Kite's thoughts on pergens: Difference between revisions

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If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a **single-split** pergen. If it has two fractions, it's a **double-split** pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called **single-pair** notation because it adds only a single pair of accidentals to conventional notation. **Double-pair** notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.

Every double-split pergen is either a **true double** or a **false double**. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = ~~2*⋅P5~~ - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.

Every double-split pergen is either a **true double** or a **false double**. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = 2⋅P5 - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.

A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.

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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).

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).

If the pergen is not explicitly false, put the pergen in its **unreduced** form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n*P8 - m*M)/~~n*m~~). The new multigen M' is the product of the outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P < G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.

If the pergen is not explicitly false, put the pergen in its **unreduced** form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n⋅P8 - m⋅M)/nm). The new multigen M' is the product of the original pergen's outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P < G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.

For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2*P8 - 3*P5)/(3*2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This __is__ explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6*G - m3. The comma splits both the octave and the fifth.

For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2⋅P8 - 3⋅P5)/(3⋅2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This __is__ explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6⋅G - m3. The comma splits both the octave and the fifth.

This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2*P8 - 4*P4)/(2*4) = (2*M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.

This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2⋅P8 - 4⋅P4)/(2⋅4) = (2⋅M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.

A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.

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Gedras can be manipulated exactly like monzos. Just as adding two intervals (a,b) and (a',b') gives us (a+a',b+b'), likewise [k,s] added to [k',s'] equals [k+k',s+s']. If the GCD of a and b is n, then (a,b) is a stack of n identical intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) is converted to [k,s], then the GCD of k and s is also n, and [k,s] = [nk',ns'] = n[k',s'].

Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n*G = P5 - 2*m3 = [7,4] - 2*[3,2] = [7,4] - [6,4] = [1,0] = A1.

Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n⋅G = P5 - 2⋅m3 = [7,4] - 2⋅[3,2] = [7,4] - [6,4] = [1,0] = A1.

Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - ~~m*P~~ = P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2*m2 = d3). The enharmonic's **count** is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v5m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:

Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - mP = P8 - 5⋅M2 = [12,7] - 5⋅[2,1] = [2,2] = 2⋅[1,1] = 2⋅m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2⋅m2 = d3). The enharmonic's **count** is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v5m2. Since P8 = 5⋅P + 2⋅E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:

P1 -- ^^M2=v3m3 -- v4 -- ^5 -- ^3M6=vvm7 -- P8C -- D^^=Ebv3 -- Fv -- G^ -- A^3=Bbvv -- C

Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^5d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:

Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5⋅m2 = [5,3] - 5⋅[1,1] = [5,3] - [5,5] = [0,-2] = -2⋅[0,1] = two descending d2's. The d2 must be upped, and E = ^5d2. Since P4 = 5⋅G - 2⋅E, G must be ^^m2. The genchain is:

P1 -- ^^m2=v3A1 -- vM2 -- ^m3 -- ^3d4=vvM3 -- P4C -- Db^^ -- Dv -- Eb^ -- Evv -- F

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

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P < G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^4M2 + ^1 = ^5M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10⋅P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10⋅G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x⋅m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5⋅P + 2⋅E, and P = ^4M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P < G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^4M2 + ^1 = ^5M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.

P1- - ^4M2=v6m3 -- vvP4 -- ^^P5 -- ^6M6=v4m7 -- P8C -- D^4=Ebv6 -- Fvv -- G^^ -- A^6=Bbv4 -- CP1 -- ^5M2=v5m3 -- P4C -- D^5=Ebv5 -- F

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.

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A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).

Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. But by using double-pair notation, we can get enharmonics that are 2nds. First we find P11/2, which equals two generators: P11/2 = 2*G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2*m6 = [1,0] = A1, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6.

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://

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

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

~~~~

~~~~The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is 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:~~~~

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

P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11~~~~</span~~>C -- E/=Fv\ -- Ab^=Av -- C^/=Db\ -- F~~ 

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.

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

==Alternate enharmonics==

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If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a single-split pergen. If it has two fractions, it's a double-split pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called single-pair notation because it adds only a single pair of accidentals to conventional notation. Double-pair notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.

