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A square superparticular, or ''square-particular'' for short, is a [[superparticular]] [[interval]] whose numerator is a square number, which is to say, a superparticular of the form
A square superparticular, or ''square-particular'' for short, is a [[superparticular]] [[interval]] whose numerator is a square number, which is to say, a superparticular of the form


$$ \frac {k^2}{k^2 - 1} = \frac {k/(k - 1)}{(k + 1)/k} $$
$$ \frac {k²}{k² - 1} = \frac {k/(k - 1)}{(k + 1)/k} $$


which is square-superparticular ''k'' for a given integer {{nowrap| ''k'' > 1 }}. A suggested shorthand for this interval is '''S''k''''' for the ''k''-th square superparticular, where the ''S'' stands for ''second-order/square superparticular''. This will be used later in this article as the notation will prove powerful in understanding the commas and implied tempered structures of [[regular temperament]]s. Note that {{nowrap| S2 {{=}} [[4/3]] }} is the first musically meaningful square-particular, as {{nowrap| S1 {{=}} 1/0 }}.
which is square-superparticular ''k'' for a given integer {{nowrap| ''k'' > 1 }}. A suggested shorthand for this interval is '''S''k''''' for the ''k''-th square superparticular, where the ''S'' stands for ''second-order/square superparticular''. This will be used later in this article as the notation will prove powerful in understanding the commas and implied tempered structures of [[regular temperament]]s. Note that {{nowrap| S2 {{=}} [[4/3]] }} is the first musically meaningful square-particular, as {{nowrap| S1 {{=}} 1/0 }}.
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=== Short proof of the superparticularity of triangle-particulars ===
=== Short proof of the superparticularity of triangle-particulars ===
$$ S(k) \cdot S(k + 1) = \frac{\frac{k}{k - 1}}{\frac{k + 2}{k + 1}} = \frac{k(k + 1)}{(k - 1)(k + 2)} = \frac{k^2 + k}{k^2 + k - 2} $$
<nowiki>$$ S(k) \cdot S(k + 1) = \frac{\frac{k}{k - 1}}{\frac{k + 2}{k + 1}} = \frac{k(k + 1)}{(k - 1)(k + 2)} = \frac{k² + k}{k² + k - 2} $$</nowiki>


Then notice that {{nowrap| ''k''<sup>2</sup> + ''k'' }} is always a multiple of 2; therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could be reasoned that it would likely be half as accurate as tempering out either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals are [[triangular number]]s, hence the alternative name ''triangle-particular''.
Then notice that {{nowrap| ''k''<sup>2</sup> + ''k'' }} is always a multiple of 2; therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could be reasoned that it would likely be half as accurate as tempering out either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals are [[triangular number]]s, hence the alternative name ''triangle-particular''.
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Tempering out S(''k'' - 1)/S(''k'' + 1) implies that (''k'' + 2)/(''k'' - 2) is divisible exactly into two halves of (''k'' + 1)/(''k'' - 1). It also implies that the intervals (''k'' + 2)/''k'' (= s) and ''k''/(''k'' - 2) (= L) are equidistant from (''k'' + 1)/(''k'' - 1) (= M) because, to make them equidistant, we need to temper out:
Tempering out S(''k'' - 1)/S(''k'' + 1) implies that (''k'' + 2)/(''k'' - 2) is divisible exactly into two halves of (''k'' + 1)/(''k'' - 1). It also implies that the intervals (''k'' + 2)/''k'' (= s) and ''k''/(''k'' - 2) (= L) are equidistant from (''k'' + 1)/(''k'' - 1) (= M) because, to make them equidistant, we need to temper out:


$$ \frac{L/M}{M/s} = \frac{ \left(\frac{k}{k-2}\right)/\left(\frac{k+1}{k-1}\right) }{ \left(\frac{k+1}{k-1}\right)/\left(\frac{k+2}{k}\right) } = \frac{Ls}{M^2} = \frac{\frac{k+2}{k-2}}{\left(\frac{k+1}{k-1}\right)^2} $$
<nowiki>$$ \frac{L/M}{M/s} = \frac{ \left(\frac{k}{k-2}\right)/\left(\frac{k+1}{k-1}\right) }{ \left(\frac{k+1}{k-1}\right)/\left(\frac{k+2}{k}\right) } = \frac{Ls}{M²} = \frac{\frac{k+2}{k-2}}{\left(\frac{k+1}{k-1}\right)²} $$</nowiki>


