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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 triples 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, 490491, 794613, 1241696, 1881041, 2773721, 3994321, 5632798, 7796461, 10612071, 14228061, 18816876, 24577433, 31737701, 40557401, 51330826, 64389781, 80106643, 98897541
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OFFSET
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3,1
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COMMENTS
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There are (nC3)^2 ways of choosing two triples out of n points with repetition.
There are 3! = 6 ways of mapping the points of one triple to the other.
However, given each triple 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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LINKS
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FORMULA
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M(n) = 6*C(n,3)^2 - C(n,3) + 1.
M(n) = 1/6*(n^6-6*n^5+13*n^4-13*n^3+7*n^2-2*n+6).
G.f.: x^3*(6+51*x+66*x^2-13*x^3+15*x^4-6*x^5+x^6)/(1-x)^7. - Colin Barker, May 02 2012
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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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MATHEMATICA
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CoefficientList[Series[(6 + 51 x + 66 x^2 - 13 x^3 + 15 x^4 - 6 x^5 + x^6) / (1 - x)^7, {x, 0, 30}], x] (* Vincenzo Librandi, Aug 14 2013 *)
LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {6, 93, 591, 2381, 7316, 18761, 42253}, 30] (* Harvey P. Dale, Mar 07 2020 *)
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PROG
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(PARI) a(n) = 6* binomial(n, 3)^2 - binomial(n, 3) + 1; \\ Michel Marcus, Aug 13 2013
(Magma) I:=[6, 93, 591, 2381, 7316, 18761, 42253]; [n le 7 select I[n] else 7*Self(n-1)-21*Self(n-2)+35*Self(n-3)-35*Self(n-4)+21*Self(n-5)-7*Self(n-6)+Self(n-7): n in [1..30]]; // Vincenzo Librandi, Aug 14 2013
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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