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#1 |
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Nov 2016
29 Posts |
I thought this was a fun coincidence and I couldn't see it mentioned elsewhere:
Taking the first differences of the sequence of Mersenne numbers ending only in 7: [127, 2047, 32767, 524287...] and of Mersenne numbers ending in 1: [31, 511, 8191, 131071...] produces secondary sequences that seem to count the number of edges in a icosohedral geodesic sphere. The first differences of 24n+3-1 (http://oeis.org/A241955) are the numbers: [120, 1920, 30720, 491520, 7864320, 125829120, 2013265920, 32212254720, 515396075520, 8246337208320, 131941395333120] The first differences of 24n+1-1 (http://oeis.org/A241888) are the numbers: [30, 480, 7680, 122880, 1966080, 31457280, 503316480, 8053063680, 128849018880, 2061584302080, 32985348833280] If we interlace these values as: [30, 120, 480, 1920, 7680, 30720, 122880, 491520, 1966080, 7864320, 31457280, 125829120, 503316480, 2013265920, 8053063680, 32212254720, 128849018880, 515396075520, 2061584302080, 8246337208320, 32985348833280, 131941395333120] We get a sequence that matches http://oeis.org/A277451 - "Number of edges in geodesic dome generated from icosahedron by recursively dividing each triangle in 4." The OEIS entry claims this matches the edges of a geodesic dome, but I only have values matching a geodesic sphere. Please let me know if you think anything does not hold. Some notes I found on geodesic spheres (from the program Jmol): "All faces are triangles. Not equilateral, but close. The geodesic sphere is constructed by starting with an icosohedron, a platonic solid with 12 vertices and 20 equilateral triangles for faces. One should split each triangular face into 4 faces by creating a new vertex at the midpoint of each edge. These midpoints are still in the plane, so they are then 'pushed out' to the surface of the enclosing sphere by normalizing their length back to 1.0. Faces + Vertices = Edges + 2 Faces: 20, 80, 320, 1280, 5120, 20480 start with 20 faces ... at each level multiply by 4 Edges: 30, 120, 480, 1920, 7680, 30720 start with 30 edges ... also multiply by 4 Vertices: 12, 42, 162, 642, 2562, 10242 start with 12 vertices and 30 edges. when you subdivide, each edge contributes one vertex 12 + 30 = 42 vertices at the next level 80 faces + 42 vertices - 2 = 120 edges at the next level" I have attached an image from http://kilodot.com/post/3690269953/c...-geodesic-grid |
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#2 |
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"Jeppe"
Jan 2016
Denmark
23×3×7 Posts |
It is not hard to see that both procedures lead to the geometric progression with initial term 30 and common ratio 4. In other words $f(n)=30\cdot 4^n$. /JeppeSN
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#3 |
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Nov 2016
29 Posts |
I did feel that my post looks a little hyperbolic/crankish when that is all that is happening here. However, I haven't found any other direct references to this sequence on google, google books, arxiv, oeis etc.
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