… and notice that the latter expression is the one we have shown is equal to S(''k'' - 1)/S(''k'' + 1) up to an offset ''k'' (→ [[S-expression/Advanced results #Mathematical derivations]]). In other words, that tempering out S(''k'' - 1)/S(''k'' + 1) results in (''k'' + 1)/(''k'' - 1) being half of (''k'' + 2)/(''k'' - 2) is an implication that it makes (''k'' + 2)/''k'', (''k'' + 1)/(''k'' - 1), and ''k''/(''k'' - 2) equidistant.  
… and notice that the latter expression is the one we have shown is equal to S(''k'' - 1)/S(''k'' + 1) up to an offset ''k'' (→ [[S-expression/Advanced results #Mathematical derivations]]). In other words, that tempering out S(''k'' - 1)/S(''k'' + 1) results in (''k'' + 1)/(''k'' - 1) being half of (''k'' + 2)/(''k'' - 2) is an implication that it makes (''k'' + 2)/''k'', (''k'' + 1)/(''k'' - 1), and ''k''/(''k'' - 2) equidistant.  
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$$
$$
{\rm S}k = {\rm S}(2k - 1) \cdot {\rm S}(2k)^2 \cdot {\rm S}(2k + 1)
{\rm S}k = {\rm S}(2k - 1) \cdot {\rm S}(2k)² \cdot {\rm S}(2k + 1)
$$
$$


