Kite's thoughts on pergens: Difference between revisions

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

If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is **split** into N parts. The interval which is split into multiple generators is the **multi-gen**. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, etc.

If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is **split** into N parts. The interval which is split into multiple generators is the **multi-gen**. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, m7, etc.

For example, the srutal temperament splits the octave in two, and its pergen name is half-octave. The pergen is written (P8/2, P5). Not only the srutal temperament, but also the srutal comma 2048/2025 is said to split the octave. The dicot temperament splits the fifth in two, and is called half-fifth, written (P8, P5/2). Porcupine is third-fourth, (P8, P4/3). Semaphore, which means "semi-fourth", is of course half-fourth.

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For example, consider 2.3.5.7 with commas 256/243 and 225/224. The first comma splits the octave into 5 parts, and makes the 5th be exactly 3/5 of the octave. The multigen must use primes 2 and 5. In this case, the pergen is (P8/5, y3), the same as Blackwood.

To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder [[@http://x31eq.com/temper/uv.html|x31eq.com/temper/uv.html]] that will find such a matrix, it's the reduced mapping. Next make a **square mapping** by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes ~~> 3~~ may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.

To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder [[@http://x31eq.com/temper/uv.html|x31eq.com/temper/uv.html]] that will find such a matrix, it's the reduced mapping. Next make a **square mapping** by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.

For rank-2, we can compute the pergen directly from the square matrix = [(x y), (0, z)]. Let the pergen be (P8/m, M/n), where M is the multigen, P is the period P8/m, and G is the generator M/n.

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G = (-y, x) / xz + nP = (-y, x) / xz + nP8/x = (nz - y, x) / xz

**The rank-2 pergen from the [(x, 0) (y, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x <= n <= x**

**The rank-2 pergen from the [(x, y) (0, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x <= n <= x**

For example, 250/243 has a mapping [(1, 2, 3) (0, -3, -5)] and a square mapping of [(1, 2) (0, -3)]. The pergen is (P8/1, (-3n-2,1)/(-3)) = (P8, (3n+2,-1)/3), with -1 <= n <= 1. No value of n reduces the fraction, so the best multigen is the one with the least cents, (2,-1) = P4. The pergen is (P8, P4/3).

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||~ 3/1 ||= 0 ||= 1 ||= -1 || ||

||~ 7/1 ||= 0 ||= 0 ||= 2 || /2 ||

Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. ~~We can let~~ ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is > 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.

Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. Let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is > 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.

=__Applications__=

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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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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:

//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

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

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

//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

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

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

//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:

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

P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- ~~P11C -- E~~/~~=Fv\ -- Ab^=Av -- C^/=Db\ -- F~~

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

P1 -- /M3=\4 -- ~~vM6~~=~~^m6~~ -- /8=\m9 -- ~~P11~~

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

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

One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. Next find 4*G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3*m2 = [0,-1] = descending d2. Thus E = ^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

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

If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is split into N parts. The interval which is split into multiple generators is the multi-gen. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, etc.

If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is split into N parts. The interval which is split into multiple generators is the multi-gen. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, m7, etc.

For example, the srutal temperament splits the octave in two, and its pergen name is half-octave. The pergen is written (P8/2, P5). Not only the srutal temperament, but also the srutal comma 2048/2025 is said to split the octave. The dicot temperament splits the fifth in two, and is called half-fifth, written (P8, P5/2). Porcupine is third-fourth, (P8, P4/3). Semaphore, which means &quot;semi-fourth&quot;, is of course half-fourth.

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For example, consider 2.3.5.7 with commas 256/243 and 225/224. The first comma splits the octave into 5 parts, and makes the 5th be exactly 3/5 of the octave. The multigen must use primes 2 and 5. In this case, the pergen is (P8/5, y3), the same as Blackwood.

To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder <a class="wiki_link_ext" href="http://x31eq.com/temper/uv.html" rel="nofollow" target="_blank">x31eq.com/temper/uv.html</a> that will find such a matrix, it's the reduced mapping. Next make a square mapping by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes ~~&gt; 3~~ may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.

To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder <a class="wiki_link_ext" href="http://x31eq.com/temper/uv.html" rel="nofollow" target="_blank">x31eq.com/temper/uv.html</a> that will find such a matrix, it's the reduced mapping. Next make a square mapping by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.

