S-expression: Difference between revisions

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== {{nowrap|S(''k'' − 1)*S''k''*S(''k'' + 1)}} (1/3-square-particulars) ==

This section concerns commas of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) = ~~( (~~k-1~~)/(~~k-2~~) )/( (~~k+2~~)/(~~k+1~~) )~~ which therefore do not (directly) involve the ''k''th harmonic.

This section concerns commas of the form {{nowrap|S(''k'' − 1) * S''k'' * S(''k'' + 1) {{=}} {{sfrac| {{sfrac|''k'' − 1|''k'' − 2}} | {{sfrac|''k'' + 2|''k'' + 1}} }} which therefore do not (directly) involve the ''k''th harmonic.

=== Significance ===

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=== Proof of simplification of 1/3-square-particulars ===

~~This section concerns commas of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) which therefore do not (directly) involve the ''k''th harmonic.~~ We can check ~~their~~ general algebraic expression for any potential simplifications:

We can check the general algebraic expression of any 1/3-square-particular for any potential simplifications:

<~~pre~~>

S(k-1) * Sk * S(k+1)

<math>\displaystyle\begin{align}

= ( (k-1~~)/(~~k-2~~) )/(~~ k/(k-1) ) * ( k/(k-1~~) )/( (~~k+1)/k ) * ( (k+1)/k ~~)/( (~~k+2~~)/(~~k+1) )

S(k-1) * S(k) * S(k+1) &= \left(\frac{\frac{k-1}{k-2}}{\frac{k}{k-1}}\right)\left(\frac{\frac{k}{k-1}}{\frac{k+1}{k}}\right)\left(\frac{\frac{k+1}{k}}{\frac{k+2}{k+1}}\right) \\

= ~~( (~~k-1~~)/(~~k-2~~) )/( (~~k+2~~)/(~~k+1~~) )~~ = (k-1)(k+1)/((k-2)(k+2)) = (k^2 - 1~~)/(~~k^2 - 4)

&= \frac{\frac{k-1}{k-2}}{\frac{k+2}{k+1}} \\

if k=3n+1 then:

&= \frac{(k-1)(k+1)}{(k-2)(k+2)} \\

S(k-1) * Sk * S(k+1) = (9n^2 + 6n~~)/(~~9n^2 + 6n - 3) = (3n^2 + 2n~~)/(~~3n^2 + 2n - 1)

&= \frac{k^2 - 1}{k^2 - 4}

if k=3n+2 then:

\end{align}</math>

S(k-1) * Sk * S(k+1) = (9n^2 + 12n + 3~~)/(~~9n^2 + 12n) = (3n^2 + 4n + 1~~)/(~~3n^2 + 4n)

if k=3n then:

If {{nowrap|''k'' {{=}} 3''n'' + 1}} then:

S(k-1) * Sk * S(k+1) = (9n^2 - 1~~)/(~~9n^2 - 4)

</~~pre~~>

In other words, what this shows is all 1/3-square-particulars of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) are superparticular iff ''k'' is throdd (not a multiple of 3), and all 1/3-square-particulars of the form S(3''k'' - 1) * S(3''k'') * S(3''k'' + 1) are throdd-particular with the numerator and denominator always being one less than a multiple of 3 (which is to say, commas of this form are throdd-particular iff ''k'' is threven and superparticular iff ''k'' is throdd).

if {{nowrap|''k'' {{=}} 3''n'' + 2}} then:

if {{nowrap|''k'' {{=}} 3''n''}} then:

In other words, what this shows is all 1/3-square-particulars of the form {{frac|S(''k'' − 1) * S''k'' * S(''k'' + 1)}} are superparticular iff ''k'' is throdd (not a multiple of 3), and all 1/3-square-particulars of the form {{nowrap|S(3''k'' − 1) * S(3''k'') * S(3''k'' + 1)}} are throdd-particular with the numerator and denominator always being one less than a multiple of 3 (which is to say, commas of this form are throdd-particular iff ''k'' is threven and superparticular iff ''k'' is throdd).