Every double-split pergen is either a true double or a false double. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = ~~2*⋅P5~~ - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.

Every double-split pergen is either a true double or a false double. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = 2⋅P5 - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.

A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.

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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).

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).

If the pergen is not explicitly false, put the pergen in its unreduced form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n*P8 - m*M)/~~n*m~~). The new multigen M' is the product of the outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P &lt; G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.

If the pergen is not explicitly false, put the pergen in its unreduced form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n⋅P8 - m⋅M)/nm). The new multigen M' is the product of the original pergen's outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P &lt; G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.

For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2*P8 - 3*P5)/(3*2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This is explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6*G - m3. The comma splits both the octave and the fifth.

For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2⋅P8 - 3⋅P5)/(3⋅2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This is explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6⋅G - m3. The comma splits both the octave and the fifth.

This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2*P8 - 4*P4)/(2*4) = (2*M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.

This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2⋅P8 - 4⋅P4)/(2⋅4) = (2⋅M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.

A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.

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Gedras can be manipulated exactly like monzos. Just as adding two intervals (a,b) and (a',b') gives us (a+a',b+b'), likewise [k,s] added to [k',s'] equals [k+k',s+s']. If the GCD of a and b is n, then (a,b) is a stack of n identical intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) is converted to [k,s], then the GCD of k and s is also n, and [k,s] = [nk',ns'] = n[k',s'].

Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n*G = P5 - 2*m3 = [7,4] - 2*[3,2] = [7,4] - [6,4] = [1,0] = A1.

Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n⋅G = P5 - 2⋅m3 = [7,4] - 2⋅[3,2] = [7,4] - [6,4] = [1,0] = A1.

Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - ~~m*P~~ = P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2*m2 = d3). The enharmonic's count is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v5m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:

Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - mP = P8 - 5⋅M2 = [12,7] - 5⋅[2,1] = [2,2] = 2⋅[1,1] = 2⋅m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2⋅m2 = d3). The enharmonic's count is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v5m2. Since P8 = 5⋅P + 2⋅E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:

P1 -- ^^M2=v3m3 -- v4 -- ^5 -- ^3M6=vvm7 -- P8C -- D^^=Ebv3 -- Fv -- G^ -- A^3=Bbvv -- C

Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^5d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:

Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5⋅m2 = [5,3] - 5⋅[1,1] = [5,3] - [5,5] = [0,-2] = -2⋅[0,1] = two descending d2's. The d2 must be upped, and E = ^5d2. Since P4 = 5⋅G - 2⋅E, G must be ^^m2. The genchain is:

P1 -- ^^m2=v3A1 -- vM2 -- ^m3 -- ^3d4=vvM3 -- P4C -- Db^^ -- Dv -- Eb^ -- Evv -- F

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

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^4M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^4M2 + ^1 = ^5M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.

For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10⋅P1 = m2. It must be downed, thus E = v10m2. Since m2 = 10⋅G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x⋅m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5⋅P + 2⋅E, and P = ^4M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^4M2 + ^1 = ^5M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.

P1- - ^4M2=v6m3 -- vvP4 -- ^^P5 -- ^6M6=v4m7 -- P8C -- D^4=Ebv6 -- Fvv -- G^^ -- A^6=Bbv4 -- CP1 -- ^5M2=v5m3 -- P4C -- D^5=Ebv5 -- F

To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.

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A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).

Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4⋅M3 = [1,2] = dd3. But by using double-pair notation, we can get enharmonics that are 2nds. First we find P11/2, which equals two generators: P11/2 = 2*G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2*m6 = [1,0] = A1, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6.

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:

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

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

~~~~

~~~~The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is 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:~~~~

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

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

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.

