Kite's thoughts on pergens: Difference between revisions

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 02:41:31 UTC</tt>.<br>

: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 06:41:17 UTC</tt>.<br>

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

: The original revision id was <tt>624544941</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 341:

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 = v<span style="vertical-align: super;">4</span>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:

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.

<span style="display: block; text-align: center;">P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11

</span><span style="display: block; text-align: center;">C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F

//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<span style="vertical-align: super;">4</span>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://

<span style="display: block; text-align: center;">//P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11//

</span><span style="display: block; text-align: center;">//C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F//

</span>

Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.

//Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.//

<span style="display: block; text-align: center;">P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11

<span style="display: block; text-align: center;">//P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11//

</span><span style="display: block; text-align: center;">C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F

</span><span style="display: block; text-align: center;">//C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F//

</span><span style="display: block; text-align: center;">

~~</span><span style="display: block; text-align: center;">~~<span style="display: block; text-align: left;">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:</span><span style="display: block; text-align: center;">P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11</span><span style="display: block; text-align: center;">C -- E\~~=Fb/~~ -- Ab^=Av -- C#\=Db/ -- F</span>

<span style="display: block; text-align: left;">The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is P11 - 4*M3 = [17,10] - [16,8] = [1,2] = dd3. For the second enharmonic, we use the second pair of accidentals: E' = \\\\dd3, and G = /M3. E = vvA1, so ^1 = 50¢ + 3.5c. E' = \\\\dd3, so /1 = 25¢ - 4.25c. E + E' = vv\\\\d3 = 2 * v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:</span><span style="display: block; text-align: left;">

<span style="display: block; text-align: left;">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.</span><span style="display: block; text-align: center;">P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11</span>

</span>

<span style="display: block; text-align: left;">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.</span>

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

</span>

//E' = \\m2, so /1 = 50¢ - 2.5c. But P11 = 1700¢ - c, so /1 should be 425¢ - c/4 - (400¢ + 4c) = 25¢ - 4.25c = (100¢ - 17c)/4 = dd3/4, and E' = \\\\dd3. E/ = Gbb\\\. G#// = Bbb\\ = Av = Ab^. So \\vm2 = 0¢.//

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 = ^<span style="vertical-align: super;">3</span>d2, and 4*G' = ^m2. Before we can divide by 4, we must quadruple all ups and downs: E = ^<span style="vertical-align: super;">12</span>d2 and 4*G' = ^<span style="vertical-align: super;">4</span>m2. The bare alt-generator is ^<span style="vertical-align: super;">4</span>[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^<span style="vertical-align: super;">4</span>m2 - 4*(^1) = m2. Thus E' = \<span style="vertical-align: super;">4</span>m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^<span style="vertical-align: super;">4</span>m2 = v<span style="vertical-align: super;">4</span>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.

<span style="display: block; text-align: center;">P1 -- v<span style="vertical-align: super;">4</span>M3 -- v<span style="vertical-align: super;">8</span>A5=^<span style="vertical-align: super;">4</span>m6-- P8

Line 389:

Line 400:

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.

==Chord names==

Using pergens, all rank-2 chords can be named using ups and downs, and if needed highs and lows as well. See the [[Ups and Downs Notation|ups and downs]] 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.

In certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. (Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.)

Given a specific temperament, the full mapping gives the notation of higher primes, and thus of any ratio. Thus JI chords can be named. For example, pajara's pergen is (P8/2, P5), half-octave, with P = vA4 or ^d5, G = P5, and E = ^^d2. The full mapping is [(2 0) (2 1) (7 -2) (8 -2)]. This tells us 7/1 = 8*P - 2*G = 4*P8 - 2*P5 = WWm7, and 7/4 = m7. 5/1 = 7*P - 2*G = 7/1 minus a half-octave. From this it follows that 5/4 = m7 - ^d5 = vM3. A 4:5:6:7 chord is written C Ev G Bb = C7(v3).

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&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 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.

== ==

==Combining pergens==

Line 403:

Line 425:

==Pergens and EDOs==

Pergens have much in common with edos. Pergens (rank-2 ~~ones)~~ assume only primes 2 and 3, edos assume only prime 2. There are an infinite number of edos, but ~~far~~ fewer than a hundred have been explored. There are an infinite number of pergens, but ~~far~~ fewer than a hundred will suffice most composers.

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.

Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset~~. However (P8/3, P5) is supported. In practice, if the generator's keyspan is very small, a partially supported pergen. For example, 22edo and 2\22 generator~~.

Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset.

How many pergens are fully supported by a given edo? Surprisingly, an infinite number! For example, 12edo supports (P8, P4/5), (P8, ~~P11~~/17), (P8, ~~WWP4~~/29), (P8, ~~W<span style="vertical~~-~~align: super;">3<~~/~~span>P4/41~~), ~~etc~~.

How many pergens are fully supported by a given edo? Surprisingly, an infinite number! There are infinite possible multigens, each one divisible by its keyspan. For example, 12edo supports, among other pergens, this series: (P8, P5/7), (P8, P12/19), (P8, WWP5/31),... (P8, (i-1,1)/n), where n = 12i+7.

How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is M's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.

How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is the multigen's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.

Given an edo, a period ~~p\N~~ and a generator ~~g\N~~, what is the pergen? The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*f is a multiple of g. If n = 1, |b| = 1.

Given an edo, a period, and a generator, what is the pergen? For 12edo, P = 12\12 and G = 1\12, it could be either (P8, P4/5) or (P8, P5/7).

//The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*(N+f) is a multiple of g + x*p, with 0 <= x < m. If n = 1, |b| = 1.//

//square mapping is [(N/Pk, 0), ( )]//

//xP % N = F//

This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. 13-edo and 18-edo are incompatible with heptatonic notation, ~~and~~ 13b-edo and 18b-edo are ~~used instead~~.

This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. The generator's keyspan depends on the multigen's keyspan, and thus on the 5th's keyspan. The latter is occasionally ambiguous, as in 13-edo and 18-edo. Since both of these edos are incompatible with heptatonic notation, 13edo's half-5th pergen is actually notated as a half-upfifth. 13b-edo and 18b-edo are listed as well.

