Kite's thoughts on pergens: Difference between revisions

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<h2>IMPORTED REVISION FROM WIKISPACES</h2>

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==Finding an example temperament==

To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4⋅P and P8. If P is 6/5, the comma is 4⋅P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4⋅P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3⋅G - P4 = (10/9)^3 / (4/3) = 250/243.

To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4⋅P and P8. If P is 6/5, the comma is 4⋅P - P8 = (6/5)4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4⋅P = (2/1) · (7/6)-4. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3⋅G - P4 = (10/9)^3 / (4/3) = 250/243.

Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is **explicitly false**. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4⋅G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4).

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Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅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://

//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//

//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.//

//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 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

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:

v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:

P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11C -- E/=Fv\ -- Ab^=Av -- 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.

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.

P1 -- v4M3 -- v8A5=^4m6-- P8

C -- Ev4 -- Ab^4 -- CP1 -- v3/M3 -- v6``//``A5=^6``//``m6=^6\\d7 -- ^3\m9 -- FC -- Ev3/ -- G#v6``//``=Ab^6``//``=Bbb^6\\ -- Db^3\ -- F

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==Alternate enharmonics==

Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12*[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12*[4,2] = [-15,-5] = -5*[3,1] = -~~5 * v~~12A2, which is an improvement but still awkward. The period is ^4m3 and the generator is v3M2.

Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v12A2, which is an improvement but still awkward. The period is ^4m3 and the generator is v3M2.

P1 -- ^4m3 -- v4M6 -- C

C -- Eb^4 -- Av4 -- C

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//These commas are called **notational commas**. They are not necessarily tempered out, they merely determine how a higher prime is mapped to a 3-limit interval, and thus how ratios containing higher primes are notated on the staff. By definition, the only commas that map to P1 are notational ones, and those that are the sum or difference of notational ones. There is widespread agreement that 5's notational comma is 81/80. But the choice of notational commas for other primes, especially 11 and 13, is somewhat arbitrary. For example, if 11's notational comma is 33/32, 11/8 is notated as a perfect 4th. But if it's 729/704, 11/8 is an augmented 4th.//

//An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - ~~2*G~~ = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.//

//An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2·G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.//

//For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.//

//(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - ~~4*G~~ = [1,2] = dd3. It must be downed, thus E = v4dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.//

//(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4·G = [1,2] = dd3. It must be downed, thus E = v4dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.//

==Alternate keyspans and stepspans==

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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).

Given a specific temperament, the full period/generator 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 2 7 8) (0 1 -2 -2)] = [(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.

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 2 0 1) (0 1 4 4)] = [(2 0) (2 1) (0 4) (1 4)]. The square mapping (the first two columns) are the same, hence the pergen is the same, but the other columns are different, hence the higher primes are mapped differently. 5/4 = M3 and 7/4 = M3 + vd5 = vm7, and 4:5:6:7 = C E G Bbv = C,v7.

== ==

==Combining pergens==

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If the enharmonic is larger than a 2nd, it may be possible to split it into several smaller enharmonics. For example, {P8/5, P5}. If 7/M = 7/5 is rounded to 1, the enharmonic is a 3rd. The enharmonic must be 5x, and also 7y + 4, so the 3-exponent = -10 = dim 3rd. Adding ups and downs, we have enharmonic = v5d3 and period = ^M2. Fortunately, d3 = m2 + m2, and the 3rd can be reduced to two 2nds. The downs must be doubled, so that the period = ^^M2, and the enharmonic = v10d3 = 2 *

v5m2. The enharmonic must be applied twice in the course of an octave: P1 - ^^M2 - ^4M3=vP4 - ^P5 - ^3M6=vvm7 - P8

As a side note, every comma implies an edo, except for those that map to P1: notational ones, and those that are the sum or difference of notational ones.

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<h2 id="toc7"><a name="Further Discussion-Finding an example temperament"></a>Finding an example temperament</h2>

To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4⋅P and P8. If P is 6/5, the comma is 4⋅P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4⋅P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3⋅G - P4 = (10/9)^3 / (4/3) = 250/243.

To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4⋅P and P8. If P is 6/5, the comma is 4⋅P - P8 = (6/5)4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4⋅P = (2/1) · (7/6)-4. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3⋅G - P4 = (10/9)^3 / (4/3) = 250/243.

Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is explicitly false. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4⋅G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4).

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Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅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:

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

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.

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 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 ~~

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:

v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:

P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11C -- E/=Fv\ -- Ab^=Av -- 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.

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.

P1 -- v4M3 -- v8A5=^4m6-- P8

C -- Ev4 -- Ab^4 -- CP1 -- v3/M3 -- v6//A5=^6//m6=^6\\d7 -- ^3\m9 -- FC -- Ev3/ -- G#v6//=Ab^6//=Bbb^6\\ -- Db^3\ -- F

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<h2 id="toc9"><a name="Further Discussion-Alternate enharmonics"></a>Alternate enharmonics</h2>

Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12*[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12*[4,2] = [-15,-5] = -5*[3,1] = -~~5 * v~~12A2, which is an improvement but still awkward. The period is ^4m3 and the generator is v3M2.

Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v12A2, which is an improvement but still awkward. The period is ^4m3 and the generator is v3M2.

P1 -- ^4m3 -- v4M6 -- C

C -- Eb^4 -- Av4 -- C

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These commas are called notational commas. They are not necessarily tempered out, they merely determine how a higher prime is mapped to a 3-limit interval, and thus how ratios containing higher primes are notated on the staff. By definition, the only commas that map to P1 are notational ones, and those that are the sum or difference of notational ones. There is widespread agreement that 5's notational comma is 81/80. But the choice of notational commas for other primes, especially 11 and 13, is somewhat arbitrary. For example, if 11's notational comma is 33/32, 11/8 is notated as a perfect 4th. But if it's 729/704, 11/8 is an augmented 4th.

An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - ~~2*G~~ = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.

An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2·G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.

For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.

(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - ~~4*G~~ = [1,2] = dd3. It must be downed, thus E = v4dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.

(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4·G = [1,2] = dd3. It must be downed, thus E = v4dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.

<h2 id="toc10"><a name="Further Discussion-Alternate keyspans and stepspans"></a>Alternate keyspans and stepspans</h2>

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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).

Given a specific temperament, the full period/generator 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 2 7 8) (0 1 -2 -2)] = [(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.

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 2 0 1) (0 1 4 4)] = [(2 0) (2 1) (0 4) (1 4)]. The square mapping (the first two columns) are the same, hence the pergen is the same, but the other columns are different, hence the higher primes are mapped differently. 5/4 = M3 and 7/4 = M3 + vd5 = vm7, and 4:5:6:7 = C E G Bbv = C,v7.

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

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

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<h2 id="toc15"><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>

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~~ ~~

If the enharmonic is larger than a 2nd, it may be possible to split it into several smaller enharmonics. For example, {P8/5, P5}. If 7/M = 7/5 is rounded to 1, the enharmonic is a 3rd. The enharmonic must be 5x, and also 7y + 4, so the 3-exponent = -10 = dim 3rd. Adding ups and downs, we have enharmonic = v5d3 and period = ^M2. Fortunately, d3 = m2 + m2, and the 3rd can be reduced to two 2nds. The downs must be doubled, so that the period = ^^M2, and the enharmonic = v10d3 = 2 *

v5m2. The enharmonic must be applied twice in the course of an octave: P1 - ^^M2 - ^4M3=vP4 - ^P5 - ^3M6=vvm7 - P8

As a side note, every comma implies an edo, except for those that map to P1: notational ones, and those that are the sum or difference of notational ones.