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

<h2 id="toc9"><a name="Further Discussion-Alternate enharmonics"></a>Alternate enharmonics</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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+: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 18:00:38 UTC</tt>.<br>
-: The original revision id was <tt>624584219</tt>.<br>
+: The original revision id was <tt>624585243</tt>.<br>
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 If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a **single-split** pergen. If it has two fractions, it's a **double-split** pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called **single-pair** notation because it adds only a single pair of accidentals to conventional notation. **Double-pair** notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.
-Every double-split pergen is either a **true double** or a **false double**. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = 2*⋅P5 - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.
+Every double-split pergen is either a **true double** or a **false double**. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = 2⋅P5 - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.
 A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.
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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&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.
-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).
+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).
-If the pergen is not explicitly false, put the pergen in its **unreduced** form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n*P8 - m*M)/n*m). The new multigen M' is the product of the outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P &lt; G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.
+If the pergen is not explicitly false, put the pergen in its **unreduced** form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P8 - m&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M)/nm). The new multigen M' is the product of the original pergen's outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P &lt; G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.
-For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2*P8 - 3*P5)/(3*2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This __is__ explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6*G - m3. The comma splits both the octave and the fifth.
+For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P8 - 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P5)/(3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This __is__ explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - m3. The comma splits both the octave and the fifth.
-This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2*P8 - 4*P4)/(2*4) = (2*M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.
+This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P4)/(2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;4) = (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.
 A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.
@@ Line 326: / Line 326: @@
 Gedras can be manipulated exactly like monzos. Just as adding two intervals (a,b) and (a',b') gives us (a+a',b+b'), likewise [k,s] added to [k',s'] equals [k+k',s+s']. If the GCD of a and b is n, then (a,b) is a stack of n identical intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) is converted to [k,s], then the GCD of k and s is also n, and [k,s] = [nk',ns'] = n[k',s'].
-Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n*G = P5 - 2*m3 = [7,4] - 2*[3,2] = [7,4] - [6,4] = [1,0] = A1.
+Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = P5 - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m3 = [7,4] - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[3,2] = [7,4] - [6,4] = [1,0] = A1.
-Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - m*P = P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2*m2 = d3). The enharmonic's **count** is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:
+Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - mP = P8 - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M2 = [12,7] - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[2,1] = [2,2] = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[1,1] = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2 = d3). The enharmonic's **count** is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. Since P8 = 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P + 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:
 &lt;span style="display: block; text-align: center;"&gt;P1 -- ^^M2=v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;m3 -- v4 -- ^5 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M6=vvm7 -- P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^^=Ebv&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt; -- Fv -- G^ -- A^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;=Bbvv -- C&lt;/span&gt;
-Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:
+Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2 = [5,3] - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[1,1] = [5,3] - [5,5] = [0,-2] = -2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[0,1] = two descending d2's. The d2 must be upped, and E = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;d2. Since P4 = 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;E, G must be ^^m2. The genchain is:
 &lt;span style="display: block; text-align: center;"&gt;P1 -- ^^m2=v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;A1 -- vM2 -- ^m3 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d4=vvM3 -- P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Db^^ -- Dv -- Eb^ -- Evv -- F&lt;/span&gt;
 To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multigen as before. Then deduce the period from the enharmonic. If the multigen was changed by unreducing, find the original generator from the period and the alternate generator.