~~[13b and 18b - needs explanation]~~

||||~ pergen ||~ supporting edos (12-31 only) ||

||= (P8, P5) ||= unsplit ||= 12, 13b, 14*, 15*, 16, 17, 18b*, 19, 20*, 21*,

||= (P8, P5) ||= unsplit ||= 12, 13, 13b, 14*, 15*, 16, 17, 18, 18b*, 19, 20*,

22, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31 ||

21*, 22, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31 ||

||~ halves ||~ ||~ ||

||= (P8/2, P5) ||= half-octave ||= 12, 14, 16, 18b, 20*, 22, 24*, 26, 28*, 30* ||

||= (P8/2, P5) ||= half-octave ||= 12, 14, 16, 18, 18b, 20*, 22, 24*, 26, 28*, 30* ||

||= (P8, P4/2) ||= half-fourth ||= 13b, 14, 15*, 18b*, 19, 20*, 23, 24, 25*, 28*, 29, 30* ||

||= (P8, P5/2) ||= half-fifth ||= 14*, 17, 18b, 20*, 21*, 24, 27, 28*, 30*, 31 ||

||= (P8, P5/2) ||= half-fifth ||= 13, 14*, 17, 18b, 20*, 21*, 24, 27, 28*, 30*, 31 ||

||= (P8/2, P4/2) ||= half-everything ||= 14, 18b, 20*, 24, 28*, 30* ||

||~ thirds ||~ ||~ ||

||= (P8/3, P5) ||= third-octave ||= 12, 15, 18b*, 21, 24*, 27, 30* ||

||= (P8/3, P5) ||= third-octave ||= 12, 15, 18, 18b*, 21, 24*, 27, 30* ||

||= (P8, P4/3) ||= third-fourth ||= 13b, 14*, 15, 21*, 22, 28*, 29, 30* ||

||= (P8, P5/3) ||= third-fifth ||= 15*, 16, 20*, 21, 25*, 26, 30*, 31 ||

||= (P8, P11/3) ||= third-11th ||= 15, 17, 21, 23, 30* ||

||= (P8, P11/3) ||= third-11th ||= 13, 15, 17, 21, 23, 30* ||

||= (P8/3, P4/2) ||= third-8ve, half-4th ||= 15, 18b*, 24, 30* ||

||= (P8/3, P5/2) ||= third-8ve, half-5th ||= 18b, 21, 24, 27, 30 ||

Line 439:

Line 462:

||= (P8/4, P5) ||= quarter-octave ||= 12, 16, 20, 24*, 28 ||

||= (P8, P4/4) ||= quarter-fourth ||= 18b*, 19, 20*, 28, 29, 30* ||

||= (P8, P5/4) ||= quarter-fifth ||= 14*, 20, 21*, 27, 28* ||

||= (P8, P5/4) ||= quarter-fifth ||= 13, 14*, 20, 21*, 27, 28* ||

||= (P8, P11/4) ||= quarter-eleventh ||= 14, 17, 20, 28*, 31 ||

||= (P8, P12/4) ||= quarter-twelfth ||= 13b, 15*, 18b, 20*, 23, 25*, 28, 30* ||

||= (P8/4, P4/2) ||= quarter-octave, half-fourth ||= 20, 24, 28 ||

||= (P8/2, M2/4) ||= half-octave, quarter-tone ||= 20, 22, 24, 26, 28 ||

||= (P8/2, M2/4) ||= half-octave, quarter-tone ||= 18, 20, 22, 24, 26, 28 ||

||= (P8/2, P4/4) ||= half-octave, quarter-fourth ||= 18b, 20*, 28, 30* ||

||= (P8/2, P5/4) ||= half-octave, quarter-fifth ||= 14, 20, 28* ||

Line 454:

Line 477:

||= (P8/3, P12/4) ||= third-octave, quarter-twelfth ||= 15, 18b, 30* ||

||= (P8/4, P4/4) ||= quarter-everything ||= 20, 28 ||

Line 477:

Line 499:

d = -r

Chord names: All rank-2 chords can be named using ups and downs, as if they were edos. For example, in half-octave, a 4:5:6 chord is C Ev G = C.v. There are multiple spellings for many chords. Whenever the enharmonic isn't an A1, even the degree of a chord note can change. It would be possible to spell the chord C Fb^ G, but there's no reason to. But in certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.

~~Half-octave with a vvM2 enharmonic: 4:5:6 = C Eb^ G. So better to have E = ^^d2.~~

Line 528:

Line 548:

<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</title></head><body><h1 id="toc0"> </h1>

<div id="toc"><h1 class="nopad">Table of Contents</h1><div style="margin-left: 1em;"><a href="#toc0"> </a></div>

<div id="toc"><h1 class="nopad">Table of Contents</h1><div style="margin-left: 1em;"><a href="#toc0"> </a></div>

<div style="margin-left: 1em;"><a href="#Definition">Definition</a></div>

<div style="margin-left: 1em;"><a href="#Definition">Definition</a></div>

<div style="margin-left: 1em;"><a href="#Derivation">Derivation</a></div>

<div style="margin-left: 1em;"><a href="#Derivation">Derivation</a></div>

<div style="margin-left: 1em;"><a href="#Applications">Applications</a></div>

<div style="margin-left: 1em;"><a href="#Applications">Applications</a></div>

<div style="margin-left: 1em;"><a href="#Further Discussion">Further Discussion</a></div>

<div style="margin-left: 1em;"><a href="#Further Discussion">Further Discussion</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Extremely large multigens">Extremely large multigens</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Extremely large multigens">Extremely large multigens</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Singles and doubles">Singles and doubles</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Singles and doubles">Singles and doubles</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Finding an example temperament">Finding an example temperament</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Finding an example temperament">Finding an example temperament</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Finding a notation for a pergen">Finding a notation for a pergen</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Finding a notation for a pergen">Finding a notation for a pergen</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Alternate enharmonics">Alternate enharmonics</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Alternate enharmonics">Alternate enharmonics</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Alternate keyspans and stepspans">Alternate keyspans and stepspans</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Alternate keyspans and stepspans">Alternate keyspans and stepspans</a></div>

<div style="margin-left: 2em;"><a href="#~~toc11~~"> </a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Chord names">Chord names</a></div>

<div style="margin-left: 2em;"><a href="#~~Further Discussion-Combining pergens">Combining pergens</a>&lt~~;/div>

<div style="margin-left: 2em;"><a href="#toc12"> </a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Pergens and EDOs">Pergens and EDOs</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Combining pergens">Combining pergens</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Misc notes">Misc notes</a></div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Pergens and EDOs">Pergens and EDOs</a></div>

</div>

<div style="margin-left: 2em;"><a href="#Further Discussion-Misc notes">Misc notes</a></div>

