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Search: id:A158121
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| A158121 |
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Given n points in the complex plane, let M(n) the number of distinct Moebius transformations that take 3 distinct points to 3 distinct points. Note that the triplets may have some or all of the points in common. |
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1
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| 6, 93, 591, 2381, 7316, 18761, 42253, 86281, 163186, 290181
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OFFSET
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3,1
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COMMENT
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There are (nC3)^2 ways of choosing two triplets out of n points with repetition.
There are 3! = 6 ways of mapping the points of one triplet to the other.
However, given each triplet pair, there is one case where each of the initial
three points is mapped to itself, resulting in the identity Moebius transformation.
There are nC3 cases of this, all but one redundant.
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REFERENCES
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Michael P. Hitchman, Geometry With an Introduction to Cosmic Topology, Jones and Bartlett Publishers, 2009, pages 59-60.
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FORMULA
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M(n) = 6*(nC3)^2 - (nC3) + 1
M(n) = 1/6(n^6-6*n^5+13*n^4-13*n^3+7*n^2-2*n+6)
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EXAMPLE
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For n=3, M(3) = 3! = 6, since there aren't any redundancies.
For n=4, M(4) = (6*4^2) - 3 = 93, since there are 3 redundant mappings.
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CROSSREFS
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Sequence in context: A009607 A009684 A009527 this_sequence A103212 A033935 A078103
Adjacent sequences: A158118 A158119 A158120 this_sequence A158122 A158123 A158124
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KEYWORD
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easy,nonn
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AUTHOR
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Matthew Lehman (matt.comicopia(AT)gmail.com), Mar 12 2009
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