-For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2 + ^1 = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.
+For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P + 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P &lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2 + ^1 = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.
 &lt;span style="display: block; text-align: center;"&gt;P1- - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 -- vvP4 -- ^^P5 -- ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 -- P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; -- Fvv -- G^^ -- A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 -- P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; -- F&lt;/span&gt;
 To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.
@@ 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&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*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.
+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://
@@ Line 350: / Line 350: @@
 //P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11//
 //C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F//
-&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 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;
+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;span style="display: block; text-align: center;"&gt;P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11&lt;/span&gt;&lt;/span&gt;&lt;span style="display: block; text-align: center;"&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; &lt;/span&gt;&lt;/span&gt;
+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.
 &lt;span style="display: block; text-align: center;"&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3 -- v&lt;span style="vertical-align: super;"&gt;8&lt;/span&gt;A5=^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m6-- P8
 &lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- Ab^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/M3 -- v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``A5=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``m6=^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\d7 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\m9 -- F&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Ev&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;/ -- G#v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Ab^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;``//``=Bbb^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;\\ -- Db^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;\ -- F&lt;/span&gt;
 ==Alternate enharmonics==
@@ Line 2,062: / Line 2,065: @@
 If a pergen has only one fraction, like (P8/2, P5) or (P8, P4/3), the pergen is a &lt;strong&gt;single-split&lt;/strong&gt; pergen. If it has two fractions, it's a &lt;strong&gt;double-split&lt;/strong&gt; pergen. A single-split pergens can result from tempering out only a single comma, although it can be created by multiple commas. A single-split pergen can be notated with only ups and downs, called &lt;strong&gt;single-pair&lt;/strong&gt; notation because it adds only a single pair of accidentals to conventional notation. &lt;strong&gt;Double-pair&lt;/strong&gt; notation uses both ups/downs and highs/lows. In general, single-pair notation is preferred, because it's simpler. However, double-pair notation may be preferred, especially if the enharmonic for single-pair notation is a 3rd or larger.&lt;br /&gt;
 &lt;br /&gt;
-Every double-split pergen is either a &lt;strong&gt;true double&lt;/strong&gt; or a &lt;strong&gt;false double&lt;/strong&gt;. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = 2*⋅P5 - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.&lt;br /&gt;
+Every double-split pergen is either a &lt;strong&gt;true double&lt;/strong&gt; or a &lt;strong&gt;false double&lt;/strong&gt;. A true double, like third-everything (P8/3, P4/3) or half-octave quarter-fourth (P8/2, P4/4), can only arise when at least two commas are tempered out, and requires double pair notation. A false double, like half-octave quarter-tone (P8/2, M2/4), can arise from a single comma, and can be notated with single pair notation. Thus a false double behaves like a single-split, and is easier to construct and easier to notate. In a false double, the multigen split automatically splits the octave as well: if M2 = 4 gen, then P8 = M9 - M2 = 2⋅P5 - 4 gen = (P5 - 2 gen) / 2. In general, if a pergen's multigen is (a,b), the octave is split into at least |b| parts.&lt;br /&gt;
 &lt;br /&gt;
 A pergen (P8/m, (a,b)/n) is a false double if and only if GCD (m,n) = |b|. The next section discusses an alternate test.&lt;br /&gt;
@@ Line 2,068: / Line 2,071: @@
 &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*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.&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;
 &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*G - M2 = (33/32)^4 / (9/8) = (-17, 2, 0, 0, 4).&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;
 &lt;br /&gt;
-If the pergen is not explicitly false, put the pergen in its &lt;strong&gt;unreduced&lt;/strong&gt; form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n*P8 - m*M)/n*m). The new multigen M' is the product of the outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P &amp;lt; G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.&lt;br /&gt;
+If the pergen is not explicitly false, put the pergen in its &lt;strong&gt;unreduced&lt;/strong&gt; form, which is always explicitly false if the pergen is a false double. The unreduced form replaces the generator with the difference between the period and the generator: (P8/m, M/n) becomes (P8/m, P8/m - M/n) = (P8/m, (n&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P8 - m&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M)/nm). The new multigen M' is the product of the original pergen's outer elements (P8 and n) minus the product of the inner elements (m and M), divided by the product of the fractions (m and n). Invert M' if descending (if P &amp;lt; G), and simplify if m and n aren't coprime. M' will have a larger fraction and/or a larger size in cents, hence the name unreduced.&lt;br /&gt;
