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As you know, the picture below is one of M.C. Escher's works.

enter image description here

I have thought a lot about what would be the rule of the function drawn on the surface of sphere? Is it a $\mathbb R\rightarrow\mathbb R^3$ or $\mathbb R^3\rightarrow\mathbb R^3$ function? Honestly, the idea that he drew it is new now and I don’t think his works be outdated ever. Thanks for sharing your knowledge about this great picture with me. :)

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    @Babak: I just remembered this. Now that I have upgraded my Mathematica (and my laptop) and learned to do animations with it, I can oblige. Please let me know, if you wanted a different kind of rotation (in case you are still interested that is - it's been close to two years :-)2014-03-23

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It seems to me that the bands are supposed to twist indefinitely when they approach the poles (as observed by Erick Wong). That effect is achieved by using loxodromes as boundary lines. The loxodromes are roughly the spherical equivalents of the logarithmic spirals in that a loxodrome makes a constant angle with all the latitudes that it crosses - very much like the logarithmic spirals that have constant angles between their tangents and the radii.

The attaced image is generated by Mathematica. I used $m=1/2$ (see the link for the meaning of this parameter), so e.g. one of the bands was generated by

ParametricPlot3D[{Cos[t+u]/Cosh[t/2],Sin[t+u]/Cosh[t/2],Tanh[t/2]},{t,-8,8},{ u,0,Pi/4},PlotPoints->{81,5}]

and the other parts came out with the same formula, but the range of the variable $u$ shifted by an integer multiple of $\pi/2$. The range of $t$ is should be symmetric, but the endpoints (here $\pm8$) are, again, a largely arbitrary choice of mine. I did a linear change of parameters to that on the Wikipage in order to make sure that all the loxodromes cover the same interval of latitudes, and also to make one half of the boundaries of the tiles to have constant latitude.

Here's the image. To get a better match with Escher's painting it might be necessary to fine-tune the value of $m$ further.

enter image description here

To see the "endless twisting" here is a close-up of the polar region. The image is necessarily quite flat now. The formula is the same, but the range of the parameter $t$ is now $4\le t\le 16$.

enter image description here

Edit: Here's an animated version

enter image description here

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    Added an animation.2014-03-23
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Its formula is $ f:D\to\mathbb{R}^3:(\phi,\theta)\mapsto(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) $ where $D=\{(\phi,\theta)\in[-\pi,\pi]\times[0,\pi]:\sin(4\phi+12\theta)>0\}$.

Here is a code in Mathematica

ParametricPlot3D[{Cos[\[Phi]] Sin[\[Theta]],    Sin[\[Theta]] Sin[\[Phi]],    Cos[\[Theta]]}, {\[Phi], -\[Pi], \[Pi]}, {\[Theta], 0, \[Pi]},   RegionFunction -> (Sin[4 #4 + 12 #5] > 0 &), Mesh -> None,   BoundaryStyle -> Black] 

and the graph

enter image description here

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    I've already came to this idea but due to the lack of Mathematica I can't produce a good picture.2012-07-05