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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:<br>
-: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 18:00:38 UTC</tt>.<br>
+: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 18:17:15 UTC</tt>.<br>
-: The original revision id was <tt>624585243</tt>.<br>
+: The original revision id was <tt>624585795</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>
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 ==Finding an example temperament==
-To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P and P8. If P is 6/5, the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - P4 = (10/9)^3 / (4/3) = 250/243.
+To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P and P8. If P is 6/5, the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P - P8 = (6/5)&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P = (2/1) · (7/6)&lt;span style="vertical-align: super;"&gt;-4&lt;/span&gt;. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - P4 = (10/9)^3 / (4/3) = 250/243.
 Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is **explicitly false**. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4).
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 Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M3 = [1,2] = dd3. 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&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m6 = [1,0] = A1, so E = vvA1 and 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = ^m6 or vM6. Next we find G = (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;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://
+//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://
 //P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11//
 //C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F//
-//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.//
 //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 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
+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:
-v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:
 &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;
+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.
-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
 &lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Ab^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/M3 -- v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``A5=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``m6=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\d7 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\m9 -- F&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/ -- G#v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Ab^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Bbb^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\ -- Db^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\ -- F&lt;/span&gt;
@@ Line 361: / Line 359: @@
 ==Alternate enharmonics==
-Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12*[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12*[4,2] = [-15,-5] = -5*[3,1] = -5 * v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;A2, which is an improvement but still awkward. The period is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 and the generator is v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M2.
+Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;A2, which is an improvement but still awkward. The period is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 and the generator is v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M2.
 &lt;span style="display: block; text-align: center;"&gt;P1 -- ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M6 -- C
 C -- Eb^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Av&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C
@@ Line 383: / Line 381: @@
 //These commas are called **notational commas**. They are not necessarily tempered out, they merely determine how a higher prime is mapped to a 3-limit interval, and thus how ratios containing higher primes are notated on the staff. By definition, the only commas that map to P1 are notational ones, and those that are the sum or difference of notational ones. There is widespread agreement that 5's notational comma is 81/80. But the choice of notational commas for other primes, especially 11 and 13, is somewhat arbitrary. For example, if 11's notational comma is 33/32, 11/8 is notated as a perfect 4th. But if it's 729/704, 11/8 is an augmented 4th.//
-//An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2*G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.//
+//An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2·G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.//
 //For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.//
-//(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.//
+//(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4·G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.//
 ==Alternate keyspans and stepspans==
@@ Line 399: / Line 397: @@
 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).
+Given a specific temperament, the full period/generator 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 2 7 8) (0 1 -2 -2)] = [(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.
+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 2 0 1) (0 1 4 4)] = [(2 0) (2 1) (0 4) (1 4)]. The square mapping (the first two columns) are the same, hence the pergen is the same, but the other columns are different, hence the higher primes are mapped differently. 5/4 = M3 and 7/4 = M3 + vd5 = vm7, and 4:5:6:7 = C E G Bbv = C,v7.
 == ==
 ==Combining pergens==
@@ Line 525: / Line 523: @@
-If the enharmonic is larger than a 2nd, it may be possible to split it into several smaller enharmonics. For example, {P8/5, P5}. If 7/M = 7/5 is rounded to 1, the enharmonic is a 3rd. The enharmonic must be 5x, and also 7y + 4, so the 3-exponent = -10 = dim 3rd. Adding ups and downs, we have enharmonic = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;d3 and period = ^M2. Fortunately, d3 = m2 + m2, and the 3rd can be reduced to two 2nds. The downs must be doubled, so that the period = ^^M2, and the enharmonic = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;d3 = 2 *