 &lt;br /&gt;
-For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2*P8 - 3*P5)/(3*2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This &lt;u&gt;is&lt;/u&gt; explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6*G - m3. The comma splits both the octave and the fifth.&lt;br /&gt;
+For example, (P8/3, P5/2) is a false double that isn't explicitly false. Its unreduced generator is (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P8 - 3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P5)/(3&lt;span class="nowrap"&gt;⋅&lt;/span&gt;2) = m3/6, and the unreduced pergen is (P8/3, m3/6). This &lt;u&gt;is&lt;/u&gt; explicitly false, thus the comma can be found from m3/6 alone. G is about 50¢, and the comma is 6&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - m3. The comma splits both the octave and the fifth.&lt;br /&gt;
 &lt;br /&gt;
-This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2*P8 - 4*P4)/(2*4) = (2*M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.&lt;br /&gt;
+This suggests an alternate true/false test: if neither the pergen nor the unreduced pergen is explicitly false, the pergen is a true double. For example, (P8/4, P4/2) isn't explicitly false. Its unreduced form has (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P8 - 4&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P4)/(2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;4) = (2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M2)/8, which simplifies to M2/4. The unreduced pergen is (P8/4, M2/4), which also isn't explicitly false, thus (P8/4, P4/2) is a true double. It requires two commas, one for each fraction. The two commas must use different higher primes, e.g. 648/625 and 49/48. Thus true doubles require commas of at least 7-limit, whereas false doubles require only 5-limit.&lt;br /&gt;
 &lt;br /&gt;
 A false double pergen's temperament can also be constructed from two commas, as if it were a true double. For example, (P8/3, P4/2) results from 128/125 and 49/48, which split the octave and the 4th respectively.&lt;br /&gt;
@@ Line 2,098: / Line 2,101: @@
 Gedras can be manipulated exactly like monzos. Just as adding two intervals (a,b) and (a',b') gives us (a+a',b+b'), likewise [k,s] added to [k',s'] equals [k+k',s+s']. If the GCD of a and b is n, then (a,b) is a stack of n identical intervals, with (a,b) = (na', nb') = n(a',b'), and if (a,b) is converted to [k,s], then the GCD of k and s is also n, and [k,s] = [nk',ns'] = n[k',s'].&lt;br /&gt;
 &lt;br /&gt;
-Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n*G = P5 - 2*m3 = [7,4] - 2*[3,2] = [7,4] - [6,4] = [1,0] = A1.&lt;br /&gt;
+Gedras greatly facilitate finding a pergen's period, generator and enharmonic(s). A given fraction of a given 3-limit interval can be approximated by simply dividing the keyspan and stepspan directly, and rounding off. This approximation will usually produce an enharmonic interval with the smallest possible keyspan and stepspan, which is the best enharmonic for notational purposes. As noted above, the smaller of two equivalent periods or generators is preferred, so fractions of the form N/2 should be rounded down, not up. For example, consider the half-fifth pergen. P5 = [7,4], and half a 5th is approximately [round(7/2), round (4/2)] = [3,2] = (5,-3) = m3. Here xE = M - n&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G = P5 - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m3 = [7,4] - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[3,2] = [7,4] - [6,4] = [1,0] = A1.&lt;br /&gt;
 &lt;br /&gt;
-Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - m*P = P8 - 5*M2 = [12,7] - 5*[2,1] = [2,2] = 2*[1,1] = 2*m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2*m2 = d3). The enharmonic's &lt;strong&gt;count&lt;/strong&gt; is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. Since P8 = 5*P + 2*E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:&lt;br /&gt;
+Next, consider (P8/5, P5). P = [12,7]/5 = [2,1] = M2. xE = P8 - mP = P8 - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;M2 = [12,7] - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[2,1] = [2,2] = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[1,1] = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2. Because x = 2, E will occur twice in the perchain, even thought the comma only occurs once (i.e. the comma = 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2 = d3). The enharmonic's &lt;strong&gt;count&lt;/strong&gt; is 2. The bare enharmonic is m2, which must be downed to vanish. The number of downs equals the splitting fraction, thus E = v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m2. Since P8 = 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P + 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;E, the period must be ^^M2, to make the ups and downs come out even. The period's (or generator's) ups or downs always equals the count. Equipped with the period and the enharmonic, the perchain is easily found:&lt;br /&gt;
 &lt;span style="display: block; text-align: center;"&gt;P1 -- ^^M2=v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;m3 -- v4 -- ^5 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;M6=vvm7 -- P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^^=Ebv&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt; -- Fv -- G^ -- A^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;=Bbvv -- C&lt;/span&gt;&lt;br /&gt;
-Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5*m2 = [5,3] - 5*[1,1] = [5,3] - [5,5] = [0,-2] = -2*[0,1] = two descending d2's. The d2 must be upped, and E = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;d2. Since P4 = 5*G - 2*E, G must be ^^m2. The genchain is:&lt;br /&gt;