<h1 id="toc1"><a name="Definition"></a><u><strong>Definition</strong></u></h1>

</div>

<h1 id="toc1"><a name="Definition"></a><u><strong>Definition</strong></u></h1>

<br />

Line 2,107:

Line 2,128:

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).<br />

<br />

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<span style="vertical-align: super;">4</span>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' = <em>d2, and G = ^\M3 or ^/d4. Here is the genchain:<br />

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. <br />

<br />

</span><span style="display: block; text-align: center;">C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F<br />

<em>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<span style="vertical-align: super;">4</span>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' = </em>d2, and G = ^\M3 or ^/d4. Here is the genchain:<em><br />

</span><span style="display: block; text-align: center;"></em>C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F<em><br />

</span><br />

Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.<br />

</em>Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.<em><br />

</span><span style="display: block; text-align: center;">C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F<br />

</span><span style="display: block; text-align: center;"></em>C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F<em><br />

</span><span style="display: block; text-align: center;"><br />

~~</span><span style="display: block; text-align: center;">~~<span style="display: block; text-align: left;">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' = ~~</em>d2~~, and G = \~~M3 or~~ /d4. Here is the genchain:</span><span style="display: block; text-align: center;">P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11</span><span style="display: block; text-align: center;">C -- E\~~=Fb/~~ -- Ab^=Av -- C#\=Db/ -- F</span><br />

<span style="display: block; text-align: left;">The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is P11 - 4*M3 = [17,10] - [16,8] = [1,2] = dd3. For the second enharmonic, we use the second pair of accidentals: E' = \\\\dd3, and G = /M3. E = vvA1, so ^1 = 50¢ + 3.5c. E' = \\\\dd3, so /1 = 25¢ - 4.25c. E + E' = vv\\\\d3 = 2 * v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:</span><span style="display: block; text-align: left;"><br />

<span style="display: block; text-align: left;">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.</span><span style="display: block; text-align: center;">P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11</span><br />

</span><br />

<br />

<span style="display: block; text-align: left;">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.</span><br />

C -- E/=F\ -- Av=Ab^ -- C/=Db\ -- F<br />

</span><br />

</em>E' = \\m2, so /1 = 50¢ - 2.5c. But P11 = 1700¢ - c, so /1 should be 425¢ - c/4 - (400¢ + 4c) = 25¢ - 4.25c = (100¢ - 17c)/4 = dd3/4, and E' = \\\\dd3. E/ = Gbb\\\. G#<em> = Bbb\\ = Av = Ab^. So \\vm2 = 0¢.</em><br />

<br />

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 = ^<span style="vertical-align: super;">3</span>d2, and 4*G' = ^m2. Before we can divide by 4, we must quadruple all ups and downs: E = ^<span style="vertical-align: super;">12</span>d2 and 4*G' = ^<span style="vertical-align: super;">4</span>m2. The bare alt-generator is ^<span style="vertical-align: super;">4</span>[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^<span style="vertical-align: super;">4</span>m2 - 4*(^1) = m2. Thus E' = \<span style="vertical-align: super;">4</span>m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^<span style="vertical-align: super;">4</span>m2 = v<span style="vertical-align: super;">4</span>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.<br />

Line 2,155:

Line 2,187:

<br />

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.<br />

<h2 id="toc11"> </h2>

<br />

<h2 id="toc12"><a name="Further Discussion-Combining pergens"></a>Combining pergens</h2>

<h2 id="toc11"><a name="Further Discussion-Chord names"></a>Chord names</h2>

<br />

Using pergens, all rank-2 chords can be named using ups and downs, and if needed highs and lows as well. See the <a class="wiki_link" href="/Ups%20and%20Downs%20Notation">ups and downs</a> 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. <br />

<br />

In certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. (Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.)<br />

<br />

Given a specific temperament, the full mapping gives the notation of higher primes, and thus of any ratio. Thus JI chords can be named. For example, pajara's pergen is (P8/2, P5), half-octave, with P = vA4 or ^d5, G = P5, and E = ^^d2. The full mapping is [(2 0) (2 1) (7 -2) (8 -2)]. This tells us 7/1 = 8*P - 2*G = 4*P8 - 2*P5 = WWm7, and 7/4 = m7. 5/1 = 7*P - 2*G = 7/1 minus a half-octave. From this it follows that 5/4 = m7 - ^d5 = vM3. A 4:5:6:7 chord is written C Ev G Bb = C7(v3). <br />

<br />

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;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 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.<br />

<br />

<h2 id="toc12"> </h2>

<h2 id="toc13"><a name="Further Discussion-Combining pergens"></a>Combining pergens</h2>

<br />

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).<br />

Line 2,164:

Line 2,207:

However, (P8/2, M2/4) + (P8, P4/2) = (P8/4, P4/2), so the sum isn't always obvious.<br />

<br />

<h2 id="~~toc13~~"><a name="Further Discussion-Pergens and EDOs"></a>Pergens and EDOs</h2>

<h2 id="toc14"><a name="Further Discussion-Pergens and EDOs"></a>Pergens and EDOs</h2>

<br />

Pergens have much in common with edos. Pergens (rank-2 ~~ones)~~ assume only primes 2 and 3, edos assume only prime 2. There are an infinite number of edos, but ~~far~~ fewer than a hundred have been explored. There are an infinite number of pergens, but ~~far~~ fewer than a hundred will suffice most composers.<br />

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.<br />

<br />

Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset~~. However (P8/3, P5) is supported. In practice, if the generator's keyspan is very small, a partially supported pergen. For example, 22edo and 2\22 generator~~.<br />

Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset.<br />

<br />

How many pergens are fully supported by a given edo? Surprisingly, an infinite number! For example, 12edo supports (P8, P4/5), (P8, ~~P11~~/17), (P8, ~~WWP4~~/29), (P8, ~~W<span style="vertical~~-~~align: super;">3<~~/~~span>P4/41~~), ~~etc~~.<br />

How many pergens are fully supported by a given edo? Surprisingly, an infinite number! There are infinite possible multigens, each one divisible by its keyspan. For example, 12edo supports, among other pergens, this series: (P8, P5/7), (P8, P12/19), (P8, WWP5/31),... (P8, (i-1,1)/n), where n = 12i+7.<br />

<br />

How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is M's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.<br />

How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is the multigen's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.<br />

<br />

Given an edo, a period ~~p\N~~ and a generator ~~g\N~~, what is the pergen? The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*f is a multiple of g. If n = 1, |b| = 1.<br />