For rank-2, we can compute the pergen directly from the square matrix = [(x y), (0, z)]. Let the pergen be (P8/m, M/n), where M is the multigen, P is the period P8/m, and G is the generator M/n.

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G = (-y, x) / xz + nP = (-y, x) / xz + nP8/x = (nz - y, x) / xz

The rank-2 pergen from the [(x, 0) (y, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x &lt;= n &lt;= x

The rank-2 pergen from the [(x, y) (0, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x &lt;= n &lt;= x

For example, 250/243 has a mapping [(1, 2, 3) (0, -3, -5)] and a square mapping of [(1, 2) (0, -3)]. The pergen is (P8/1, (-3n-2,1)/(-3)) = (P8, (3n+2,-1)/3), with -1 &lt;= n &lt;= 1. No value of n reduces the fraction, so the best multigen is the one with the least cents, (2,-1) = P4. The pergen is (P8, P4/3).

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</table>

Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. ~~We can let~~ ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.

Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. Let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.

<h1 id="toc3"><a name="Applications"></a>Applications</h1>

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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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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:

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

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

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

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

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

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

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:

~~ ~~

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

P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- ~~P11C -- E/=Fv\ -- Ab^=Av -- C^/=Db\ -- F~~<~~br />~~

~~<br~~ />

Using vM6/2 for 2*G gives a different but equally valid notation: M6/2 = [9,5]/2 = [4,2] = M3, and M6 - 2*(M3) = [1,1] = m2, and E' = \\m2 and G = /M3 or \4.</span~~>~~ 

~~P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11 ~~

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

</span~~&gt~~;~~ ~~

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

~~ ~~

~~<br~~ />

One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. Next find 4*G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3*m2 = [0,-1] = descending d2. Thus E = ^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

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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>