+Most single-split pergens are dealt with similarly. For example, (P8, P4/5) has a bare generator [5,3]/5 = [1,1] = m2. The bare enharmonic is P4 - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2 = [5,3] - 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[1,1] = [5,3] - [5,5] = [0,-2] = -2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;[0,1] = two descending d2's. The d2 must be upped, and E = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;d2. Since P4 = 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G - 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;E, G must be ^^m2. The genchain is:&lt;br /&gt;
 &lt;br /&gt;
 &lt;span style="display: block; text-align: center;"&gt;P1 -- ^^m2=v&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;A1 -- vM2 -- ^m3 -- ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d4=vvM3 -- P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- Db^^ -- Dv -- Eb^ -- Evv -- F&lt;/span&gt;&lt;br /&gt;
 To find the single-pair notation for a false double pergen, find an explicitly false form of the pergen, and find the generator and enharmonic from the fractional multigen as before. Then deduce the period from the enharmonic. If the multigen was changed by unreducing, find the original generator from the period and the alternate generator.&lt;br /&gt;
 &lt;br /&gt;
-For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10*P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10*G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x*m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5*P + 2*E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P &amp;lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2 + ^1 = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.&lt;br /&gt;
+For example, (P8/5, P4/2) isn't explicitly false, so we must unreduce it to (P8/5, m2/10). The bare alternate generator G' is [1,1]/10 = [0,0] = P1. The bare enharmonic is m2 - 10&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P1 = m2. It must be downed, thus E = v&lt;span style="vertical-align: super;"&gt;10&lt;/span&gt;m2. Since m2 = 10&lt;span class="nowrap"&gt;⋅&lt;/span&gt;G' + E, G' is ^1. The octave plus or minus some number of enharmonics must equal 5 periods, thus (P8 + x&lt;span class="nowrap"&gt;⋅&lt;/span&gt;m2) must be divisible by 5, and ([12,7] + x[1,1]) mod 5 must be 0. The smallest (least absolute value) x that satisfies this equation is -2, and P8 = 5&lt;span class="nowrap"&gt;⋅&lt;/span&gt;P + 2&lt;span class="nowrap"&gt;⋅&lt;/span&gt;E, and P = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2. Next, find the original half-fourth generator. P = P8/5 ~ 240¢, and G = P4/2 ~250¢. Because P &amp;lt; G, G' is not P - G but G - P, and G is not P - G' but P + G', which equals ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2 + ^1 = ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2. The alternate generator is usually simpler than the original generator, and the alternate multigen is usually more complex than the original multigen.&lt;br /&gt;
 &lt;span style="display: block; text-align: center;"&gt;P1- - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;m3 -- vvP4 -- ^^P5 -- ^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;M6=v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m7 -- P8&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt; -- Fvv -- G^^ -- A^&lt;span style="vertical-align: super;"&gt;6&lt;/span&gt;=Bbv&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt; -- C&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;P1 -- ^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;M2=v&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;m3 -- P4&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;C -- D^&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt;=Ebv&lt;span style="vertical-align: super;"&gt;5&lt;/span&gt; -- F&lt;/span&gt;&lt;br /&gt;
 To find the double-pair notation for a true double pergen, find each pair from each half of the pergen. Each pair has its own enharmonic. For (P8/2, P4/2), the split octave implies P = vA4 and E = ^^d2, and the split 4th implies G = /M2 and E' = \\m2.&lt;br /&gt;
@@ Line 2,113: / Line 2,116: @@
 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&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*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;
+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;
@@ Line 2,122: / Line 2,125: @@
 &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;span style="display: block; text-align: center;"&gt;&lt;br /&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;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;
-&lt;span style="display: block; text-align: center;"&gt;P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11&lt;/span&gt;&lt;/span&gt;&lt;span style="display: block; text-align: center;"&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; &lt;/span&gt;&lt;/span&gt;&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;
 &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;
+&lt;br /&gt;
 &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;
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 &lt;!-- ws:start:WikiTextHeadingRule:61:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc15"&gt;&lt;a name="Further Discussion-Misc notes"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:61 --&gt;Misc notes&lt;/h2&gt;
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-Pergens were discovered by Kite Giedraitis in 2017, and developed with the help of Praveen Venkataramana. Earlier drafts of this article can be found at &lt;!-- ws:start:WikiTextUrlRule:3073: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:3073 --&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: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;
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