Given an edo, a period, and a generator, what is the pergen? For 12edo, P = 12\12 and G = 1\12, it could be either (P8, P4/5) or (P8, P5/7). <br />

<em>The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*(N+f) is a multiple of g + x*p, with 0 &lt;= x &lt; m. If n = 1, |b| = 1.</em><br />

<em>square mapping is [(N/Pk, 0), ( )]</em><br />

<br />

This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. 13-edo and 18-edo are incompatible with heptatonic notation, ~~and~~ 13b-edo and 18b-edo are ~~used instead~~.~~<br />~~

This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. The generator's keyspan depends on the multigen's keyspan, and thus on the 5th's keyspan. The latter is occasionally ambiguous, as in 13-edo and 18-edo. Since both of these edos are incompatible with heptatonic notation, 13edo's half-5th pergen is actually notated as a half-upfifth. 13b-edo and 18b-edo are listed as well.<br />

~~<br />~~

~~[13b and 18b - needs explanation]~~<br />

<br />

Line 2,194:

Line 2,238:

<td style="text-align: center;">unsplit<br />

</td>

22, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31<br />

21*, 22, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31<br />

</td>

</tr>

Line 2,211:

Line 2,255:

<td style="text-align: center;">half-octave<br />

</td>

</td>

</tr>

Line 2,227:

Line 2,271:

<td style="text-align: center;">half-fifth<br />

</td>

</td>

</tr>

Line 2,251:

Line 2,295:

<td style="text-align: center;">third-octave<br />

</td>

</td>

</tr>

Line 2,275:

Line 2,319:

<td style="text-align: center;">third-11th<br />

</td>

</td>

</tr>

Line 2,355:

Line 2,399:

<td style="text-align: center;">quarter-fifth<br />

</td>

</td>

</tr>

Line 2,387:

Line 2,431:

<td style="text-align: center;">half-octave, quarter-tone<br />

</td>

</td>

</tr>

Line 2,474:

Line 2,518:

<br />

~~<br />~~

<h2 id="toc15"><a name="Further Discussion-Misc notes"></a>Misc notes</h2>

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

<br />

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><br />

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><br />

<br />

Line 2,495:

Line 2,538:

d = -r<br />

<br />

Chord names: All rank-2 chords can be named using ups and downs, as if they were edos. For example, in half-octave, a 4:5:6 chord is C Ev G = C.v. There are multiple spellings for many chords. Whenever the enharmonic isn't an A1, even the degree of a chord note can change. It would be possible to spell the chord C Fb^ G, but there's no reason to. But in certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.<br />

<br />

~~Half-octave with a vvM2 enharmonic: 4:5:6 = C Eb^ G. So better to have E = ^^d2.<br />~~