@@ Line 1: / Line 1: @@
 <h2>IMPORTED REVISION FROM WIKISPACES</h2>
 This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br>
-: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 06:56:04 UTC</tt>.<br>
+: This revision was by author [[User:TallKite|TallKite]] and made on <tt>2018-01-08 17:31:45 UTC</tt>.<br>
-: The original revision id was <tt>624545475</tt>.<br>
+: The original revision id was <tt>624584219</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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 A **pergen** (pronounced "peer-gen") is a way of identifying a regular temperament solely by its period and generator(s). For any temperament, there are many possible periods and generators. For the pergen, they are chosen to use the fewest, and smallest, prime factors possible. Fractions are allowed, e.g. half-octave, but avoided if possible.
-If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is **split** into N parts. The interval which is split into multiple generators is the **multi-gen**. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, etc.
+If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is **split** into N parts. The interval which is split into multiple generators is the **multi-gen**. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, m7, etc.
 For example, the srutal temperament splits the octave in two, and its pergen name is half-octave. The pergen is written (P8/2, P5). Not only the srutal temperament, but also the srutal comma 2048/2025 is said to split the octave. The dicot temperament splits the fifth in two, and is called half-fifth, written (P8, P5/2). Porcupine is third-fourth, (P8, P4/3). Semaphore, which means "semi-fourth", is of course half-fourth.
@@ Line 63: / Line 63: @@
 For example, consider 2.3.5.7 with commas 256/243 and 225/224. The first comma splits the octave into 5 parts, and makes the 5th be exactly 3/5 of the octave. The multigen must use primes 2 and 5. In this case, the pergen is (P8/5, y3), the same as Blackwood.
-To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder [[@http://x31eq.com/temper/uv.html|x31eq.com/temper/uv.html]] that will find such a matrix, it's the reduced mapping. Next make a **square mapping** by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes &gt; 3 may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.
+To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder [[@http://x31eq.com/temper/uv.html|x31eq.com/temper/uv.html]] that will find such a matrix, it's the reduced mapping. Next make a **square mapping** by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.
 For rank-2, we can compute the pergen directly from the square matrix = [(x y), (0, z)]. Let the pergen be (P8/m, M/n), where M is the multigen, P is the period P8/m, and G is the generator M/n.
@@ Line 72: / Line 72: @@
 G = (-y, x) / xz + nP = (-y, x) / xz + nP8/x = (nz - y, x) / xz
-&lt;span style="display: block; text-align: center;"&gt;**&lt;span style="font-size: 110%;"&gt;The rank-2 pergen from the [(x, 0) (y, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x &lt;= n &lt;= x&lt;/span&gt;**
+&lt;span style="display: block; text-align: center;"&gt;**&lt;span style="font-size: 110%;"&gt;The rank-2 pergen from the [(x, y) (0, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x &lt;= n &lt;= x&lt;/span&gt;**
 &lt;/span&gt;
 For example, 250/243 has a mapping [(1, 2, 3) (0, -3, -5)] and a square mapping of [(1, 2) (0, -3)]. The pergen is (P8/1, (-3n-2,1)/(-3)) = (P8, (3n+2,-1)/3), with -1 &lt;= n &lt;= 1. No value of n reduces the fraction, so the best multigen is the one with the least cents, (2,-1) = P4. The pergen is (P8, P4/3).
@@ Line 107: / Line 107: @@
 ||~ 3/1 ||= 0 ||= 1 ||= -1 ||   ||
 ||~ 7/1 ||= 0 ||= 0 ||= 2 || /2 ||
-Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. We can let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.
+Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. Let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.
 =__Applications__=
@@ Line 287: / Line 287: @@
 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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 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&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.
-//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
+//P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11//
-C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F
+//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
+//P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11//
-C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F
+//C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F//
