-gen, the conditions are edo = Mx and edo = Ny ± T. For {P8/2, P4/2}, the two conditions are mutually exclusive: the edo must be both even and odd. Therefore there must be two accidental pairs, each with its own enharmonic interval. In the main table, this pergen is notated with both ups/downs and highs/lows. Since the 8ve and 4th are split, the 5th is too. Each interval has its own genchain. One of these is notated with ups/downs, another with highs/lows, and the third with both. The 3 possible ways of allocating the two accidental pairs are all listed. Furthermore, ups/downs can be exchanged for highs/lows.&lt;br /&gt;
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-For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;2, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;4. The octave genchain would be C - E#^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;=Abbv&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt; - C. Seeing the same pitch represented as both E and A is rather disconcerting. For this reason, enharmonics that are unisons or 2nds are preferred.&lt;br /&gt;
+For {P8/2, P5/3}, the edo = 2x = 3y ± 1. The edo must be even, thus y must be odd. Possible edos are 2, 4, 8, 10, 14, 16, 20, 22, 26, 28... The main table has ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;2, which implies 26-edo. Most of the other edos aren't practical. 10 and 20 imply the m3. 22-edo suggests a d&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;4. The octave genchain would be C - E#^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;=Abbv&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt; - C.&lt;br /&gt;
-&lt;br /&gt;
-[&lt;em&gt;Question: if the edo is 14, is the enharmonic 2 A1's = AA1?&lt;/em&gt;]&lt;br /&gt;
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 Even if a pergen with two fractions &lt;u&gt;can&lt;/u&gt; be notated with a single accidental pair, a second pair may be preferred. Again, analogous to 22-edo, a notation that causes familiar chords to be misspelled is not very welcoming.&lt;br /&gt;