<br />

@@ 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 02:41:31 UTC</tt>.<br>
+: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 06:41:17 UTC</tt>.<br>
-: The original revision id was <tt>624537647</tt>.<br>
+: The original revision id was <tt>624544941</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 341: / Line 341: @@
 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 = 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:
+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.
-&lt;span style="display: block; text-align: center;"&gt;P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11
-&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F
+//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;span style="display: block; text-align: center;"&gt;//P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11//
+&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;//C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F//
 &lt;/span&gt;
-Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.
+//Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.//
-&lt;span style="display: block; text-align: center;"&gt;P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11
+&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;//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&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;The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is P11 - 4*M3 = [17,10] - [16,8] = [1,2] = dd3. For the second enharmonic, we use the second pair of accidentals: E' = \\\\dd3, and G = /M3. E = vvA1, so ^1 = 50¢ + 3.5c. E' = \\\\dd3, so /1 = 25¢ - 4.25c. E + E' = vv\\\\d3 = 2 * v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:&lt;/span&gt;&lt;span style="display: block; text-align: left;"&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;/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/=Fv\ -- 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;
+//E' = \\m2, so /1 = 50¢ - 2.5c. But P11 = 1700¢ - c, so /1 should be 425¢ - c/4 - (400¢ + 4c) = 25¢ - 4.25c = (100¢ - 17c)/4 = dd3/4, and E' = \\\\dd3. E/ = Gbb\\\. G#// = Bbb\\ = Av = Ab^. So \\vm2 = 0¢.//
 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;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
@@ Line 389: / Line 400: @@
 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.
+==Chord names==
+Using pergens, all rank-2 chords can be named using ups and downs, and if needed highs and lows as well. See the [[Ups and Downs Notation|ups and downs]] 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.
+In certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. (Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.)
+Given a specific temperament, the full mapping gives the notation of higher primes, and thus of any ratio. Thus JI chords can be named. For example, pajara's pergen is (P8/2, P5), half-octave, with P = vA4 or ^d5, G = P5, and E = ^^d2. The full mapping is [(2 0) (2 1) (7 -2) (8 -2)]. This tells us 7/1 = 8*P - 2*G = 4*P8 - 2*P5 = WWm7, and 7/4 = m7. 5/1 = 7*P - 2*G = 7/1 minus a half-octave. From this it follows that 5/4 = m7 - ^d5 = vM3. A 4:5:6:7 chord is written C Ev G Bb = C7(v3).
+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;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 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.
 == ==
 ==Combining pergens==
@@ Line 403: / Line 425: @@
 ==Pergens and EDOs==
-Pergens have much in common with edos. Pergens (rank-2 ones) assume only primes 2 and 3, edos assume only prime 2. There are an infinite number of edos, but far fewer than a hundred have been explored. There are an infinite number of pergens, but far fewer than a hundred will suffice most composers.
+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.
-Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset. However (P8/3, P5) is supported. In practice, if the generator's keyspan is very small, a partially supported pergen. For example, 22edo and 2\22 generator.
+Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset.
-How many pergens are fully supported by a given edo? Surprisingly, an infinite number! For example, 12edo supports (P8, P4/5), (P8, P11/17), (P8, WWP4/29), (P8, W&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;P4/41), etc.
+How many pergens are fully supported by a given edo? Surprisingly, an infinite number! There are infinite possible multigens, each one divisible by its keyspan. For example, 12edo supports, among other pergens, this series: (P8, P5/7), (P8, P12/19), (P8, WWP5/31),... (P8, (i-1,1)/n), where n = 12i+7.
-How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is M's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.
+How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is the multigen's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.
-Given an edo, a period p\N and a generator g\N, what is the pergen? The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*f is a multiple of g. If n = 1, |b| = 1.
+Given an edo, a period, and a generator, what is the pergen? For 12edo, P = 12\12 and G = 1\12, it could be either (P8, P4/5) or (P8, P5/7).
+//The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*(N+f) is a multiple of g + x*p, with 0 &lt;= x &lt; m. If n = 1, |b| = 1.//
+//square mapping is [(N/Pk, 0), ( )]//
+//xP % N = F//
-This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. 13-edo and 18-edo are incompatible with heptatonic notation, and 13b-edo and 18b-edo are used instead.
+This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. The generator's keyspan depends on the multigen's keyspan, and thus on the 5th's keyspan. The latter is occasionally ambiguous, as in 13-edo and 18-edo. Since both of these edos are incompatible with heptatonic notation, 13edo's half-5th pergen is actually notated as a half-upfifth. 13b-edo and 18b-edo are listed as well.
-[13b and 18b - needs explanation]
 ||||~ pergen ||~ supporting edos (12-31 only) ||
-||= (P8, P5) ||= unsplit ||= 12, 13b, 14*, 15*, 16, 17, 18b*, 19, 20*, 21*,
+||= (P8, P5) ||= unsplit ||= 12, 13, 13b, 14*, 15*, 16, 17, 18, 18b*, 19, 20*,
-, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31 ||
+*, 22, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31 ||
 ||~ halves ||~   ||~   ||
-||= (P8/2, P5) ||= half-octave ||= 12, 14, 16, 18b, 20*, 22, 24*, 26, 28*, 30* ||
+||= (P8/2, P5) ||= half-octave ||= 12, 14, 16, 18, 18b, 20*, 22, 24*, 26, 28*, 30* ||
 ||= (P8, P4/2) ||= half-fourth ||= 13b, 14, 15*, 18b*, 19, 20*, 23, 24, 25*, 28*, 29, 30* ||
-||= (P8, P5/2) ||= half-fifth ||= 14*, 17, 18b, 20*, 21*, 24, 27, 28*, 30*, 31 ||
+||= (P8, P5/2) ||= half-fifth ||= 13, 14*, 17, 18b, 20*, 21*, 24, 27, 28*, 30*, 31 ||
 ||= (P8/2, P4/2) ||= half-everything ||= 14, 18b, 20*, 24, 28*, 30* ||
 ||~ thirds ||~   ||~   ||
-||= (P8/3, P5) ||= third-octave ||= 12, 15, 18b*, 21, 24*, 27, 30* ||
+||= (P8/3, P5) ||= third-octave ||= 12, 15, 18, 18b*, 21, 24*, 27, 30* ||
 ||= (P8, P4/3) ||= third-fourth ||= 13b, 14*, 15, 21*, 22, 28*, 29, 30* ||
 ||= (P8, P5/3) ||= third-fifth ||= 15*, 16, 20*, 21, 25*, 26, 30*, 31 ||
-||= (P8, P11/3) ||= third-11th ||= 15, 17, 21, 23, 30* ||
+||= (P8, P11/3) ||= third-11th ||= 13, 15, 17, 21, 23, 30* ||
 ||= (P8/3, P4/2) ||= third-8ve, half-4th ||= 15, 18b*, 24, 30* ||