 &lt;span style="display: block; text-align: center;"&gt;
-&lt;/span&gt;&lt;span style="display: block; text-align: center;"&gt;
 &lt;span style="display: block; text-align: left;"&gt;The bare generator is m6/2 = [8,5]/2 = [4,2] = M3, and the bare enharmonic is P11 - 4*M3 = [17,10] - [16,8] = [1,2] = dd3. For the second enharmonic, we use the second pair of accidentals: E' = \\\\dd3, and G = /M3. E = vvA1, so ^1 = 50¢ + 3.5c. E' = \\\\dd3, so /1 = 25¢ - 4.25c. E + E' = vv\\\\d3 = 2 * v\\m2 = 0¢. E' - E = [0,2] = 2 * ^\\d2. Here is the genchain:&lt;/span&gt;
+&lt;span style="display: block; text-align: 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;span style="display: block; text-align: center;"&gt;P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11C -- E/=Fv\ -- Ab^=Av -- C^/=Db\ -- F&lt;/span&gt;
-&lt;span style="display: block; text-align: left;"&gt;Using vM6/2 for 2*G gives a different but equally valid notation: M6/2 = [9,5]/2 = [4,2] = M3, and M6 - 2*(M3) = [1,1] = m2, and E' = \\m2 and G = /M3 or \4.&lt;/span&gt;
-&lt;span style="display: block; text-align: center;"&gt;P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11&lt;/span&gt;
-C -- E/=F\ -- Av=Ab^ -- C/=Db\ -- F
-&lt;/span&gt;
-E' = \\m2, so /1 = 50¢ - 2.5c. But P11 = 1700¢ - c, so /1 should be 425¢ - c/4 - (400¢ + 4c) = 25¢ - 4.25c = (100¢ - 17c)/4 = dd3/4, and E' = \\\\dd3. E/ = Gbb\\\. G# //= Bbb\\ = Av = Ab^. So \\vm2 = 0¢.//
 One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. Next find 4*G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3*m2 = [0,-1] = descending d2. Thus E = ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d2, and 4*G' = ^m2. Before we can divide by 4, we must quadruple all ups and downs: E = ^&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;d2 and 4*G' = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2. The bare alt-generator is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 - 4*(^1) = m2. Thus E' = \&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3.
 &lt;span style="display: block; text-align: center;"&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3 -- v&lt;span style="vertical-align: super;"&gt;8&lt;/span&gt;A5=^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m6-- P8
@@ Line 566: / Line 555: @@
 A &lt;strong&gt;pergen&lt;/strong&gt; (pronounced &amp;quot;peer-gen&amp;quot;) is a way of identifying a regular temperament solely by its period and generator(s). For any temperament, there are many possible periods and generators. For the pergen, they are chosen to use the fewest, and smallest, prime factors possible. Fractions are allowed, e.g. half-octave, but avoided if possible.&lt;br /&gt;
 &lt;br /&gt;
-If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is &lt;strong&gt;split&lt;/strong&gt; into N parts. The interval which is split into multiple generators is the &lt;strong&gt;multi-gen&lt;/strong&gt;. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, etc.&lt;br /&gt;
+If a rank-2 temperament uses the primes 2 and 3 in its comma(s), or in its prime subgroup (i.e. doesn't explicitly exclude the octave or the fifth), then the period can be expressed as the octave 2/1, or some fraction of an octave. Furthermore, the generator can usually be expressed as some 3-limit interval, or some fraction of such an interval. The fraction is always of the form 1/N, thus the octave and/or the 3-limit interval is &lt;strong&gt;split&lt;/strong&gt; into N parts. The interval which is split into multiple generators is the &lt;strong&gt;multi-gen&lt;/strong&gt;. The 3-limit multi-gen is referred to not by its ratio but by its conventional name, e.g. P5, M6, m7, etc.&lt;br /&gt;
 &lt;br /&gt;
 For example, the srutal temperament splits the octave in two, and its pergen name is half-octave. The pergen is written (P8/2, P5). Not only the srutal temperament, but also the srutal comma 2048/2025 is said to split the octave. The dicot temperament splits the fifth in two, and is called half-fifth, written (P8, P5/2). Porcupine is third-fourth, (P8, P4/3). Semaphore, which means &amp;quot;semi-fourth&amp;quot;, is of course half-fourth.&lt;br /&gt;
@@ Line 819: / Line 808: @@
 For example, consider 2.3.5.7 with commas 256/243 and 225/224. The first comma splits the octave into 5 parts, and makes the 5th be exactly 3/5 of the octave. The multigen must use primes 2 and 5. In this case, the pergen is (P8/5, y3), the same as Blackwood.&lt;br /&gt;
 &lt;br /&gt;