-v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. The enharmonic must be applied twice in the course of an octave: P1 - ^^M2 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3=vP4 - ^P5 - ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M6=vvm7 - P8
 As a side note, every comma implies an edo, except for those that map to P1: notational ones, and those that are the sum or difference of notational ones.
@@ Line 2,071: / Line 2,066: @@
 &lt;!-- ws:start:WikiTextHeadingRule:45:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc7"&gt;&lt;a name="Further Discussion-Finding an example temperament"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:45 --&gt;Finding an example temperament&lt;/h2&gt;
   &lt;br /&gt;
-To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P and P8. If P is 6/5, the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P - P8 = (6/5)^4 / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P = (2/1) * (7/6)^-4. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - P4 = (10/9)^3 / (4/3) = 250/243.&lt;br /&gt;
+To find an example of a temperament with a specific pergen, we must find the comma(s) the temperament tempers out. To construct a comma that creates a single-split pergen, find a ratio for P or G that contains only one higher prime, with exponent ±1, of appropriate cents to add up to approximately the octave or the multigen. The comma is the difference between the stacked ratios and the larger interval. For example, (P8/4, P5) requires a P of about 300¢. The comma is the difference between 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P and P8. If P is 6/5, the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P - P8 = (6/5)&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; / (2/1) = 648/625. If P is 7/6, the comma is P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P = (2/1) · (7/6)&lt;span style="vertical-align: super;"&gt;-4&lt;/span&gt;. Neither 13/11 nor 32/27 would work for P, too many and too few higher primes respectively. (P8, P4/3) requires a G of about (498¢)/3 = 166¢, perhaps 10/9. The comma is 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - P4 = (10/9)^3 / (4/3) = 250/243.&lt;br /&gt;
 &lt;br /&gt;
 Finding the comma(s) for a double pergen is trickier. As previously noted, if a pergen's multigen is (a,b), the octave is split into at least |b| parts. Therefore if a pergen (P8/m, (a,b)/n) has m = |b|, it is &lt;strong&gt;explicitly false&lt;/strong&gt;. If so, proceed as if the octave were unsplit: (P8/2, M2/4) requires G ~ 50¢, perhaps 33/32, and the comma is 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4).&lt;br /&gt;
@@ Line 2,118: / Line 2,113: @@
 Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M3 = [1,2] = dd3. 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&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m6 = [1,0] = A1, so E = vvA1 and 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = ^m6 or vM6. Next we find G = (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G)/2, using either ^m6 or vM6.&lt;br /&gt;
 &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. &lt;em&gt;Here is the genchain:&lt;/em&gt;&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. &lt;em&gt;Here is the genchain:&lt;/em&gt;&lt;br /&gt;
 &lt;em&gt;P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11&lt;/em&gt;&lt;br /&gt;
 &lt;em&gt;C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F&lt;/em&gt;&lt;br /&gt;
 &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;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;em&gt;P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11&lt;/em&gt;&lt;br /&gt;
 &lt;em&gt;C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F&lt;/em&gt;&lt;br /&gt;
 &lt;br /&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&lt;br /&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;br /&gt;
-v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:&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;
+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;
-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;
 &lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Ab^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/M3 -- v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:025:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:025 --&gt;A5=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:026:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:026 --&gt;m6=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\d7 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\m9 -- F&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/ -- G#v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:027:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:027 --&gt;=Ab^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;&lt;!-- ws:start:WikiTextRawRule:028:``//`` --&gt;//&lt;!-- ws:end:WikiTextRawRule:028 --&gt;=Bbb^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\ -- Db^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\ -- F&lt;/span&gt;&lt;br /&gt;
@@ Line 2,136: / Line 2,129: @@
 &lt;!-- ws:start:WikiTextHeadingRule:49:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc9"&gt;&lt;a name="Further Discussion-Alternate enharmonics"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:49 --&gt;Alternate enharmonics&lt;/h2&gt;
   &lt;br /&gt;
-Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12*[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12*[4,2] = [-15,-5] = -5*[3,1] = -5 * v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;A2, which is an improvement but still awkward. The period is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 and the generator is v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M2.&lt;br /&gt;
+Sometimes the enharmonic found by rounding off the gedra can be greatly improved by rounding off differently. For example, (P8/3, P4/4) unreduces to (P8/3, WWM6/12), a false double. The bare alternate generator is WWM6/12 = [33,19]/12 = [3,2] = m3. The bare enharmonic is [33,19] - 12·[3,2] = [-3,-5] = a quintuple-diminished 6th! This would make for a very confusing notation. However, [33,19]/12 can be rounded very inaccurately all the way up to [4,2] = M3. The enharmonic becomes [33,19] - 12·[4,2] = [-15,-5] = -5·[3,1] = -5·v&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;A2, which is an improvement but still awkward. The period is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 and the generator is v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M2.&lt;br /&gt;