 ||= (P8/3, P5/2) ||= third-8ve, half-5th ||= 18b, 21, 24, 27, 30 ||
@@ Line 439: / Line 462: @@
 ||= (P8/4, P5) ||= quarter-octave ||= 12, 16, 20, 24*, 28 ||
 ||= (P8, P4/4) ||= quarter-fourth ||= 18b*, 19, 20*, 28, 29, 30* ||
-||= (P8, P5/4) ||= quarter-fifth ||= 14*, 20, 21*, 27, 28* ||
+||= (P8, P5/4) ||= quarter-fifth ||= 13, 14*, 20, 21*, 27, 28* ||
 ||= (P8, P11/4) ||= quarter-eleventh ||= 14, 17, 20, 28*, 31 ||
 ||= (P8, P12/4) ||= quarter-twelfth ||= 13b, 15*, 18b, 20*, 23, 25*, 28, 30* ||
 ||= (P8/4, P4/2) ||= quarter-octave, half-fourth ||= 20, 24, 28 ||
-||= (P8/2, M2/4) ||= half-octave, quarter-tone ||= 20, 22, 24, 26, 28 ||
+||= (P8/2, M2/4) ||= half-octave, quarter-tone ||= 18, 20, 22, 24, 26, 28 ||
 ||= (P8/2, P4/4) ||= half-octave, quarter-fourth ||= 18b, 20*, 28, 30* ||
 ||= (P8/2, P5/4) ||= half-octave, quarter-fifth ||= 14, 20, 28* ||
@@ Line 454: / Line 477: @@
 ||= (P8/3, P12/4) ||= third-octave, quarter-twelfth ||= 15, 18b, 30* ||
 ||= (P8/4, P4/4) ||= quarter-everything ||= 20, 28 ||
@@ Line 477: / Line 499: @@
 d = -r
-Chord names: All rank-2 chords can be named using ups and downs, as if they were edos. For example, in half-octave, a 4:5:6 chord is C Ev G = C.v. There are multiple spellings for many chords. Whenever the enharmonic isn't an A1, even the degree of a chord note can change. It would be possible to spell the chord C Fb^ G, but there's no reason to. But in certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.
-Half-octave with a vvM2 enharmonic: 4:5:6 = C Eb^ G. So better to have E = ^^d2.
@@ Line 528: / Line 548: @@
 <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">&lt;html&gt;&lt;head&gt;&lt;title&gt;pergen&lt;/title&gt;&lt;/head&gt;&lt;body&gt;&lt;!-- ws:start:WikiTextHeadingRule:31:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc0"&gt;&lt;!-- ws:end:WikiTextHeadingRule:31 --&gt; &lt;/h1&gt;
-  &lt;!-- ws:start:WikiTextTocRule:61:&amp;lt;img id=&amp;quot;wikitext@@toc@@normal&amp;quot; class=&amp;quot;WikiMedia WikiMediaToc&amp;quot; title=&amp;quot;Table of Contents&amp;quot; src=&amp;quot;/site/embedthumbnail/toc/normal?w=225&amp;amp;h=100&amp;quot;/&amp;gt; --&gt;&lt;div id="toc"&gt;&lt;h1 class="nopad"&gt;Table of Contents&lt;/h1&gt;&lt;!-- ws:end:WikiTextTocRule:61 --&gt;&lt;!-- ws:start:WikiTextTocRule:62: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#toc0"&gt; &lt;/a&gt;&lt;/div&gt;
+  &lt;!-- ws:start:WikiTextTocRule:63:&amp;lt;img id=&amp;quot;wikitext@@toc@@normal&amp;quot; class=&amp;quot;WikiMedia WikiMediaToc&amp;quot; title=&amp;quot;Table of Contents&amp;quot; src=&amp;quot;/site/embedthumbnail/toc/normal?w=225&amp;amp;h=100&amp;quot;/&amp;gt; --&gt;&lt;div id="toc"&gt;&lt;h1 class="nopad"&gt;Table of Contents&lt;/h1&gt;&lt;!-- ws:end:WikiTextTocRule:63 --&gt;&lt;!-- ws:start:WikiTextTocRule:64: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#toc0"&gt; &lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:62 --&gt;&lt;!-- ws:start:WikiTextTocRule:63: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Definition"&gt;Definition&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:64 --&gt;&lt;!-- ws:start:WikiTextTocRule:65: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Definition"&gt;Definition&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:63 --&gt;&lt;!-- ws:start:WikiTextTocRule:64: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Derivation"&gt;Derivation&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:65 --&gt;&lt;!-- ws:start:WikiTextTocRule:66: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Derivation"&gt;Derivation&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:64 --&gt;&lt;!-- ws:start:WikiTextTocRule:65: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Applications"&gt;Applications&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:66 --&gt;&lt;!-- ws:start:WikiTextTocRule:67: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Applications"&gt;Applications&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:65 --&gt;&lt;!-- ws:start:WikiTextTocRule:66: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Further Discussion"&gt;Further Discussion&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:67 --&gt;&lt;!-- ws:start:WikiTextTocRule:68: --&gt;&lt;div style="margin-left: 1em;"&gt;&lt;a href="#Further Discussion"&gt;Further Discussion&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:66 --&gt;&lt;!-- ws:start:WikiTextTocRule:67: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Extremely large multigens"&gt;Extremely large multigens&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:68 --&gt;&lt;!-- ws:start:WikiTextTocRule:69: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Extremely large multigens"&gt;Extremely large multigens&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:67 --&gt;&lt;!-- ws:start:WikiTextTocRule:68: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Singles and doubles"&gt;Singles and doubles&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:69 --&gt;&lt;!-- ws:start:WikiTextTocRule:70: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Singles and doubles"&gt;Singles and doubles&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:68 --&gt;&lt;!-- ws:start:WikiTextTocRule:69: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Finding an example temperament"&gt;Finding an example temperament&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:70 --&gt;&lt;!-- ws:start:WikiTextTocRule:71: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Finding an example temperament"&gt;Finding an example temperament&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:69 --&gt;&lt;!-- ws:start:WikiTextTocRule:70: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Finding a notation for a pergen"&gt;Finding a notation for a pergen&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:71 --&gt;&lt;!-- ws:start:WikiTextTocRule:72: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Finding a notation for a pergen"&gt;Finding a notation for a pergen&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:70 --&gt;&lt;!-- ws:start:WikiTextTocRule:71: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Alternate enharmonics"&gt;Alternate enharmonics&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:72 --&gt;&lt;!-- ws:start:WikiTextTocRule:73: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Alternate enharmonics"&gt;Alternate enharmonics&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:71 --&gt;&lt;!-- ws:start:WikiTextTocRule:72: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Alternate keyspans and stepspans"&gt;Alternate keyspans and stepspans&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:73 --&gt;&lt;!-- ws:start:WikiTextTocRule:74: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Alternate keyspans and stepspans"&gt;Alternate keyspans and stepspans&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:72 --&gt;&lt;!-- ws:start:WikiTextTocRule:73: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#toc11"&gt; &lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:74 --&gt;&lt;!-- ws:start:WikiTextTocRule:75: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Chord names"&gt;Chord names&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:73 --&gt;&lt;!