-To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder &lt;a class="wiki_link_ext" href="http://x31eq.com/temper/uv.html" rel="nofollow" target="_blank"&gt;x31eq.com/temper/uv.html&lt;/a&gt; that will find such a matrix, it's the reduced mapping. Next make a &lt;strong&gt;square mapping&lt;/strong&gt; by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes &amp;gt; 3 may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.&lt;br /&gt;
+To find a temperament's pergen, first find the period-generator mapping. This is a matrix with a column for each prime in the subgroup, and a row for each period/generator. Not all such mappings will work, the matrix must be in row echelon form. Graham Breed's website has a temperament finder &lt;a class="wiki_link_ext" href="http://x31eq.com/temper/uv.html" rel="nofollow" target="_blank"&gt;x31eq.com/temper/uv.html&lt;/a&gt; that will find such a matrix, it's the reduced mapping. Next make a &lt;strong&gt;square mapping&lt;/strong&gt; by discarding columns, usually the columns for the highest primes. But lower primes may need to be discarded, as in the previous (P8/5,y3) example, to ensure that the diagonal has no zeros. Lower primes may also be discarded to minimize splitting, see the Breedsmic example below. Then invert the matrix to get the monzos for each period/generator. Add/subtract periods from the generator to get alternate generators. If the interval becomes descending, invert it. For rank-3, add/subtract both periods and generators from the 2nd generator to get more alternates. Choose among the alternates to minimize the splitting and the cents.&lt;br /&gt;
 &lt;br /&gt;
 For rank-2, we can compute the pergen directly from the square matrix = [(x y), (0, z)]. Let the pergen be (P8/m, M/n), where M is the multigen, P is the period P8/m, and G is the generator M/n.&lt;br /&gt;
@@ Line 828: / Line 817: @@
 G = (-y, x) / xz + nP = (-y, x) / xz + nP8/x = (nz - y, x) / xz&lt;br /&gt;
 &lt;br /&gt;
-&lt;span style="display: block; text-align: center;"&gt;&lt;strong&gt;&lt;span style="font-size: 110%;"&gt;The rank-2 pergen from the [(x, 0) (y, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x &amp;lt;= n &amp;lt;= x&lt;/span&gt;&lt;/strong&gt;&lt;br /&gt;
+&lt;span style="display: block; text-align: center;"&gt;&lt;strong&gt;&lt;span style="font-size: 110%;"&gt;The rank-2 pergen from the [(x, y) (0, z)] square mapping is (P8/x, (nz-y,x)/xz), with -x &amp;lt;= n &amp;lt;= x&lt;/span&gt;&lt;/strong&gt;&lt;br /&gt;
 &lt;/span&gt;&lt;br /&gt;
 For example, 250/243 has a mapping [(1, 2, 3) (0, -3, -5)] and a square mapping of [(1, 2) (0, -3)]. The pergen is (P8/1, (-3n-2,1)/(-3)) = (P8, (3n+2,-1)/3), with -1 &amp;lt;= n &amp;lt;= 1. No value of n reduces the fraction, so the best multigen is the one with the least cents, (2,-1) = P4. The pergen is (P8, P4/3).&lt;br /&gt;
@@ Line 1,092: / Line 1,081: @@
 &lt;/table&gt;
-Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. We can let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &amp;gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.&lt;br /&gt;
+Again, period = P8 and gen1 = P5/2. Gen2 = (-3,-1,2)/2. To add gen1 to gen2, add a double gen1 to the 2nd multigen, the multigen2. A double half-fifth is a fifth = (-1,1,0), and this gives us (-4,0,2)/2 = (-2,0,1) = 7/4. The fraction disappears, the multigen becomes the gen, and we can add/subtract the period and the gen1 directly. Subtracting an octave and inverting makes gen2 = 8/7 = r2. Adding an octave and subtracting 4 half-fifths makes 64/63 = r1. The pergen is (P8, P5/2, r1) = half-fifth with red. Let ^1 = 64/63, and the pergen is (P8, P5/2, ^1), half-fifth with ups. This is far better than (P8, P5/2, gg7/4). The pergen sometimes uses a larger prime in place of a smaller one in order to avoid splitting gen2, but only if the smaller prime is &amp;gt; 3. In other words, the first priority is to have as few higher primes (colors) as possible, next to have as few fractions as possible, finally to have the higher primes be as small as possible.&lt;br /&gt;
 &lt;br /&gt;
 &lt;!-- ws:start:WikiTextHeadingRule:37:&amp;lt;h1&amp;gt; --&gt;&lt;h1 id="toc3"&gt;&lt;a name="Applications"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:37 --&gt;&lt;u&gt;Applications&lt;/u&gt;&lt;/h1&gt;
@@ Line 2,073: / Line 2,062: @@
 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 2,124: / Line 2,113: @@
 A false-double pergen can use double-pair notation, which is found as if it were a true double. Double-pair notation is often preferable when the single-pair enharmonic is not a unison or a 2nd, as with (P8/2, P4/3).&lt;br /&gt;
 &lt;br /&gt;