 &lt;span style="display: block; text-align: center;"&gt;P1 -- ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m3 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M6 -- C&lt;br /&gt;
 C -- Eb^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Av&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;br /&gt;
@@ Line 2,158: / Line 2,151: @@
 &lt;em&gt;These commas are called &lt;strong&gt;notational commas&lt;/strong&gt;. They are not necessarily tempered out, they merely determine how a higher prime is mapped to a 3-limit interval, and thus how ratios containing higher primes are notated on the staff. By definition, the only commas that map to P1 are notational ones, and those that are the sum or difference of notational ones. There is widespread agreement that 5's notational comma is 81/80. But the choice of notational commas for other primes, especially 11 and 13, is somewhat arbitrary. For example, if 11's notational comma is 33/32, 11/8 is notated as a perfect 4th. But if it's 729/704, 11/8 is an augmented 4th.&lt;/em&gt;&lt;br /&gt;
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-&lt;em&gt;An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2*G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.&lt;/em&gt;&lt;br /&gt;
+&lt;em&gt;An alternate enharmonic will arise if the notational comma changes. For example, 11's notational comma can be either 33/32, with 11/8 notated as a P4, or 729/704, with 11/8 notated as an A4. The keyspan of all 11-limit intervals will reflect this choice of notational comma. For (P8, P5/2), G ~ 350¢. If G = 11/9, the (vanishing, not notational) comma is P5 - 2·G = 243/242. For the first notational comma, 11/9 is a m3, and the comma is an A1. For the 2nd, 11/9= M3, and the comma is a d1.&lt;/em&gt;&lt;br /&gt;
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 &lt;em&gt;For single-comma pergens, the enharmonic should equal the comma's mapping. For example, (P8, P5/2) might arise from 243/242, which splits the 5th into two 11/9 halves.&lt;/em&gt;&lt;br /&gt;
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-&lt;em&gt;(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4*G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.&lt;/em&gt;&lt;br /&gt;
+&lt;em&gt;(P8, P11/4) has a bare generator [17,10]/4 = [4,2] = M3. The bare enharmonic is P11 - 4·G = [1,2] = dd3. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;dd3, and G = ^M3. The enharmonic is unfortunately not a unison or 2nd. Note that the generator's stepspan could have been rounded up instead of down, making G = [4,3] = d4. This would make E = [-1,2] = d43. Rounding down is clearly preferable! In general, rounding down is better, because the smaller of two equivalent generators or periods is preferred. However, there are exceptions.&lt;/em&gt;&lt;br /&gt;
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 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;
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-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;
+Given a specific temperament, the full period/generator 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 2 7 8) (0 1 -2 -2)] = [(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;
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-A different temperament may result in the same pergen with the same enharmonic, but may still produce a different name for the same chord. For example, injera (2.3.5.7 with 81/80 and 50/49, or rryy&amp;amp;gT) is also half-octave. However, the tipping point for the d2 enharmonic is at 700¢, and while pajara favors a fifth wider than that, injera favors a fifth narrower than that. Hence ups and downs are exchanged, and E = vvd2, and P = ^A4 = vd5. The mapping is [(2 0) (2 1) (0 4) (1 4)]. The square mapping (the first two columns) are the same, hence the pergen is the same, but the other columns are different, hence the higher primes are mapped differently. 5/4 = M3 and 7/4 = M3 + vd5 = vm7, and 4:5:6:7 = C E G Bbv = C,v7.&lt;br /&gt;
+A different temperament may result in the same pergen with the same enharmonic, but may still produce a different name for the same chord. For example, injera (2.3.5.7 with 81/80 and 50/49, or rryy&amp;amp;gT) is also half-octave. However, the tipping point for the d2 enharmonic is at 700¢, and while pajara favors a fifth wider than that, injera favors a fifth narrower than that. Hence ups and downs are exchanged, and E = vvd2, and P = ^A4 = vd5. The mapping is [(2 2 0 1) (0 1 4 4)] = [(2 0) (2 1) (0 4) (1 4)]. The square mapping (the first two columns) are the same, hence the pergen is the same, but the other columns are different, hence the higher primes are mapped differently. 5/4 = M3 and 7/4 = M3 + vd5 = vm7, and 4:5:6:7 = C E G Bbv = C,v7.&lt;br /&gt;
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-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:3146: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:3146 --&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:3135: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:3135 --&gt;&lt;br /&gt;
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-If the enharmonic is larger than a 2nd, it may be possible to split it into several smaller enharmonics. For example, {P8/5, P5}. If 7/M = 7/5 is rounded to 1, the enharmonic is a 3rd. The enharmonic must be 5x, and also 7y + 4, so the 3-exponent = -10 = dim 3rd. Adding ups and downs, we have enharmonic = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;d3 and period = ^M2. Fortunately, d3 = m2 + m2, and the 3rd can be reduced to two 2nds. The downs must be doubled, so that the period = ^^M2, and the enharmonic = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;d3 = 2 *&lt;br /&gt;
-v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. The enharmonic must be applied twice in the course of an octave: P1 - ^^M2 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3=vP4 - ^P5 - ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M6=vvm7 - P8&lt;br /&gt;
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 As a side note, every comma implies an edo, except for those that map to P1: notational ones, and those that are the sum or difference of notational ones.&lt;br /&gt;