-- ws:start:WikiTextTocRule:74: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Combining pergens"&gt;Combining pergens&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:75 --&gt;&lt;!-- ws:start:WikiTextTocRule:76: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#toc12"&gt; &lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:74 --&gt;&lt;!-- ws:start:WikiTextTocRule:75: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Pergens and EDOs"&gt;Pergens and EDOs&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:76 --&gt;&lt;!-- ws:start:WikiTextTocRule:77: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Combining pergens"&gt;Combining pergens&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:75 --&gt;&lt;!-- ws:start:WikiTextTocRule:76: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Misc notes"&gt;Misc notes&lt;/a&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:77 --&gt;&lt;!-- ws:start:WikiTextTocRule:78: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Pergens and EDOs"&gt;Pergens and EDOs&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:76 --&gt;&lt;!-- ws:start:WikiTextTocRule:77: --&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:78 --&gt;&lt;!-- ws:start:WikiTextTocRule:79: --&gt;&lt;div style="margin-left: 2em;"&gt;&lt;a href="#Further Discussion-Misc notes"&gt;Misc notes&lt;/a&gt;&lt;/div&gt;
-&lt;!-- ws:end:WikiTextTocRule:77 --&gt;&lt;!-- ws:start:WikiTextHeadingRule:33:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc1"&gt;&lt;a name="Definition"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:33 --&gt;&lt;u&gt;&lt;strong&gt;Definition&lt;/strong&gt;&lt;/u&gt;&lt;/h1&gt;
+&lt;!-- ws:end:WikiTextTocRule:79 --&gt;&lt;!-- ws:start:WikiTextTocRule:80: --&gt;&lt;/div&gt;
+&lt;!-- ws:end:WikiTextTocRule:80 --&gt;&lt;!-- ws:start:WikiTextHeadingRule:33:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc1"&gt;&lt;a name="Definition"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:33 --&gt;&lt;u&gt;&lt;strong&gt;Definition&lt;/strong&gt;&lt;/u&gt;&lt;/h1&gt;
   &lt;br /&gt;
 &lt;br /&gt;
@@ Line 2,107: / Line 2,128: @@
 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' = &lt;em&gt;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. &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;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;
+&lt;em&gt;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;em&gt;&lt;br /&gt;
+&lt;span style="display: block; text-align: center;"&gt;&lt;/em&gt;P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11&lt;em&gt;&lt;br /&gt;
+&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;&lt;/em&gt;C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F&lt;em&gt;&lt;br /&gt;
 &lt;/span&gt;&lt;br /&gt;
-Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.&lt;br /&gt;
+&lt;/em&gt;Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.&lt;em&gt;&lt;br /&gt;
-&lt;span style="display: block; text-align: center;"&gt;P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11&lt;br /&gt;
+&lt;span style="display: block; text-align: center;"&gt;&lt;/em&gt;P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11&lt;em&gt;&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;/em&gt;C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F&lt;em&gt;&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;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;The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is P11 - 4*M3 = [17,10] - [16,8] = [1,2] = dd3. For the second enharmonic, we use the second pair of accidentals: E' = \\\\dd3, and G = /M3. E = vvA1, so ^1 = 50¢ + 3.5c. E' = \\\\dd3, so /1 = 25¢ - 4.25c. E + E' = vv\\\\d3 = 2 * v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:&lt;/span&gt;&lt;span style="display: block; text-align: left;"&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;
+&lt;/span&gt;&lt;br /&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/=Fv\ -- Ab^=Av -- C^/=Db\ -- F&lt;/span&gt;&lt;br /&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;br /&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;
+&lt;/em&gt;E' = \\m2, so /1 = 50¢ - 2.5c. But P11 = 1700¢ - c, so /1 should be 425¢ - c/4 - (400¢ + 4c) = 25¢ - 4.25c = (100¢ - 17c)/4 = dd3/4, and E' = \\\\dd3. E/ = Gbb\\\. G#&lt;em&gt; = Bbb\\ = Av = Ab^. So \\vm2 = 0¢.&lt;/em&gt;&lt;br /&gt;
+&lt;br /&gt;
+&lt;br /&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;
 &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;br /&gt;
@@ Line 2,155: / Line 2,187: @@
   &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;
-&lt;!-- ws:start:WikiTextHeadingRule:53:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc11"&gt;&lt;!-- ws:end:WikiTextHeadingRule:53 --&gt; &lt;/h2&gt;
+&lt;br /&gt;
-  &lt;!-- ws:start:WikiTextHeadingRule:55:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc12"&gt;&lt;a name="Further Discussion-Combining pergens"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:55 --&gt;Combining pergens&lt;/h2&gt;
+&lt;!-- ws:start:WikiTextHeadingRule:53:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc11"&gt;&lt;a name="Further Discussion-Chord names"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:53 --&gt;Chord names&lt;/h2&gt;
+  &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;
+&lt;br /&gt;
+In certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. (Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.)&lt;br /&gt;
+&lt;br /&gt;
+Given a specific temperament, the full mapping gives the notation of higher primes, and thus of any ratio. Thus JI chords can be named. For example, pajara's pergen is (P8/2, P5), half-octave, with P = vA4 or ^d5, G = P5, and E = ^^d2. The full mapping is [(2 0) (2 1) (7 -2) (8 -2)]. This tells us 7/1 = 8*P - 2*G = 4*P8 - 2*P5 = WWm7, and 7/4 = m7. 5/1 = 7*P - 2*G = 7/1 minus a half-octave. From this it follows that 5/4 = m7 - ^d5 = vM3. A 4:5:6:7 chord is written C Ev G Bb = C7(v3). &lt;br /&gt;
+&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 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;
+&lt;br /&gt;
+&lt;!-- ws:start:WikiTextHeadingRule:55:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc12"&gt;&lt;!-- ws:end:WikiTextHeadingRule:55 --&gt; &lt;/h2&gt;
+ &lt;!-- ws:start:WikiTextHeadingRule:57:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc13"&gt;&lt;a name="Further Discussion-Combining pergens"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:57 --&gt;Combining pergens&lt;/h2&gt;
   &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;
@@ Line 2,164: / Line 2,207: @@
 However, (P8/2, M2/4) + (P8, P4/2) = (P8/4, P4/2), so the sum isn't always obvious.&lt;br /&gt;
 &lt;br /&gt;
-&lt;!-- ws:start:WikiTextHeadingRule:57:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc13"&gt;&lt;a name="Further Discussion-Pergens and EDOs"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:57 --&gt;Pergens and EDOs&lt;/h2&gt;
+&lt;!-- ws:start:WikiTextHeadingRule:59:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc14"&gt;&lt;a name="Further Discussion-Pergens and EDOs"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:59 --&gt;Pergens and EDOs&lt;/h2&gt;
   &lt;br /&gt;
-Pergens have much in common with edos. Pergens (rank-2 ones) assume only primes 2 and 3, edos assume only prime 2. There are an infinite number of edos, but far fewer than a hundred have been explored. There are an infinite number of pergens, but far 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;