-Even single-split pergens may benefit from double-pair notation. For example, (P8, P11/4) has an enharmonic that's a 3rd: P11/4 = [17,10]/4 = [4,2] = M3, and P11 - 4*M3 = [1,2] = dd3. But by using double-pair notation, we can get enharmonics that are 2nds. First we find P11/2, which equals two generators: P11/2 = 2*G = [17,10]/2 = [8,5] = m6. The bare enharmonic is P11 - 2*m6 = [1,0] = A1, so E = vvA1 and 2*G = ^m6 or vM6. Next we find G = (2*G)/2, using either ^m6 or vM6.&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;
 &lt;br /&gt;
-&lt;em&gt;But ^m6 has an up sign, and there's no such thing as half an up. The answer is to double all ups and downs: P11/2 = ^^m6 or vvM6, and E = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;A1. The bare generator is ^^m6/2 = ^^[8,5]/2 = ^[4,2] = ^M3, and the bare enharmonic is ^^m6 - 2*(^M3) = [0,1] = d2. For the second enharmonic, we use the second pair of accidentals: E' =&lt;/em&gt; d2, and G = ^\M3 or ^/d4. Here is the genchain:&lt;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;
-P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11&lt;br /&gt;
+&lt;em&gt;P1 -- ^\M3=^/d4 -- ^^m6=vvM6 -- v\A8=v/m9 -- P11&lt;/em&gt;&lt;br /&gt;
-C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F&lt;br /&gt;
+&lt;em&gt;C -- E^\=Fb^/ -- Ab^^=Avv -- C#v\=Dbv/ -- F&lt;/em&gt;&lt;br /&gt;
 &lt;br /&gt;
-Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.&lt;br /&gt;
+&lt;em&gt;Using vvM6/2 for 2*G gives a different but equally valid notation: vvM6/2 = vv[9,5]/2 = v[4,2] = vM3, and vvM6 - 2*(vM3) = [1,1] = m2, and E' = \\m2 and G = v/M3 or v\4.&lt;/em&gt;&lt;br /&gt;
-P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11&lt;br /&gt;
+&lt;em&gt;P1 -- v/M3=v\4 -- vvM6=^^m6 -- ^/8=^\m9 -- P11&lt;/em&gt;&lt;br /&gt;
-C -- Ev/=Fv\ -- Avv=Ab^^ -- C^/=Db^\ -- F&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;/span&gt;&lt;span style="display: block; text-align: center;"&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;
-&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;
-&lt;span style="display: block; text-align: center;"&gt;P1 -- \M3=/d4 -- ^m6=vM6 -- \A8=/m9 -- P11C -- E/=Fv\ -- Ab^=Av -- C^/=Db\ -- F&lt;/span&gt;&lt;br /&gt;
-&lt;br /&gt;
-&lt;span style="display: block; text-align: left;"&gt;Using vM6/2 for 2*G gives a different but equally valid notation: M6/2 = [9,5]/2 = [4,2] = M3, and M6 - 2*(M3) = [1,1] = m2, and E' = \\m2 and G = /M3 or \4.&lt;/span&gt;&lt;br /&gt;
-&lt;span style="display: block; text-align: center;"&gt;P1 -- /M3=\4 -- vM6=^m6 -- /8=\m9 -- P11&lt;/span&gt;&lt;br /&gt;
-C -- E/=F\ -- Av=Ab^ -- C/=Db\ -- F&lt;br /&gt;
-&lt;/span&gt;&lt;br /&gt;
-E' = \\m2, so /1 = 50¢ - 2.5c. But P11 = 1700¢ - c, so /1 should be 425¢ - c/4 - (400¢ + 4c) = 25¢ - 4.25c = (100¢ - 17c)/4 = dd3/4, and E' = \\\\dd3. E/ = Gbb\\\. G# &lt;em&gt;= Bbb\\ = Av = Ab^. So \\vm2 = 0¢.&lt;/em&gt;&lt;br /&gt;
-&lt;br /&gt;
-&lt;br /&gt;
-&lt;br /&gt;
 One might think with (P8/3, P11/4), that one pair would be needed to split the octave, and another two pairs to split the 11th, making three in all. But proceeding as before, we find only two pairs are needed. First unreduce to get (P8/3, m3/12). m3/12 is the alternate generator G'. Next find 4*G' = m3/3 = [3,2]/3 = [1,1] = m2. Next, the bare enharmonic: m3 - 3*m2 = [0,-1] = descending d2. Thus E = ^&lt;span style="vertical-align: super;"&gt;3&lt;/span&gt;d2, and 4*G' = ^m2. Before we can divide by 4, we must quadruple all ups and downs: E = ^&lt;span style="vertical-align: super;"&gt;12&lt;/span&gt;d2 and 4*G' = ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2. The bare alt-generator is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;[1,1]/4 = ^[0,0] = ^1, and the bare 2nd enharmonic is ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 - 4*(^1) = m2. Thus E' = \&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 and G' = ^/1.The period can be deduced from 4*G': P8/3 = (m10 - m3)/3 = (m10)/3 - 4*G' = P4 - ^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m2 = v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3. From the unreducing, we know that G - P = P11/4 - P8/3 = m3/12, so G = P11/4 = P8/3 + m3/12 = v4M3 + ^/1 = v3/M3.&lt;br /&gt;
 &lt;span style="display: block; text-align: center;"&gt;P1 -- v&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;M3 -- v&lt;span style="vertical-align: super;"&gt;8&lt;/span&gt;A5=^&lt;span style="vertical-align: super;"&gt;4&lt;/span&gt;m6-- P8&lt;br /&gt;
@@ Line 2,512: / Line 2,490: @@
 &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:3082: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:3082 --&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: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;
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