20200502, 10:20  #1 
Feb 2012
Prague, Czech Republ
2^{2}·41 Posts 
Divisibility function
\(\forall N \ne 0,\ x \in [1,\ N]: \ sin \, x \pi +\imath \, sin \frac{N \pi}{x} = 0 \leftrightarrow xN.
\) I'm not aware of any practical value, but I like the pictures, produced in Maxima by draw2d( nticks=5000, grid=true, xrange=[1,1], yrange=[1,1], terminal='pngcairo, parametric(sin(%pi*x),sin(%pi*n/x),x,1, n) ), n=<insert N here>; edit: fixed tex formatting Last fiddled with by jnml on 20200502 at 10:30 
20200502, 10:22  #2 
Feb 2012
Prague, Czech Republ
2^{2}·41 Posts 
6 to 9

20200502, 10:23  #3 
Feb 2012
Prague, Czech Republ
2^{2}·41 Posts 
10 to 13

20200502, 10:25  #4 
Feb 2012
Prague, Czech Republ
10100100_{2} Posts 
14 to 17 and that's all.

20200502, 11:52  #5 
Dec 2012
The Netherlands
2^{2}·3·5^{3} Posts 
Look up Lissajous curves.

20200502, 12:37  #6 
Feb 2012
Prague, Czech Republ
244_{8} Posts 
I don't think the function in #1 qualifies as a Lissajous curve.
The reason I see is that the argument of the sinus function of the complex part is not a proportional to \(t\), but to its inverse. In fact, that's what makes the sum a divisibility function, which the Lissajous curve  AFAICT  is not. 
20200502, 12:41  #7 
Feb 2017
Nowhere
3802_{10} Posts 
Related, but Lissajous figures are parameterized by x = sin(L_{1}(t)), y = sin(L_{2}(t)) (or cosine) where L_{1} and L_{2} are linear functions.
The parameterization here is of the form x = sin(L_{1}(t)), y = sin(1/L_{2}(t)), in a range of tvalues that insures L_{2}(t) is not 0. 
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