=== Table of 1/3-square-particulars ===

@@ Line 783: / Line 783: @@
 == {{nowrap|S(''k'' − 1)*S''k''*S(''k'' + 1)}} (1/3-square-particulars) ==
-This section concerns commas of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) which therefore do not (directly) involve the ''k''th harmonic.
+This section concerns commas of the form {{nowrap|S(''k'' − 1) * S''k'' * S(''k'' + 1) {{=}} {{sfrac|&nbsp;{{sfrac|''k'' − 1|''k'' − 2}}&nbsp;|&nbsp;{{sfrac|''k'' + 2|''k'' + 1}}&nbsp;}} which therefore do not (directly) involve the ''k''th harmonic.
 === Significance ===
@@ Line 795: / Line 795: @@
 === Proof of simplification of 1/3-square-particulars ===
-This section concerns commas of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) which therefore do not (directly) involve the ''k''th harmonic. We can check their general algebraic expression for any potential simplifications:
+We can check the general algebraic expression of any 1/3-square-particular for any potential simplifications:
-<pre>
-S(k-1) * Sk * S(k+1)
+<math>\displaystyle\begin{align}
- = ( (k-1)/(k-2) )/( k/(k-1) ) * ( k/(k-1) )/( (k+1)/k ) * ( (k+1)/k )/( (k+2)/(k+1) )
+S(k-1) * S(k) * S(k+1) &= \left(\frac{\frac{k-1}{k-2}}{\frac{k}{k-1}}\right)\left(\frac{\frac{k}{k-1}}{\frac{k+1}{k}}\right)\left(\frac{\frac{k+1}{k}}{\frac{k+2}{k+1}}\right) \\
- = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) = (k-1)(k+1)/((k-2)(k+2)) = (k^2 - 1)/(k^2 - 4)
+&= \frac{\frac{k-1}{k-2}}{\frac{k+2}{k+1}} \\
-if k=3n+1 then:
+&= \frac{(k-1)(k+1)}{(k-2)(k+2)} \\
-S(k-1) * Sk * S(k+1) = (9n^2 + 6n)/(9n^2 + 6n - 3) = (3n^2 + 2n)/(3n^2 + 2n - 1)
+&= \frac{k^2 - 1}{k^2 - 4}
-if k=3n+2 then:
+\end{align}</math>
-S(k-1) * Sk * S(k+1) = (9n^2 + 12n + 3)/(9n^2 + 12n) = (3n^2 + 4n + 1)/(3n^2 + 4n)
-if k=3n then:
+If {{nowrap|''k'' {{=}} 3''n'' + 1}} then:
-S(k-1) * Sk * S(k+1) = (9n^2 - 1)/(9n^2 - 4)
-</pre>
+<math>S(k-1) * Sk * S(k+1) = \frac{9n^2 + 6n}{9n^2 + 6n - 3} = \frac{3n^2 + 2n}{3n^2 + 2n - 1}</math>
-In other words, what this shows is all 1/3-square-particulars of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) are superparticular iff ''k'' is throdd (not a multiple of 3), and all 1/3-square-particulars of the form S(3''k'' - 1) * S(3''k'') * S(3''k'' + 1) are throdd-particular with the numerator and denominator always being one less than a multiple of 3 (which is to say, commas of this form are throdd-particular iff ''k'' is threven and superparticular iff ''k'' is throdd).
+if {{nowrap|''k'' {{=}} 3''n'' + 2}} then:
+<math>S(k-1) * Sk * S(k+1) = \frac{9n^2 + 12n + 3}{9n^2 + 12n} = \frac{3n^2 + 4n + 1}{3n^2 + 4n}</math>
+if {{nowrap|''k'' {{=}} 3''n''}} then:
+<math>S(k-1) * Sk * S(k+1) = \frac{9n^2 - 1}{9n^2 - 4}</math>
+In other words, what this shows is all 1/3-square-particulars of the form {{frac|S(''k'' − 1) * S''k'' * S(''k'' + 1)}} are superparticular iff ''k'' is throdd (not a multiple of 3), and all 1/3-square-particulars of the form {{nowrap|S(3''k'' − 1) * S(3''k'') * S(3''k'' + 1)}} are throdd-particular with the numerator and denominator always being one less than a multiple of 3 (which is to say, commas of this form are throdd-particular iff ''k'' is threven and superparticular iff ''k'' is throdd).
 === Table of 1/3-square-particulars ===