S-expression: Difference between revisions
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== S(k − 1)* | == {{nowrap|S(''k'' − 1)*S''k''*S(''k'' + 1)}} (1/3-square-particulars) == | ||
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. | 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. | ||
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4. Square-particulars, {{frac|1|2}}-square-particulars (a.k.a. [[triangle-particular]]s), and {{frac|1|3}}-square-particulars are part of a more general sequence with interesting properties: [[1/n-square-particular|1/''n''-square-particular]]s. | 4. Square-particulars, {{frac|1|2}}-square-particulars (a.k.a. [[triangle-particular]]s), and {{frac|1|3}}-square-particulars are part of a more general sequence with interesting properties: [[1/n-square-particular|1/''n''-square-particular]]s. | ||
=== Proof of simplification of | === Proof of simplification of 1/3-square-particulars === | ||
We can check the general algebraic expression of any 1/3-square-particular for any potential simplifications: | We can check the general algebraic expression of any 1/3-square-particular for any potential simplifications: | ||