I prefer a much bruter method to show the semiparticulars' superparticularity
Demonstration of semiparticulars' superparticularity
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If k = 4n:
[math]\displaystyle{
\begin {align}
{\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n)/{\rm S}(4n + 2) \\
&= \frac {((4n) + 3)(4n)^2}{((4n) - 1)((4n) + 2)^2 } \\
&= \frac {(4n + 3)(4n)^2}{(4n - 1)(4n + 2)^2} \\
&= \frac {(4n + 3)(2n)^2}{(4n - 1)(2n + 1)^2} \\
&= \frac {16n^3 + 12n^2}{16n^3 + 12n^2 - 1}
\end {align}
}[/math]
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If k = 4n + 1:
[math]\displaystyle{
\begin {align}
{\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n + 1)/{\rm S}(4n + 3) \\
&= \frac {((4n + 1) + 3)(4n + 1)^2}{((4n + 1) - 1)((4n + 1) + 2)^2 } \\
&= \frac {(4n + 4)(4n + 1)^2}{4n(4n + 3)^2} \\
&= \frac {(n + 1)(4n + 1)^2}{n(4n + 3)^2} \\
&= \frac {16n^3 + 24n^2 + 9n + 1}{16n^3 + 24n^2 + 9n}
\end {align}
}[/math]
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If k = 4n - 1:
[math]\displaystyle{
\begin {align}
{\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n - 1)/{\rm S}(4n + 1) \\
&= \frac {((4n - 1) + 3)(4n - 1)^2}{((4n - 1) - 1)((4n - 1) + 2)^2 } \\
&= \frac {(4n + 2)(4n - 1)^2}{(4n - 2)(4n + 1)^2} \\
&= \frac {(2n + 1)(4n - 1)^2}{(2n - 1)(4n + 1)^2} \\
&= \frac {32n^3 - 6n + 1}{32n^3 - 6n - 1}
\end {align}
}[/math]
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If k = 4n - 2:
[math]\displaystyle{
\begin {align}
{\rm S}k/{\rm S}(k + 2) &= {\rm S}(4n - 2)/{\rm S}(4n) \\
&= \frac {((4n - 2) + 3)(4n - 2)^2}{((4n - 2) - 1)((4n - 2) + 2)^2 } \\
&= \frac {(4n + 1)(4n - 2)^2}{(4n - 3)(4n)^2} \\
&= \frac {(4n + 1)(2n - 1)^2}{(4n - 3)(2n)^2} \\
&= \frac {16n^3 - 12n^2 + 1}{16n^3 - 12n^2}
\end {align}
}[/math]
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So we see for k = 4n, 4n + 1, and 4n - 2, a coefficient of 22 is canceled out, whereas for the k = 4n - 1, a coefficient of 2 is canceled out. FloraC (talk) 18:31, 7 February 2022 (UTC)