 &lt;br /&gt;
-Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset. However (P8/3, P5) is supported. In practice, if the generator's keyspan is very small, a partially supported pergen. For example, 22edo and 2\22 generator.&lt;br /&gt;
+Just as edos are said to support temperaments, they can support pergens. If a temperament is supported, its pergen is too, and vice versa. An edo can't suppoprt a pergen if the split is impossible. For example, all odd-numbered edos are incompatible with half-octave pergens. A pergen is somewhat unsupported by an edo if the period and generator can only generate a subset of the edo. For example, (P8, P5) is somewhat unsupported by 15edo, because any chain-of-5ths scale could only make a 5-edo subset.&lt;br /&gt;
 &lt;br /&gt;
-How many pergens are fully supported by a given edo? Surprisingly, an infinite number! For example, 12edo supports (P8, P4/5), (P8, P11/17), (P8, WWP4/29), (P8, W&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;P4/41), etc.&lt;br /&gt;
+How many pergens are fully supported by a given edo? Surprisingly, an infinite number! There are infinite possible multigens, each one divisible by its keyspan. For example, 12edo supports, among other pergens, this series: (P8, P5/7), (P8, P12/19), (P8, WWP5/31),... (P8, (i-1,1)/n), where n = 12i+7.&lt;br /&gt;
 &lt;br /&gt;
-How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is M's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.&lt;br /&gt;
+How many edos support a given pergen? Again, an infinite number. For (P8/m, M/n) to be supported by N-edo, N must be divisible by m, and k by n, where k is the multigen's N-edo keyspan. To be fully supported, N/m and k/n must be coprime.&lt;br /&gt;
 &lt;br /&gt;
-Given an edo, a period p\N and a generator g\N, what is the pergen? The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*f is a multiple of g. If n = 1, |b| = 1.&lt;br /&gt;
+Given an edo, a period, and a generator, what is the pergen? For 12edo, P = 12\12 and G = 1\12, it could be either (P8, P4/5) or (P8, P5/7). &lt;br /&gt;
+&lt;em&gt;The octave fraction n is N/p. Let the edo's 5th be f\N. To find the multigen M, we must find a monzo (a,b) such that a*N + b*(N+f) is a multiple of g + x*p, with 0 &amp;lt;= x &amp;lt; m. If n = 1, |b| = 1.&lt;/em&gt;&lt;br /&gt;
+&lt;em&gt;square mapping is [(N/Pk, 0), ( )]&lt;/em&gt;&lt;br /&gt;
+&lt;em&gt;xP % N = F&lt;/em&gt;&lt;br /&gt;
 &lt;br /&gt;
-This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. 13-edo and 18-edo are incompatible with heptatonic notation, and 13b-edo and 18b-edo are used instead.&lt;br /&gt;
+This table lists all pergens up to quarter-splits, with all edos that support them. Partial support is indicated with an asterisk. The generator's keyspan depends on the multigen's keyspan, and thus on the 5th's keyspan. The latter is occasionally ambiguous, as in 13-edo and 18-edo. Since both of these edos are incompatible with heptatonic notation, 13edo's half-5th pergen is actually notated as a half-upfifth. 13b-edo and 18b-edo are listed as well.&lt;br /&gt;
-&lt;br /&gt;
-[13b and 18b - needs explanation]&lt;br /&gt;
 &lt;br /&gt;
@@ Line 2,194: / Line 2,238: @@
          &lt;td style="text-align: center;"&gt;unsplit&lt;br /&gt;
 &lt;/td&gt;
-         &lt;td style="text-align: center;"&gt;12, 13b, 14*, 15*, 16, 17, 18b*, 19, 20*, 21*,&lt;br /&gt;
+         &lt;td style="text-align: center;"&gt;12, 13, 13b, 14*, 15*, 16, 17, 18, 18b*, 19, 20*, &lt;br /&gt;
-, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31&lt;br /&gt;
+*, 22, 23, 24*, 25*, 26, 27, 28*, 29, 30*, 31&lt;br /&gt;
 &lt;/td&gt;
      &lt;/tr&gt;
@@ Line 2,211: / Line 2,255: @@
          &lt;td style="text-align: center;"&gt;half-octave&lt;br /&gt;
 &lt;/td&gt;
-         &lt;td style="text-align: center;"&gt;12, 14, 16, 18b, 20*, 22, 24*, 26, 28*, 30*&lt;br /&gt;
+         &lt;td style="text-align: center;"&gt;12, 14, 16, 18, 18b, 20*, 22, 24*, 26, 28*, 30*&lt;br /&gt;
 &lt;/td&gt;
      &lt;/tr&gt;
@@ Line 2,227: / Line 2,271: @@
          &lt;td style="text-align: center;"&gt;half-fifth&lt;br /&gt;
 &lt;/td&gt;
-         &lt;td style="text-align: center;"&gt;14*, 17, 18b, 20*, 21*, 24, 27, 28*, 30*, 31&lt;br /&gt;
+         &lt;td style="text-align: center;"&gt;13, 14*, 17, 18b, 20*, 21*, 24, 27, 28*, 30*, 31&lt;br /&gt;
 &lt;/td&gt;
      &lt;/tr&gt;
@@ Line 2,251: / Line 2,295: @@
          &lt;td style="text-align: center;"&gt;third-octave&lt;br /&gt;
 &lt;/td&gt;
-         &lt;td style="text-align: center;"&gt;12, 15, 18b*, 21, 24*, 27, 30*&lt;br /&gt;
+         &lt;td style="text-align: center;"&gt;12, 15, 18, 18b*, 21, 24*, 27, 30*&lt;br /&gt;
 &lt;/td&gt;
      &lt;/tr&gt;
@@ Line 2,275: / Line 2,319: @@
          &lt;td style="text-align: center;"&gt;third-11th&lt;br /&gt;
 &lt;/td&gt;
-         &lt;td style="text-align: center;"&gt;15, 17, 21, 23, 30*&lt;br /&gt;
+         &lt;td style="text-align: center;"&gt;13, 15, 17, 21, 23, 30*&lt;br /&gt;
 &lt;/td&gt;
      &lt;/tr&gt;
@@ Line 2,355: / Line 2,399: @@
          &lt;td style="text-align: center;"&gt;quarter-fifth&lt;br /&gt;
 &lt;/td&gt;
-         &lt;td style="text-align: center;"&gt;14*, 20, 21*, 27, 28*&lt;br /&gt;
+         &lt;td style="text-align: center;"&gt;13, 14*, 20, 21*, 27, 28*&lt;br /&gt;
 &lt;/td&gt;
      &lt;/tr&gt;
@@ Line 2,387: / Line 2,431: @@
          &lt;td style="text-align: center;"&gt;half-octave, quarter-tone&lt;br /&gt;
 &lt;/td&gt;
-         &lt;td style="text-align: center;"&gt;20, 22, 24, 26, 28&lt;br /&gt;
+         &lt;td style="text-align: center;"&gt;18, 20, 22, 24, 26, 28&lt;br /&gt;
 &lt;/td&gt;
      &lt;/tr&gt;
@@ Line 2,474: / Line 2,518: @@
 &lt;br /&gt;
 &lt;br /&gt;
-&lt;br /&gt;
+&lt;!-- ws:start:WikiTextHeadingRule:61:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc15"&gt;&lt;a name="Further Discussion-Misc notes"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:61 --&gt;Misc notes&lt;/h2&gt;
-&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;
   &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;
+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:3098: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:3098 --&gt;&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;
@@ Line 2,495: / Line 2,538: @@
 d = -r&lt;br /&gt;
 &lt;br /&gt;
-Chord names: All rank-2 chords can be named using ups and downs, as if they were edos. For example, in half-octave, a 4:5:6 chord is C Ev G = C.v. There are multiple spellings for many chords. Whenever the enharmonic isn't an A1, even the degree of a chord note can change. It would be possible to spell the chord C Fb^ G, but there's no reason to. But in certain pergens, one spelling isn't always clearly better. For example, in half-fourth, C E G A^ and C E G Bbv are the same chord, and either spelling might be used. This exact same ambiguity occurs in 24-edo. Even in 12-edo, there are chords with ambiguous spellings. B D F Ab = Bdim7, and B D F# G# = Bmin6. But without the 5th, the chord could be spelled either B D Ab or B D G#.&lt;br /&gt;
 &lt;br /&gt;
-Half-octave with a vvM2 enharmonic: 4:5:6 = C Eb^ G. So better to have E = ^^d2.&lt;br /&gt;
 &lt;br /&gt;
 &lt;br /&gt;