Kite's thoughts on pergens: 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>2018-01-08 02:31~~:02~~ UTC</tt>.

: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 02:41:31 UTC</tt>.

: The original revision id was <tt>~~624537469~~</tt>.

: The original revision id was <tt>624537647</tt>.

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

The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.

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P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11

C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F

The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is m6 - 2*(M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = \M3 or /d4. Here is the genchain:P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11C -- E\=Fb/ -- Ab^=Av -- C#\=Db/ -- F

Using vM6/2 for 2*G gives a different but equally valid notation: M6/2 = [9,5]/2 = [4,2] = M3, and M6 - 2*(M3) = [1,1] = m2, and E' = \\m2 and G = /M3 or \4.P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11

C -- E/=F\ -- Av=Ab^ -- C/=Db\ -- F

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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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 = ^3d2, and 4*G' = ^m2. Before we can divide by 4, we must quadruple all ups and downs: E = ^12d2 and 4*G' = ^4m2. The bare alt-generator is ^4[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^4m2 - 4*(^1) = m2. Thus E' = \4m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^4m2 = v4M3. 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.

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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).

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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v4A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = ^\M3 or ^/d4. 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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v4A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = d2, and G = ^\M3 or ^/d4. Here is the genchain:

P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11

C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F

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P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11

C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F

The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is m6 - 2*(M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = d2, and G = \M3 or /d4. Here is the genchain:P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11C -- E\=Fb/ -- Ab^=Av -- C#\=Db/ -- F

Using vM6/2 for 2*G gives a different but equally valid notation: M6/2 = [9,5]/2 = [4,2] = M3, and M6 - 2*(M3) = [1,1] = m2, and E' = \\m2 and G = /M3 or \4.P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11

C -- E/=F\ -- Av=Ab^ -- C/=Db\ -- F

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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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 = ^3d2, and 4*G' = ^m2. Before we can divide by 4, we must quadruple all ups and downs: E = ^12d2 and 4*G' = ^4m2. The bare alt-generator is ^4[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^4m2 - 4*(^1) = m2. Thus E' = \4m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^4m2 = v4M3. 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.

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<h2 id="toc14"><a name="Further Discussion-Misc notes"></a>Misc notes</h2>

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 <a href="http://xenharmonic.wikispaces.com/pergen+names">http://xenharmonic.wikispaces.com/pergen+names</a>

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 <a href="http://xenharmonic.wikispaces.com/pergen+names">http://xenharmonic.wikispaces.com/pergen+names</a>

@@ 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>2018-01-08 02:31:02 UTC</tt>.<br>
+: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 02:41:31 UTC</tt>.<br>
-: The original revision id was <tt>624537469</tt>.<br>
+: The original revision id was <tt>624537647</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 348: / Line 348: @@
 &lt;span style="display: block; text-align: center;"&gt;P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11
 &lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F
+&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;
+&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;&lt;span style="display: block; text-align: left;"&gt;The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is m6 - 2*(M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = \M3 or /d4. Here is the genchain:&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E\=Fb/ -- Ab^=Av -- C#\=Db/ -- F&lt;/span&gt;
+&lt;span style="display: block; text-align: left;"&gt;Using vM6/2 for 2*G gives a different but equally valid notation: M6/2 = [9,5]/2 = [4,2] = M3, and M6 - 2*(M3) = [1,1] = m2, and E' = \\m2 and G = /M3 or \4.&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11&lt;/span&gt;
+C -- E/=F\ -- Av=Ab^ -- C/=Db\ -- 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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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. 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.
@@ Line 2,103: / Line 2,107: @@
 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;
 &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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = //d2, and G = ^\M3 or ^/d4. 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. But by using double-pair notation, we can get enharmonics that are 2nds. First 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, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6. But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = &lt;em&gt;d2, and G = ^\M3 or ^/d4. Here is the genchain:&lt;br /&gt;
 &lt;span style="display: block; text-align: center;"&gt;P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11&lt;br /&gt;
 &lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F&lt;br /&gt;
@@ Line 2,110: / Line 2,114: @@
 &lt;span style="display: block; text-align: center;"&gt;P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11&lt;br /&gt;
 &lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F&lt;br /&gt;
+&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;&lt;br /&gt;
+&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;&lt;span style="display: block; text-align: left;"&gt;The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is m6 - 2*(M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' = &lt;/em&gt;d2, and G = \M3 or /d4. Here is the genchain:&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E\=Fb/ -- Ab^=Av -- C#\=Db/ -- F&lt;/span&gt;&lt;br /&gt;
+&lt;span style="display: block; text-align: left;"&gt;Using vM6/2 for 2*G gives a different but equally valid notation: M6/2 = [9,5]/2 = [4,2] = M3, and M6 - 2*(M3) = [1,1] = m2, and E' = \\m2 and G = /M3 or \4.&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11&lt;/span&gt;&lt;br /&gt;
+C -- E/=F\ -- Av=Ab^ -- C/=Db\ -- F&lt;br /&gt;
 &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 proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. 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. 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;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-Misc notes"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:59 --&gt;Misc notes&lt;/h2&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:3057: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:3057 --&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:3075: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:3075 --&gt;&lt;br /&gt;
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