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{| class="wikitable center-1"
{| class="wikitable center-1"
|-
|-
! Comma
! rowspan="2" | Comma<br>(by size)
! S-expressions
! colspan="4" | S-expressions
|-
! colspan="2" |Main
! colspan="2" |Secondary
|-
|-
| [[28/27]]
| [[28/27]]
| S7⋅S8, S4/S6
| S7⋅S8
|S4/S6
|
|
|-
|-
| [[36/35]]
| [[36/35]]
| S6, S8⋅S9
| S6
|S8⋅S9
|
|
|-
|-
| [[64/63]]
| [[64/63]]
| S8, S6/S9, S4/(S6⋅S7), (S4⋅S5⋅S6)/S3
| S8
|S6/S9
|S4/(S6⋅S7)
|(S4⋅S5⋅S6)/S3
|-
|-
| [[81/80]]
| [[81/80]]
| S9, S6/S8
| S9
|S6/S8
|
|
|-
| [[245/243]]
| S7/S9
|
|
|S10/(S12/S14)
|-
|-
| [[176/175]]
| [[176/175]]
| S8/S10, S22⋅S23⋅S24
| S8/S10
|S22⋅S23⋅S24
|
|(S26⋅S27)²/S351
|-
|[[15625/15552]]
|S25²⋅S26
|
|S15⋅(S25/S27)
|
|-
|-
|[[225/224]]
|[[225/224]]
|S15, S25*S26*S27
|S15
|S25⋅S26⋅S27
|
|
|-
|-
| [[243/242]]
| [[243/242]]
| S9/S11, S15/S55, S15/(S22/S24)
| S9/S11
|
|S15/S55
|S15/(S22/S24)
|-
|-
| [[325/324]]
| [[325/324]]
| S25⋅S26, S10/S12
| S10/S12
|S25⋅S26
|
|
|-
|[[352/351]]*
|S12/(S9/S11)
|S11⋅S12/S9
|
|(S8/S9)/(S64⋅S65)
|-
| [[364/363]]*
| S14/(S11/S13)
|(S13⋅S14)/S11
|S24⋅S25⋅S26/S22
|(S9/S11)/S27
|-
|[[385/384]]
|S33⋅S34⋅S35
|
|
|(S8/S9)/(S64²/S65)
|-
|[[Septischisma|<small>33554432​/33480783</small>]]*
|(S8/S9)²/S15
|
|(S16/S18)³/(S19/S20)
|S49⋅S55⋅(S64²⋅S65)²
|-
|-
| [[540/539]]
| [[540/539]]
| S12/S14, (S9⋅S10)/S7, (S6/S7)/(S8/S10)
| S12/S14
|
|(S9⋅S10)/S7
|(S6/S7)/(S8/S10)
|-
|[[4000/3993]]
|S10/S11
|
|(S12/S14)/S99
|S25⋅(S64⋅S65)/S55
|-
|-
| [[676/675]]
| [[676/675]]
| S26, S13/S15, S49*(S64*S65)^2*S99
| S13/S15
|S26
|
|S49⋅(S64⋅S65)²⋅S99
|-
|[[32805/32768]]*
|
|
|S15/(S8/S9)
|S19/(S16/S18)²
|-
|-
|[[1001/1000]]
|[[1001/1000]]
|sqrt(S26*S49*S99), S49*S64*S65*S99
|
|Cp10
|√{{Overline|S26⋅S49⋅S99}}
|S49⋅(S64⋅S65)⋅S99
|-
|[[4459/4455]]
|
|
|S49⋅(S64⋅S65)
|(S64⋅S65)/S26⋅S99
|-
|-
|[[1216/1215]]
|[[1216/1215]]
|S16/S18, S64*S65*S76*S77
|S16/S18
|
|
|(S64⋅S65)⋅(S76⋅S77)
|-
|[[10985/10976]]
|S13/S14
|
|(S64⋅S65²)⋅S99
|
|-
|-
| [[1225/1224]]
| [[1225/1224]]
| S35, S49⋅S50
| S35
|S49⋅S50
|
|
|-
|[[41503/41472]]*
|
|
|S49⋅S55
|S65⋅(S76⋅S77²)
|-
|<small>[[131072/130977]]</small>
|S64²⋅S65
|
|S32/S63
|
|-
|[[43904/43875]]
|S14/S15
|
|S49⋅S64
|S56⋅(S76⋅S77)
|-
|-
|[[2080/2079]]
|[[2080/2079]]
|S64⋅S65, S78⋅S79⋅S80, sqrt(S26/S49/S99)
|S64⋅S65
|S78⋅S79⋅S80
|√{{Overline|S26/(S49⋅S99)}}
|
|-
|-
| [[2601/2600]]
| [[2601/2600]]
| S51, S17/(S25⋅S26)
| S51
|
|S17/(S25⋅S26)
|
|-
|-
| [[3025/3024]]
| [[3025/3024]]
| S55, S22/S24, (S25/S27)⋅S99
| S55
|S22/S24
|(S25/S27)⋅S99
|
|-
|-
|[[3136/3135]]
|[[3136/3135]]
|S56, S96*S97*S98
|S56
|S96⋅S97⋅S98
|
|
|-
|-
|[[4000/3993]]
|[[4225/4224]]
|S10/S11, (S12/S14)/S99, S25*(S64*S65)/S55
|S65
|
|(S25/S27)⋅S351
|
|-
|-
|[[4375/4374]]
|[[4375/4374]]
|S25/S27, S55/S99
|S25/S27
|-
|
|[[4459/4455]]
|S55/S99
|S49*S64*S65, S64+S65-S26+S99
|S65/S351
|-
|-
|[[6656/6655]]
|[[6656/6655]]
|S64*S65/S55, S64*S351/S99
|
|
|(S64⋅S65)/S55
|S64⋅S351/S99
|-
|-
| [[9801/9800]]
| [[9801/9800]]
| S99, S33/S35
| S99
|S33/S35
|
|
|-
|-
|[[10241/10240]]
|[[10241/10240]]*
|S14*S18*S19/S12/S16, S76*S77/S64
|
|
|(S76⋅S77)/S64
|((S18⋅S19)/S16)/(S12/S14)
|-
|-
|[[10985/10976]]
|[[10648/10647]]*
|S13/S14, (S64*S65^2)*S99
|
|
|S99/S351
|S55⋅S65
|-
|-
| [[25921/25920]]
| [[25921/25920]]
| S161, S46/S48
| S161
|-
|
|[[32805/32768]]
|S46/S48
|S15/(S8/S9), S19/(S16/S18)^2
|
|-
|[[41503/41472]]
|S49*S55, S65*S76*S77^2
|-
|[[43904/43875]]
|S14/S15, S49*S64, S56*S76*S77
|-
|-
| <small>[[123201/123200]]</small>
| <small>[[123201/123200]]</small>
| S351, S78/S80, S25/S27/S65
| S351
|S78/S80
|
|(S25/S27)/S65
|}
|}
'''*important:''' commas marked with an asterisk appear commonly but do not appear elsewhere on this page, so the extra slots are used as extra equivalent expressions.


{{Note| Examples that can ''easily'' (with one or two algebraic rewriting steps) be shown to result from the aforementioned [[#A useful general rule|useful general rule]] are not included. }}
{{Note| Examples that can ''easily'' (with one or two algebraic rewriting steps) be shown to result from the aforementioned [[#A useful general rule|useful general rule]] are not included. }}
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